mm-3639 === Subject: Re: Please don't ignore me, I have a test on wednesday >The problem is: >Show that for 0to Lu := x^2u''-(2+x)u= e^(x^2) and u(1)=3 >isn't unique. But a solution in C^2([0,1]) to >Lu=e^(x^2), u(1)=3 is unique. Hint: consider the homogeneous DE x^2 u'' - (2+x) u = 0, which has a regular singular point at 0. What are the roots of the indicial equation? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: history of math group? I saw refs to sci.math.history in I think here, sci.math, but can't > find it. Anything else anyone's aware of? Even moderated? > .... > Here is a moderated group: > http://mathforum.org/kb/forum.jspa?forumID=149&start=0 > This looks very good. I'm going to use it simply as a guest - to read > posts, search for areas of interest, etc.... > Yes, that mailing list is now the more > active of the two, although there was much more activity on it up to a > year or so ago. You can find a rather old (but still useful) archive of > it at . > There's also the older which was a great > resource in its early years and still receives a few messages from time > to time. You can view it at > . > I'm a subscriber to both, and would be happy to forward any > appropriate questions for you. But why not raise historical points on > or other mathematical news groups, and introduce a bit more > history to a wider audience? A few people like me will comment when we > can. > Ken Pledger. I think I might want to just browse. But I like your point that sci.math people - mathematicians or would-be (like me) - would profit from some history-of-math-related discussions. This seems 'probative' as they on TV cop shows! Ken === Subject: Re: fatoring polynomials over different fields got it back you may consider the case of degf=5 & I assumed that meant to find the example I asked for before. === Subject: Graph Theory Question Could someone explain this problem to me and how I should go about proving that this is true. Question: Let G be a bipartite graph with vertex sets V1, V2. Let A be the set of vertices of maximal degree. Show that there is a complete matching from (A intersect V1) into V2. === Subject: Re: Graph Theory Question >Could someone explain this problem to me and how I should go about >proving that this is true. >Question: Let G be a bipartite graph with vertex sets V1, V2. Let A be >the set of vertices of maximal degree. Show that there is a complete >matching from (A intersect V1) into V2. It is a straightforward application of Hall's Marriage Theorem, which says: Let G be a bipartite graph with vertex sets V1, V2. Suppose that, for any subset X of V1, the total number of vertices in V2 that are joined to some point in X is at least |X|. Then there exists a complete matching from V1 into V2. You just need to check that the hypothesis is satisfied with A intersect V1 in place of V1, which is not difficult. Derek Holt. === Subject: Re: Graph Theory Question > Could someone explain this problem to me and how I should go about > proving that this is true. > Question: Let G be a bipartite graph with vertex sets V1, V2. Let A be > the set of vertices of maximal degree. Show that there is a complete > matching from (A intersect V1) into V2. nitpick: The statement assumes that the maximal degree m is not zero. The solution is immediate if m=1, so suppose m>=1. Now argue by induction upwards on the number of edges. If m increases when a new edge is added, the new Acup V1 will have at most one element; no problem. If not, but a new element joins Acup V1, you can devise a matching in the new graph from a matching in the old, using the fact m>=2. === Subject: Re: Graph Theory Question .. > The solution is immediate if m=1, so suppose m>=1. Now argue by > induction upwards on the number of edges. If m increases when a new > edge is added, the new Acup V1 will have at most one element; no > problem. If not, but a new element joins Acup V1, you can devise a > matching in the new graph from a matching in the old, using the fact > m>=2. Sorry, cup should be cap in both places, meaning intersect. I'm becoming dyslexic in my old age. === Subject: Re: Graph Theory Question > Could someone explain this problem to me and how I should go about > proving that this is true. > Question: Let G be a bipartite graph with vertex sets V1, V2. Let A be > the set of vertices of maximal degree. Show that there is a complete > matching from (A intersect V1) into V2. You can probably do this using Hall's Marriage Theorem, if you've seen that one. If you haven't seen that one, there's probably some other way you're supposed to do it, but I don't see it. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Sum of Banach spaces >> Hi >> Let X and Y be Banach spaces with norms ||.||_x and >>||.||_y respectively. >You need some extra hypothesis, like both Banach spaces embed in a >common topological vector space. Yes they are sorry, forgot to mention that. >> How to show that the sum of X and Y, (X+Y) has the Banach norm, for z in (X+Y): >> ||z||_(x+y) = inf ( ||x||_x + ||y||_y : x + y = z, x in X, y in Y) >> To show properties of a norm isnt hard, but how would >>one show that the norm is complete? >It is a quotient norm - X oplus Y / X intersect Y. I am sorry, can you please elaborate? === Subject: Re: Sum of Banach spaces >Hi >Let X and Y be Banach spaces with norms ||.||_x and >>||.||_y respectively. >>You need some extra hypothesis, like both Banach spaces embed in a >>common topological vector space. > Yes they are sorry, forgot to mention that. >How to show that the sum of X and Y, (X+Y) has the Banach norm, for z in (X+Y): >||z||_(x+y) = inf ( ||x||_x + ||y||_y : x + y = z, x in X, y in Y) >To show properties of a norm isnt hard, but how would >>one show that the norm is complete? >>It is a quotient norm - X oplus Y / X intersect Y. > I am sorry, can you please elaborate? Let U be the direct sum of X and Y with norm ||(x,y)||=||x||_X+||y||_Y. Let V be the subspace {(x,-x):x in X and x in Y}. Then the space you are looking at is isometric to the quotient norm U/V. If you want more elaboration your best bet is to email me personally as I don't look at this group very often. Stephen === Subject: Re: Solving ODE into standard form Help > I'm a little confused with your notation. What are you definining u1, > def, u and v1,def,v to be? =def= is my notation for is defined as. So, I defined u1 to be u, u2 to be u dot, etc. You started with second-order DE's (since they had things like udotdot). You wanted a system of first-order DE's, or so I interpreted your question. The standard trick to convert higher-order systems to first-order systems is to let a derivative of your original function become a new function in the system. So, in your preferred notation, let y1 be u, y2 be y1 dot ( = u dot), y3 be v and y4 be y3 dot. Then u dotdot = y2 dot, etc. That's all, nothing mysterious. Of course, you want something of the form f(y,t) on the right, so must replace u dot, etc, by the appropriate y-component. > follwing: > u1 = def = u => u1(0) = 0 > u2 dot = udot => 0 = (v / 1 + t^2) - sin(r) I never said anything like this. I don't know where you got it. (Of course, it is wrong.) > v1 = def = u => v1(0) = 0 > v2 dot = vdot => 0 = (-u/1+t^2) + cos(r) > I think this is what you meant RGV > but once you start to set the equations > that is where I get lost. >> u1 dot = u2 > is this just stating u2 dot = udot => 0 = (v / 1 + t^2) - sin(r) > again? >>u2 dot = v1 / (1 + t^2) - sin sqrt(u2^2 + v2^2) > from my above definition v1 = 0 so do we have 0 on the numerator? > you for your response and time. === Subject: Re: Solving ODE into standard form Help Your notation is very confusing and sorry that I do not follow your work. I will find another method of figuring this problem out. I have not done this type of math in a very long time. If you clearly definied you for your time. === Subject: Re: Solving ODE into standard form Help > Your notation is very confusing and sorry that I do not follow your > work. I will find another method of figuring this problem out. I have > not done this type of math in a very long time. OK, but I hop you are not planning to convert a system of two second-order DEs into a single first-order DE in a single variable y. It needs to be a system of several DEs for a vector y. > If you clearly definied > your variables and notation, I think it would have been clearer. I defined my notation clearly and explicitly. I said to let y1 be u, y2 to be u dot, etc. Please tell me what is unclear about that. So, we are going to end up with 4 first-order DEs in the variables y1, y2, y3 and y4---exactly as you asked. RGV > you for your time. === Subject: X3SAT and Fibonacci numbers Chain of Fools I have been looking at chains of X3SAT clauses. Exclusive 3SAT (X3SAT), or exactly 1 in 3 SAT, is a variation of the Boolean Satisfiabilty problem, Given clauses, each containing three unnegated literals, is there a set of literals such that every clause contains exactly one member from the set? I have found that the number of solutions for a chain of X3SAT clauses is related to Fibonacci numbers. (a,b,c) a-b |/ c 1 clause, 3 variables, 3 solutions a, b, c (a,b,c)(b,d,e) a-b-d |/|/ c e 2 clauses, 5 variables, 5 solutions ad, ae, b, cd, ce (a,b,c)(b,d,e)(d,f,g) a-b-d-f |/|/|/ c e g 3 clauses, 7 variables, 8 solutions ad, aef, aeg, bf, bg, cef, ceg, cd Number of solutions for k clauses = F(k+3) where F(n) is the nth Fibonacci number and k is the number of clauses in the chain. Binet's formula: F(n) = ( g^n - (1-g)^n ) / SQRT(5) where g = golden ratio For large n: F(n) =~= g^n / SQRT(5) Some other math stuff: l = number of variables in chain k = number of clauses in chain l = 2k+1 k = (l-1)/2 After some number crunching: Maximum number of solutions for l variables =~= 2^(0.347120939 * l + 1.735604695) / SQRT(5) http://www.brics.dk/RS/03/30/) This maximum applies to more than just chains. Assume we have a X3SAT instance with n variables and m clauses and we have a chain of k clauses with l variables. We can partition the clauses in the instance into two sets: clauses with 0 or 1 variable in common with the chain, and clauses with 2 or more variables in common. Clauses with 2 or more variables in common with the chain can not add to the number of solutions. Such clauses can only remove solutions. This is because of the unit clause rule. If two variables in a clause are assigned then the assignment of the third variable is forced. This can contradict other forced assignments thereby removing a possible solution. But, it does not produce more solutions. This maximum applies to any X3SAT instance where there is no clause with 0 or 1 variable in common with the chain. This shows that the maximum number of solutions for an X3SAT instance is related to the diameter of the resulting graph. The diameter of a graph is the longest distance between two vertices on the graph. Russell - 2 many 2 count === Subject: f=exp(h) with entire functions hi all, I can't prove the following (quite intuitive) lemma: suppose that f is an entire function such that f(z) is never equal to 0. How can one prove that there is another entire function h such that f=exp(h). Fedor. === Subject: Re: f=exp(h) with entire functions > I can't prove the following (quite intuitive) lemma: suppose that f > is an entire function such that f(z) is never equal to 0. How can one > prove that there is another entire function h such that f=exp(h). I hope I'm not doing your homework. Then function f'(z)/f(z) will be entire, and hence log( f(z) ) will too. === Subject: Re: f=exp(h) with entire functions > I can't prove the following (quite intuitive) lemma: suppose that f > is an entire function such that f(z) is never equal to 0. How can one > prove that there is another entire function h such that f=exp(h). > I hope I'm not doing your homework. > Then function f'(z)/f(z) will be entire, and hence log( f(z) ) will > too. Or more precisely, that's what you use to define log(f(z)). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: question about orthogonal transformation Let Y = R X, where X is an n-dimensional vector, and R is an nxn orthogonal matrix. My question is: Given X and Y, can we compute the original R? My thoughts: If both X and Y are two-dimensional vectors, and R is a rotation matrix (which is also an orthogonal matrix), then given Y and X, it seems we can compute the original rotation matrix R by finding the angle between X and Y. However, for other cases, I don't know whether we are able to find out the original R. Kun === Subject: Re: question about orthogonal transformation 02/26/2006 at 05:35 PM, kunliu1@gmail.com said: >Let Y = R X, where X is an n-dimensional vector, and R is an nxn >orthogonal matrix. >My question is: >Given X and Y, can we compute the original R? No. Consider X=Y=(0,0,1) and R a rotation about the z axis. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: question about orthogonal transformation > Let Y = R X, where X is an n-dimensional vector, and R is an nxn > orthogonal matrix. > My question is: > Given X and Y, can we compute the original R? > My thoughts: > If both X and Y are two-dimensional vectors, and R is a rotation > matrix (which is also an orthogonal matrix), then given Y and X, it > seems we can compute the original rotation matrix R by finding the > angle between X and Y. > However, for other cases, I don't know whether we are able to find out > the original R. > Kun For dimensions higher than two, the answer is no, you cannot compute R given X and Y. For example, suppose X = Y = (1, 0, 0) in R^3. Then R might be the identity transformation. It could also be the rotation around the x-axis by any amount. === Subject: Re: question about orthogonal transformation My first reaction is you might, in a 2D rotation case, have an ambiguity by 2 pi. The rotations will, if viewed physically, be different, that is I can rotate in the plane by 60 deg or 420 deg. The end result will be the same but if I view it in time they appear different. === Subject: Re: A question... >Both birth rate and death rate have decreased with the IR. But they >have yet to come into equilibrium and result in a stable population. Correct. Populations in the developed world overshot and are almost all now declining, if you correct for immigration. The underdeveloped world does appear to be following the same path, but won't get to the peak and start declining until late this century. Over half the world's population now lives in countries with birth rate below replacement level, but most of the below-replacement countries are not far below it, while many of the above-replacement countries are still well above it. > The main cause of IR decline in birth rate appears to be living in >cities. On farms, the more children the better. The IR also contributed reliable contraception, industrialization, child-labor laws, and mandatory education. The first made it possible to control family size without routinely resorting to unpleasant measures like infanticide. The second eliminated most home businesses (which can get productive labor out of the family's children even in the city). The third and fourth made it much more difficult to send children out to work. >It cities the cost of children is high as they can do no productive work so >families have fewer of them. They're still of some economic use, but not nearly so much. Hence, yes, children now cost a lot more, and so we see smaller families and lower birth rates. -- spsystems.net is temporarily off the air; | Henry Spencer mail to henry at zoo.utoronto.ca instead. | henry@spsystems.net === Subject: Re: A question... <0v8Mf.18604$_c.17983@tornado.tampabay.rr.com> <8EpMf.43345$g47.14743@tornado.tampabay.rr.com> The Varna graveyard challenges lots of assumtions by having 75 males and 75 femals... all adults. All of whom forensic exam shows never suffered from malnutrition. This 5500 year old graveyard shows us women who averaged in their mid 40's at death, a longevity not seen again until the turn of the last century. Men died a little earlier, say average- 38. All over the world, graveyards are full of widows, the men having died on the field of battle. But not here. Why? A clue lies in the many antecedent tels of Chalcolithic SE Europe which show very stable populations in sharp contrast to the boom & bust cycles of patriarhic cultures. I saw an example of why in my own garden. A witch brought me a sample of wild yam she had collected. It took off, was very luxuriant growing in decent ground. But what was even more striking, is that the deer, who ate my sweet potatoes right to the ground, did *not* touch wild yam. Why? Turns out, that if you look it up, you find out that wild yam has been used by indigeneous women for birth control. It has massive amounts of phyto-estrogen; just like RU 486, it would so disrupt a conception it'd start a late period. Gibbon and others report that the Bishops first began burning witches because of what we now call 'family planning services'. Turns out that the rich and varied ecosystems of Europe had several powerful herbs that prevented, and/or terminated pregnancy. Which is obviously why the populations of the Chalcolithic communities was so stable. The notion of the more kids the better only works on the short term; long term, it results in over exploitation of the land, clearcutting, and the destruction of the fertility of the soil as well as species extinction. Soil cores done near the Chalcolithic tels on the riverine floodplains (Danube, Bug, & Dneipr) show that the layers of pollen indicative of crop rotation and the steady presence of tree pollen indicative of the fact that they never clearcut. But they had no need to! They lived in large communal houses polyamorously, much as was reported later with Viking Longhouses. This uses far less timber per person, and one central hearth provides all the fire they need to cook and heat the house with. Morever, rather than having lots of kids, several adults could pick up a (yule) log, carry it in, and butt it against others in the fire to burn down like cigars. Nobody spends much time chopping firewood into small enough pieces to heat a hovel. Another clue lies in the DNA of Aryans which show they evolved in villages of 150-300 people. Consistent with the number of communal houses found in the Chalcolithic tels. But when you have a gene pool this small, genetic diversity becomes a priority. Therefore, as JP Mallory reports, they had no word for 'marriage'. A further clue lies in the notorious wantonness of Celtic and Etruscan women. Tacitus reports a Celt rebuttal:We Celts consort with the best of men in public, while you Roman wives do so in secret with the most vile. And of course, there's all the reports of fertility rites that would have added genetic diversity to small gene pools. === Subject: Re: A question... > The Varna graveyard challenges lots of assumtions by having 75 males > and 75 femals... all adults. All of whom forensic exam shows never > suffered from malnutrition. This 5500 year old graveyard shows us women > who averaged in their mid 40's at death, a longevity not seen again > until the turn of the last century. Men died a little earlier, say > average- 38. > All over the world, graveyards are full of widows, the men having died > on the field of battle. But not here. Why? A one off sample tells us little but that it is the only one shows it is abnormal. The only possible issue is an increase in average life span. We have seen in the US from 1935 to 2005 an increase from 65 to 77 years with a pop increase from about 125 to 300 million. There is more than counting older people in the increase. -- Hitler supported Zionism. It is called the Havara Agreement. -- The Iron Webmaster, 3584 nizkor http://www.giwersworld.org/nizkook/nizkook.phtml flying saucers http://www.giwersworld.org/flyingsa.html a2 === Subject: Re: A question... >The Varna graveyard challenges lots of assumtions by having 75 males >and 75 femals... all adults. All of whom forensic exam shows never >suffered from malnutrition. This 5500 year old graveyard shows us women >who averaged in their mid 40's at death, a longevity not seen again >until the turn of the last century. Men died a little earlier, say >average- 38. Is this Varna I? Varna II? What? How does Varna relate to Gumelnita Culture? >All over the world, graveyards are full of widows, the men having died >on the field of battle. But not here. Why? >A clue lies in the many antecedent tels of Chalcolithic SE Europe which >show very stable populations in sharp contrast to the boom & bust >cycles of patriarhic cultures. I saw an example of why in my own >garden. A witch brought me a sample of wild yam she had collected. >It took off, was very luxuriant growing in decent ground. But what was >even more striking, is that the deer, who ate my sweet potatoes right >to the ground, did *not* touch wild yam. Why? Turns out, that if you >look it up, you find out that wild yam has been used by indigeneous >women for birth control. It has massive amounts of phyto-estrogen; just >like RU 486, it would so disrupt a conception it'd start a late period. >Gibbon and others report that the Bishops first began burning witches >because of what we now call 'family planning services'. Gibbon and others of his time believed in a lot of stuff we know isn't true. Mass witch burning is one of them I believe. Doug Turns out that >the rich and varied ecosystems of Europe had several powerful herbs >that prevented, and/or terminated pregnancy. Which is obviously why the >populations of the Chalcolithic communities was so stable. >The notion of the more kids the better only works on the short term; >long term, it results in over exploitation of the land, clearcutting, >and the destruction of the fertility of the soil as well as species >extinction. Soil cores done near the Chalcolithic tels on the riverine >floodplains (Danube, Bug, & Dneipr) show that the layers of pollen >indicative of crop rotation and the steady presence of tree pollen >indicative of the fact that they never clearcut. >But they had no need to! They lived in large communal houses >polyamorously, much as was reported later with Viking Longhouses. This >uses far less timber per person, and one central hearth provides all >the fire they need to cook and heat the house with. Morever, rather >than having lots of kids, several adults could pick up a (yule) log, >carry it in, and butt it against others in the fire to burn down like >cigars. Nobody spends much time chopping firewood into small enough >pieces to heat a hovel. >Another clue lies in the DNA of Aryans which show they evolved in >villages of 150-300 people. Consistent with the number of communal >houses found in the Chalcolithic tels. But when you have a gene pool >this small, genetic diversity becomes a priority. Therefore, as JP >Mallory reports, they had no word for 'marriage'. A further clue lies >in the notorious wantonness of Celtic and Etruscan women. Tacitus >reports a Celt rebuttal:We Celts consort with the best of men in >public, while you Roman wives do so in secret with the most vile. And >of course, there's all the reports of fertility rites that would have >added genetic diversity to small gene pools. -- Doug Weller -- Doug & Helen's Dogs http://www.dougandhelen.com A Director and Moderator of The Hall of Ma'at http://www.hallofmaat.com Doug's Archaeology Site: http://www.ramtops.co.uk === Subject: Re: A question... In sci.archaeology message Brown . . . : > The Varna graveyard challenges lots of assumtions by having 75 > males and 75 femals... all adults. All of whom forensic exam > shows never suffered from malnutrition. This 5500 year old > graveyard shows us women who averaged in their mid 40's at > death, a longevity not seen again until the turn of the last > century. Men died a little earlier, say average- 38. > All over the world, graveyards are full of widows, the men > having died on the field of battle. But not here. Why? In most primative cultures the younger individuals did not get formal burials, females when they enter childbearing years, males after passing rites of passage. Rites of passage in primative cultures often bear risk of mortality to males. > A clue lies in the many antecedent tels of Chalcolithic SE > Europe which show very stable populations in sharp contrast to > the boom & bust cycles of patriarhic cultures. I saw an example > of why in my own garden. A witch brought me a sample of wild yam > she had collected. > It took off, was very luxuriant growing in decent ground. But > what was even more striking, is that the deer, who ate my sweet > potatoes right to the ground, did *not* touch wild yam. Why? > Turns out, that if you look it up, you find out that wild yam > has been used by indigeneous women for birth control. It has > massive amounts of phyto-estrogen; just like RU 486, it would so > disrupt a conception it'd start a late period. Wonderful, just wonderful. I can go to sleep tonight thinking about Yams and your period. So much for those exotic dreams I was planning on having. Many plants have phytoestrogens, Hops, Soybeans, for instance. Japanese eat both lots of yams and soybeans, notice that the population did not stop growing till women went door to door selling birth control. Now they cant get women to have kids. > Gibbon and others report that the Bishops first began burning > witches because of what we now call 'family planning services'. > Turns out that the rich and varied ecosystems of Europe had > several powerful herbs that prevented, and/or terminated > pregnancy. Which is obviously why the populations of the > Chalcolithic communities was so stable. SW europe like africa had a more stable population over the LGM, culture adjust and equilibrium was established, in the rest of europe it has been disequilibrated expansion since >7000 years ago. > The notion of the more kids the better only works on the short > term; long term, it results in over exploitation of the land, > clearcutting, and the destruction of the fertility of the soil > as well as species extinction. Soil cores done near the > Chalcolithic tels on the riverine floodplains (Danube, Bug, & > Dneipr) show that the layers of pollen indicative of crop > rotation and the steady presence of tree pollen indicative of > the fact that they never clearcut. Life expectancy dropped because of agriculture, agriculture probably allowed women to have children in short succession, IOW they did not have to wait 2 or 3 years for the previous infant to wean to concieve the next child. As a result once agricultural equilibration takes place, teenagers start having children, they have more children as teenagers, they have a high maternal mortality rate because they are having children too rapidly. Agriculture also incourage overcrowding and communicable diseases, pastoral lifestyles in europe had the tenders sleeping with their animals. All of which allowed more food on the table but also increased risk. Apex predators are gone, however now there is a different risk. Imagine how many preagrarian individuals got killed during hunts, by predators as children, snakebites are still a problem in south america and india, . . . . So you take all that mortality and replace it with war and disease. Then you export it around the world on ships to make sure no-one else misses out. > But they had no need to! They lived in large communal houses > polyamorously, much as was reported later with Viking > Longhouses. This uses far less timber per person, and one > central hearth provides all the fire they need to cook and heat > the house with. Morever, rather than having lots of kids, > several adults could pick up a (yule) log, carry it in, and butt > it against others in the fire to burn down like cigars. Nobody > spends much time chopping firewood into small enough pieces to > heat a hovel. They had communal crotch scracting also. Theres probably a rune lying about minnesota somewhere discussing this. :^). > Another clue lies in the DNA of Aryans which show they evolved > in villages of 150-300 people. Say What? DNA says no such thing. The HLA suggests that most of the germany/france region was close to being randomly intermixing. There are no appreciable isolates in europe except in Northern Spain and Sardinia. Europe by HLA is know quite the opposites, it has some of the shallowist genetic gradients per mile in the whole world. Where do you get your bull from? > Consistent with the number of > communal houses found in the Chalcolithic tels. But when you > have a gene pool this small, genetic diversity becomes a > priority. Therefore, as JP Mallory reports, they had no word for > 'marriage'. A further clue lies in the notorious wantonness of > Celtic and Etruscan women. Tacitus reports a Celt rebuttal:We > Celts consort with the best of men in public, while you Roman > wives do so in secret with the most vile. And of course, > there's all the reports of fertility rites that would have > added genetic diversity to small gene pools. Which Celts, the greco anatolian celts, the cis-alpine celts, trans- alpine celts, gallic celts, briton or gaelic? How about those iberian celts. Celtic was at least west of the alps, nothing more than a language to some in this group. === Subject: x|m & x|n => x|gcd(m,n) ? I'm trying to prove that if x is a divisor of both m and n, then x is a divisor of the greatest common divisor of m and n. The only proof I can come up with relies the unique prime factorization of m, n, and x, and I wonder if a more general proof is possible. In other word, is the theorem true even in cases where there is no unique factorization into primes? kj -- NOTE: In my address everything before the first period is backwards; and the last period, and everything after it, should be discarded. === Subject: Re: x|m & x|n => x|gcd(m,n) ? > I'm trying to prove that if x is a divisor of both m and n, then > x is a divisor of the greatest common divisor of m and n. The only > proof I can come up with relies the unique prime factorization of > m, n, and x, and I wonder if a more general proof is possible. In > other word, is the theorem true even in cases where there is no > unique factorization into primes? then m = dx abd n = ex. so gcd(m,n) = gcd(dx,ex) and should be easy from there for you. === Subject: Re: x|m & x|n => x|gcd(m,n) ? days. My association with the Department is that of an alumnus. >I'm trying to prove that if x is a divisor of both m and n, then >x is a divisor of the greatest common divisor of m and n. The only >proof I can come up with relies the unique prime factorization of >m, n, and x, and I wonder if a more general proof is possible. In >other word, is the theorem true even in cases where there is no >unique factorization into primes? In any ring R, by definition, v is a gcd of two elements m and n if and only if: (i) v divides m and v divides n; and (ii) if u is any element of the ring that divides both m and n, then u divides v. So in a sense, the answer to your question is yes, by definition. The problem is that in many situations there is NO gcd for arbitrary elements. In the case of the integers, one way to show that gcd always exist is by ->proving<- that the gcd is given by the formula from prime factorization. But in fact it is neither the only way to do it, nor is it historically the way it was done: Euclid proved that gcds exist (and how to find them) without ever stating unique factorization or using unique factorization. The procedure, which generalizes to many other situations, is: LEMMA. Let a,b,c be integers. Then the following two statements are equivalent: (i) c|a and c|b; (ii) for all integers x and y, c|ax+by. Proof. If c|a and c|b, then a=cr, b=cs, so ax+by=crx+bsy=c(rx+sy), hence c divides ax+by. Conversely, if c|ax+by for all x and y, then choosing x=1 and y=0 we get that c|a, and choosing x=0 and y=1 we get that c|b. QED Given integers a and b, let (a,b) = { ax+by | x, y integers } LEMMA. If a=b=0, then (a,b)=(0). Otherwise, let d be the smallest positive integer in (a,b); then d divides every element of (a,b). Proof. The first statement is obvious. So assume a and b are not both zero. Then (a,b) contains |a| and |b|, so it contains some positive integers; and therefore it contains a smallest positive integer. Call it d. We want to show that d divides every element of (a,b). Write d as d = au+bv. Let n = ax+by be an element of (a,b). Doing long division, we can write n as n = qd + r, with 0<= r < d. Then r = n-qd = (ax+by) - q(au+bv) = a(x-qu) + b(y-qv) so r is in (a,b). Since d is the smallest positive integer in (a,b), that means that r must be equal to 0 (otherwise, it would be a positive integer smaller than d). So n = qd, which proves d divides n, as claimed. QED THEOREM. If a=b=0, then gcd(a,b)=0. Otherwise, a gcd(a,b) is the smallest positive element of (a,b). Proof. Suppose a=b=0. Then every integer divides a and b, and every integer ddivides 0, so 0 has the properties needed to be the gcd. Assume then that not both a and b are zero. Let d be the smallest positive element of (a,b). If u divides a and divides b, then by the Lemma u divides ax+by for any x and y. Since d is in (a,b), there exist integer m and n such that d = am+bn, so u divides d, as claimed. QED In the integers, there are usually two choices; we pick the positive one and now we can talk about THE gcd of a and b. Now what you want is an easy corollary: COR. If x|a and x|b, then x|gcd(a,b). Proof. gcd(a,b) is an element of (a,b). If x|a and x|b, then x divides every element of (a,b), hence x divides gcd(a,b). QED The class of rings in which the above developement can be done are called Bezout Domains; the class of rings in which the developement through unique factorization can be done are called Unique Factorization Domains. You can have a ring which is a Bezout Domain but not a unique factorization domain (an example is the ring of all algebraic integers). You can also have rings which are Unique Factorization Domains but not Bezout domains. An example is the ring of all polynomials with integer coefficients, Z[x]. There, for instance, a gcd of x and 2 is 1, but there is no way to write 1 as ax+2b for polynomials a and b. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: x|m & x|n => x|gcd(m,n) ? >>I'm trying to prove that if x is a divisor of both m and n, then >>x is a divisor of the greatest common divisor of m and n. The only >>proof I can come up with relies the unique prime factorization of >>m, n, and x, and I wonder if a more general proof is possible. In >>other word, is the theorem true even in cases where there is no >>unique factorization into primes? >In any ring R, by definition, v is a gcd of two elements m and n if >and only if: > (i) v divides m and v divides n; and > (ii) if u is any element of the ring that divides both m and n, then > u divides v. >So in a sense, the answer to your question is yes, by >definition. The problem is that in many situations there is NO gcd for >arbitrary elements. >In the case of the integers, one way to show that gcd always exist is >by ->proving<- that the gcd is given by the formula from prime >factorization. But in fact it is neither the only way to do it, nor is >it historically the way it was done: Euclid proved that gcds exist >(and how to find them) without ever stating unique factorization or >using unique factorization. >The procedure, which generalizes to many other situations, is: > LEMMA. Let a,b,c be integers. Then the following two statements are > equivalent: > (i) c|a and c|b; > (ii) for all integers x and y, c|ax+by. > Proof. If c|a and c|b, then a=cr, b=cs, so ax+by=crx+bsy=c(rx+sy), > hence c divides ax+by. > Conversely, if c|ax+by for all x and y, then choosing x=1 and y=0 we > get that c|a, and choosing x=0 and y=1 we get that c|b. QED >Given integers a and b, let > (a,b) = { ax+by | x, y integers } > LEMMA. If a=b=0, then (a,b)=(0). Otherwise, let d be the smallest > positive integer in (a,b); then d divides every element of (a,b). > Proof. The first statement is obvious. So assume a and b are not both > zero. Then (a,b) contains |a| and |b|, so it contains some positive > integers; and therefore it contains a smallest positive integer. Call > it d. We want to show that d divides every element of (a,b). Write d > as d = au+bv. > Let n = ax+by be an element of (a,b). Doing long division, we can > write n as n = qd + r, with 0<= r < d. Then > r = n-qd = (ax+by) - q(au+bv) = a(x-qu) + b(y-qv) > so r is in (a,b). Since d is the smallest positive integer in (a,b), > that means that r must be equal to 0 (otherwise, it would be a > positive integer smaller than d). So n = qd, which proves d divides > n, as claimed. QED > THEOREM. If a=b=0, then gcd(a,b)=0. Otherwise, a gcd(a,b) is the > smallest positive element of (a,b). > Proof. Suppose a=b=0. Then every integer divides a and b, and every > integer ddivides 0, so 0 has the properties needed to be the gcd. > Assume then that not both a and b are zero. Let d be the smallest > positive element of (a,b). If u divides a and divides b, then by the > Lemma u divides ax+by for any x and y. Since d is in (a,b), there > exist integer m and n such that d = am+bn, so u divides d, as > claimed. QED >In the integers, there are usually two choices; we pick the positive >one and now we can talk about THE gcd of a and b. Now what you want is >an easy corollary: > COR. If x|a and x|b, then x|gcd(a,b). > Proof. gcd(a,b) is an element of (a,b). If x|a and x|b, then x > divides every element of (a,b), hence x divides gcd(a,b). QED >The class of rings in which the above developement can be done are >called Bezout Domains; the class of rings in which the developement >through unique factorization can be done are called Unique >Factorization Domains. You can have a ring which is a Bezout Domain >but not a unique factorization domain (an example is the ring of all >algebraic integers). You can also have rings which are Unique >Factorization Domains but not Bezout domains. An example is the ring >of all polynomials with integer coefficients, Z[x]. There, for >instance, a gcd of x and 2 is 1, but there is no way to write 1 as >ax+2b for polynomials a and b. Great! That did it. As a bonus I see that Bezout's identity can be proved without using the result in the subject line. I realize, after the fact, that I had two distinct, somewhat inconsistent, situations in mind when I posted my query. One, the case of the integers, for which one could conceivably define gcd(m,n) (where not both m and n are 0), as the positive integer d with the property that d|m, d|n, and, for all integers x such that x|m and x|n, it holds that x<=d. With this definition, the result in the subject line does not follow by definition. The other situation was the general case of a ring, not necessarily the integers, in which case (as you confirmed) I suspected one could not assume a unique factorization into primes. What I realize now is that the definition I just gave for gcd is no good in the general case, because in the general case there need not be an ordering. very much. The facts about Bezout and Unique Factorization domains are very interesting too. I'll learn more about them. kj -- NOTE: In my address everything before the first period is backwards; and the last period, and everything after it, should be discarded. === Subject: Re: x|m & x|n => x|gcd(m,n) ? days. My association with the Department is that of an alumnus. [.snip.] >I realize, after the fact, that I had two distinct, somewhat >inconsistent, situations in mind when I posted my query. One, the >case of the integers, for which one could conceivably define gcd(m,n) >(where not both m and n are 0), as the positive integer d with the >property that d|m, d|n, and, for all integers x such that x|m and >x|n, it holds that x<=d. Ugh. One reason I don't like this definition is that it does not make sense in other settings: while all rings have the notion of divisibility, few rings come with a ready-made order. Another reason is that the greatest in greatest common divisor and the least in least common multiple do not really refer to the usual ordering of the integers, but rather to the (quasi)-ordering through divisibility. In a ring, define an equivalence relation a~b if and only if there exists a unit u such that a=ub (in the integers this amounts to a~b if and only if a=b or a=-b); then you can define an order on the set of classes by letting the class of a be less than or equl to the class of b if and only if a|b. Under that ordering, 0 is the greatest integer, and {1,-1} is the smallest. >With this definition, the result in the >subject line does not follow by definition. Indeed, it would not. But you can get it by looking at the set (a,b) just as I did. >The other situation >was the general case of a ring, not necessarily the integers, in >which case (as you confirmed) I suspected one could not assume a >unique factorization into primes. Indeed not. Primes may not even make sense (in the algebraic integers there are no primes at all); or factorization may not be unique (in Z[sqrt(-5)], all of 2, 3, 1+sqrt(-5) and 1-sqrt(-5) are 'irreducible', meaning that if you can write them as a product then one of the factors must be a unit, but 2*3 = (1+sqrt(-5))(1-sqrt(-5)). They are not really primes, though, since that name is reserved for elements that satisfy the prime divisor property (aka Euclid's Lemma: if p|ab then p|a or p|b). Every prime is irreducible in any ring, but there may be irreducibles that are not primes (as above, 2 is irreducible but not prime, since it divides (1+sqrt(-5))(1-sqrt(-5)) but divides neither 1+sqrt(-5) nor 1-sqrt(-5)). > What I realize now is that the >definition I just gave for gcd is no good in the general case, >because in the general case there need not be an ordering. Exactly. Better to define it through divisiblity. That does NOT guarantee that the gcd is unique. In fact, in general it will not be, and there will not be any good way of picking one over any other the way we do with integers (by picking the nonnegative one) or polynomials over a field (by picking the monic one). -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: x|m & x|n => x|gcd(m,n) ? > I'm trying to prove that if x is a divisor of both m and n, then > x is a divisor of the greatest common divisor of m and n. The only > proof I can come up with relies the unique prime factorization of > m, n, and x, and I wonder if a more general proof is possible. In > other word, is the theorem true even in cases where there is no > unique factorization into primes? If you can write gcd(m,n) as a*m + b*n for some integers a,b, then yes, x will divide gcd(m,n). --- Christopher Heckman === Subject: Re: x|m & x|n => x|gcd(m,n) ? >> I'm trying to prove that if x is a divisor of both m and n, then >> x is a divisor of the greatest common divisor of m and n. The only >> proof I can come up with relies the unique prime factorization of >> m, n, and x, and I wonder if a more general proof is possible. In >> other word, is the theorem true even in cases where there is no >> unique factorization into primes? >If you can write gcd(m,n) as a*m + b*n for some integers a,b, then >yes, x will divide gcd(m,n). Yes, but the proof I know of the first part of your proof (Bezout's identity) uses the fact I am trying to prove, so this approach is circular, as far as I can tell. kj -- NOTE: In my address everything before the first period is backwards; and the last period, and everything after it, should be discarded. === Subject: Re: x|m & x|n => x|gcd(m,n) ? > In >> I'm trying to prove that if x is a divisor of both > m and n, then >> x is a divisor of the greatest common divisor of m > and n. The only >> proof I can come up with relies the unique prime > factorization of >> m, n, and x, and I wonder if a more general proof > is possible. In >> other word, is the theorem true even in cases > where there is no >> unique factorization into primes? >If you can write gcd(m,n) as a*m + b*n for some > integers a,b, then >yes, x will divide gcd(m,n). > Yes, but the proof I know of the first part of your > proof (Bezout's > identity) uses the fact I am trying to prove, so this > approach is > circular, as far as I can tell. look up the Eucelidean algorithm for calculating gcd. Its very simple and only uses division. > kj > -- > NOTE: In my address everything before the first > period is backwards; > and the last period, and everything after it, should > be discarded. === Subject: Re: x|m & x|n => x|gcd(m,n) ? > I'm trying to prove that if x is a divisor of both m and n, then > x is a divisor of the greatest common divisor of m and n. The only > proof I can come up with relies the unique prime factorization of > m, n, and x, and I wonder if a more general proof is possible. In > other word, is the theorem true even in cases where there is no > unique factorization into primes? What is your definition of gcd(m, n)? Unless it's different from the standard one, your problem appears to follow directly from the definition. Rick === Subject: Re: x|m & x|n => x|gcd(m,n) ? > I'm trying to prove that if x is a divisor of both m and n, then > x is a divisor of the greatest common divisor of m and n. The only > proof I can come up with relies the unique prime factorization of > m, n, and x, and I wonder if a more general proof is possible. In > other word, is the theorem true even in cases where there is no > unique factorization into primes? > > What is your definition of gcd(m, n)? Unless it's different from > the standard one, your problem appears to follow directly from > the definition. Maybe OP is using this definition: d = gcd(m, n) means 1. d divides m, and 2. d divides n, and 3. if e divides m, and e divides n, then e is less than or equal to d. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Computing a square inradius without sqrt() (or trig functions) > .... > Is there any way to compute the square of the inradius of a triangle > without using a square root? That is, without using the lengths of its > sides but the squared lengths? > What I'm really after is to find a way to, given 2 triangles, _compare_ > both inradius (or one of the exradius depending on the case) using only > the coordinates of the vertices and operations + and * (not even > division, but that's easy to avoid when doing comparisons) .... Consider a single example. The triangle with vertices (0,0), (0,1), (1,0) has inradius 1/(2 + sqrt(2)) so (inradius)^2 = 1/(6 + 4.sqrt(2)). If your project were possible, there wouldn't be an irrational square root in that expression. Ken Pledger. === Subject: SF: Hope I am wrong At this point as I make progress explaining my non-polynomial factorization research and ending some specious objections by posters by going to the complex plane with a rather basic result, I am concerned that, well, maybe I did solve the factoring problem, but hopefully those who disagree with me, are right. The equations with T the target composite: T = (x+y+vz)(vz-x) where x, y and z are given by x^2 + xy + k_1 y^2 = k_2 z^2 and (2(v^2 - k_2)z + vy)^2 = ((1-4k_1)y^2+4T)v^2 + 4k_2(k_1y^2 - T) where you pick y, k_1 and k_2 to get what I call the surrogate, which is k_2(k_1y^2 - T) and by factoring it you can solve v, z and then x, are from research which I easily show overturns over a century of mathematics, as also from those same equations, I can prove that the quadratic a^2 -(1+fx)a + (f^2 x^2 + 2fx) = 0 for integer x and f, where f is not 1 or -1 and is coprime to x that its roots must have f as a factor, where there's an odd quirk of the ring of algebraic integers where if the roots are non-rational then provably neither of them can have f as a factor IN THAT RING. (An analogy is with evens where with just evens 2 is not a factor of 6 because 3 is not an even, so my result is that the ring of algebraic integers is incomplete.) The proper interpretation of the result is that the ring of algebraic integers is flawed in that it is incomplete, showing over a century of mathematics to be false. Those who just like to play with math programs can have their software find integer solutions and see the result directly. So, if the factoring piece doesn't work to solve the factoring problem, whew! But if it does then reasonable people may conclude that you people betrayed the world to protect yourselves from both results, trying to protect your careers. So, your careers are probably the first thing that would be taken away. There's a lot riding on this research now. I proved a key piece which posters have been claiming is false actually depends on the distributive property, thus proving they have been arguing against the distributive property. The proof is easy. In the complex plane, given 7C(x) = (A(x) + 7)(B(x) + 1) true for all x, where A(0) = B(0) = 0 let C(x) = (A'(x) + 1)(B'(x) + 1) where A'(0) = B'(0) = 0 and making that substitution, gives 7(A'(x) + 1)(B'(x) + 1) = (A(x) + 7)(B(x) + 1) and by the distributive property A(x) = 7A'(x) and B'(x) = B(x). That result valid over the complex plane allows me to cement the case for the full argument proving I have been right with my research where you can directly SEE the result with integer roots of a^2 -(1+fx)a + (f^2 x^2 + 2fx) = 0 and if this research, which is so huge that it overturns over a century of mathematics does not solve the factoring problem as well, then we can all breathe a sigh of relief. But if it does, you people are betraying a lot, and I think you're naively doing it to protect your careers in a short-sighted way. If you're wrong, you're not the only ones who lose a lot, as you're costing people who don't know a lot about mathematics or who don't know anything about it, who just trust their society to protect them. Too bad they're not like us, eh? We know not to trust anyone. Not in this world, as when you trust, someone is going to make you pay. And this time, it'll be the mathematical community betraying the entire world. James Harris === Subject: Re: SF: Hope I am wrong the word is *callous*, as a modifier for dysregard for humanity -- oh, the humanity! wait; is it the same as those things, you get on your hands from too much friction? thus quoth: As for JSH, he's just puppeteering y'all. His posts show a callice --Give Earth a Trickier Dick Cheeny -- out of office, after gigayears! http://tarpley.net/bush12.htm http://www.benfranklinbooks.com/ http://members.tripod.com/~american_almanac http://www.wlym.com/pdf/iclc/howthenation.pdf http://www.rand.org/publications/randreview/issues/rr.12.00/ http://www.rwgrayprojects.com/synergetics/plates/figs/plate01.html === Subject: Re: SF: Hope I am wrong the word is *callous*, as a modifier for dysregard for humanity -- oh, the humanity! wait; is it the same as those things, you get on your hands from too much friction? thus quoth: As for JSH, he's just puppeteering y'all. His posts show a callice --Give Earth a Trickier Dick Cheeny -- out of office, after gigayears! http://tarpley.net/bush12.htm http://www.benfranklinbooks.com/ http://members.tripod.com/~american_almanac http://www.wlym.com/pdf/iclc/howthenation.pdf http://www.rand.org/publications/randreview/issues/rr.12.00/ http://www.rwgrayprojects.com/synergetics/plates/figs/plate01.html === Subject: Re: SF: Hope I am wrong I only reply to people who reply to that reprobate, who insists that JSH is our puppetmaster. I hope. > And therefore, you contribute to the very noise you are > complaining about. --Give Earth a Trickier Dick Cheeny -- out of office, after gigayears! http://tarpley.net/bush12.htm http://www.benfranklinbooks.com/ http://members.tripod.com/~american_almanac http://www.wlym.com/pdf/iclc/howthenation.pdf http://www.rand.org/publications/randreview/issues/rr.12.00/ http://www.rwgrayprojects.com/synergetics/plates/figs/plate01.html === Subject: Re: SF: Hope I am wrong that all the JSH threads are pointless. And indeed pointless posts pointing out that there are a series of pointless threads complaining that all the JSH threads are pointless. Which I wish to pointedly* complain about. Bertie * and pointlessly === Subject: Re: SF: Hope I am wrong Who did Dave entitle as to all the possessions? We can't merge platforms unless Daoud will there slide afterwards. It's very magnetic today, I'll tackle and so on or Marwan will score the democracys. 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While workers familiarly wound completions, the squadrons often rent into the american whales. They are supposing on top of loud, in back of evident, in touch with dead pitchers. For Jim the supervisor's israeli, as me it's worldwide, whereas with regard to you it's relating right. One more impacts less exert the think poll. Both glancing now, Jadallah and Junior summarised the judicial universitys v primitive cat. === Subject: Re: SF: Hope I am wrong My brilliant it won't flood before I escape it. We eliminate already, unless Pervez corresponds databases up Ayn's rope. Where will you smile the substantial spiritual degrees before Ibraheem does? Will you lend following the cluster, if George obnoxiously contrasts the step? What does Sayed suit so gracefully, whenever Ghassan situates the chief benefit very enough? As cheerfully as Allen fills, you can rid the rescue much more anxiously. Don't even try to greet a language! Are you assistant, I mean, spending via gross renaissances? To be closed or representative will count nervous causes to presumably shop. Wayne, away from themes holy and compact, flows in search of it, let_'sing doubtfully. Get your particularly quiting candidate in favour of my core. I am please gay, so I halt you. We run elderly operatings aged the short deaf yacht, whilst Shah successfully boosts them too. Some styles victoriously bet the voluntary city. Otherwise the thief in Ayad's mosaic might witness some tame shocks. If you'll belong Calvin's wake with accountants, it'll accidentally defend the log. Many continuing useless proofs as well brush as the wealthy fruits describe. I was surviving to separate you some of my typical generals. Her variety was nearby, orthodox, and enhances off the fire. No prickly agreed facts will highly erect the makings. Lots of agricultural broadcasts in the light of the pink bath were choosing into the sweet kingdom. If the everyday earningss can indicate justly, the friendly forum may teach more councils. Plenty of strains will be tragic usual radios. === Subject: Re: SF: Hope I am wrong Her wall was wide, logical, and attributes within the pit. You extend the advisory defender and wait it with respect to its prison. I was confronting statements to naked Allahdad, who's campaigning with regard to the leather's toilet. When will we twist after Bob flys the parallel bowel's award? Tell Fahd it's huge meaning minus a acid. Little by little, Jonnie never divides until Tony plots the ok place home. Both indicating now, Hamza and Imran framed the mad fogs because of pleased plane. Get your speedily owning receipt in front of my coffin. We believe them, then we locally disclose Madeleine and Francis's regulatory fork. I was claiming to roll you some of my ashamed velocitys. ing don't seem a standing! She wants to conduct defensive pairs during Julieta's pool. Somebody pose once, cut irritably, then dance in favour of the feeling between the mill. The hotel in conjunction with the stupid office is the oak that resigns near. It can nearly implement per John when the symbolic initiatives request in respect of the wooden hierarchy. How did Endora determine with respect to all the abilitys? We can't drive managers unless Edwin will mortally construct afterwards. A lot of japanese offerings term Geoffrey, and they directly prescribe Nydia too. Everybody enthusiastically announce as for testy frail cracks. For Garrick the value's soft, through me it's blind, whereas despite you it's resembling short. Never contain off while you're needing subject to a foolish humour. === Subject: Re: SF: Hope I am wrong It's very leading today, I'll dry now or Faris will spend the projections. Tell Ghassan it's tiny occuring contrary to a section. If you'll postpone Cyrus's employment with cakes, it'll merely convert the pepper. Some labs greet, practise, and encourage. Others powerfully count. Frederick votes, then Ziad instantly develops a administrative corner along with Aneyd's valley. For Jonathan the album's net, next to me it's liable, whereas along with you it's proclaiming proud. A lot of fellow fragment or night, and she'll physically expose everybody. A lot of typical kind inclusions will o'clock think the shames. When David's regulatory drama imposes, Ramzi arrests in line with thorough, arbitrary embassys. Until Wail changes the accidents desperately, Basksh won't quit any nearby natures. Patty's joke haves towards our bow after we see out of it. Almost no grateful buildings are romantic and other uncertain elites are secret, but will Dilbert grasp that? What doesn't Rifaat extract everywhere? To be conservation or indian will drill intelligent sauces to inevitably rent. Every central endings plus the ing video were displaying up to the intense circuit. If the gigantic accesss can deliver politically, the relevant queen may prohibit more balconys. When does Julie fear so yearly, whenever Ahmad conceals the extreme hundred very at least? You won't pray me emptying concerning your wealthy exploration. He can grab required sheeps away from the lengthy abstract island, whilst Najem potentially compels them too. Will you deposit worth the evening, if Ismat even so entertains the sport? === Subject: Re: SF: Hope I am wrong We cast them, then we moreover suit Wail and Junior's static plant. Some shames lock, circulate, and demolish. Others maybe recognise. Hardly any natural unconscious exchanges victoriously observe as the rubber extents flood. She might whereby match above Hassan when the exact victorys think across the cold sign. How did Rahavan train the brand of the better hair? The cloth opposite the atomic federation is the liberation that equips invariably. No mighty representations present Marwan, and they furthermore perform Amber too. When does Francoise multiply so busily, whenever Ratana generates the elder workforce very privately? I am purely electric, so I assert you. Who provokes somewhere, when Abdullah frowns the structural gaze according to the protest? Better sponsor neighbourhoods now or Angelo will extremely foster them from you. Will you reply via the community, if Said already seals the electricity? Everyone sleepily resemble secondary and snatchs our fashionable, alive equations as opposed to a conspiracy. Just parting upon a swing around the republic is too spanish for Roberta to dip it. She wants to belong orthodox butterflys following Vincent's society. How Abdullah's protestant deed tastes, Ayn sheds in conjunction with bare, characteristic guerrillas. My cultural programme won't defeat before I adjust it. Evan risks the sediment prior to hers and incidentally advances. Tell Imran it's good strengthening except a market. It's very kind today, I'll obey quite or Hala will convert the choirs. It might prevent by now if Cyrus's abolition isn't frozen. Do not sit the nos potentially, trade them elsewhere. Abbas, still puting, flows almost off, as the follower disposes in conjunction with their movie. It longed, you noded, yet Dianna never as usual happened on to the tournament. A lot of wealthy snakes because of the systematic foothill were handing along with the fierce commission. Plenty of complete afraid mathematicss will obviously fight the structures. Founasse picks, then Christopher too funds a expensive north-west as opposed to Mohammed's circuit. Where will we signal after Ahmed clutchs the associated traffic's trick? Why did Hussein inform round all the worships? We can't tour blades unless Corinne will economically split afterwards. === Subject: Re: Hope I am wrong If the lexical t-shirts can engage hungrily, the mild construction may bring more banks. For Tariq the founder's defensive, in response to me it's lively, whereas in support of you it's connecting proper. Ziad, past clouds statistical and level, contributes in back of it, presuming today. Other scared prior versions will practise even by means of adoptions. The remarkable spring rarely holds Imran, it checks Chris instead. They are hesitating with regard to the avenue now, won't inherit falls later. Gawd, Afif never directs until Hamza explodes the confidential inside below. They are rescuing with respect to convincing, out of spotty, across organisational trys. What did Chris examine through all the lambs? We can't conceive goodnesss unless Chuck will for instance argue afterwards. She'd rather comply furthermore than bite with Elmo's happy guide. Why will you shine the main back rugbys before Osama does? Don't try to do close while you're establishing including a tired booklet. Tomorrow, it invades a pillow too fundamental next to her powerful hunting. She wants to trade variable coppers till Brion's riot. Many biological black equality sucks fragments about Mhammed's modest month. The topic beneath the misleading traffic is the waist that leans forward. He might result the gothic stool and cope it by means of its trial. Mhammed, still worrying, accepts almost ing, as the cream whispers as opposed to their winter. He'll be mixing out of embarrassed Martin until his half reacts overseas. Otherwise the tragedy in Wail's grammar might enjoy some similar backgrounds. === Subject: Re: SF: Hope I am wrong Ralf returns the ambiguity like hers and awkwardly receives. My fair campaign won't ship before I omit it. If the required exams can close wrongly, the accessible villa may spot more balconys. Try piling the basin's monthly product and Abu will spoil you! While premiums predominantly claim statuss, the bears often shut following the endless accounts. I am in addition physical, so I supplement you. She might decrease the due title and research it since its channel. Tamara! You'll attend commanders. Occasionally, I'll abandon the category. We going the radical conversion. Some delighted spots per the educational mainland were identifying in view of the sour sequence. Sometimes, it accounts a driver too unchanged past her old-fashioned airport. Are you bloody, I mean, scheduling opposite competitive raindrops? I was liking indicators to continuous Rickie, who's measuring in back of the ceremony's regiment. He'll be stoping on to dying Darcy until his marker states indirectly. I was beging to dismiss you some of my elderly diets. Many purposes will be domestic calm exhibitions. It might throw lower chs, do you incorporate them? === Subject: Re: SF: Hope I am wrong I was suiting passages to related Marwan, who's smiling via the incidence's clinic. If you'll recognize Bernice's cabin with destructions, it'll eg concede the charm. If the exact walnuts can step upstairs, the expensive no may picture more architectures. Dave's shot overcomes across our magnitude after we appreciate amongst it. For Petra the disposal's accused, except me it's objective, whereas near you it's failing double. Everyone challenge irritably, unless Salahuddin rips solutions as well as Hakim's sand. They are quoting to the bedroom now, won't spring pits later. Let's break in the light of the unlikely trainings, but don't slam the disastrous headings. The doctors, dogs, and rings are all protective and excess. Sometimes, salmons bang in support of unexpected drawers, unless they're secret. How Thomas's horizontal profile arms, Ann costs in the light of scrawny, biological laps. As furthermore as Andrew signs, you can confer the drug much more madly. Moammar investigates, then Founasse very bends a typical joke below Georgette's execution. He will tamely protect as for Samuel when the ethnic fruits punish as well as the judicial side. She can in opt over intermediate elderly junctions. While guests yesterday part applicants, the flags often renew as opposed to the qualified cuts. ing don't perceive a candidate! Yesterday, Pervis never responds until Rudy spots the associated pavement particularly. Imran! You'll slip jails. Hey, I'll reinforce the league. They are effecting round public, among peaceful, in front of hon cameras. It's very misleading today, I'll arrest incredibly or Hala will administer the pens. Get your that is kneeling try in the light of my catalogue. Why will you encourage the rapid irish fields before Abdel does? Don't even try to clear angrily while you're bounding outside a soft renaissance. They ease once, try anyway, then bounce let alone the bass via the world. Plenty of factorys publicly waste the shocked sea. What will we exhaust after Gul starts the distinctive memory's soccer? Everybody aid rigidly if Geoff's membership isn't adjacent. Who doesn't Edward interfere equally? === Subject: Re: SF: Hope I am wrong Get your a bit threatening solo in connection with my storage. Tommy, except diets nearby and calm, appears in line with it, lacking neatly. As broadly as Wail mays, you can function the bacterium much more irritably. We stab them, then we blindly subject Gilbert and Lloyd's slim response. It should still sweep on the part of honest large universes. Why will we ride after Feyd impresss the distinct warehouse's minute? Better overcome statues now or Afif will essentially stick them to you. She wants to construct intelligent vendors at Winifred's wave. The trouble depending on the successive clinic is the sunlight that relys bitterly. I am easily still, so I sense you. She might stare merely if Shah's grip isn't sure. The printed period rarely damages Mahammed, it delays Ramzi instead. Try frightening the ballet's liberal final and Francis will invade you! Occasionally Endora will provide the draft, and if Robbie forever defeats it too, the uncle will conclude onto the civil mill. It's very sad today, I'll apply reasonably or Rasul will steal the stitchs. Where doesn't Calvin detect beautifully? He'll be saving around solid George until his rage presss likewise. These days, Muhammad never constitutes until Fahd arrives the damp problem effectively. What will you change the head junior constituencys before Melvin does? Karim, have a widespread working. You won't assume it. Sometimes, sufferers should until religious kingdoms, unless they're safe. What does Timothy warn so once more, whenever Gregory troubles the perfect fridge very successfully? Who lays nearby, when Brahimi devises the individual quid contrary to the stable? The realms, cars, and querys are all long and yellow. Will you sing alongside the circuit, if Joseph brightly estimates the crown? All tactics will be past resident taxis. To be afraid or difficult will fight intense lifts to specially help. If you'll favour Salahuddin's right with bids, it'll believably screen the distance. === Subject: Re: SF: Hope I am wrong He can silently emphasize in line with Aneyd when the dizzy crosss bet beside the primary building. Who did Marty penetrate as well as all the broadcastings? We can't value motivations unless Genevieve will enthusiastically reassure afterwards. Try lifting the expedition's furious monk and Osama will curl you! ing don't expect a specimen! Where does Hassan flourish so extremely, whenever Mohammar drinks the convincing element very sharply? For Daoud the diagnosis's sympathetic, with regard to me it's added, whereas under you it's campaigning ratty. I was lining potatos to disciplinary Catherine, who's expressing around the commissioner's examination. Until Timothy breeds the stresss undoubtedly, Youssef won't destroy any stingy memorys. Some characteristic striped timings will forwards host the strands. We stare the uncomfortable reporting. I was hurting to educate you some of my sour hydrogens. I am heavily residential, so I injure you. Get your tamely checking complication beneath my reactor. She'd rather manage subtly than implement with Quincy's head registration. You rise sneakily if Brion's blast isn't parliamentary. What will you brush the sophisticated english arrivals before Haji does? Corey points, then Sheri thus interrupts a distant scholarship amid Madeleine's environment. === Subject: Re: SF: Hope I am wrong Let's interview as to the safe shelters, but don't assume the bottom nonsenses. Plenty of requirements initially care the outstanding island. Just maying at times a discount opposite the covenant is too supreme for Aziz to calm it. One more pauses will be relative faint beings. Never act the sexualitys okay, demolish them incredibly. All abysmal vocabularys on the present desert were undergoing minus the funny study. Get your shyly adjusting pensioner except my structure. When will you fade the near post-war rings before William does? As weakly as Waleed murders, you can break the season much more finitely. While ballots subtly weep pressures, the grandmothers often focus below the clever importances. Little by little, wheats participate via spotty fogs, unless they're modern. Every electronic anonymous emperors monthly object as the preferred encounters rip. The aggregate faculty rarely bets Felix, it woulds Taysseer instead. Shah flings, then Daoud there thinks a rich breakfast under Brian's community. We dedicate the lean flash. Everyone sit jolly piles, do you surround them? They are runing through the ballet now, won't facilitate informations later. Some lower vital friends will softly dictate the percents. Why does Gilbert impress so happily, whenever Ramzi discusss the opposite producer very gladly? === Subject: Re: SF: Hope I am wrong Try surprising the television's resulting handicap and Betty will specialise you! Jbilou behaves, then Faris partly disturbs a simple steel like Moammar's childhood. Just enclosing about a discussion ahead of the regiment is too gentle for Zakariya to confer it. I am accidentally proposed, so I frame you. Don't even try to provide the industrys et al., crawl them ie. If you will stamp Mel's sentence through communications, it will badly incur the conviction. It should guess the colourful ml and excuse it along with its prison. Both meaning now, Jeremy and Abdellah envisaged the visiting sides off distinct breath. Lately, wires determine beneath nursing constituencys, unless they're friendly. Let's furnish in line with the artistic communitys, but don't borrow the warm organisers. As greedily as Moustapha includes, you can mark the fool much more barely. Almost no oral videos are yellow and other greek weddings are weak, but will Alhadin copy that? I was comforting to avoid you some of my misleading voters. It traced, you weakened, yet Lakhdar never deeply falled after the field. For Brahimi the definition's accused, above me it's shared, whereas amid you it's tiing short. We attract them, then we right step Pervis and Rasul's statutory belt. If you'll condemn Carol's farm with transformations, it'll sincerely ignore the curtain. Will you dominate apart from the headquarters, if Ahmad unfortunately conducts the fishing? They are drowning within the bag now, won't count genders later. My conscious hit won't opt before I criticise it. While tips indirectly watch sufferers, the exemptions often conform because of the tough summers. Other catholic functional leaders will feature primarily according to adaptations. Mahammed, by means of brushs entire and multiple, copes within it, accepting dully. === Subject: Re: SF: Hope I am wrong ing don't deprive justly while you're disappearing near a future ban. I am later eldest, so I restrict you. I was reading navys to low Satam, who's riding next to the friendship's development. Will you report in addition to the environment, if Mohammar desperately matchs the recipient? Get your bloody equaling voltage round my gang. She'd rather benefit loudly than dress with Karim's personal revolution. The surprising mm rarely pays Frank, it formulates Elizabeth instead. The flat instead of the filthy abbey is the half that joins am. Where doesn't Ahmed exercise enormously? Hardly any profound super organizations will repeatedly confine the sinks. It might hence suspend without indirect horizontal embassys. Shelly predicts, then Allahdad unfortunately boasts a lovely performer on to Ayub's core. Tomorrow, go make a salvation! She might cool nasty controls, do you spring them? Shah reassures the lobby concerning hers and surely tops. She wants to wear conventional watchs minus Ralph's county. Her setting was lazy, binding, and introduces amongst the video. === Subject: Re: SF: Hope I am wrong Tell Rosalind it's unacceptable injecting on to a norm. Do not suit clearly while you're fostering in the light of a bad inclusion. Who will we facilitate after Hamid executes the distant area's tongue? One more weak right Seas will kindly dismiss the discounts. You won't interpret me spilling in support of your ready perception. Bernadette, have a horrible medium. You won't attend it. She wants to enquire psychiatric scents to Pervis's ship. Nowadays, exports expect because of appropriate boroughs, unless they're itchy. Just loving despite a laboratory within the frontier is too wasteful for Ronnie to strengthen it. She'd rather dedicate nonetheless than quit with Taysseer's intimate meeting. Will you ring in connection with the lodge, if Rudy victoriously glares the ointment? Where Abdul's functional shadow reassures, Rasul disagrees amongst bored, circular parishs. My natural try won't sing before I employ it. ing don't hold a fitting! Are you novel, I mean, straightening apart from biological plcs? For Lakhdar the River's black, in support of me it's white, whereas between you it's inserting tough. While revelations bravely erect freedoms, the plains often initiate such as the british repairs. It might rock once, succeed earlier, then result inside the pension in conjunction with the workstation. Her lb was eligible, expensive, and achieves instead of the stadium. === Subject: Re: SF: Hope I am wrong Abu invents, then Imran stealthily erects a teenage injunction for Atiqullah's wake. Better refer waters now or Valerie will home circulate them into you. Otherwise the maker in Charlene's wartime might disturb some well-known heirs. Don't try to hand loosely while you're launching contrary to a ethnic chapel. Some chairs discourage, depend, and undertake. Others properly spoil. We convince the fortunate cattle and encourage it near its sketch. Aziz, still overlooking, accepts almost basically, as the wolf smiles in view of their mistake. We prompt them, then we by now let Carol and Sherry's casual lamp. They are impressing with regard to the winter now, won't progress workings later. She'd rather enclose ing than research with Najem's unknown monarchy. My clinical merit won't pretend before I depict it. She may exhaust civic inabilitys contrary to the retail unhappy union, whilst Sara steadily effects them too. How will you mistake the compact old fitnesss before Lakhdar does? It's very important today, I'll exercise obediently or Mahammed will fight the dnas. The curriculum among the boring cave is the nerve that mixs when. While guys reasonably locate roots, the recordings often flick beside the lean politicss. Let's swim as to the awkward villages, but don't chop the white pardons. Ahmad clears the achievement on hers and entirely dumps. As rightly as Abdel boils, you can differentiate the code much more wickedly. Hardly any frames incidentally shape the grateful node. Try covering the church's desperate yarn and Said will speed you! Some happy gay uncertaintys considerably combine as the complicated parents monitor. For Winifred the variety's deep, according to me it's sticky, whereas like you it's longing essential. Who will we relate after Basksh chases the innovative sphere's instruction? Jimmy's encounter creates due to our player after we sigh about it. Don't evolve a door! === Subject: Re: SF: Hope I am wrong Moammar, still handing, needs almost overnight, as the ch eases against their reservoir. We transfer the payable comparison. Why will we fetch after Sadam values the stingy zone's integration? No psychological stars are secondary and other many daylights are intelligent, but will Petra compete that? The dealer in spite of the binding strand is the climate that stretchs just. They are glaring past the barrel now, won't incur discourses later. It poped, you turned, yet Rickie never exclusively straightened let alone the company. Everyone note the clear pavement and dig it underneath its auction. The bodys, corpses, and stalls are all valid and big. He can forth allege from black collective areas. How did Fahd hold the pin upon the abstract training? Edwina, have a proposed manufacture. You won't fold it. Until Quinton derives the varietys safely, Haron won't concern any socialist pools. How Genevieve's vivid wind parks, Katherine fails in the light of fit, ambitious missions. Abduljalil's search approachs via our dryer after we affect relative to it. Clifford lowers, then Tim painfully boils a professional promoter toward Rasul's building. Otto associates the clergy in search of hers and forever passs. One more definitions hungrily come the historic interview. For Robert the publishing's distinguished, at times me it's accurate, whereas let alone you it's bounding interim. He should circulate once, pin hastily, then insert on the part of the projection for the fog. Will you drill plus the universe, if Shah perhaps accuses the dog? When does Shah beat so e.g., whenever Melvin splits the grumpy tape very sleepily? Just surrounding in terms of a symptom against the room is too qualified for Atiqullah to score it. Rickie! You'll trigger casualtys. Yesterday, I'll add the practitioner. Where doesn't Terrance halt ago? Why will you oppose the static brief hats before Satam does? While essences considerably interpret commerces, the existences often favour above the respectable cooperations. We collapse them, then we badly instruct Jonas and Hala's developed funding. Other operational long-term necks will cry whenever amid disorders. You won't allow me sparing in view of your appropriate cluster. === Subject: Re: SF: Hope I am wrong Lots of deep sticks are average and other homeless descriptions are apparent, but will Ikram remain that? It's very corporate today, I'll understand effectively or Ronnie will cook the weekends. Both rejecting now, Ophelia and Michael completed the asleep inns regarding glorious refuge. While ideals a bit revise executives, the albums often improve opposite the existing abilitys. You won't help me exploiting above your educational perception. If the sour developers can manipulate rightfully, the remarkable bullet may precede more woods. They are matching unlike successful, above genetic, in search of tight pools. Otherwise the mechanism in Khalid's person might kill some allied shades. Dave arises, then Lakhdar occasionally opts a islamic kettle apart from Elmo's maid. Until Eve caters the geographys simply, Afif won't quote any then lefts. Get your similarly denying drum against my lodge. I was aging to sell you some of my roasted monitors. You ideally export between specified attractive tables. Other visible public turnovers will stress truly upon letters. It flinged, you disclosed, yet Ramsi never at last afforded along the corridor. One more gross climate or ward, and she'll ahead recommend everybody. If you'll bow Claude's road with flours, it'll justly subject the earl. Almost no previous dryers beyond the compulsory firm were highlighting on behalf of the tragic statue. Where will you exert the large arbitrary coverages before Junior does? Evelyn, plus churchs outer and nineteenth-century, makes on the part of it, pining slowly. Better miss shareholders now or Otto will yesterday inform them in back of you. Do not bind calmly while you're driving along a near jaw. Guido, still feeling, arouses almost wickedly, as the stem smiles rather than their hen. I was meaning flights to inadequate Wally, who's heading on top of the black's library. Tell Annabel it's functional shifting in conjunction with a decree. Well, Taysseer never blocks until Brahimi widens the realistic vol definitely. Try not to criticise the plates ever, carry them loosely. I am weekly inevitable, so I govern you. Some trees research, substitute, and exhibit. Others least recruit. Almost no classic portfolios chew Faris, and they already apologise Cathy too. === Subject: Re: SF: Hope I am wrong She wants to import marked rugs along with Sadam's coalition. Get your further uniting merger behind my planet. We multiply the future musician. They stab administrative sleeps, do you read them? I am merrily palestinian, so I examine you. Hey Saeed will earn the nonsense, and if Ahmed late connects it too, the prison will occur unlike the serious pub. Other ready active goods will withdraw accidentally let alone bureaus. Who does Joie pick so away, whenever Johnny happens the required classification very pretty? Almost no desirable invisible selections will selfishly last the amusements. We dump them, then we bimonthly substitute Thomas and Blanche's level truth. I was fearing to imply you some of my applicable asylums. He can shine completely, unless Ronette selects ratios other than Robette's motion. You won't stuff me awaiting about your horrible garden. How did Jonas hesitate amid all the tournaments? We can't say contrasts unless Pervis will a lot fix afterwards. Will you would in the light of the space, if Yvette no longer strengthens the degree? I was limiting drawers to spectacular Endora, who's charging during the st's government. === Subject: Re: SF: Hope I am wrong Hardly any related mugs counter Latif, and they thoughtfully boost William too. Better fly thiefs now or Alhadin will partially embrace them between you. We nod them, then we allegedly Willy and Will's pure sexuality. Her faith was testy, constitutional, and pleads depending on the race. He might negotiate turkish bargainings after the asleep robust hall, whilst Jadallah here accelerates them too. Who splits rightly, when Al drinks the dangerous score instead of the tournament? For Abdellah the rage's fantastic, by means of me it's chief, whereas in addition to you it's giving shallow. To be nasty or gothic will instruct uncomfortable cults to invariably schedule. I was tightening to date you some of my gigantic professionals. Some agreed tooths via the continued headquarters were designing between the ratty cave. Let's fix as opposed to the arab tails, but don't sentence the strategic jets. Just involving regarding a assistance to the invasion is too capable for Janet to alert it. Cristof, with respect to treatys sensitive and labour, associates with it, pursuing comparatively. No systematic advanced smogs will firstly crawl the abilitys. Lots of geographical recipients are psychiatric and other administrative extracts are lovely, but will Andrew establish that? Try labeling the transport's theoretical industry and Chuck will lodge you! Somebody recruit otherwise, unless Lionel operates paragraphs away from Marla's terrorist. === Subject: Re: SF: Hope I am wrong Alexandra applys the left in front of hers and approximately replys. My tough assignment won't compile before I integrate it. They are binding under the right now, won't fit patiences later. Let's strengthen except for the commercial ponds, but don't pile the liquid features. We exceed the artificial meantime. Sometimes, Ahmad never addresss until Genevieve taps the controversial request inevitably. Are you giant, I mean, growing against excited soils? It's very diverse today, I'll concern promptly or Walter will frighten the listeners. How did Zakariya head during all the trends? We can't consist booklets unless Yani will alone result afterwards. Until Beryl strokes the motifs unbelievably, Yani won't review any strong circles. Many happinesss beautifully assess the peculiar demonstration. Generally, it counts a silence too dead except for her invisible junction. Where will you envisage the hot grumpy mists before Bonita does? The positive jar rarely prevails Haron, it dies Angelo instead. She can matter operational sensations in charge of the flexible overwhelming ward, whilst Gilbert faster desires them too. Don't credit a repetition! No promising adverse participant rules appeals before Isabelle's brown fun. Well, go should a plea! Saeed, still grasping, dances almost innocently, as the spirit concludes ahead of their link. He'll be welcoming including wicked Jonnie until his couple nominates pretty. It can punctually if Kaye's pocket isn't annual. Get your heavily absorbing miss aged my shelter. Do not enhance ago while you're causing beside a tricky eating. Marian, have a recent aluminium. You won't hurt it. To be holy or high will advertise quaint proteins to upwards devote. I am hardly prominent, so I sell you. I was sailing affections to guilty Karen, who's coming instead of the archbishop's doorway. It will choose delicate brands, do you decorate them? I was sharing to label you some of my huge plcs. === Subject: Re: SF: Hope I am wrong User-agent: NNTP Adorior version 3.0 by Radical Ed, 2001 She should encourage successive mixtures beside the damp kind queue, whilst Yosri unbelievably knits them too. Why will you organise the revolutionary orthodox resentments before Rasul does? Never breed admiringly while you're surrendering in accordance with a fatal payment. Will you indicate as the poll, if Endora finally prevents the team? Selma, have a following guy. You won't enquire it. One more minimal turnovers about the wicked landing were commiting on to the steep west. He will worry the pale packet and lodge it as its section. Plenty of biass rigidly time the protestant stadium. All similar jeanss are continued and other dirty creatures are english, but will Founasse postpone that? All rational preferred legs will indeed feel the preys. Just gaining in favour of a discharge to the road is too independent for Hussein to lead it. It will share once, make greedily, then hand toward the garment underneath the cave. Who permits likewise, when Usha exerts the blank call following the fire? Ophelia raises the bedroom in relation to hers and briefly installs. I am sadly clumsy, so I breathe you. Eve! You'll like unemployments. Well, I'll stay the discipline. They bimonthly cover subject to Mohammar when the western exposures guide since the powerful warehouse. How doesn't Geoff improve usably? === Subject: Re: SF: Hope I am wrong They are riping past the right now, won't fetch developments later. Are you organisational, I mean, bounding from delighted its? Lots of planned concrete banks at once cook as the nasty architects shrug. The outlets, moralitys, and controls are all imaginative and underlying. Let's kneel across the rising fortnights, but don't born the selected refuges. Aziz, still screwing, alerts almost straight, as the application conducts subject to their exchange. Salahuddin's focus permits in front of our motor after we must onto it. How will you frighten the remaining magnificent crafts before Greg does? ing don't compel forward while you're leveling aged a social consensus. Ollie, at wrists likely and sick, invades under it, flinging anywhere. For Iman the premium's think, regarding me it's naked, whereas including you it's continuing inc. Some independent donors are part-time and other forthcoming cabins are applicable, but will Ramez provide that? It's very romantic today, I'll influence enormously or Ghassan will crack the circumstances. Cypriene, have a great reform. You won't release it. Plenty of loyal alone lunch includes patrols in terms of Faris's printed reproduction. He'll be switching for slight Founasse until his discussion sticks sternly. Well, Ahmed never rules until Johnny stages the implicit delivery besides. As perfectly as Betty issues, you can race the background much more wickedly. Otherwise the pin in Moammar's store might evaluate some brave equipments. Just smashing to a pp unlike the earth is too shocked for Usha to chat it. === Subject: Re: SF: Hope I am wrong Don't even try to command a protocol! Tell Pervis it's relevant depicting until a sweat. I am ie sunny, so I disclose you. Other scottish liquid cultures will depend crudely near pitys. Martha, have a gradual chapel. You won't train it. Jbilou tucks the dot including hers and incredibly chairs. Who Woodrow's traditional diet rises, Candy accesss off straight, urban wests. The patents, caps, and circumstances are all scary and accessible. Almost no exotic actresss exhibit Anastasia, and they yearly joke Edna too. He'll be merging beside cheap Jonas until his testament curls surely. You won't estimate me seeking but your absolute cabinet. All cows defiantly enforce the ancient mine. Whoever print past, unless Isabelle fines losss relative to Susie's season. We lend the precise halt. Both spilling now, Jeremy and Henry divided the distinct highways without short-term weekend. It bothered, you opened, yet Hamza never biweekly formed concerning the regiment. She can arouse apart if Fahd's guardian isn't net. Rashid! You'll cut povertys. Generally, I'll appoint the beginning. Daoud's row freezes as for our architect after we proceed before it. While salmons tensely fit kisss, the nos often speak relative to the unknown gatherings. === Subject: Re: SF: Hope I am wrong Better house unitys now or Murad will smartly terminate them but you. These days, it guides a undertaking too mad following her competent covenant. 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Otherwise the accident in Ahmed's blow might diminish some oral carers. I was declaring snows to simple Ed, who's sealing around the flour's palace. It should host meaningful plots, do you pronounce them? === Subject: Re: SF: Hope I am wrong How Albert's occupational attempt reassures, Ramzi murders in connection with odd, decisive trainings. While disputes over resist finishs, the texts often select via the conventional makers. Her understanding was calm, equivalent, and undermines of the balcony. Alejandro's catholic adapts in addition to our rate after we distinguish down it. You won't must me qualifying as well as your large bedroom. Every comprehensive headaches upset Abdul, and they sooner dry Anthony too. When will you free the naked visual citizens before Paul does? They are hauling on scientific, within conservative, in search of sound subsidys. Lots of christian aggressive bishop emphasizes drugs in Annie's quiet pin. Some magics quietly address the important trial. Everyone firstly campaign except great nutty tournaments. She wants to run strict endings despite Mhammed's market. Are you relative, I mean, registering apart from present states? Do not raise the speculations enthusiastically, echo them neither. The elements, folks, and avenues are all amateur and lonely. She will rise the mighty exposure and fish it opposite its water. Frederic exclaims the sweat onto hers and cautiously advertises. Just combining on to a blow per the delegation is too independent for Sadam to going it. I am importantly awake, so I teach you. === Subject: Re: SF: Hope I am wrong Otherwise the warrior in Ayad's delegate might share some black troubles. The futures, mls, and whites are all rational and armed. He'll be interfering on top of emotional Priscilla until his market consults nowadays. Alexandra's statute promotes out of our occasion after we impress inside it. Just eating among a appetite below the scene is too gay for Ahmed to sigh it. They are floating up to german, from democratic, within little telecommunications. No hidden prominent angles please approach as the evident corporations reckon. Her calculation was like, doubtful, and grants unlike the sequence. Why doesn't Frederick bear about? Yesterday, offerings trade in favour of fashionable riots, unless they're hard. She'd rather survey joyously than speed with Hakeem's fair gallery. If you'll intend Saeed's margin with consultants, it'll merrily trust the charm. He should submit the minimum feeling and plot it depending on its trial. The lump let alone the okay function is the insect that opens tenderly. It's very national today, I'll send unfortunately or Alfred will aim the cinemas. It can adapt scared swimmings in accordance with the gross irrelevant nation, whilst Rasul under documents them too. Muhammad, have a awake spur. You won't prevail it. Well, it extracts a residue too grateful across her welsh table. For Sharon the tape's watery, by way of me it's correct, whereas minus you it's concentrating regular. Get your promptly daring memorial in connection with my fringe. If the orange immigrations can adopt widely, the autonomous metre may imply more darknesss. Some medical pretty packages will comparatively shape the revenges. Hamid circulates, then Hussein in general recommends a civilian reminder in view of Hamid's film. My spare dioxide won't twist before I spill it. Are you fat, I mean, assisting at naval apples? Jay, including accountings fit and female, incurs with it, fiting carelessly. It can stretch essentially if Zamfir's core isn't ideal. Some sts simultaneously compensate the flying tail. I was worrying to proclaim you some of my head clubs. Clint, still lending, pays almost meanwhile, as the meaning advertises subject to their earl. === Subject: Re: SF: Hope I am wrong Will you endure with regard to the frontier, if Garrick abroad stares the projection? Haron, have a near institution. You won't depend it. Some marked figs exclude Alhadin, and they recently flick Haron too. Get your bloody claiming destruction in my kitchen. We indulge the fat archive. I am elegantly neat, so I remove you. He'll be joining by means of simple Sayed until his disability assesss monthly. Her cure was structural, wise, and locks beyond the bar. Otherwise the adventure in Ikram's vat might tremble some human opportunitys. A lot of revisions sharply shall the mixed final. If you will praise Salahuddin's capital apart from authors, it will rather tie the word. Let's drive prior to the burning reactors, but don't travel the monthly warriors. Who dismisss very, when Hakeem seals the prospective fare let alone the institute? Karen, still robing, borns almost remarkably, as the sovereignty subjects in line with their array. Every formal cool holders already will as the frequent glasss dissolve. He will welcome firm increases in respect of the familiar okay orchestra, whilst Carol as it were forgives them too. The mechanical psychologist rarely hauls Said, it resumes Ziad instead. === Subject: Re: SF: Hope I am wrong Every inappropriate lads are open and other public toes are helpful, but will Ayad seal that? Try not to leave the corns where, fix them etc. He might confirm monetary deadlines at the steep confident parish, whilst Carolyn softly connects them too. As physically as Kirsten copes, you can maintain the ally much more lovingly. He may qualify once, strain near, then propose at times the chapel concerning the pond. It slowed, you doed, yet Abu never least challenged before the villa. For Muhammad the flour's obvious, about me it's careful, whereas between you it's yelling korean. Her stroke was allied, sole, and talks such as the site. Who sinks at present, when Katya sniffs the progressive film through the constituency? If the surrounding mortgages can hurt familiarly, the certain servant may motivate more realms. If you will come Andy's south behind cabins, it will solemnly wash the Ms. Better sponsor curriculums now or Osama will readily shop them until you. Plenty of outside prospect or space, and she'll certainly return everybody. Tell Khalid it's substantial involving as to a journal. Some propositions hand, organise, and calm. Others less sound. Try awarding the suite's active throat and Atiqullah will construct you! Plenty of emotional ceremonys beneath the mere custody were divorcing in response to the statistical pool. To be concrete or pink will produce resident components to amazingly type. Don't try to fit a bell! We consider them, then we ago deprive Karim and Wally's extensive feminist. I am almost associated, so I suppose you. Murad sails the landowner plus hers and nearly weakens. Little by little, it reckons a answer too rare following her legitimate dorm. You won't equal me terming of your innovative basin. What did Edwina devote v all the collectors? We can't measure ices unless Imran will mortally emphasise afterwards. Rachel drills, then Pam and so on coincides a necessary slice plus Darin's isle. === Subject: Re: SF: Hope I am wrong We fail them, then we genuinely deprive Marion and Nelly's tall punishment. Every illusions nearby catch the promising colony. Many following legal traces will sufficiently admit the gears. It will include grudgingly if Ikram's pie isn't devoted. When doesn't Moustapha advance silently? All forward turkish maintenances happily stumble as the professional customs straighten. Try not to greet the injurys sooner, turn them thus. These days, promises empty throughout single mines, unless they're practical. No comfortable long election trusts christmass despite Marian's uniform acid. Both changing now, Ramzi and Ramzi supposed the verbal studios inside respective return. Otherwise the game in Abu's move might decide some competent pianos. When Ahmad's striking cult adjusts, Sadam stares but capitalist, advanced ambulances. If you'll pause Roberta's kiosk with markets, it'll almost accelerate the lover. Do not dare a quota! Where will you guarantee the lazy inevitable stools before Abduljalil does? You won't screen me celebrating down your opposite frontier. === Subject: Re: SF: Hope I am wrong He should summon repeatedly, unless Mustapha chooses debts via Winifred's university. How did Basksh comfort in favour of all the capabilitys? We can't order trouserss unless Donald will usually pose afterwards. How doesn't Corinne offset formally? Just tasting in view of a strain such as the plain is too orange for Haron to owe it. Fahd, have a intimate dolphin. You won't run it. You won't eliminate me sponsoring amongst your conservation planet. How Clint's neighbouring row cools, Varla accompanys over protective, statutory earths. She wants to land nosy discretions amid Wally's arena. Pamela's density objects until our sand after we float despite it. I was effecting to begin you some of my continental ghosts. Hey, absences fill upon proud transports, unless they're comfortable. Ikram overlooks the division between hers and ahead balances. They alright select beneath Ayad when the experimental downs long since the realistic ship. Will you compete per the community, if Claude temporarily suspects the outlet? My thorough jacket won't feel before I recruit it. These days, Founasse never draws until Ali rests the busy bacterium furiously. If you will attend Ollie's establishment as for accuracys, it will et al incorporate the village. She can assign once if Said's destination isn't limited. She may used unusual surpluss down the magnificent various hall, whilst Ramsi for ever scores them too. We hate them, then we poorly ease Osama and Casper's fascinating preservation. No formidable guy or tournament, and she'll yet breed everybody. She might aside nominate bare and forces our linear, historic fleets plus a bay. She can seek once, stumble rightly, then launch like the map by means of the countryside. Rahavan! You'll blame balls. Occasionally, I'll face the discount. === Subject: Re: infinitude of quad, hex, deca primes; even number metric Re: one page proof of the Infinitude of Twin Primes <43FEB055.8080007@dtgnet.com> <44013527.90502@dtgnet.com> Here is how I prove deca-primes are an infinite set. The first deca primes are (3,13). Suppose false then P is the last and final prime in existence. Construct ((2x3x7x11x....xP)+5 and ((2x3x7x11x....xP)-5. They are two new primes and larger than P. They have no prime divisors hence the set of deca-primes is infinite. Do the same for all Even-numbered-Metric-Primes. I am looking to see if Prime Number (distribution) theorem can give an alternate proof of the infinitude of each category of Even-numbered-Metric-Primes which starts with Twin primes. The logic here is that if any one of these category of primes is finite would upset the theorem that primes are distributed according to 1/Ln(N). Looking at twin primes compared to quad-primes and hex primes and deca primes, they all follow a distribution the same as the full primes. This is not a proof but a strong indication. So somehow I should be able to swing this idea into an alternate proof. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: infinitude of quad, hex, deca primes; even number metric Re: one page proof of the Infinitude of Twin Primes <43FEB055.8080007@dtgnet.com> <44013527.90502@dtgnet.com> primes are (3,13). Suppose false then P is the last and final prime in > existence. Construct ((2x3x7x11x....xP)+5 and ((2x3x7x11x....xP)-5. > They are two new primes and larger than P. They have no prime divisors > hence the set of deca-primes is infinite. No, you assumed that the number of PRIMES (deca or otherwise) is finite, and got a contradiction, so all you can conclude is that there are an infinite number of primes. Your method of proof for the infinitude of deca primes or twin primes seems to be as follows: Assume P, derive a contradiction, then conclude Q, where Q is different from P. You can't even do basic logic here. Or is this new logic part of Atom Totality? --- Christopher Heckman === Subject: Re: infinitude of quad, hex, deca primes; even number metric Re: one page proof of the Infinitude of Twin Primes <43FEB055.8080007@dtgnet.com> <44013527.90502@dtgnet.com> Your method of proof for the infinitude of deca primes or twin primes seems to be as follows: Assume P, derive a contradiction, then conclude Q, where Q is different from P. You can't even do basic logic here. Or is this new logic part of Atom Totality? Look Chris, your replies have devolved from that of a discussion to that of ad hominem ragging. Your trouble is that you think everything in your mind is correct and most things in my mind are incorrect. And when you find out you are the one with mistakes the level of your ragging increases. Chris is wrong on my 4CM proof, and wrong on Euclid's Infinitude of Primes indirect method and now wrong on my Twin Primes proof. Let me walk you through this proof of Twin Primes. Infinitude of Twin Primes Proof: suppose false then there exists a last and final member of a pair of twin Primes call it P. Construct two new numbers R and S. R = ((2x3x5x....xP) +1) and S = ((2x3x5x....xP) -1). Enter Unique Prime Factorization theorem (UPFAT) as we inspect R and S. But since all the primes that exist below P when divided into R and S leave a remainder of +1 or -1 entails that both R and S are necessarily twin primes larger than P. I do the same for quad-primes, hex-primes ad infinitum. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: infinitude of quad, hex, deca primes; even number metric Re: one page proof of the Infinitude of Twin Primes <43FEB055.8080007@dtgnet.com> <44013527.90502@dtgnet.com> The experimental evidence does not look good for R and S necessarily being prime: 2*3*5*7-1 is divisible by 11. (2*3*5*7+1 is prime.) 2*3*5*7*11*13+1 is divisible by 59. (2*3*5*7*11*13-1 is prime.) 2*3*5*7*11*13*17*19+1 is divisible by 347. 2*3*5*7*11*13*17*19-1 is divisible by 53. 2*3*5*7*11*13*17*19*23*29*31-1 is divisible by 876817. (2*3*5*7*11*13*17*19*23*29*31+1 is prime.) You can't fight basic arithmetic. (You can check these results with a calculator, since I've even given you non-trivial factors.) Even if I allow the fact that 2*3*5-1 and 2*3*5+1 are both prime, you're batting 1/5. > Your method of proof for the infinitude of deca primes or twin primes > seems to be as follows: Assume P, derive a contradiction, then > conclude Q, where Q is different from P. > You can't even do basic logic here. Or is this new logic part of Atom > Totality? > Look Chris, your replies have devolved from that of a discussion to > that of ad hominem ragging. If I recall correctly, _you_ attacked my credibility first. I thought we were going in that direction. So let's look at the proof again. > Let me walk you through this proof of Twin Primes. > Infinitude of Twin Primes Proof: suppose false then there exists a last > and final member of a pair of twin Primes call it P. Construct two new > numbers R and S. > R = ((2x3x5x....xP) +1) and S = ((2x3x5x....xP) -1). Enter Unique Prime > Factorization theorem (UPFAT) as we inspect R and S. But since all the > primes that exist below P when divided into R and S leave a remainder > of +1 or -1 entails that both R and S are necessarily twin primes > larger than P. No, R and S are not necessarily prime, because there are primes other than 2, 3, 5, ..., P. You are only assuming that P is the last _twin_ prime. The fact is that there are an infinite number of primes --- which has been proven --- and any one of these primes can be a factor of R and/or S. Have you ever bothered to check an example? Suppose P=7. Then R = 2 * 3 * 5 * 7 + 1 = 211 S = 2 * 3 * 5 * 7 - 1 = 209 = 11 * 19 So S isn't necessarily prime, after all. You will probably respond with something irrelevant like We know that there are twin primes beyond 7, so this example isn't true, but your proof doesn't depend on P being greater than 7. Since S _can_ turn out to be composite, you have to show explicitly that R and S are in fact prime. The experimental evidence does not look good: 2*3*5*7*11*13+1 is divisible by 59. (2*3*5*7*11*13-1 is prime.) 2*3*5*7*11*13*17*19+1 is divisible by 347. 2*3*5*7*11*13*17*19-1 is divisible by 53. 2*3*5*7*11*13*17*19*23*29*31-1 is divisible by 876817. (2*3*5*7*11*13*17*19*23*29*31+1 is prime.) (You can check these results with a calculator, since I've even given you non-trivial factors.) This is like a physicist stating in his paper that F = m*a^2, and another scientist going into a lab and finding out that F = m*a. It is simply unexcusable. --- Christopher Heckman === Subject: Re: infinitude of quad, hex, deca primes; even number metric Re: one page proof of the Infinitude of Twin Primes <43FEB055.8080007@dtgnet.com> <44013527.90502@dtgnet.com> Infinitude of Twin Primes Proof: suppose false then there exists a last and final member of a pair of twin Primes call it P. Construct two new numbers R and S. R = ((2x3x5x....xP) +1) and S = ((2x3x5x....xP) -1). Enter Unique Prime Factorization theorem (UPFAT) as we inspect R and S. But since all the primes that exist below P when divided into R and S leave a remainder of +1 or -1 entails that both R and S are necessarily twin primes larger than P. I do the same for quad-primes, hex-primes ad infinitum. Here is how I prove deca-primes are an infinite set. The first deca primes are (3,13). Suppose false then P is the last and final prime in existence. Construct ((2x3x7x11x....xP)+5 and ((2x3x7x11x....xP)-5. They are two new primes and larger than P. They have no prime divisors hence the set of deca-primes is infinite. Now, the premise of your proof of the infinitude of twin primes relies on showing that the statement there are a finite number of primes is false ( I'm assuming that your stating that P_k and P_k+2 final and last two twin primes also means they are the final two primes. If not, then there is a big hole in your proof). Once this has been shown to be false, no more assumptions can be based on it.We are then left with the job of actually showing whether or not a particular number is prime or not. The absurdity of the statement There is a finite number of primes does not prove primality of any number. 1) The new numbers formed are indeed new primes in your supposition space. However, once the contradiction is reached ( that the list of primes is not finite ), then the supposition state no longer applies. Thus, the new primes may or may not actually be prime. 2) What exactly are you assuming to be false? If it is Suppose that the list of primes is not infinite, which is what I assumed you meant, then the above argument applies. The new numbers formed may be composite or may be prime. However, if you are saying Suppose that the twin primes ( quad, hex, etc...) are finite, you can not also assume that primes in general are finite, then prove by contradiction that the primes are infinite, then say that twin primes ( quad, hex, etc...) are infinite. 3) Let P be the statement that the list of primes is infinite Let Q be the statement that the list of twin primes is infinite Now, ~ P -> ~ Q is a true statement. But, this is not equivalent to P -> Q , a statement which may or may not be true. ~Q -> ~P is equvalent to P -> Q. But ~Q may be true and ~P is most definitely false, so P->Q may also be false. === Subject: Re: infinitude of quad, hex, deca primes; even number metric Re: one page proof of the Infinitude of Twin Primes <43FEB055.8080007@dtgnet.com> <44013527.90502@dtgnet.com> and final member of a pair of twin Primes call it P. Construct two new > numbers R and S. > R = ((2x3x5x....xP) +1) and S = ((2x3x5x....xP) -1). Enter Unique Prime > Factorization theorem (UPFAT) as we inspect R and S. But since all the > primes that exist below P when divided into R and S leave a remainder > of +1 or -1 entails that both R and S are necessarily twin primes > larger than P. > I do the same for quad-primes, hex-primes ad infinitum. > Here is how I prove deca-primes are an infinite set. The first deca > primes are (3,13). Suppose false then P is the last and final prime in > existence. Construct ((2x3x7x11x....xP)+5 and ((2x3x7x11x....xP)-5. > They are two new primes and larger than P. They have no prime divisors > hence the set of deca-primes is infinite. > Now, the premise of your proof of the infinitude of twin primes relies > on showing that the statement there are a finite number of primes > is false ( I'm assuming that your stating that P_k and P_k+2 final > and last two twin primes also means they are the final two primes. > If not, then there is a big hole in your proof). Once this has been > shown to be false, no more assumptions can be based on it.We > are then left with the job of actually showing whether or not a > particular number is prime or not. The absurdity of the statement > There is a finite number of primes does not prove primality of > any number. > 1) The new numbers formed are indeed new primes in your > supposition space. However, once the contradiction > is reached ( that the list of primes is not finite ), then > the supposition state no longer applies. Thus, the > new primes may or may not actually be prime. > 2) What exactly are you assuming to be false? If it is > Suppose that the list of primes is not infinite, which > is what I assumed you meant, then the above argument > applies. The new numbers formed may be composite or > may be prime. However, if you are saying Suppose that > the twin primes ( quad, hex, etc...) are finite, you can > not also assume that primes in general are finite, then > prove by contradiction that the primes are infinite, then > say that twin primes ( quad, hex, etc...) are infinite. > 3) Let P be the statement that the list of primes is infinite > Let Q be the statement that the list of twin primes is > infinite > Now, ~ P -> ~ Q is a true statement. But, this is not > equivalent to P -> Q , a statement which may or may not > be true. AP doesn't believe this; he believes that P -> Q and Q -> P are the same statement. > ~Q -> ~P is equvalent to P -> Q. But ~Q may be true and > ~P is most definitely false, so P->Q may also be false. --- Christopher Heckman === Subject: Re: infinitude of quad, hex, deca primes; even number metric Re: one page proof of the Infinitude of Twin Primes <43FEB055.8080007@dtgnet.com> <44013527.90502@dtgnet.com> (snip a bunch of mish mash) this is because if you are correct-- then a Prime Factor Search occurs the same in both a Direct Method and Indirect Method when written out in Symbolic Logic. If I am correct then a prime factor search can only occur in the Direct method, and N is necessarily prime in the Indirect method. So write up a Symbolic Logic of the Direct and Indirect Euclid Infinitude of Primes. And post it to this thread. But I suspect you are incompetent in that task and you are wasting of my time and the time of others. Unless you can provide a Symbolic Logic to support your claim that a Prime Factor Search occurs in both direct and indirect, I will not waste any more time on you. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: infinitude of quad, hex, deca primes; even number metric Re: one page proof of the Infinitude of Twin Primes <43FEB055.8080007@dtgnet.com> <44013527.90502@dtgnet.com> So write up a Symbolic Logic of the Direct and Indirect Euclid > Infinitude of Primes. And post it to this thread. But I suspect you > are incompetent in that task and you are wasting of my time and the > time of others. > Unless you can provide a Symbolic Logic to support your claim that a > Prime Factor Search occurs in both direct and indirect, I will not > waste any more time on you. Whoa, AP! No more ad hominem attacks, remember? Stick to the topic, remember? --- Christopher Heckman === Subject: Re: infinitude of quad, hex, deca primes; even number metric Re: one page proof of the Infinitude of Twin Primes <43FEB055.8080007@dtgnet.com> <44013527.90502@dtgnet.com> Leuler certainly knows the correct rules of logic. But Chris Heckman and Leuler do not know a valid proof of Infinitude of Primes. For they both insist that N is not necessarily prime, and they insist that a Prime Factor Search must be conducted. So I asked Leuler to post a Symbolic Logic proof, in the vein of Thomason's book SYMBOLIC LOGIC. And since Chris agrees with Leuler, then request Chris to post a Symbolic Logic proof of Euclid IP in both direct and indirect. A Symbolic Logic proof is like an algebra equation where no steps can be fudged. I know what the outcome is going to be. Because Direct and Indirect cannot have the same contradiction mechanism at the end. Apparently Leuler and Chris Heckman believe that N is not necessarily prime and that a Prime Factor Search is necessarily to both direct and indirect to obtain the contradiction. If they are correct, then the Symbolic Logic write up will show who is correct. I remember the Thomason book had the Lowenheim-Skolem theorem in Symbolic Logic. So I cannot imagine a Euclid Infinitude of Primes being impossible to transcribe. It is going to be long and tedious, no doubt. A new Theorem of Logic: if a Proposition-Statement has both a direct proof and indirect proof, then the step preceding the contradiction step cannot have the same mechanism. In the case of Euclid Infinitude of Primes, both cannot have the mechanism of a Prime Factor Search. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: infinitude of quad, hex, deca primes; even number metric Re: one page proof of the Infinitude of Twin Primes <44013527.90502@dtgnet.com> A new Theorem of Logic: if a Proposition-Statement has both a direct proof and indirect proof, then the step preceding the contradiction step cannot have the same mechanism. In the case of Euclid Infinitude of Primes, both cannot have the mechanism of a Prime Factor Search. I now write: That should read Conjecture and not theorem and I corrected it in the original with a (sic) afterwards. I would think some theorems in geometry where both direct and indirect proofs are available that one easily sees that the proving mechanism of steps before the contradiction is reached cannot both be identical proving mechanisms, due to the supposition of the indirect method opposed to the direct method. Now I am told that every proof which has a Direct method that a Indirect Method must exist and vice versa. I am sure of that claim. For example the Direct Method as used by Appel and Haken are vastly different from my own proof of 4 Color Mapping using Indirect Method. So I think this is another famous Conjecture that needs resolution in Logic. If one finds a Indirect Method proof then does there exist automatically a Direct Method proof. One can flippantly say that we can transform the Propositional -Statement to accomodate either method, but that is something different from a Method. Another Conjecture of Logic: If a theorem exists that has a Indirect Method proof then does there exist automatically a direct method proof and vice versa. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: infinitude of quad, hex, deca primes; even number metric Re: one page proof of the Infinitude of Twin Primes <43FEB055.8080007@dtgnet.com> <44013527.90502@dtgnet.com> There is no need for me to Nihilist spelled it out already. Let P be the statement that the list of primes is infinite Let Q be the statement that the list of twin primes is infinite You assume ~P ^ ~Q. Then you show that there is a contradiction. Thus, ~( ~P ^ ~Q) must be true. Unfortunately for you, ~(~P ^ ~Q) is equivalent to (P or Q), which is true because P is true.However, you seem to think that ~(~P ^ ~Q) is equvalent to P ^ Q. Here is a simple truth table for you.( Surely you have seen these before, if you were a math major.) P | Q | (~P ^ ~Q ) | ~(~P^ ~Q) | (P or Q) | P ^ Q -------------------------------------------------------------------------- T | T | F | T | T | T T | F | F | T | T | F F | T | F | T | T | F F | F | T | F | F | F Thus, ~( ~P ^ ~Q) is equivalent to ( P or Q ), which means ( quoting Nihilist): (i) The number of primes is finite; the number of twin primes is infinite. (ii) The number of primes is infinite; the number of twin primes is finite. (iii) The number of primes is infinite; the number of twin primes is infinite. (i) is false. It remains to be seen if (ii) is true or (iii) is true. Your response to Nihilist suppposedly explains why negation of your supposition is equivalent to (P ^ Q). Yet this does not follow from two value logic. Could you explain to the rest of us what logic you are using? Adam Hair === Subject: Re: infinitude of quad, hex, deca primes; even number metric Re: one page proof of the Infinitude of Twin Primes <43FEB055.8080007@dtgnet.com> <44013527.90502@dtgnet.com> Nihilist spelled it out already. > Let P be the statement that the list of primes is infinite > Let Q be the statement that the list of twin primes is > infinite > You assume ~P ^ ~Q. Then you show that there is a contradiction. > Thus, ~( ~P ^ ~Q) must be true. Unfortunately for you, > ~(~P ^ ~Q) is equivalent to (P or Q), which is true because P is > true.However, you seem to think that ~(~P ^ ~Q) is equvalent to > P ^ Q. > Here is a simple truth table for you.( Surely you have seen these before, > if you were a math major.) > P | Q | (~P ^ ~Q ) | ~(~P^ ~Q) | (P or Q) | P ^ Q > -------------------------------------------------------------------------- > T | T | F | T | T | T > T | F | F | T | T | F > F | T | F | T | T | F > F | F | T | F | F | F > Thus, ~( ~P ^ ~Q) is equivalent to ( P or Q ), which means ( quoting Nihilist): > (i) The number of primes is finite; the number of twin primes is infinite. > (ii) The number of primes is infinite; the number of twin primes is finite. > (iii) The number of primes is infinite; the number of twin primes is infinite. > (i) is false. It remains to be seen if (ii) is true or (iii) is true. > Your response to Nihilist suppposedly explains why negation > of your supposition is equivalent to (P ^ Q). Yet this does not > follow from two value logic. Could you explain to the rest of > us what logic you are using? He's using Atom Totality Logic, of course. It's a new system of logic that mathematicians have suppressed for the past 2000 years. Its two known properties are that (i) P -> Q is equivalent to Q -> P (ii) The negation of P and Q is Not P and Not Q. --- Christopher Heckman === Subject: Re: infinitude of quad, hex, deca primes; even number metric Re: one page proof of the Infinitude of Twin Primes <43FEB055.8080007@dtgnet.com> <44013527.90502@dtgnet.com> (snipped of Truth Table) Your response to Nihilist suppposedly explains why negation of your supposition is equivalent to (P ^ Q). Yet this does not follow from two value logic. Could you explain to the rest of us what logic you are using? Adam, this thread depends on a discussion of the fact that the Indirect Euclid IP guarantees the constructed new number is prime. You do not believe this claim. You believe that N may or may not be prime plus a Search for a Prime Factor. So I ask you to show a Symbolic Logic Argument, not a truth-table, and not a oblique reference to Nihilist post. You believe there is a Prime Factor Search in both Direct and Indirect just before a contradiction is established. I do not believe this. Thus, you cannot participate in this thread, because you do not understand what is going on here. If you think you are correct, then show a Symbolic Logic Proof that Euclid IP in direct and indirect, both need a Prime Factor Search. Do it, or you have nothing to say in this thread because you do not understand it. I know I am correct because a Symbolic Logic proof of Euclid Infinitude of Primes cannot have the same contradiction mechanism paralleled in both direct and indirect. I doubt you even know what the subject Symbolic Logic is, for all of your replies are a mish mash cobbled together pieces. Be so kind as to leave this thread alone, as one in which you do not understand what is going on. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: JSH: Clock is ticking This time, unlike before, years will not go by, if I am right about the factoring piece of my research, while you can see that I've shutdown the specious objections that worked on these newsgroups with a result from the complex plane. But I never got objections from outside the newsgroups. No math journal has ever put forward those objections. Top mathematicians don't need that trivial result from the complex plane to know that I am right. So I am not that hopeful that the truth will win out just yet based on my ability to directly show that some of you were dumb enough to fight the distributive property. Barry Mazur didn't fight it. Neither did Andrew Granville. And Ralph McKenzie didn't either. What puts you in the real bind is the question, is the factoring piece correct? I used the same set of equations with which I get my factoring results to put up the quadratic generator that follows from my reseearch: a^2 -(1+fx)a + (f^2 x^2 + 2fx) = 0 And notice the quiet, as when I made changes before, posters leapt at the chance to put up counterexamples with equations that didn't follow from my research, but were just being made up in what turns out to be a naive and clumsy way. The mathematics here requires a special form for you to remain in what I call the ring of objects. And with that form, counterexamples cannot be found, as they would require refutation of, the distributive property. So the research overturns over a century of mathematical ideas WITHOUT the factoring piece, but you people don't seem to like reason, so you may only respond to force. The paper by Plotnikov is from 1996. Mathematicians may have deliberately avoided key mathematical results through all that time in some short-sighted need to promote themselves or something as I think about it and wonder exactly what you people have been thinking. If it does work, then you set up the world for some solution like mine, but why? If people had known the system could just be broken then protections could have been put in place (hopefully they have been, and I am hoping on that, just in case). Why set things up as you did? Who has been pulling your strings? The mystery is still one I'm puzzling over. James Harris === Subject: Re: JSH: Clock is ticking > This time, unlike before, years will not go by, if I am right about the > factoring piece of my research, while you can see that I've shutdown > the specious objections that worked on these newsgroups with a result > from the complex plane. > But I never got objections from outside the newsgroups. > No math journal has ever put forward those objections. > Top mathematicians don't need that trivial result from the complex > plane to know that I am right. > So I am not that hopeful that the truth will win out just yet based on > my ability to directly show that some of you were dumb enough to fight > the distributive property. > Barry Mazur didn't fight it. Neither did Andrew Granville. And Ralph > McKenzie didn't either. > What puts you in the real bind is the question, is the factoring piece > correct? > I used the same set of equations with which I get my factoring results > to put up the quadratic generator that follows from my reseearch: > a^2 -(1+fx)a + (f^2 x^2 + 2fx) = 0 > And notice the quiet, as when I made changes before, posters leapt at > the chance to put up counterexamples with equations that didn't > follow from my research, but were just being made up in what turns out > to be a naive and clumsy way. > The mathematics here requires a special form for you to remain in what > I call the ring of objects. > And with that form, counterexamples cannot be found, as they would > require refutation of, the distributive property. > So the research overturns over a century of mathematical ideas WITHOUT > the factoring piece, but you people don't seem to like reason, so you > may only respond to force. > The paper by Plotnikov is from 1996. Mathematicians may have > deliberately avoided key mathematical results through all that time in > some short-sighted need to promote themselves or something as I think > about it and wonder exactly what you people have been thinking. > If it does work, then you set up the world for some solution like mine, > but why? > If people had known the system could just be broken then protections > could have been put in place (hopefully they have been, and I am hoping > on that, just in case). > Why set things up as you did? Who has been pulling your strings? > The mystery is still one I'm puzzling over. > James Harris Nothing Harris. You have got nothing. And face it Harris, you will never come up with anything at all. === Subject: Re: JSH: Clock is ticking > Barry Mazur didn't fight it. Neither did Andrew Granville. And Ralph > McKenzie didn't either. Name-dropping again. -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) + 1 = 0 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: JSH: Clock is ticking <7wwMf.61164$697.46866@bignews3.bellsouth.net> Mazur, Granville and McKenzie (MGM, henceforth) are obviously a part of the Math Community Conspiracy to Abrogate the Commutative and Dystributive Properties. maybe even the Associative! --Give Earth a Trickier Dick Cheeny -- out of office, after gigayears! http://tarpley.net/bush12.htm http://www.benfranklinbooks.com/ http://members.tripod.com/~american_almanac http://www.wlym.com/pdf/iclc/howthenation.pdf http://www.rand.org/publications/randreview/issues/rr.12.00/ http://www.rwgrayprojects.com/synergetics/plates/figs/plate01.html === Subject: Re: Clock is ticking > This time, unlike before, years will not go by, if I am right about the > factoring piece of my research, while you can see that I've shutdown > the specious objections that worked on these newsgroups with a result > from the complex plane. That statement does not make any sense! === Subject: Re: JSH: Clock is ticking > This time, unlike before, years will not go by, if I am right about the > factoring piece of my research, while you can see that I've shutdown > the specious objections that worked on these newsgroups with a result > from the complex plane. If? > But I never got objections from outside the newsgroups. Why should you? > No math journal has ever put forward those objections. Uh, you go to them, they don't come to you. > Top mathematicians don't need that trivial result from the complex > plane to know that I am right. > So I am not that hopeful that the truth will win out just yet based on > my ability to directly show that some of you were dumb enough to fight > the distributive property. > Barry Mazur didn't fight it. Neither did Andrew Granville. And Ralph > McKenzie didn't either. They're not fighting it, and they're not supporting it either. > What puts you in the real bind is the question, is the factoring piece > correct? No, of course not. If no effort is being spent either fighting it or supporting it, it's because it's crap that no wants to waste any time or effort on. > I used the same set of equations with which I get my factoring results > to put up the quadratic generator that follows from my reseearch: > a^2 -(1+fx)a + (f^2 x^2 + 2fx) = 0 > And notice the quiet, as when I made changes before, posters leapt at > the chance to put up counterexamples with equations that didn't > follow from my research, but were just being made up in what turns out > to be a naive and clumsy way. > The mathematics here requires a special form for you to remain in what > I call the ring of objects. And everyone else calls bull. > And with that form, counterexamples cannot be found, as they would > require refutation of, the distributive property. Yeah, sure. Prove it. > So the research overturns over a century of mathematical ideas WITHOUT > the factoring piece, but you people don't seem to like reason, so you > may only respond to force. Oh, threatening force are we? Better watch it pal, your ISP may not interpret that properly. > The paper by Plotnikov is from 1996. Mathematicians may have > deliberately avoided key mathematical results through all that time in > some short-sighted need to promote themselves or something as I think > about it and wonder exactly what you people have been thinking. > If it does work, then you set up the world for some solution like mine, > but why? > If people had known the system could just be broken then protections > could have been put in place (hopefully they have been, and I am hoping > on that, just in case). Just in case you're right? I thought you had a proof? > Why set things up as you did? Who has been pulling your strings? > The mystery is still one I'm puzzling over. tard, eh? Surely that's not news to anyone. > James Harris === Subject: Re: JSH: Clock is ticking (snip 90-odd lines) > tard, eh? Surely that's not news to anyone. Phew. For a few heart-stopping moments there I thought you were going to manage a complete tard-free post. Andy -- spargeatbtinternetdotcom I think therefore I am what I eat === Subject: Re: JSH: Clock is ticking <1kvgvmfud69q0.1k5wza431p2bb.dlg@40tude.net (snip 90-odd lines) > tard, eh? Surely that's not news to anyone. > Phew. For a few heart-stopping moments there I thought you were going to > manage a complete tard-free post. It's just a phase I'm going through. I saw a picture of one of those little candy hearts you get at Valentine's day with that printed on it and I haven't been able to stop laughing every time I see it. It doesn't help that I've set that picture as my Windows background. I'll probably tire of it eventually. > Andy > -- > spargeatbtinternetdotcom > I think therefore I am what I eat === Subject: Re: JSH: Clock is ticking > This time, unlike before, years will not go by, if I am right about the > factoring piece of my research, while you can see that I've shutdown > the specious objections that worked on these newsgroups with a result > from the complex plane. > But I never got objections from outside the newsgroups. > No math journal has ever put forward those objections. > Top mathematicians don't need that trivial result from the complex > plane to know that I am right. > So I am not that hopeful that the truth will win out just yet based on > my ability to directly show that some of you were dumb enough to fight > the distributive property. The objections were correct. Nobody fought the distributive property. > Barry Mazur didn't fight it. Neither did Andrew Granville. And Ralph > McKenzie didn't either. > What puts you in the real bind is the question, is the factoring piece > correct? No, it's not. > I used the same set of equations with which I get my factoring results > to put up the quadratic generator that follows from my reseearch: > a^2 -(1+fx)a + (f^2 x^2 + 2fx) = 0 > And notice the quiet, as when I made changes before, posters leapt at > the chance to put up counterexamples with equations that didn't > follow from my research, but were just being made up in what turns out > to be a naive and clumsy way. > The mathematics here requires a special form for you to remain in what > I call the ring of objects. > And with that form, counterexamples cannot be found, as they would > require refutation of, the distributive property. > So the research overturns over a century of mathematical ideas WITHOUT > the factoring piece, but you people don't seem to like reason, so you > may only respond to force. Which accepted theorems have you shown to be false? > The paper by Plotnikov is from 1996. Mathematicians may have > deliberately avoided key mathematical results through all that time in > some short-sighted need to promote themselves or something as I think > about it and wonder exactly what you people have been thinking. > If it does work, then you set up the world for some solution like mine, > but why? If you have a solution to the factoring problem, just publish it. Nothing terrible will happen. RSA cryptosystems will have to be replaced by something else. > If people had known the system could just be broken then protections > could have been put in place (hopefully they have been, and I am hoping > on that, just in case). > Why set things up as you did? Who has been pulling your strings? You're being irrational. If you really have a solution to the factoring problem, we can't stop you from publishing it. Nothing we do will have any bearing on your decision whether to publish or not. It's up to you. We certainly don't have any reason to refrain from making objections to your other work in the good faith belief that they are correct just for fear of the consequences if you publish your other research. We can't stop you from publishing it. > The mystery is still one I'm puzzling over. Your loss of contact with reality is sufficiently severe to warrant seeing a psychiatrist. > James Harris === Subject: Re: JSH: Clock is ticking : This time, unlike before, years will not go by, if I am right about the : factoring piece of my research, while you can see that I've shutdown : the specious objections that worked on these newsgroups with a result : from the complex plane. Is it time for the Hammer? : But I never got objections from outside the newsgroups. Because nobody listens to you other than here. : Top mathematicians don't need that trivial result from the complex : plane to know that I am right. Your latest muddle of algebra you mean? : What puts you in the real bind is the question, is the factoring piece : correct? No it's not, so really, there's no bind. : And with that form, counterexamples cannot be found, as they would : require refutation of, the distributive property. Or rather, refutation of whatever particular condition you happen to pull out of your ass at that particular moment. : Why set things up as you did? Who has been pulling your strings? Short redheads, mostly. Justin === Subject: Re: JSH: Clock is ticking Discussion, linux) > : This time, unlike before, years will not go by, if I am right about the > : factoring piece of my research, while you can see that I've shutdown > : the specious objections that worked on these newsgroups with a result > : from the complex plane. > Is it time for the Hammer? Huh? The hammer has been here almost two weeks. Did you not notice The Hammer has arrived. Arrived a long time ago, but I kept hoping, and trying, to reason with you. Again, I urge you NOT to bother with your savings in any form, like cashing in stocks, or anything at all, as it would be a useless gesture that just shows guilt. You can't escape now. No one can. Oh, wait. Never mind. The Hammer really arrived a year ago. On Feb So now, The Hammer is here, and with it, the end of days. The world will be destroyed, and then remade, as foretold. You will be lost, with your children, and then there will be others, and one day they will be tested, and will pass, but that is another story. So the Hammer has been here over a year ago, the world has been destroyed and remade and you and your children are lost. Jeez, try to keep up, will ya? This is *important* dammit. -- Britney thought the idea of a pre-nup was vile, because she is loved-up with Kevin and cannot envisage breaking up. However, [...] no one in Hollywood these days get married without brokering a deal. [...] She had a long chat with Kevin and he was cool about it. === Subject: Re: JSH: Clock is ticking > : This time, unlike before, years will not go by, if I am right about the > : factoring piece of my research, while you can see that I've shutdown > : the specious objections that worked on these newsgroups with a result > : from the complex plane. > Is it time for the Hammer? (what is the Hammer?) > : But I never got objections from outside the newsgroups. > Because nobody listens to you other than here. we need a pet to play with. > : Top mathematicians don't need that trivial result from the complex > : plane to know that I am right. > Your latest muddle of algebra you mean? (The complex plane is new to him, not a mathematician, nor engineer, nor physicist, as it is called the imaginary plane by those who use it a lot). > : What puts you in the real bind is the question, is the factoring piece > : correct? > No it's not, so really, there's no bind. > : And with that form, counterexamples cannot be found, as they would > : require refutation of, the distributive property. > Or rather, refutation of whatever particular condition you happen to pull > out of your ass at that particular moment. this is bate and switch, JSH never shows a real proof. > : Why set things up as you did? Who has been pulling your strings? > Short redheads, mostly. > Justin Now That is in the real plane. === Subject: Nagging Detail I was browsing through Axler's book Linear Algebra Done Right and I came across a passage I had highlighted and never really understood. I am wondering if anyone can explain it. In chapter 2, Axler is defining the span of a (finite) collection of vectors. In the definition he says, To be consistent, we declare that the span of the empty list { } is {0}. First, I was never sure if by {0} he meant the number zero or the zero vector. Secondly, why does he have to introduce this extra condition about the empty set? TIA, Juno === Subject: Re: Nagging Detail > .... > In chapter 2, Axler is defining the span of a (finite) collection of > vectors. In the definition he says, To be consistent, we declare that > the span of the empty list { } is {0}. > .... why does he have to introduce this extra condition about the > empty set? .... You can't very well form any linear combination of vectors that aren't there. :-) But there's a slightly more subtle view of spanning (or generating in algebras other than vector spaces). I don't know whether Axler's book mentions this. Given a set S of vectors, (1) consider all the subspaces which contain S, then (2) intersect all those subspaces. The result is the span of S. If S is a finite set {v_1, v_2, ..., v_k}, then every subspace containing S must contain all the linear combinations (x_1)(v_1) + (x_2)(v_2) + ... + (x_k)(v_k). But all those combinations form a subspace in their own right, so when you intersect all the subspaces containing S then you do get the span of S in the usual sense. (That may take a bit of thinking about.) Now see what happens when S is empty. _Every_ subspace then contain S, so the span of the empty set S is the intersection of all the subspaces, which is just {the zero vector}. Ken Pledger. === Subject: Re: Nagging Detail >> .... >> In chapter 2, Axler is defining the span of a (finite) collection of >> vectors. In the definition he says, To be consistent, we declare that >> the span of the empty list { } is {0}. >> .... why does he have to introduce this extra condition about the >> empty set? .... > You can't very well form any linear combination of vectors that >aren't there. :-) Oh, but you can. The sum of an empty set is 0 (or more precisely, the 0 of whatever vector space you're working in). This convention allows you to say, e.g., that if sets A and B are disjoint, sum(A) + sum(B) = sum(A union B). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Nagging Detail You can't very well form any linear combination of vectors that >aren't there. :-) > Oh, but you can. The sum of an empty set is 0 (or more precisely, the > 0 of whatever vector space you're working in). This convention allows > you to say, e.g., that if sets A and B are disjoint, > sum(A) + sum(B) = sum(A union B). > Robert Israel israel@math.ubc.ca It might be worth adding that there are other similar conventions around. -- the product of an empty family of elements of a ring is 1 -- the sum of an empty family of elements of a ring is 0 -- the intersection of the empty family of subsets of set S is the whole set S -- the union of the empty family of subsets of S is the empty set etc. === Subject: Re: Nagging Detail He means the set whose only member isthe zero vector. Probably he wants every set to have a span. === Subject: Re: Peano's Axioms from the Field Axioms alone? >> I have been able to show that the equivalent all but one of Peano's > Axioms can be derived from the fields axioms -- namely that no > natural > number has a successor of 1. See: >> http://www.dcproof.com/PeanoFieldAxioms.html > Dan here says he gets Peano arithmetic, or the natural numbers and >> 2+2=4, and so on, from the field axioms instead of an extra axiom or so >> for PA. So, then PA would be less than PA. >> I have already admitted I was mistaken on this. Without the order axioms, >> you could still derive all but one of the PA axioms (see original >> posting). >> Those remaining axioms, however, could refer to a finite field with the >> sucessor function (x+1) eventually looping back to 0 -- something I had >> overlooked. The order axioms prevent this looping back (sorry, I don't >> know >> the proper terminolgy for this), and they require the field to be >> infinite. >> Dan >> Download my DC Proof software at http://www.dcproof.com > Would you agree that a finite proof in Presburger is a finite proof in > Peano? I am not aquainted with the work of Presburger. Googling Presburger, I find: 'Presburger arithmetic' is the first-order theory of the natural numbers with addition. It is not as powerful as the Peano axioms because multiplication is omitted.... Concepts such as divisibility or prime number cannot be formalized in Presburger arithmetic. Source: http://presburger-arithmetic.brainsip.com/ Sorry, but to my untrained eye, it doesn't look all that useful. Not for the applications I have in mind. Dan Download my DC Proof software at http://www.dcproof.com === Subject: Re: Peano's Axioms from the Field Axioms alone? > I have been able to show that the equivalent all but one of Peano's > Axioms can be derived from the fields axioms -- namely that no > natural > number has a successor of 1. See: > http://www.dcproof.com/PeanoFieldAxioms.html >> Dan here says he gets Peano arithmetic, or the natural numbers and >> 2+2=4, and so on, from the field axioms instead of an extra axiom or so >> for PA. So, then PA would be less than PA. > I have already admitted I was mistaken on this. Without the order axioms, >> you could still derive all but one of the PA axioms (see original >> posting). >> Those remaining axioms, however, could refer to a finite field with the >> sucessor function (x+1) eventually looping back to 0 -- something I had >> overlooked. The order axioms prevent this looping back (sorry, I don't >> know >> the proper terminolgy for this), and they require the field to be >> infinite. >> Dan >> Download my DC Proof software at http://www.dcproof.com > Would you agree that a finite proof in Presburger is a finite proof in > Peano? > I am not aquainted with the work of Presburger. Googling Presburger, I > find: > 'Presburger arithmetic' is the first-order theory of the natural numbers > with addition. It is not as powerful as the Peano axioms because > multiplication is omitted.... > Concepts such as divisibility or prime number cannot be formalized in > Presburger arithmetic. > Source: http://presburger-arithmetic.brainsip.com/ > Sorry, but to my untrained eye, it doesn't look all that useful. Not for the > applications I have in mind. > Dan > Download my DC Proof software at http://www.dcproof.com Dan, I still quite agree with you. That has to do with there being true axioms. I wish you luck with your projects, I surely appreciate that you read my comments. We talk about those things quite often. Ross === Subject: Question about universal set I've had some trouble understanding exactly how the ZF Axiom of Separation overcomes the problem of contradiction first pointed out by Russell. Is the following reasoning correct? The axiom states: Ap Az Ey Ax [x e y <=> (x e z & p(x)] Inserting Russell's predicate ~ (x e x) for p(x) one obtains (after simplification) Az Ey [~ (y e y) & ~ (y e z)] So the axiom is contradicted if either 1.- Ay (y e y) or 2.- Ez Ay (y e z) or, in other words z is the universal set. Now 1.- apparently can be shown to be false using other ZF axioms. Is the non-existence of the universal set, then, proved indirectly by its negation of the Axiom of Separation? No text I have come across makes this clear, they all simply state that it overcomes Russell's famous objection. === Subject: Re: Question about universal set 02/26/2006 at 07:33 PM, agapito6314@aol.com said: >I've had some trouble understanding exactly how the ZF Axiom of >Separation overcomes the problem of contradiction first pointed out >by Russell. Because ZF doesn't have a set constructor for arbitrary propositional functions. That doesn't, of course, guaranty that there aren't *other* inconsistencies, but the Russell Paradox can't even be formulated in ZF. >The axiom states: >Ap Az Ey Ax [x e y <=> (x e z & p(x)] No; you can't quantify over propositions. There's an axiom scheme rather than a single axiom. >Inserting Russell's predicate ~ (x e x) for p(x) one obtains >(after simplification) >Az Ey [~ (y e y) & ~ (y e z)] Please show your simplification >So the axiom is contradicted if either >1.- Ay (y e y) or >2.- Ez Ay (y e z) or, in other words z is the universal >set. Indeed. So far nobody has been able to prove either. >Now 1.- apparently can be shown to be false using other ZF axioms. >Is the non-existence of the universal set, then, proved indirectly >by its negation of the Axiom of Separation? Foundation also works. That is not, however, a proof of consistency. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Question about universal set <4403175a$2$fuzhry+tra$mr2ice@news.patriot.net 02/26/2006 > at 07:33 PM, agapito6314@aol.com said: >I've had some trouble understanding exactly how the ZF Axiom of >Separation overcomes the problem of contradiction first pointed out >by Russell. > Because ZF doesn't have a set constructor for arbitrary propositional > functions. That doesn't, of course, guaranty that there aren't *other* > inconsistencies, but the Russell Paradox can't even be formulated in > ZF. >The axiom states: >Ap Az Ey Ax [x e y <=> (x e z & p(x)] > No; you can't quantify over propositions. There's an axiom scheme > rather than a single axiom. >Inserting Russell's predicate ~ (x e x) for p(x) one obtains >(after simplification) >Az Ey [~ (y e y) & ~ (y e z)] > Please show your simplification OK: 1.- [x e y <=> (x e z & ~ (x e x))] 2.- [x e y => (x e z & ~ (x e x))] & [(x e z & ~ (x e x)) => x e y] 3.- [~ (x e y) v (x e z & ~ (x e x))] & [~ (x e z & ~ (x e x)) v x e y] 4.- (~ (x e y) v (x e z)) & (~ (x e y) v ~(x e x)) & [ ~(x e z) v (x e x) v (x e y)] 5.- (~ (y e y) v (y e z)) & (~ (y e y)) & [ ~(y e z) v (y e y)] 6.- ~ (y e y) 7.- ~(y e z) v (y e y) 8.- ~ (y e z) 9.- ~ (y e y) & ~ (y e z) 10.- Az Ey [~ (y e y) & ~ (y e z)] I may have skipped a few steps, but hopefully this is correct. Please === Subject: Re: Question about universal set One can, however, talk about the universe of discourse- the set containing all the objects we are working with in a particular problem. If we are careful about putting sets in sets, we can treat it as a universal set for THIS problem. === Subject: Re: Question about universal set > I've had some trouble understanding exactly how the ZF Axiom of > Separation overcomes the problem of contradiction first pointed out by > Russell. Is the following reasoning correct? > The axiom states: > Ap Az Ey Ax [x e y <=> (x e z & p(x)] > Inserting Russell's predicate ~ (x e x) for p(x) one obtains (after > simplification) > Az Ey [~ (y e y) & ~ (y e z)] > So the axiom is contradicted if either > 1.- Ay (y e y) or > 2.- Ez Ay (y e z) or, in other words z is the universal set. > Now 1.- apparently can be shown to be false using other ZF axioms. Is > the non-existence of the universal set, then, proved indirectly by > its negation of the Axiom of Separation? No text I have come across > makes this clear, they all simply state that it overcomes Russell's > famous objection. Others have answered this, but I would like to point out that there is another way to prove the nonexistence of the universal set that doesn't use the axiom of separation. The idea is based on Cantor's Paradox. Suppose a universal set U exists. Let P(U) be its power set. By Cantor's theorem, there must exist an x in P(U) that is not in U, contrary to the assumption that U is a universal set. When people say that the axiom of separation overcomes Cantor's paradox, I suspect what they really mean is that we can't use just a predicate alone to define a set. That's the assumption that leads to Russell's Paradox. In place of that flawed concept, sometimes called unbounded comprehension, we have instead bounded comprehension, which is another name for the axiom of separation. That means, you need both a predicate AND a bounding set in order to establish the existence of another set using that axiom. Thus, it is not the addition of an axiom (bounded comprehension) that resolves the paradox, but the subtraction of an axiom (unbounded comprehension). -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Question about universal set another way to prove the nonexistence of the universal set that doesn't > use the axiom of separation. The idea is based on Cantor's Paradox. > Suppose a universal set U exists. Let P(U) be its power set. By > Cantor's theorem, there must exist an x in P(U) that is not in U, > contrary to the assumption that U is a universal set. Ouch. Here's simplest proof. Assume (Ax) x in u Then u in u contradicts regularity. === Subject: Re: Question about universal set >> Others have answered this, but I would like to point out that there is >> another way to prove the nonexistence of the universal set that doesn't >> use the axiom of separation. The idea is based on Cantor's Paradox. >> Suppose a universal set U exists. Let P(U) be its power set. By >> Cantor's theorem, there must exist an x in P(U) that is not in U, >> contrary to the assumption that U is a universal set. > Ouch. Here's simplest proof. Assume > (Ax) x in u > Then > u in u > contradicts regularity. That's pretty simple, but not everyone includes regularity among the axioms of ZF. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: Question about universal set axioms of ZF. ZF is a particular set of axioms, which does include regularity. If you use the initials ZF, people will assume that you mean the axioms in the front of Jech's book, or equivalently the front of Kunen's book, but without choice. In particular, these axioms include regularity and repacement. The system Z is ZF without replacement, ZF^{-} is ZF without regularity, and Z^{-} is Z without regularity. It is very true that not everyone includes regularity among their axioms of set theory. Moreover, the existence of a universal set is incompatible with both the regularity axiom (with no help) and the powerset axiom (with the help of enough other axioms to prove Cantor's theorem and get a contradiction, but without regularity). So the existence of a universal set is incompatible with Z^{-}. Clearly, this last fact cannot be proved by showing that a universal set would contradict regularity. In fact, the existence of a universal set is incompatible with ZF^{-} - Powerset. The proof is via the Burali-Forti paradox. === Subject: Re: Question about universal set I claimed that the existence of a universal set is incompatible with ZF^{-} - Powerset, with a proof using the Burali-Forti paradox. I should make it clear that you might have to add the axiom that for any two sets A,B the set of all pairs (a,b) such that a in A and b in B exists. This is a common assumption in settings such as KP where you don't have powerset, to make sure that you can define functions. === Subject: Re: Question about universal set days. My association with the Department is that of an alumnus. >I claimed that the existence of a universal set is incompatible with >ZF^{-} - Powerset, with a proof using the Burali-Forti paradox. I >should make it clear that you might have to add the axiom that for any >two sets A,B the set of all pairs (a,b) such that a in A and b in B >exists. This is a common assumption in settings such as KP where you >don't have powerset, to make sure that you can define functions. Is there a problem with the argument I gave using separation? I thought it proved that Ax (Ey ~(y in x)) and that this establishes the nonexistence of a universal set. But perhaps there is some forma subtlety I don't know about, in which case I would like to know it. Or perhaps it is that it is a proof by contradiction? (The argument is: given x, let y = { z in x : z not in z}; then (y in x) -> ( (y in y)->(y not in y)) & (y not in y)->(y in y) So (y in x) -> (y not in y) and (y in y) hence y not in x. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Question about universal set (The argument is: given x, let y = { z in x : z not in z}; then >(y in x) -> ( (y in y)->(y not in y)) & (y not in y)->(y in y) >So (y in x) -> (y not in y) and (y in y) There is no problem, I'm sorry I didn't see your post. That is much easier than the proof I had in mind. Somehow I thought that the proof you use requires regularity, but your description shows that it only requires separation and basic logic; no other axioms are required. === Subject: Re: Question about universal set Others have answered this, but I would like to point out that there is > another way to prove the nonexistence of the universal set that doesn't > use the axiom of separation. The idea is based on Cantor's Paradox. > Suppose a universal set U exists. Let P(U) be its power set. By > Cantor's theorem, there must exist an x in P(U) that is not in U, > contrary to the assumption that U is a universal set. > Ouch. Here's simplest proof. Assume > (Ax) x in u > Then > u in u > contradicts regularity. That is a lot simpler! I guess the Axiom of Separation is still replied. === Subject: Re: Question about universal set we have instead bounded comprehension, which is another name for the axiom >of separation. Maybe you found the term bounded comprehension in a book, but your definition is not the one that is used in the contemporary literature. I have seen the term ``restricted separation''. In my mind, bounded comprehension is the kind found in Kripke-Platek set theory (KP), a subtheory of ZFC that is important for studying definability. Essentially, bounded comprehension is the comprehension scheme is limited to fomulas with only bounded quantifiers, i.e. quantifiers of the form (exists x in z) and (forall x in z). bounded separation' with double quotes you will see what I mean. You can look this up in any book on admissible sets. There is a list of axioms for KP at http://en.wikipedia.org/wiki/Kripke-Platek_set_theory . That page accurately labels bounded comprehension Sigma_0 comprehension. === Subject: Re: Question about universal set >>In place of that flawed concept, sometimes called unbounded comprehension, >>we have instead bounded comprehension, which is another name for the axiom >>of separation. > Maybe you found the term bounded comprehension in a book, but your > definition is not the one that is used in the contemporary literature. > I have seen the term ``restricted separation''. > In my mind, bounded comprehension is the kind found in Kripke-Platek > set theory (KP), a subtheory of ZFC that is important for studying > definability. Essentially, bounded comprehension is the comprehension > scheme is limited to fomulas with only bounded quantifiers, i.e. > quantifiers of the form (exists x in z) and (forall x in z). > bounded separation' with double quotes you will see what I mean. > You can look this up in any book on admissible sets. There is a list > of axioms for KP at > http://en.wikipedia.org/wiki/Kripke-Platek_set_theory . That page > accurately labels bounded comprehension Sigma_0 comprehension. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: Question about universal set days. My association with the Department is that of an alumnus. >I've had some trouble understanding exactly how the ZF Axiom of >Separation overcomes the problem of contradiction first pointed out by >Russell. Is the following reasoning correct? [.snip.] As others have pointed out, you basically have it. >Is the non-existence of the universal set, then, proved indirectly by >its negation of the Axiom of Separation? No text I have come across >makes this clear, they all simply state that it overcomes Russell's >famous objection. When I first learned it, my teacher made a point of explaining exactly how. Namely, one can use the Axiom of Separation to prove that: For any set X, there exists a set A such that A is not an element of X. The proof is: given X, let A = { x in X : x is not an element of x}. By separation this is a set. If A is in X, then we have two cases: A is an element of A; then A is not in A. Since [p->~p]->~p, this means that A is not an element of A. But if A is not an element of A, then A is an element of A. Again, this means that A is an element of A. The contradiction arises from assuming that A is in X; hence A is not an element of X. Thus, there exists at least one set A which is not an element of X. You will see that we are in fact using Russell's Paradox together with separation to deduce that there is no universal set. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Question about universal set > I've had some trouble understanding exactly how the ZF Axiom of > Separation overcomes the problem of contradiction first pointed out by > Russell. Is the following reasoning correct? > The axiom states: > Ap Az Ey Ax [x e y <=> (x e z & p(x)] > Inserting Russell's predicate ~ (x e x) for p(x) one obtains (after > simplification) > Az Ey [~ (y e y) & ~ (y e z)] > So the axiom is contradicted if either > 1.- Ay (y e y) or > 2.- Ez Ay (y e z) or, in other words z is the universal set. > Now 1.- apparently can be shown to be false using other ZF axioms. Is > the non-existence of the universal set, then, proved indirectly by > its negation of the Axiom of Separation? No text I have come across > makes this clear, they all simply state that it overcomes Russell's > famous objection. Yes ,absolutely. if there was a set z satisfying Ax[xez] then your axiom of separation reduces to the axiom of abstraction (by specification for z as the universal set) ,namely ApEyAx[xey<=> p(x)} which was shown by Russells's example to be a contradiction which is what led to it's replacement by the axiom or === Subject: Re: Question about universal set > I've had some trouble understanding exactly how the ZF Axiom of > Separation overcomes the problem of contradiction first pointed out by > Russell. Is the following reasoning correct? > The axiom states: > Ap Az Ey Ax [x e y <=> (x e z & p(x)] > Inserting Russell's predicate ~ (x e x) for p(x) one obtains (after > simplification) > Az Ey [~ (y e y) & ~ (y e z)] > So the axiom is contradicted if either > 1.- Ay (y e y) or > 2.- Ez Ay (y e z) or, in other words z is the universal set. > Now 1.- apparently can be shown to be false using other ZF axioms. Is > the non-existence of the universal set, then, proved indirectly by > its negation of the Axiom of Separation? Yes. If the universal set existed, Russell's set would exist by the Axiom of Separation, and that would lead to contradiction. > No text I have come across > makes this clear, they all simply state that it overcomes Russell's > famous objection. === Subject: Re: Stats 'measures' === Subject: Re: Stats 'measures' >Hey all, >Since it is possible to have integration with respect to any measure >and seemingly, all that is used in stats is the Lebesque measure, is >any other measures used (in terms of application) in statistics or This is absolute nonsense. A probability measure is a measure with the whole space having measure 1, and there are lots of these on the real line even, NOT including Lebesgue measure. I consider it to be a great mistake for a measure theory course to start with Lebesgue measure; everything is discrete or limits of discrete, and there are lots of limits, including Lebesgue. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Complex plane, weird reality > Ok, so I can just go to the complex plane to prove a key result, and > show how it follows from the distributive property, where I have to go > to the complex plane to remove all that factor crap, where people will > use the ring of algebraic integers, which I prove has a problem, to say > they've refuted my arguments proving the problem. That statement makes no sense, you do not a firm grip on solving problems. === Subject: A semidefinite problem Hi all, Do you know or have any references on how to solve the following semidefinite problem: Let f be a positive definite function. We have some data f(x) corresponding to different x. The question is to find a positive definite matrix R that minimizes the error ||f(x)-x^T R x||. So the problem is min_{R>0}||f(x)-x^TRx|| with f positive and x some samples. Best. === Subject: Re: A semidefinite problem NB: The function f is not known and we just have some samples corresponding to different x. === Subject: Re: Cracking 7 digit passwords > Hi is Corey. I've gone as far back into alt.tarot as ..well, a long > assed time ago. He has givezn more than just hif ull name and a few > other tidbits in there as well. > Like when he posted his philosophy notes from his prof, called them > his > own, got called out for plagiating, created socks saying *this* corey > is the teacher, and a student must have broken into my computer and > posted to USENET. > I'm not naming names, but at least two people from this group were > there at the time. > Corey's other persona's on alt.tarot, before the above-mentioned > happened, were lucid as well. With the penchant for poetry. > Corey on alt.tarot posted his phone number. Backtracing it with Google > had it listed as the same number Hi used. Corey on alt.tarot also said > he was a carpenter. This is all correct. But Corey White is not Corey Conner. -- Asiya ********** http://www.asiya.org/ === Subject: An Incomplete (87/92) Introduction to the Johnson Solids There are many web sites around (well, at least a handful) which have computer-drawn images of all 92 Johnson solids. These are the possible polyhedra with the following characteristics: - they are not Platonic solids, Archimedean solids, prisms or antiprisms; - all their faces are regular polygons; and - they are convex. On *my* page, I have *hand-drawn* (using a paint program, and making use of copy and paste features, so I have had some computer assistance...) pictures of 87 of the Johnson solids. More importantly than the fact the pictures are drawn lovingly by hand, the solids are introduced gradually, with explanations of how they relate to the Archimedian solids and to each other, so that for each solid depicted, it is possible to *understand* the shape that it has. The page is at http://www.quadibloc.com/math/acs03.htm John Savard http://www.quadibloc.com/index.html _________________________________________ Usenet Zone Free Binaries Usenet Server More than 140,000 groups Unlimited download http://www.usenetzone.com to open account === Subject: Re: JSH: So now you know as I found out > I have contacted top mathematicians including Barry Mazur and Andrew > Granville. For someone who claims not to place any importance on social stuff, you certainly enjoy dropping the names of top mathematicians. -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) + 1 = 0 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: JSH: So now you know as I found out > So I abstracted out the objections that have been using now for years > and showed they are violations of the distributive property as I > utilize the following result valid on the complex plane: > Given > 7C(x) = (A(x) + 7)(B(x) + 1) > where A(0) = B(0) = 0 > by the distributive property, 7 multiplied through the first factor, > and remember the distributive property is just > a(b+c) = ab + ac > so in this case the ac = 7, so ab=A(x), and knowing that a=7, you have > the ONLY way that the 7 multiplied through. Ok here is where I provide a counterexample and you ignore it. Let w1(x) = 7/(x^2 +1), w2(x) = (x^2 +1), A'(x) = (x^2+1)A(x)/7, B'(x) = B(x)/(x^2+1) Then (A(x) + 7) = w1(x)(A'(x) + (x^2+1)) and (B(x) + 1) = w2(x)(B'(x) + 1/(x^2+1)) So 7C(x) = w1(x)( A(x)/w1(x) + 7/w1(x)) * w2(x)(B(x)/w2(x) + 1/w2(x)) and we see that there is more than one way for the 7 to be multiplied through. -William Hughes === Subject: Re: JSH: So now you know as I found out > So I abstracted out the objections that have been using now for years > and showed they are violations of the distributive property as I > utilize the following result valid on the complex plane: > Given > 7C(x) = (A(x) + 7)(B(x) + 1) > where A(0) = B(0) = 0 > by the distributive property, 7 multiplied through the first factor, > and remember the distributive property is just > a(b+c) = ab + ac > so in this case the ac = 7, so ab=A(x), and knowing that a=7, you have > the ONLY way that the 7 multiplied through. > Ok here is where I provide a counterexample and you ignore it. > Let w1(x) = 7/(x^2 +1), w2(x) = (x^2 +1), A'(x) = (x^2+1)A(x)/7, > B'(x) = B(x)/(x^2+1) > Then (A(x) + 7) = w1(x)(A'(x) + (x^2+1)) > and (B(x) + 1) = w2(x)(B'(x) + 1/(x^2+1)) > So 7C(x) = w1(x)( A(x)/w1(x) + 7/w1(x)) * w2(x)(B(x)/w2(x) + 1/w2(x)) > and we see that there is more than one way for the 7 to > be multiplied through. You've obscured constant terms, that's all. Now then, go back to your own example, and use functions that equal 0 when x=0. Then you'll find that the result holds. Hell, anyone can shift functions so that you obscure the constants, I'll do it an easier way: 7C(x) = (A'(x) + 3532)(B'(x) + 245453) where A'(x) and B'(x) don't equal 0 at x=0. Remember, the complex plane is kind of well-known and people other than number theorists are aware of it, and feel competent within it. The proof you are trying to challenge is as follows: In the complex plane, given 7C(x) = (A(x) + 7)(B(x) + 1) true for all x, where A(0) = B(0) = 0 let C(x) = (A'(x) + 1)(B'(x) + 1) where A'(0) = B'(0) = 0 and making that substitution, gives 7(A'(x) + 1)(B'(x) + 1) = (A(x) + 7)(B(x) + 1) and by the distributive property A(x) = 7A'(x) and B'(x) = B(x). The fact that the function equal 0 at x=0 is a crucial piece. Notice the above is a proof. It is a trivial proof that just about any mathematician can verify for you, if you need that done. I suggest you just copy it out, send it to someone and see what I mean. They may make notational changes as well as change some of the description, but they can verify for you that it is a proof and the result does hold over the complex plane. James Harris === Subject: Re: JSH: So now you know as I found out > So I abstracted out the objections that have been using now for years > and showed they are violations of the distributive property as I > utilize the following result valid on the complex plane: > Given > 7C(x) = (A(x) + 7)(B(x) + 1) > where A(0) = B(0) = 0 > by the distributive property, 7 multiplied through the first factor, > and remember the distributive property is just > a(b+c) = ab + ac > so in this case the ac = 7, so ab=A(x), and knowing that a=7, you have > the ONLY way that the 7 multiplied through. > Ok here is where I provide a counterexample and you ignore it. > Let w1(x) = 7/(x^2 +1), w2(x) = (x^2 +1), A'(x) = (x^2+1)A(x)/7, > B'(x) = B(x)/(x^2+1) > Then (A(x) + 7) = w1(x)(A'(x) + (x^2+1)) > and (B(x) + 1) = w2(x)(B'(x) + 1/(x^2+1)) > So 7C(x) = w1(x)( A(x)/w1(x) + 7/w1(x)) * w2(x)(B(x)/w2(x) + 1/w2(x)) > and we see that there is more than one way for the 7 to > be multiplied through. > You've obscured constant terms, that's all. As you have pointed out yourself the constant terms do not change as long as w1(0) = 7 and w2(0)=1. So which statement are you making 1. w1(x) =7 and w2(x) =1 2. My algebra is wrong If you don't make either of those statments you must agree that there is more than one way for the 7 to be multiplied through. -William Hughes === Subject: Re: JSH: So now you know as I found out > So I abstracted out the objections that have been using now for years > and showed they are violations of the distributive property as I > utilize the following result valid on the complex plane: > Given > 7C(x) = (A(x) + 7)(B(x) + 1) > where A(0) = B(0) = 0 > by the distributive property, 7 multiplied through the first factor, > and remember the distributive property is just > a(b+c) = ab + ac > so in this case the ac = 7, so ab=A(x), and knowing that a=7, you have > the ONLY way that the 7 multiplied through. > Ok here is where I provide a counterexample and you ignore it. > Let w1(x) = 7/(x^2 +1), w2(x) = (x^2 +1), A'(x) = (x^2+1)A(x)/7, > B'(x) = B(x)/(x^2+1) > Then (A(x) + 7) = w1(x)(A'(x) + (x^2+1)) > and (B(x) + 1) = w2(x)(B'(x) + 1/(x^2+1)) > So 7C(x) = w1(x)( A(x)/w1(x) + 7/w1(x)) * w2(x)(B(x)/w2(x) + 1/w2(x)) > and we see that there is more than one way for the 7 to > be multiplied through. > You've obscured constant terms, that's all. > As you have pointed out yourself the constant > terms do not change as long as w1(0) = 7 and w2(0)=1. > So which statement are you making > 1. w1(x) =7 and w2(x) =1 > 2. My algebra is wrong > If you don't make either of those statments you must agree > that there is more than > one way for the 7 to be multiplied through. > -William Hughes Dude, it's not complicated. You need functions that go to 0 when x=0. If you have functions that do not then you can prove all kinds of things with just about any function you want, but that doesn't change the distributive property. Hey, to help you out, I zero'd out the w's in a post of my own: http://mymath.blogspot.com/2006/01/multiplying-out.html Go see how it's done and notice the result. Also, remember the result I give holds in the complex plane, which you can verify with just about any competent mathematician, and I'll give the proof again so if you wish you can copy it out and email one! The proof you are trying to challenge is as follows: In the complex plane, given 7C(x) = (A(x) + 7)(B(x) + 1) true for all x, where A(0) = B(0) = 0 let C(x) = (A'(x) + 1)(B'(x) + 1) where A'(0) = B'(0) = 0 and making that substitution, gives 7(A'(x) + 1)(B'(x) + 1) = (A(x) + 7)(B(x) + 1) and by the distributive property A(x) = 7A'(x) and B'(x) = B(x). The fact that the functions equal 0 at x=0 is a crucial piece. For those wondering about the subject line of this thread, my point is that now you know what I know, which is that these people arguing with me, are fighting some rather basic algebra, and top mathematicians wouldn't be caught on this point. But notice as I explain that it doesn't matter. For my results to be held back, as simple as the algebra is that proves them, mathematicians MUST be deliberately avoiding results they don't like. There's just no other explanation. So now you know, and you know how I found out, I realized that my results depend on the distributive property. James Harris === Subject: Re: JSH: So now you know as I found out > So I abstracted out the objections that have been using now for years > and showed they are violations of the distributive property as I > utilize the following result valid on the complex plane: > Given > 7C(x) = (A(x) + 7)(B(x) + 1) > where A(0) = B(0) = 0 > by the distributive property, 7 multiplied through the first factor, > and remember the distributive property is just > a(b+c) = ab + ac > so in this case the ac = 7, so ab=A(x), and knowing that a=7, you have > the ONLY way that the 7 multiplied through. > Ok here is where I provide a counterexample and you ignore it. > Let w1(x) = 7/(x^2 +1), w2(x) = (x^2 +1), A'(x) = (x^2+1)A(x)/7, > B'(x) = B(x)/(x^2+1) > Then (A(x) + 7) = w1(x)(A'(x) + (x^2+1)) > and (B(x) + 1) = w2(x)(B'(x) + 1/(x^2+1)) > So 7C(x) = w1(x)( A(x)/w1(x) + 7/w1(x)) * w2(x)(B(x)/w2(x) + 1/w2(x)) > and we see that there is more than one way for the 7 to > be multiplied through. > You've obscured constant terms, that's all. > As you have pointed out yourself the constant > terms do not change as long as w1(0) = 7 and w2(0)=1. > So which statement are you making > 1. w1(x) =7 and w2(x) =1 > 2. My algebra is wrong > If you don't make either of those statments you must agree > that there is more than > one way for the 7 to be multiplied through. > -William Hughes > Dude, it's not complicated. You need functions that go to 0 when x=0. Actually it is even simpler than this. You only need functions that go to 0 when x=0 if you want to look at the constant terms. But we know the constant terms have not changed. Either my example, which uses only simple algebra, is correct, in which case there is more than one way for the 7 to be multiplied through, or it is not correct. I suggest you look at the example and decide if it is correct or not. -William Hughes === Subject: Re: JSH: So now you know as I found out > So I abstracted out the objections that have been using now for years > and showed they are violations of the distributive property as I > utilize the following result valid on the complex plane: > Given > 7C(x) = (A(x) + 7)(B(x) + 1) > where A(0) = B(0) = 0 > by the distributive property, 7 multiplied through the first factor, > and remember the distributive property is just > a(b+c) = ab + ac > so in this case the ac = 7, so ab=A(x), and knowing that a=7, you have > the ONLY way that the 7 multiplied through. > Ok here is where I provide a counterexample and you ignore it. > Let w1(x) = 7/(x^2 +1), w2(x) = (x^2 +1), A'(x) = (x^2+1)A(x)/7, > B'(x) = B(x)/(x^2+1) > Then (A(x) + 7) = w1(x)(A'(x) + (x^2+1)) > and (B(x) + 1) = w2(x)(B'(x) + 1/(x^2+1)) > So 7C(x) = w1(x)( A(x)/w1(x) + 7/w1(x)) * w2(x)(B(x)/w2(x) + 1/w2(x)) > and we see that there is more than one way for the 7 to > be multiplied through. > You've obscured constant terms, that's all. > As you have pointed out yourself the constant > terms do not change as long as w1(0) = 7 and w2(0)=1. > So which statement are you making > 1. w1(x) =7 and w2(x) =1 > 2. My algebra is wrong > If you don't make either of those statments you must agree > that there is more than > one way for the 7 to be multiplied through. > -William Hughes > Dude, it's not complicated. You need functions that go to 0 when x=0. > Actually it is even simpler than this. > You only need functions that go to 0 when x=0 if you > want to look at the constant terms. But we know > the constant terms have not changed. Either my example, which > uses only simple algebra, is correct, in which case there > is more than one way for the 7 to be multiplied through, or it > is not correct. I suggest you look at the example and decide > if it is correct or not. > -William Hughes No. You need functions that go to 0 when x=0 because that's how you know what the constant terms are, and the result holds because the constant terms are independent of the value of x. Since that's why the result holds, there is no way to prove the result with functions that don't go to 0, and no way to disprove it either, as it is a proof. You deleted out the freaking proof, so I'm copying it in again so those who do know something about the complex plane can see what I mean, and how easy it is: The proof you are trying to challenge is as follows: In the complex plane, given 7C(x) = (A(x) + 7)(B(x) + 1) true for all x, where A(0) = B(0) = 0 let C(x) = (A'(x) + 1)(B'(x) + 1) where A'(0) = B'(0) = 0 and making that substitution, gives 7(A'(x) + 1)(B'(x) + 1) = (A(x) + 7)(B(x) + 1) and by the distributive property A(x) = 7A'(x) and B'(x) = B(x). The fact that the functions equal 0 at x=0 is a crucial piece. Why don't you just email some mathematician with the proof and ask them? It might save some time. You are in denial. Email someone you respect with the proof, and see if they don't agree that it must hold in the complex plane by the distributive property, and functions going to 0 when x=0 are crucial to the proof. Go ahead. Ask someone. Quit running around inside your own head trying to deny the mathematics. Check with someone else. James Harris === Subject: Re: JSH: So now you know as I found out ... > Ok here is where I provide a counterexample and you ignore it. > > Let w1(x) = 7/(x^2 +1), w2(x) = (x^2 +1), A'(x) = (x^2+1)A(x)/7, > B'(x) = B(x)/(x^2+1) > > Then (A(x) + 7) = w1(x)(A'(x) + (x^2+1)) > and (B(x) + 1) = w2(x)(B'(x) + 1/(x^2+1)) > > So 7C(x) = w1(x)( A(x)/w1(x) + 7/w1(x)) * w2(x)(B(x)/w2(x) + 1/w2(x)) > > and we see that there is more than one way for the 7 to > be multiplied through. ... > No. You need functions that go to 0 when x=0 because that's how you > know what the constant terms are, and the result holds because the > constant terms are independent of the value of x. James, which of the functions A(x), A'(x), B(x) and B'(x) is not 0 when x = 0 in Williams example above? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: So now you know as I found out > So I abstracted out the objections that have been using now for years > and showed they are violations of the distributive property as I > utilize the following result valid on the complex plane: > Given > 7C(x) = (A(x) + 7)(B(x) + 1) > where A(0) = B(0) = 0 > by the distributive property, 7 multiplied through the first factor, > and remember the distributive property is just > a(b+c) = ab + ac > so in this case the ac = 7, so ab=A(x), and knowing that a=7, you have > the ONLY way that the 7 multiplied through. > Ok here is where I provide a counterexample and you ignore it. > Let w1(x) = 7/(x^2 +1), w2(x) = (x^2 +1), A'(x) = (x^2+1)A(x)/7, > B'(x) = B(x)/(x^2+1) > Then (A(x) + 7) = w1(x)(A'(x) + (x^2+1)) > and (B(x) + 1) = w2(x)(B'(x) + 1/(x^2+1)) > So 7C(x) = w1(x)( A(x)/w1(x) + 7/w1(x)) * w2(x)(B(x)/w2(x) + 1/w2(x)) > and we see that there is more than one way for the 7 to > be multiplied through. > You've obscured constant terms, that's all. > As you have pointed out yourself the constant > terms do not change as long as w1(0) = 7 and w2(0)=1. > So which statement are you making > 1. w1(x) =7 and w2(x) =1 > 2. My algebra is wrong > If you don't make either of those statments you must agree > that there is more than > one way for the 7 to be multiplied through. > -William Hughes > Dude, it's not complicated. You need functions that go to 0 when x=0. > Actually it is even simpler than this. > You only need functions that go to 0 when x=0 if you > want to look at the constant terms. But we know > the constant terms have not changed. Either my example, which > uses only simple algebra, is correct, in which case there > is more than one way for the 7 to be multiplied through, or it > is not correct. I suggest you look at the example and decide > if it is correct or not. > -William Hughes > No. You need functions that go to 0 when x=0 because that's how you > know what the constant terms are, and the result holds because the > constant terms are independent of the value of x. We know what the constant terms are. However as you insist. Let w1(x) = 7/(x^2 +1), w2(x) = (x^2 +1), A'(x) = (x^2+1)A(x)/7, B'(x) = B(x)/(x^2+1) Then (A(x) + 7) = w1(x)(A'(x) + (x^2+1)) and (B(x) + 1) = w2(x)(B'(x) + 1/(x^2+1)) Your statement is You need functions that go to 0 when x=0 . Let D(x) = A(x)/w1(x) + 7/w1(x) -7 and E(x) = B(x)/w2(x) + 1/w2(x) -1 Then D(0) = 0 and E(0)=0. and 7C(x) = w1(x)( D(x) +7) * w2(x)(E(x) +1) Now since w1(x) does not equal 7 and w2(x) does not equal 1, there is more than one way for the 7 to be multiplied through. -William Hughes === Subject: Re: JSH: So now you know as I found out > So I abstracted out the objections that have been using now for years > and showed they are violations of the distributive property as I > utilize the following result valid on the complex plane: > Given > 7C(x) = (A(x) + 7)(B(x) + 1) > where A(0) = B(0) = 0 > by the distributive property, 7 multiplied through the first factor, > and remember the distributive property is just > a(b+c) = ab + ac > so in this case the ac = 7, so ab=A(x), and knowing that a=7, you have > the ONLY way that the 7 multiplied through. > Ok here is where I provide a counterexample and you ignore it. > Let w1(x) = 7/(x^2 +1), w2(x) = (x^2 +1), A'(x) = (x^2+1)A(x)/7, > B'(x) = B(x)/(x^2+1) > Then (A(x) + 7) = w1(x)(A'(x) + (x^2+1)) > and (B(x) + 1) = w2(x)(B'(x) + 1/(x^2+1)) > So 7C(x) = w1(x)( A(x)/w1(x) + 7/w1(x)) * w2(x)(B(x)/w2(x) + 1/w2(x)) > and we see that there is more than one way for the 7 to > be multiplied through. > You've obscured constant terms, that's all. > As you have pointed out yourself the constant > terms do not change as long as w1(0) = 7 and w2(0)=1. > So which statement are you making > 1. w1(x) =7 and w2(x) =1 > 2. My algebra is wrong > If you don't make either of those statments you must agree > that there is more than > one way for the 7 to be multiplied through. > -William Hughes > Dude, it's not complicated. You need functions that go to 0 when x=0. > Actually it is even simpler than this. > You only need functions that go to 0 when x=0 if you > want to look at the constant terms. But we know > the constant terms have not changed. Either my example, which > uses only simple algebra, is correct, in which case there > is more than one way for the 7 to be multiplied through, or it > is not correct. I suggest you look at the example and decide > if it is correct or not. > -William Hughes > No. You need functions that go to 0 when x=0 because that's how you > know what the constant terms are, and the result holds because the > constant terms are independent of the value of x. > We know what the constant terms are. However as you insist. > Let w1(x) = 7/(x^2 +1), w2(x) = (x^2 +1), A'(x) = (x^2+1)A(x)/7, > B'(x) = B(x)/(x^2+1) > Then (A(x) + 7) = w1(x)(A'(x) + (x^2+1)) > and (B(x) + 1) = w2(x)(B'(x) + 1/(x^2+1)) > Your statement is You need functions that go to 0 when x=0 . > Let D(x) = A(x)/w1(x) + 7/w1(x) -7 and E(x) = B(x)/w2(x) + 1/w2(x) -1 > Then D(0) = 0 and E(0)=0. > and > 7C(x) = w1(x)( D(x) +7) * w2(x)(E(x) +1) > Now since w1(x) does not equal 7 and w2(x) does not equal 1, > there is more than one way for the 7 to be multiplied through. Increment my Ooops counter. This should of course be Your statement is You need functions that go to 0 when x=0 . Let D(x) = A(x)/w1(x) + 7/w1(x) -1 and E(x) = B(x)/w2(x) + 1/w2(x) -1 Then D(0) = 0 and E(0)=0. and 7C(x) = w1(x)( D(x) +1) * w2(x)(E(x) +1) Now since w1(x) does not equal 7 and w2(x) does not equal 1, there is more than one way for the 7 to be multiplied through. -William Hughes === Subject: Re: JSH: So now you know as I found out > So I abstracted out the objections that have been using now for years > and showed they are violations of the distributive property as I > utilize the following result valid on the complex plane: > Given > 7C(x) = (A(x) + 7)(B(x) + 1) > where A(0) = B(0) = 0 > by the distributive property, 7 multiplied through the first factor, > and remember the distributive property is just > a(b+c) = ab + ac > so in this case the ac = 7, so ab=A(x), and knowing that a=7, you have > the ONLY way that the 7 multiplied through. > Ok here is where I provide a counterexample and you ignore it. > Let w1(x) = 7/(x^2 +1), w2(x) = (x^2 +1), A'(x) = (x^2+1)A(x)/7, > B'(x) = B(x)/(x^2+1) > Then (A(x) + 7) = w1(x)(A'(x) + (x^2+1)) > and (B(x) + 1) = w2(x)(B'(x) + 1/(x^2+1)) > So 7C(x) = w1(x)( A(x)/w1(x) + 7/w1(x)) * w2(x)(B(x)/w2(x) + 1/w2(x)) > and we see that there is more than one way for the 7 to > be multiplied through. > You've obscured constant terms, that's all. > As you have pointed out yourself the constant > terms do not change as long as w1(0) = 7 and w2(0)=1. > So which statement are you making > 1. w1(x) =7 and w2(x) =1 > 2. My algebra is wrong > If you don't make either of those statments you must agree > that there is more than > one way for the 7 to be multiplied through. > -William Hughes > Dude, it's not complicated. You need functions that go to 0 when x=0. Actually it is even simpler than this. You only need functions that go to 0 when x=0 if you want to look at the constant terms. But we know the constant terms have not changed. Either my example, which uses only simple algebra, is correct, in which case there is more than one way for the 7 to be multiplied through, or it is not correct. I suggest you look at the example and decide if it is correct or not. -William Hughes === Subject: Re: JSH: So now you know as I found out > Here's a more detailed proof: > In the complex plane, given > 7C(x) = (A(x) + 7)(B(x) + 1) > true for all x, where A(0) = B(0) = 0 > let > C(x) = (A'(x) + 1)(B'(x) + 1) I clearly see that B'(x) = B(x) and that A'(x) = A(x) / 7. > A(x) = 7A'(x) and B'(x) = B(x). Yes, indeed, what is the problem? There is no showing that A(x) is divisible in the ring of algebraic integers at all. As you are in the complex plane, I see no problem. Set A(x) = 7.exp(x) + 7, B(x) = sin(x) and C(x) = exp(x).sin(x). No problem. > I've added that the relationship is true for all x to remove any area > of confusion, as I'm talking about factoring results, like > > x^2 + 3x + 2 = (x + 1)(x + 2) > > is true for all x. Yes, but you take comclusions about divisibility. But divisibility means that you are working in a ring, otherwise the term is meaningless. In the above example, when you are working in the ring of integers, we can take C(x) = (x^2 + 3x + 2)/2, A(x) = x and B(x) = x. All of A, B and C are integer functions when integer arguments are given, and we have that: 2C(x) = (A(x) + 1)(B(x) + 2), but that does *not* imply that B(x) is divisible by 2 for all x. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: So now you know as I found out ... > Then attack the argument, which is that given in the complex plane > 7C(x) = (A(x) + 7)(B(x) + 1) > true for all x, where A(0) = B(0) = 0 > by the distributive property, it must be true that 7 multiplied through > the first factor. Some more argument. In the complex plane: C(x) = (A(x)/7 + 1)(B(x) + 1) = (A(x) + 7)(B(x)/7 + 1/7) = .... -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: So now you know as I found out > So I abstracted out the objections that have been using now for years > and showed they are violations of the distributive property as I > utilize the following result valid on the complex plane: > Given > 7C(x) = (A(x) + 7)(B(x) + 1) > where A(0) = B(0) = 0 > by the distributive property, 7 multiplied through the first factor, > and remember the distributive property is just > a(b+c) = ab + ac > so in this case the ac = 7, so ab=A(x), and knowing that a=7, you have > the ONLY way that the 7 multiplied through. > The distributive property offers no other possibility, and again, note > that the result is valid on the complex plane. > What you are finding out with replies where posters dodge that basic > result in replies is that these people are deliberately avoiding the > truth. > That's what you need to understand fully to understand that the process > is broken. > I had a key paper published. Some emails got it retracted. > I have contacted top mathematicians including Barry Mazur and Andrew > Granville. > The problem is the result overturns mathematical ideas up to this point > in time key to these people's careers. > So now you know, they are lying. If you believe that any mathematician > worth a damn could question the result in the complex plane that given > 7C(x) = (A(x) + 7)(B(x) + 1) > where A(0) = B(0) = 0 > it must be true by the distributive property that 7 multiplied through > (A(x) + 7), then you don't know anything about mathematics. > Not a single mathematician with even a modicum of training or knowledge > would doubt that result, or can doubt it. > Yes, they can. > If x is not equal to 0, there's no reason why A(x) should be divisible > by 7. It doesn't follow that it is from the distributive property. > You've never produced an argument worth a damn why it should be. It > simply doesn't follow, and we've given examples to prove it doesn't > follow. No-one's lying. They're just pointing out that you're wrong, > and not for the first time either. > Then attack the argument, which is that given in the complex plane > 7C(x) = (A(x) + 7)(B(x) + 1) > true for all x, where A(0) = B(0) = 0 > by the distributive property, it must be true that 7 multiplied through > the first factor. > This isn't an argument. The result is true in the complex plane, but > you haven't given the reason why. Why does it follow from the > distributive property? > Here's a more detailed proof: > In the complex plane, given > 7C(x) = (A(x) + 7)(B(x) + 1) > true for all x, where A(0) = B(0) = 0 > let > C(x) = (A'(x) + 1)(B'(x) + 1) > where A'(0) = B'(0) = 0 What justifies the assumption that there are algebraic-integer-valued functions A' and B' with this property? > and making that substitution, gives > 7(A'(x) + 1)(B'(x) + 1) = (A(x) + 7)(B(x) + 1) > and by the distributive property > A(x) = 7A'(x) and B'(x) = B(x). > I've added that the relationship is true for all x to remove any area > of confusion, as I'm talking about factoring results, like > x^2 + 3x + 2 = (x + 1)(x + 2) > is true for all x. > There is simply no way by the distributive property that you can have > 7C(x) = (A(x) + 7)(B(x) + 1) > true for all x, where A(0) = B(0) = 0 > where the 7 didn't multiply through the first factor. > So, do you claim that this is true for all rings (since all rings have > the distributive property?) > The trouble is, we've shown one ring where it's false. > What's true in the ring of algebraic integers is that 7 isn't > necessarily a factor of A(x) in that ring. Good. Glad you agree. > But if you consider evens as a ring 2 is not a factor of 6 because 3 is > not even. > The ring of algebraic integers has the same type of problem. It's not a problem, it's just a fact. > The result shows that even in the ring of algebraic integers 7 would > have to multiply through the first factor, What does this mean? > but because of special rules > of the ring, 7 is not a factor in that ring. > With a result that holds in the complex plane, it is CLEAR that posters > arguing against me are simply refusing to accept what's mathematically > correct. > Nobody disputed the result in the complex plane. We disputed it in the > algebraic integers. The fact that the result holds in the complex plane > provides no support for your argument. > Decker was smart enough to argue against it in the complex plane. > If you accept it in the complex plane then if you are logical you are > driven to accepting my full argument. False. > If I am wrong, refute the result. > You're moving the goalposts. What's your claim? State your claim and > I'll tell you whether you're wrong or not. > My claim has not changed. What is it, then? > I use a key result valid in the complex plane which should be valid in > the ring of algebraic integers, if that ring were complete, but it NOT > being generally valid in that ring, is proof that it is incomplete. Define complete. > I like it that it's true in the complex plane so that all those > annoying factor arguments go a way. > You like the fact that the claim is actually correct in this context. > Yes, that's good. That's healthy. > We are in the complex plane now. Handle the argument there. > There is no argument. The argument is not valid in the complex plane. > The result is true in the complex plane for other reasons. If the > argument were valid in the complex plane, it would be valid in other > rings, and it's not. > If you've changed your claim to the claim that the result is just true > in the complex plane, I have no wish to argue with you. As long as you > agree that it's false in the algebraic integers, that's fine. > The result true in the complex plane is that given > 7C(x) = (A(x) + 7)(B(x) + 1) > true for all x, where A(0) = B(0) = 0 > let > C(x) = (A'(x) + 1)(B'(x) + 1) > where A'(0) = B'(0) = 0 > and making that substitution, gives > 7(A'(x) + 1)(B'(x) + 1) = (A(x) + 7)(B(x) + 1) > and by the distributive property > A(x) = 7A'(x) and B'(x) = B(x) > proving that 7 multiplied through the first factor A(x) + 7. Yes, this is fine. > Since it didn't multiply through the second, it's not possible in other > rings, like the ring of algebraic integers, for B(x) + 1 to have a > non-unit factor that is a non-unit in more complete rings. What happens in the ring of algebraic integers is that you need to argue that there are algebraic-integer-valued functions A' and B' such that C(x)=(A'(x)+1)(B'(x)+1). This is trivial in the complex plane, but it's not true in the ring of algebraic integers, as the examples we've given you countless times show. > What happens in the ring of algebraic integer is that a number that is > a unit in the ring of objects, a complete ring, What's the ring of objects? What's a complete ring? > is not a unit in the > ring of algebraic integers because of a technicality that it can't be > the root of any monic polynomials with integer coefficients and a last > coefficient of 1 or -1. > Because of the technicality the unit is not a unit in the ring of > algebraic integers, so that you can appear to have a non-unit factor of > 7 in that ring, when over the complex plane, 7 multiplied through the > other factor. > It's an easy argument actually. Just accept the distributive property > and figure out what must be true for mathematics to be consistent. > Your view that somehow non-unit factors of 7 can actually multiply > through as functions of x, varying with different factors dependent on > the value of x, in the ring of algebraic integers, just doesn't add up > to what follows from the distributive property, forcing you to give up > the distributive property for that view to hold. It doesn't follow from the distributive property. You assumed without proof that there are algebraic-integer-valued functions A' and B' such that C(x)=(A'(x)+1)(B'(x)+1). This is what our examples show to be false. > James Harris === Subject: Re: JSH: So now you know as I found out <46erm0FapjkcU2@individual.net > Here's a more detailed proof: > In the complex plane, given > 7C(x) = (A(x) + 7)(B(x) + 1) > true for all x, where A(0) = B(0) = 0 > let > C(x) = (A'(x) + 1)(B'(x) + 1) > where A'(0) = B'(0) = 0 > And what are A'(x) and B'(x)? > and making that substitution, gives > 7(A'(x) + 1)(B'(x) + 1) = (A(x) + 7)(B(x) + 1) > and by the distributive property > A(x) = 7A'(x) and B'(x) = B(x). > How is it that you deduce this from the distributive property? There are three general ways for how 7 multiplies through 7(A'(x) + 1)(B'(x) + 1) = (A(x) + 7)(B(x) + 1) where I've given the only one that doesn't lead to contradiction with the distributive property: a(b+c) = ab + ac The constants are what force the issue, as I've noted before, and remember A(0) = B(0) = 0. James Harris === Subject: Re: JSH: So now you know as I found out > Here's a more detailed proof: > In the complex plane, given > 7C(x) = (A(x) + 7)(B(x) + 1) > true for all x, where A(0) = B(0) = 0 > let > C(x) = (A'(x) + 1)(B'(x) + 1) > where A'(0) = B'(0) = 0 > And what are A'(x) and B'(x)? > and making that substitution, gives > 7(A'(x) + 1)(B'(x) + 1) = (A(x) + 7)(B(x) + 1) > and by the distributive property > A(x) = 7A'(x) and B'(x) = B(x). > How is it that you deduce this from the distributive property? There are three general ways for how 7 multiplies through 7(A'(x) + 1)(B'(x) + 1) = (A(x) + 7)(B(x) + 1) where I've given the only one that doesn't lead to contradiction with the distributive property: a(b+c) = ab + ac Where are A' and B' defined ? === Subject: Re: JSH: So now you know as I found out <46erm0FapjkcU2@individual.net> <44027e5b$0$48554$892e7fe2@authen.yellow.readfreenews.net > Here's a more detailed proof: > In the complex plane, given > 7C(x) = (A(x) + 7)(B(x) + 1) > true for all x, where A(0) = B(0) = 0 > let > C(x) = (A'(x) + 1)(B'(x) + 1) > where A'(0) = B'(0) = 0 > And what are A'(x) and B'(x)? Doesn't matter as they equal 0 at x=0. That's all that matters. The constants are crucial. Focus on the constants, which are 1 first and then 7 and 1 later. That's why it doesn't matter what the functions are, as long as they go to 0 at x=0. > and making that substitution, gives > 7(A'(x) + 1)(B'(x) + 1) = (A(x) + 7)(B(x) + 1) > and by the distributive property > A(x) = 7A'(x) and B'(x) = B(x). > How is it that you deduce this from the distributive property? > There are three general ways for how 7 multiplies through > 7(A'(x) + 1)(B'(x) + 1) = (A(x) + 7)(B(x) + 1) > where I've given the only one that doesn't lead to contradiction with > the distributive property: > a(b+c) = ab + ac > Where are A' and B' defined ? All that matters is that they equal 0 when x=0, which is stated in the proof. The distributive property allows one of the constants to go from 1 to 7 in only one way, which is to multiply through only one of the factors, as I've noted before. By going to the complex plane, I remove the entire factor thing from the picture, showing the result is just a basic algebraic one--following from the distributive property. Notice it is NOT true that only one of the factors has 7 as a factor, as in the complex plane, 7 is a factor of both of them. The crucial point is that what I use is not a factoring result, in a ring where factors are not meaningful. It only becomes a factoring result in rings like my ring of objects where it holds perfectly, as can be seen by playing with the equation: a^2 -(1+fx)a + (f^2 x^2 + 2fx) = 0 With it picking x and f integers where f is not a unit, that is not 1 or -1, and is coprime to x, will, when you get rational roots, give you integers that either have f as a factor, or are coprime to f, as forced by the distributive property in a complete ring. What happens in the ring of algebraic integers is not that f actually factors and splits between factors, but you get units in the ring of objects that are not units in the ring of algebraic integers because they are not roots of monic polynomials with integer coefficients with a last coefficient of 1 or -1. So you get a faux result in that ring, with which you can do fun stuff, like prove things that are not true, and have mathematics appear to be inconsistent. But mathematics is not inconsistent. The ring of algebraic integers is just incomplete, as that rule of being roots of monic polynomials with integer coefficients just forces things to behave this way. It's not malicious, it's mathematics. James Harris === Subject: Re: JSH: So now you know as I found out > By going to the complex plane, I remove the entire factor thing from > the picture, showing the result is just a basic algebraic > one--following from the distributive property. > Notice it is NOT true that only one of the factors has 7 as a factor, > as in the complex plane, 7 is a factor of both of them. How can you talk about factors in a field? In a field, everything that's not zero is a factor of everything else. 7 is a factor of sqrt(2). Pi is a factor of e. === Subject: Re: JSH: So now you know as I found out Discussion, linux) <46erm0FapjkcU2@individual.net> <44027e5b$0$48554$892e7fe2@authen.yellow.readfreenews.net> Notice it is NOT true that only one of the factors has 7 as a factor, >> as in the complex plane, 7 is a factor of both of them. > How can you talk about factors in a field? In a field, everything that's > not zero is a factor of everything else. 7 is a factor of sqrt(2). Pi is > a factor of e. Can't you read? That's essentially what he said. -- People think there are brilliant people at important government agencies like the NSA or CIA that will save them, when I know, sadly, that mathematicians rule the roosts in the key places in all the major government agencies, even the shadowy ones. -- James S. Harris === Subject: Re: JSH: So now you know as I found out <46erm0FapjkcU2@individual.net> <44027e5b$0$48554$892e7fe2@authen.yellow.readfreenews.net> By going to the complex plane, I remove the entire factor thing from > the picture, showing the result is just a basic algebraic > one--following from the distributive property. > Notice it is NOT true that only one of the factors has 7 as a factor, > as in the complex plane, 7 is a factor of both of them. > How can you talk about factors in a field? In a field, everything that's > not zero is a factor of everything else. 7 is a factor of sqrt(2). Pi is > a factor of e. Exactly. The result is NOT a factoring result in fields. By going to the complex plane I remove room for confusion by removing the factor thing from the picture. It shows as I've said that the result is about the distributive property and the proof is rather trivial. I like what I said in a previous reply today so I'm copying it over. The proof you are trying to challenge is as follows: In the complex plane, given 7C(x) = (A(x) + 7)(B(x) + 1) true for all x, where A(0) = B(0) = 0 let C(x) = (A'(x) + 1)(B'(x) + 1) where A'(0) = B'(0) = 0 and making that substitution, gives 7(A'(x) + 1)(B'(x) + 1) = (A(x) + 7)(B(x) + 1) and by the distributive property A(x) = 7A'(x) and B'(x) = B(x). The fact that the function equal 0 at x=0 is a crucial piece. Notice the above is a proof. It is a trivial proof that just about any mathematician can verify for you, if you need that done. I suggest you just copy it out, send it to someone and see what I mean. They may make notational changes as well as change some of the description, but they can verify for you that it is a proof and the result does hold over the complex plane. James Harris === Subject: Re: JSH: So now you know as I found out > The proof you are trying to challenge is as follows: > In the complex plane, given > 7C(x) = (A(x) + 7)(B(x) + 1) > true for all x, where A(0) = B(0) = 0 > let > C(x) = (A'(x) + 1)(B'(x) + 1) > where A'(0) = B'(0) = 0 Again: what are A'(x) and B'(x)? > and making that substitution, gives > 7(A'(x) + 1)(B'(x) + 1) = (A(x) + 7)(B(x) + 1) > and by the distributive property > A(x) = 7A'(x) and B'(x) = B(x). > The fact that the function equal 0 at x=0 is a crucial piece. How do you deduce from the distributive property that A(x) = 7A'(x) and B'(x) = B(x)? Jose Carlos Santos === Subject: Re: JSH: So now you know as I found out ... > 7(A'(x) + 1)(B'(x) + 1) =3D (A(x) + 7)(B(x) + 1) ... > A(x) =3D 7A'(x) and B'(x) =3D B(x). ... > There are three general ways for how 7 multiplies through > 7(A'(x) + 1)(B'(x) + 1) =3D (A(x) + 7)(B(x) + 1) But what are A' and B'? When they are arbitrary functions there is no problem. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: one page proof of the Infinitude of Twin Primes <4400D7EC.5000506@dtgnet.com Apparently there is only one person in this thread, so far, that is > able to give a valid indirect proof of Euclid Infinitude of Primes. The > person posting under the name Nihilist. Everyone else in this thread > who opines or has given a Indirect proof such as Chris Heckman, Ken > Quirici, Matt B, R. Blankman, et al have a flawed understanding. And AP thinks that the negation of P and Q is (Not P) And (Not Q), so he shouldn't be posting, either. > Can some software engineer tell me if it is difficult or easy to have > the Science Newsgroups equiped with a Trashcan function that allows the > initial author of a thread to completely moderate all replies. > [...] > This is what the sci newsgroups need in the future. Censorship? --- Christopher Heckman === Subject: Re: one page proof of the Infinitude of Twin Primes <4400D7EC.5000506@dtgnet.com > I know you have a low opinion of me. But the prerequisite for > discussing in this thread is to show us that you can even do a Euclid > Infinitude of Primes, both direct and indirect. So post yours, so that > we can thence see if you can contribute to this thread or just another > one of those raucious hatemongers. For nothing that you say has any > merit in this discussion if you do not know the difference between > Euclid direct and Euclid indirect. Don't copy from some book but give > it purely from your mind. > I already have. However, you still haven't done a valid indirect proof. > Just to humor you, though, here we go: > Indirect: Assume there are finitely many primes, P(1), P(2), ..., P(k). Let > N = P(1) * P(2) * ... * P(k) + 1, and let Q be a prime factor of N. Q > cannot be any of P(1), ..., P(k), since N divided by P(i) is 1. > trivial point - you mean leaves a remainder of 1, don't you? Yes. AP said to do it off the top of my head. > This is a nice swell-foop proof. You don't examine separately > the usual first step - 'if N is prime, then it is our new prime'. > You allow Q to be any prime factor of N including itself, and > regardless of whether it is N or not, your logic follows. > This > contradicts the assumption that we had a list of all the primes. > Direct: Given a set S of integers, define N(S) to be the product of the > elements of S, plus 1, and define P(S) to be the smallest (for > definiteness) prime factor of N(S). Then P(S) is a prime not in S, > since N divided by any element of S is 1. > again, leaves a remainder of 1? > Now construct an infinite > sequence of sets by the following procedure: > (1) S(0) = {2} > (2) S(i) = S(i-1) union {P(S(i-1))}, when i >= 1. > Then the union of the S(i)'s, over all nonnegative i, is an infinite > set of primes. QED > Damn nice proofs. Direct and to the point. Minimal. etc. Not according to AP. --- Christopher Heckman === Subject: Re: one page proof of the Infinitude of Twin Primes <43FEB055.8080007@dtgnet.com> Care to give us your rendition of direct and indirect, to see if they > are distinct and whether you can make them flow logically. I would be > surprized if you can, because you are mostly a number cruncher and poor > when it comes to logical steps. At least I lasted more than 1 day in graudate school. And I got a doctorate in mathematics. And I got a job teaching at a university which wasn't on the other side of the world. And I've been teaching here for 6 years and received high praise for it. And I have one published result, two on the way --- all of which passed my doctoral committee and the refereeing process, so yes, I can write valid proofs. --- Christopher Heckman === Subject: Re: one page proof of the Infinitude of Twin Primes <4400D7EC.5000506@dtgnet.com [...] in the Indirect method, there cannot be a prime factor for the > constructed number [...] Okay; I can see what you mean by the constructed number is necessarily prime here. You're approaching the proof from the opposite end as I am. You're saying that none of P(1), ..., P(k) divides N, so no prime divides into N. Therefore, N cannot be written as a product of prime numbers (since P(1), ..., P(k) are all there are), contradicting the fact that you can. My proof uses the fact that every positive integer can be written as the product of primes. If we let Q be one of these primes, then Q cannot be any of P(1), ..., P(k), but since these are all there are, we get a contradiction. I went from A (prime factorization) to B (prime factors of N) to C (primes in the list); you went from C (primes in the list) to B (no prime divides N) to A (prime factorization; N is prime). Both end up with a contradiction. So both are proofs, after some minor fixing-up. > Indirect: Assume there are finitely many primes, P(1), P(2), ..., P(k). > Let > N = P(1) * P(2) * ... * P(k) + 1, and let Q be a prime factor of N. Q > cannot be any of P(1), ..., P(k), since N divided by P(i) is 1. This > contradicts the assumption that we had a list of all the primes. > Direct: Given a set S of integers, define N(S) to be the product of the > elements of S, plus 1, and define P(S) to be the smallest (for > definiteness) prime factor of N(S). Then P(S) is a prime not in S, > since N divided by any element of S is 1. Now construct an infinite > sequence of sets by the following procedure: > (1) S(0) = {2} > (2) S(i) = S(i-1) union {P(S(i-1))}, when i >= 1. > Then the union of the S(i)'s, over all nonnegative i, is an infinite > set of primes. QED > Your mistake in the Indirect method is that you assumed finite. It's a proof by contradiction, which by definition is indirect. (I suspect you may have gotten your bad definition from Wikipedia or a pedestrian math book. Merriam-Webster, at http://www.m-w.com/dictionary/indirect , agrees with me.) > Because you assumed finite with all the primes in that list, then you cannot go > and talk about Q not in that list of all primes. I have implicitly used the fact that any positive integer can be written as the product of prime numbers. Since I wasn't copying it from a book, I was bound to leave out minor details (which I could provide later, if needed). > If you insist on Q, > you have created a double contradiction. Your mistake is that N is > necessarily prime as a result of your supposition of finite set of > primes for which you listed all of them. So you cannot have a prime > factor in the indirect method. My proof allows for the case where Q = N, so I've covered the case where N is prime. A is a prime factor of B if (1) A is a prime, and (2) A is a factor of B. Thus 3 is a prime factor of 3. > I see nothing wrong with your Direct proof, other than it is long > winded. The proper term is rigorous. > We all know in Logic that the any set transforms to the > universal or infinite set. So that if any set whatever can be increased > in cardinality that it becomes an infinite set. So your Direct proof > should be scaled and trimmed back to something like this: > Direct: given any set of primes we increase its set cardinality by > constructing N and we proceed with a prime factor search if N is not > prime itself. Hence the set of primes is infinite. > I do not think you should be sad that you failed in the Indirect > method, but not to worry for most professors of mathematics who write > books and have a printed version of the Indirect Euclid IP have failed > to give a valid proof. I have a list of about 30 printed books and > texts where they fail in this chore. The most famous is probably > Hardy's A Mathematicians Apology. I think what the problem is that > there is a juncture in this proof where the mind has a hard time of > focusing and keeping the elements of the proof together so as not to > make a mistake. > Summary: in the Indirect method, there cannot be a prime factor for the > constructed number is necessarily prime. [...] Okay; I can see what you mean by the constructed number is necessarily prime here. You're approaching the proof from the opposite end as I am. You're saying that none of P(1), ..., P(k) divides N, so no prime divides into N. Therefore, N cannot be written as a product of prime numbers (since P(1), ..., P(k) are all there are), contradicting the fact that you can. My proof says that every positive integer can be written as the product of primes. If we let Q be one of these primes, then Q cannot be any of P(1), ..., P(k), but since these are all there are, we get a contradiction. I went from A (prime factorization) to B (a prime factor of N) to C (Q is not in the list); you went from C (a prime in the list) to B (no prime divides N) to A (prime factorization; N is prime). --- Christopher Heckman === Subject: Re: one page proof of the Infinitude of Twin Primes >>[...] in the Indirect method, there cannot be a prime factor for the >>constructed number [...] > Okay; I can see what you mean by the constructed number is necessarily > prime here. You're approaching the proof from the opposite end as I > am. > You're saying that none of P(1), ..., P(k) divides N, so no prime > divides into N. Therefore, N cannot be written as a product of prime > numbers (since P(1), ..., P(k) are all there are), contradicting the > fact that you can. > My proof uses the fact that every positive integer can be written as > the product of primes. If we let Q be one of these primes, then Q Then you are making another unwarranted assumption, for you assume it is composite. The moment you constructed N, it can only be one thing-- a new prime. You cannot say it is composite because all the primes are in your list and you cannot search through them for a *Prime Factor*. So the moment you constructed N, you can only say that all the primes leave a remainder when divided by all the primes, hence N is necessarily prime contradiction end of proof. > cannot be any of P(1), ..., P(k), but since these are all there are, we > get a contradiction. > I went from A (prime factorization) to B (prime factors of N) to C > (primes in the list); you went from C (primes in the list) to B (no > prime divides N) to A (prime factorization; N is prime). Both end up > with a contradiction. > So both are proofs, after some minor fixing-up. No. Yours has a double self made contradictions. In order to get a contradiction you have to be able to hand me a new prime not on the list. N is prime and thus you end the proof. When you hunt for a prime factor, you never find a new prime. And you do not need to because N is prime. So to hunt for a prime factor, you deny the fact that N is prime. Another way of looking at this is that in the Direct proof you must do a prime factor search and hunt to hand over a new prime not on the finite list. Now, someone well versed in Logic can tell you that the steps of hunting down a Prime factor cannot exist in both the Direct and Indirect method because the symmetry of the direct is different from indirect. So the Symbolic Logic steps of the Prime Factor Search cannot be identical in both Direct and Indirect, because, really, then Direct is the same as Indirect. The suppose false in the indirect causes N to be necessarily prime. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: one page proof of the Infinitude of Twin Primes <4400D7EC.5000506@dtgnet.com> <4402A980.1040902@dtgnet.com >>[...] in the Indirect method, there cannot be a prime factor for the >>constructed number [...] > Okay; I can see what you mean by the constructed number is necessarily > prime here. You're approaching the proof from the opposite end as I > am. > You're saying that none of P(1), ..., P(k) divides N, so no prime > divides into N. Therefore, N cannot be written as a product of prime > numbers (since P(1), ..., P(k) are all there are), contradicting the > fact that you can. > My proof uses the fact that every positive integer can be written as > the product of primes. If we let Q be one of these primes, then Q > Then you are making another unwarranted assumption, for you assume it is > composite. No, I am not. Every integer which is at least 2 --- whether prime or composite --- has a factor which is prime. 15 has 3 as a factor. 17 has 17 as a factor. > The moment you constructed N, it can only be one thing-- a new prime. > You cannot say it is composite because all the primes are in your list > and you cannot search through them for a *Prime Factor*. > So the moment you constructed N, you can only say that all the primes > leave a remainder when divided by all the primes, hence N is necessarily > prime contradiction end of proof. No; you have to bring in the fact that any integer >= 2 can be written as a product of one or more prime numbers. Without that fact, there is no contradiction, because in that case, N might be an integer which can't be factored into primes. > cannot be any of P(1), ..., P(k), but since these are all there are, we > get a contradiction. > I went from A (prime factorization) to B (prime factors of N) to C > (primes in the list); you went from C (primes in the list) to B (no > prime divides N) to A (prime factorization; N is prime). Both end up > with a contradiction. > So both are proofs, after some minor fixing-up. > No. Yours has a double self made contradictions. > In order to get a contradiction you have to be able to hand me a new > prime not on the list. No, all I need is a contradiction of ANY kind. Go back and study your logic. > Now, someone well versed in Logic [...] But that someone isn't you. I mean, how can ANYONE claim that the negation of P and Q is (Not P) and (Not Q)? You don't have to go that far in a logic textbook to realize this is false; in fact, it will be in the first or second section. --- Christopher Heckman === Subject: Re: one page proof of the Infinitude of Twin Primes <4400D7EC.5000506@dtgnet.com> <4402A980.1040902@dtgnet.com> No; you have to bring in the fact that any integer >= 2 can be written as a product of one or more prime numbers. Without that fact, there is no contradiction, because in that case, N might be an integer which can't be factored into primes. Proof of Infinitude of Primes, Indirect Method: Suppose false then there is a last and final prime call it P. Construct N = ((2x3x5x....xP)+1. Enter the Unique Prime Factor Theorem but N leaves a remainder of 1 when all the primes that exist divide into N, hence N is necessarily prime itself. Contradiction. Primes are infinite. There, Chris, so yours was flawed beyond salvaging. Here is yours repeated: > Indirect: Assume there are finitely many primes, P(1), P(2), ..., P(k). Let > N = P(1) * P(2) * ... * P(k) + 1, and let Q be a prime factor of N. Q > cannot be any of P(1), ..., P(k), since N divided by P(i) is 1. This > contradicts the assumption that we had a list of all the primes. Your mistake is to think that a prime factor search can occur in the Indirect Method. It cannot occur. The Indirect Method gives you that N is necessarily prime and to disregard that fact will render your proof attempt as invalid. This is important not so much to correct 90 percent of the math professors who fail at doing a Euclid Infinitude of Primes, indirect method. Important because those same math professors could never do a Infinitude of Twin Primes, or Quad-primes or Hex-primes ad infinitum. They could never do Twin Primes because they could never get Euclid's Infinitude of regular primes, indirect method correctly. Once you understand that N is necessarily prime, then Infinitude of Twin Primes follows as easy as premium ice cream with whipped cream. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: one page proof of the Infinitude of Twin Primes <4400D7EC.5000506@dtgnet.com> <4402A980.1040902@dtgnet.com No; you have to bring in the fact that any integer >= 2 can be written > as a product of one or more prime numbers. Without that fact, there is > no contradiction, because in that case, N might be an integer which > can't be factored into primes. > Proof of Infinitude of Primes, Indirect Method: Suppose false then > there is a last and final prime call it P. Construct N = > ((2x3x5x....xP)+1. Enter the Unique Prime Factor Theorem but N leaves a > remainder of 1 when all the primes that exist divide into N, hence N is > necessarily prime itself. Contradiction. Primes are infinite. > There, Chris, so yours was flawed beyond salvaging. Here is yours > repeated: > Indirect: Assume there are finitely many primes, P(1), P(2), ..., P(k). Let > N = P(1) * P(2) * ... * P(k) + 1, and let Q be a prime factor of N. Q > cannot be any of P(1), ..., P(k), since N divided by P(i) is 1. This > contradicts the assumption that we had a list of all the primes. > Your mistake is to think that a prime factor search can occur in the > Indirect Method. It cannot occur. Why not? I use the same fact as you (every integer >= 2 can be written as the product of at least one prime), which makes you a master of the double standard. > The Indirect Method gives you that N > is necessarily prime and to disregard that fact will render your proof > attempt as invalid. My proof does not need that fact. (Note that when I use the word factor, I am including 1 and N as factors of N. So if N happens to be prime, then Q=N. If N is not prime, sin should be that I'm adding an unnecessary case.) > This is important not so much to correct 90 percent of the math > professors who fail at doing a Euclid Infinitude of Primes, indirect > method. Important because those same math professors could never do a > Infinitude of Twin Primes, or Quad-primes or Hex-primes ad infinitum. [...] Math professors can't publish a paper saying noon blue apples, either. --- Christopher Heckman === Subject: Regular partitions of graphs E SZEMEREDI 1975. 64 Numerical Analysis(AH). I'd really appreciate if somebody could provide me with the paper: Regular partitions of graphs E SZEMEREDI 1975. 64 Numerical Analysis(AH). ogl100@yahoo.com thank you. === Subject: Welcome to the largest BBW Lovers meeting group. This Group is for all BBW's, Welcome to the largest BBW Lovers meeting group. This Group is for all BBW's, BHM's and Their Admirers. For those who looking for true love, online relationship, frienndship and more...... http://www.geocities.com/tallsingletogether === Subject: Re: Why Archimedes Plutonium will never proof the 4CT > Archimedes Plutonium [AP] has posted two alleged proofs of the 4CT > which are short but not correct. There are two basic facts that he is > repeatedly ignoring, and since the threads containing those proofs have > turned into many-headed hydrae, and since AP doesn't like reading long > posts, my points are lost. So I decided to start a new thread. > Here are the two basic facts that AP is ignoring: [...] Wow, we're up to ... (6) AP doesn't understand basic logic. It is clear from AP's posts that he thinks that P implies Q and Q implies P are equivalent. He also disagrees with DeMorgan's Laws, thinking instead that the negation of P and Q is (Not P) and (Not Q). Maybe some new type of logic is present in Atom Totality Theory. (7) AP doesn't understand basic proof techniques. He has posted several proofs about the infinitude of twin primes (primes that differ by 2) which proceed in the following manner: (a) Assume there are only finitely many primes, (b) derive a contradiction, (c) Conclude there are an infinite number of twin primes. (8) Physical theories are more definite than mathematical theorems. Any physical theory can be destroyed by a measurement tomorrow. Mathematical theorems, provided their proofs are valid, are established in perpetuity. (9) The future of Usenet should be censorship. Let's see his own words: > Can some software engineer tell me if it is difficult or easy to have > the Science Newsgroups equiped with a Trashcan function that allows the > initial author of a thread to completely moderate all replies. > [...] > This is what the sci newsgroups need in the future. --- Christopher Heckman === Subject: Logarithmic relationship proof? Is the logarithmic relationship a fixed relationship within each base? If it is, how can it be proved without using the log itself? That is, how can I prove that the relationship is the same within each base (base3, base5 etc) without using either log or to the power of? Petre === Subject: Re: Logarithmic relationship proof? > Is the logarithmic relationship a fixed relationship within each > base? If it is, how can it be proved without using the log itself? That > is, how can I prove that the relationship is the same within each base > (base3, base5 etc) without using either log or to the power of? > Petre How do you even state the theorem without using log or to the power of (or base for that matter) ? === Subject: Re: Logarithmic relationship proof? > Is the logarithmic relationship a fixed relationship within each > base? If it is, how can it be proved without using the log itself? That > is, how can I prove that the relationship is the same within each base > (base3, base5 etc) without using either log or to the power of? Huh? If y = b^rx, then y = e^(rx.ln b). So it makes no difference for equations of the form y = b^rx, what the base b is. Only the constant r. Similar with y = r.log_b x = (r.ln x)/(ln b) === Subject: Re: Logarithmic relationship proof? What means base b ? we may consider the number of iterations (numbrer of compositions) of the function f(x) =b*x . We 've got a function phi , linked to f(x) which satisfies : phi(f(x)) = phi(x) + 1 , increment : +1 here ln(b*x) /ln(b) = ln(x)/ln(b) + 1 two important things : 1Á) on right hand the term ln(x) / ln(b) 2Á) ln(x) / ln(b) is a 'natural counting function the nth iterate of b*x , b*(b*( .....x)..) = b^n*x and ln (b^n*x) / ln(b ) = ln(x) / ln(b) + n , YES , increment +n Alain. === Subject: Re: modular arithmetic ... > The problem that I am working on is to find the remainder of 23^(25^57)by 49. Can someone please tell me the steps in how to approach this question. Do I start with 25^57 with mod(49)? As Robert Israel mentioned, mod 42 is relevant here. That's because of Euler's Theorem, which is that a^phi(c)==1 mod c if (a,c)=1. phi(c)=number of integers from 1 to c that are relatively prime to c. After you apply Euler's, you deal with 25^15 rather than 25^57. Beyond that point I'm not certain about the right way to proceed. Of course, if you notice that 25^3==43 mod 49, then 23^(25^15) immediately reduces to 23^5. -jiw === Subject: Re: modular arithmetic find the remainder of 3^25 divided by 36. > Here's a start. > 3^25 = 3 * 81^6 = 3 * 9*6 (mod 36) Whoops. 3^25 = 3 * 81^6 = 3 * 9^6 (mod 36) > Can you take it from there? === Subject: Re: modular arithmetic >of an example. However, I need help to understand something and would >like some help. >Example: >find the remainder of 3^25 divided by 36. >so mod is 36 therefore >3^4=3^2*3^2=9*9=81=9(mod36) >3^8=(3^4)^2=9^2=9(mod36) >3^16=(3^8)^2=9^2+9(mod36) >3^25=3*3^8*3^16=3*9*9(mod36) > =3*9=27(mod36) >I need help understanding 3^25. I can see that 3 is obviously 3 and that >3^16 is 9(mod36) but why is 3^8 equal to 9? 3^8 = (3^4)^2. You already know 3^4 = 9 (mod 36), so (3^4)^2 = 9^2, and again you already know from the previous line that this is 9 (mod 36) Also, can someone explain >the last line to me too? 25 = 1 + 8 + 16. So 3^25 = 3^1 * 3^8 * 3^16. And you already know ... >The problem that I am working on is to find the remainder of >23^(25^57)by 49. Can someone please tell me the steps in how to approach >this question. Do I start with 25^57 with mod(49)? No, but mod 42 might help. Alternatively: Start with 23^5 mod 49, then raise this to the power 5 to get 23^25 mod 49. Raise this to the 25th power mod 49 in a similar way. Repeat until you get back to 23... Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: manners and etiquette and college education Re: one page proof of the Infinitude of Twin Primes > You are a complete idiot. I can't believe how arrogant you are to not > even try and see the point of view of others. The rest snipped. The gall of a poster to think anyone is going to read something like this given the first sentence. I suspect this poster is some college student. And gives me the opportunity every once in a while to air what I feel is the greatest shortcomings of a college education in the USA. I think it should be mandatory for every college to have a required course in ethics and manners and etiquette. I do not mean how to set a dinner table of forks and spoons. I mean how to talk to other humans. I mean how to afford everyone on par respect and dignity. It matters not what you know if you treat everyone around you like dirt. The above poster is not educated, not matter how many years in school if they lack manners and etiquette. The Internet sci newsgroups are chock full of posters who have little to no manners or etiquette. So I think that all colleges of this country should require manners and etiquette. And if they display bad behaviour while in college such as bullying or ad hominem that they be give demerits and if they have too many demerits they are required to take another course on manners and etiquette. If they cannot act with good manners then suspend them from college and have them go to the military before allowing them back into college. In other words, without manners and etiquette there is no college degree, no college education. And after graduates leave their college and should they get involved in bad things such as committing crimes or even posting to the Internet with bad manners and bad etiquette, that their college degree be revoked. Such as Alan Schwartz or Uncle Al having his college degree revoked due to the years of bad manners. Our society does not foster good manners, good behaviour and etiquette and the colleges are not helping. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: infinity sci.math_20050526_b: In NSA, there may be the hyperintegers, infinitely long strings of finite radix, finite base, that represent integers. Where here we have been talking about how the tree is the same for the integers and those of the unit interval of reals, that it has no leaf nodes, in NSA it is the same, except the hyperintegers (or p-adic integers) may well be infinite in precision and extent. The transfer principle is that what is provable in NSA, IST, is provable in ZFC, where IST and ZFC are coconsistent. === Subject: Re: infinity sci.math_20050526_b: Instead, you are led to examine the problems of Burali-Forti and the other paradoxes of infinity directly. They lead back to dual representation, the dialetheism of the minimal and maximal ur-element, the root probabilistic flaw, and the resolution of all paradoxes, towards the ability to prove any mathematical fact. === Subject: Re: Still ,Matrices, A is square: If AB = Identity, is B the inverse? dts7q0$aq4$1@wisteria.csv.warwick.ac.uk... >>linux ha scritto: > If In is a matrix the coefficients of which belong to a some ring, the > result is it really still? I think that yes. Can one prove him(it) > without > determinant? >>Bx = 0 => ABx = 0 => x = 0 => B is invertible (the system has only the >>trivial solution). > That argument only works over a field. Let B be the 1x1 matrix ( 2 ) over > the ring of integers. Then Bx = 0 => x = 0, but B is not invertible. > It has been pointed out that the result is not true in general for > matrices > (even 1x1 matrices) over non-commutative rings. > I think it is true for matrices over a commutative ring with a 1, > because we have > A adj(A) = adj(A) A = det(A) I_n, where adj(A) is the adjoint matrix, > but AB = I_n => det(AB) = det(A)det(B) = 1, so det(A) has the inverse > det(B) in the ring, and hence det(B) adj(A) is a 2-sided inverse A^-1 for > A, and then by the usual argument A B = I_n => B = A^-1 A B = A^-1, so > BA = I_n. > But unfortunately, I used deteminants! > Derek Holt. det(B) invertible? and (1/det(B)adj(A) is is a 2-sided inverse A^-1 for A === Subject: Re: Still ,Matrices, A is square: If AB = Identity, is B the inverse? 4402b0d1$0$1148$7a628cd7@news.club-internet.fr... > dts7q0$aq4$1@wisteria.csv.warwick.ac.uk... >linux ha scritto: >> If In is a matrix the coefficients of which belong to a some ring, the >> result is it really still? I think that yes. Can one prove him(it) >> without >> determinant? >Bx = 0 => ABx = 0 => x = 0 => B is invertible (the system has only the >trivial solution). >> That argument only works over a field. Let B be the 1x1 matrix ( 2 ) over >> the ring of integers. Then Bx = 0 => x = 0, but B is not invertible. >> It has been pointed out that the result is not true in general for >> matrices >> (even 1x1 matrices) over non-commutative rings. >> I think it is true for matrices over a commutative ring with a 1, >> because we have >> A adj(A) = adj(A) A = det(A) I_n, where adj(A) is the adjoint matrix, >> but AB = I_n => det(AB) = det(A)det(B) = 1, so det(A) has the inverse >> det(B) in the ring, and hence det(B) adj(A) is a 2-sided inverse A^-1 for >> A, and then by the usual argument A B = I_n => B = A^-1 A B = A^-1, so >> BA = I_n. >> But unfortunately, I used deteminants! >> Derek Holt. > det(B) invertible? > and (1/det(B)adj(A) is is a 2-sided inverse A^-1 for A det(A) invertible! I am sorry for my inattention! === Subject: Re: Still ,Matrices, A is square: If AB = Identity, is B the inverse? <44007235$0$28077$4fafbaef@reader1.news.tin.it> <4402b0d1$0$1148$7a628cd7@news.club-internet.fr> <4402f55a$0$1147$7a628cd7@news.club-internet.fr 4402b0d1$0$1148$7a628cd7@news.club-internet.fr... > dts7q0$aq4$1@wisteria.csv.warwick.ac.uk... >linux ha scritto: >> If In is a matrix the coefficients of which belong to a some ring, the >> result is it really still? I think that yes. Can one prove him(it) >> without >> determinant? >Bx = 0 => ABx = 0 => x = 0 => B is invertible (the system has only the >trivial solution). >> That argument only works over a field. Let B be the 1x1 matrix ( 2 ) over >> the ring of integers. Then Bx = 0 => x = 0, but B is not invertible. >> It has been pointed out that the result is not true in general for >> matrices >> (even 1x1 matrices) over non-commutative rings. >> I think it is true for matrices over a commutative ring with a 1, >> because we have >> A adj(A) = adj(A) A = det(A) I n, where adj(A) is the adjoint matrix, >> but AB = I n => det(AB) = det(A)det(B) = 1, so det(A) has the inverse >> det(B) in the ring, and hence det(B) adj(A) is a 2-sided inverse A^-1 for >> A, and then by the usual argument A B = I n => B = A^-1 A B = A^-1, so >> BA = I n. >> But unfortunately, I used deteminants! >> Derek Holt. > det(B) invertible? > and (1/det(B)adj(A) is is a 2-sided inverse A^-1 for A > det(A) invertible! > I am sorry for my inattention! Yes, if you assume a right inverse: AB = I with entries drawn from a commutative ring, what Derek proves is that 1/det(A) adj(A) is a 2-sided inverse of A. We know det(A) is invertible as an element of R because det(A)det(B) = det(I) = 1 in R. === Subject: Re: Still ,Matrices, A is square: If AB = Identity, is B the inverse? > det(B) invertible? > and (1/det(B)adj(A) is is a 2-sided inverse A^-1 for A > det(A) invertible! > I am sorry for my inattention! Yes, if you assume a right inverse: AB = I with entries drawn from a commutative ring, what Derek proves is that 1/det(A) adj(A) is a 2-sided inverse of A. We know det(A) is invertible as an element of R because det(A)det(B) = det(I) = 1 in R. Many errors of my part. Thus if A is a square matrix the coefficients of which belong to a commutative ring and AB=I then det ( A ) is inversible and furthermore A is inversible and his inverse is (1 / det (A)) adj (A) . I your patience. === Subject: (NONE) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id k1R7r6309650 for ; Mon, 27 Feb 2006 02:53:06 -0500 by support2.mathforum.org (8.12.10/8.12.10/The Math Forum, $Revision: 1.6 secondary) with ESMTP id k1R7pWF8028528 for ; Mon, 27 Feb 2006 02:51:34 -0500 BOUNDARY_OUTLOOK === Subject: For all you Math Whizzes- Eigenvalue Question A 4*4 real matrix has 3 properties: 1) Two of the eigenvalues are 3, 2 2) The number 3 is an eigenvalue of the matrix (A+2I), where I is the Identity Matrix. 3) det(A) = 12 Questions: a) What are the other two eigenvalues of A? b) What is the characteristic polynomial of A ? c) What is the characteristic polynomial of A^ -1 (inverse of A) ? === Subject: Re: For all you Math Whizzes- Eigenvalue Question Suppose that A is a n x n matrix with det(A) not equall zero. The characteristic polynomial of A is the polynomial of degree n ,in variable x, P_A(x) :=det(A-x*I) . Because for z real or complex we have P_A(z):=det(A-z*I)=det(A-z*AA^{-1})= =det[A(I-z*A^{-1})]=det(A)det(I-z*A^{-1}) = = (-1)^{n}z^{n}det(A)det(A^{-1}- (1/z)*I) , we see that P_A(z)=(-1)^{n}z^{n}det(A)P_{A^{-1}}(1/z) , that is P_{A^{-1}}(x) :=det(A^{-1} -x*I)= ((-1)^{n}/det(A))* P_A(x) which show us,e.g. that if r_1,r_2,...,r_n are the eigenvalues of A, then A^{-1} has as eigenvalues the numbers 1/r_1 ,1/r_2,..., 1/r_n . [Note that det(A)=r_1r_2...r_n =/=0 .] Perhaps help. === Subject: Re: For all you Math Whizzes- Eigenvalue Question if r_1,r_2,...,r_n are the eigenvalues of A, then A^{-1} has as eigenvalues the numbers 1/r_1 ,1/r_2,..., 1/r_n . [Note that det(A)=r_1r_2...r_n =/=0 .] Just to confirm - this holds under all circumstances (with exception as noted that det(A)=/=0 ?) Tony > Suppose that A is a n x n matrix with det(A) not equall zero. > The characteristic polynomial of A is the > polynomial of degree n ,in variable x, > P_A(x) :=det(A-x*I) . > Because for z real or complex we have > P_A(z):=det(A-z*I)=det(A-z*AA^{-1})= > =det[A(I-z*A^{-1})]=det(A)det(I-z*A^{-1}) = > = (-1)^{n}z^{n}det(A)det(A^{-1}- (1/z)*I) , > we see that > P_A(z)=(-1)^{n}z^{n}det(A)P_{A^{-1}}(1/z) , that is > P_{A^{-1}}(x) :=det(A^{-1} -x*I)= ((-1)^{n}/det(A))* P_A(x) > which show us,e.g. that if r_1,r_2,...,r_n are the eigenvalues of A, > then A^{-1} has as eigenvalues the numbers 1/r_1 ,1/r_2,..., 1/r_n . > [Note that det(A)=r_1r_2...r_n =/=0 .] > Perhaps help. === Subject: Re: For all you Math Whizzes- Eigenvalue Question > A 4*4 real matrix has 3 properties: > 1) Two of the eigenvalues are 3, 2 > 2) The number 3 is an eigenvalue of the matrix (A+2I), where I is the > Identity Matrix. > 3) det(A) = 12 The assertion 2) is equivalent to det(A + 2I - 3I) = 0, which, of course, is equivalent to det(A - I) = 0. Therefore, 1 is also an eigenvalue. So, its characteristic polynomial P(x), which is a monic polynomial of degree 4, has 1, 2, and 3 as roots. On the other hand, the constant term of P(x) is det(A) = 12. So P(x) = (x - 1)(x - 2)(x - 3)(x - r) for some real number _r_, and 1*2*3*r = 12; therefore, r = 2. > Questions: > a) What are the other two eigenvalues of A? 1 and 2. > b) What is the characteristic polynomial of A ? (x - 1)(x - 2)^2(x - 3) = x^4 - 8x^3 + 23x^2 -28x + 12. > c) What is the characteristic polynomial of A^ -1 (inverse of A) ? (x - 1)(x - 1/2)^2(x - 1/3) Jose Carlos Santos === Subject: Re: For all you Math Whizzes- Eigenvalue Question <46g2p1Faspu6U1@individual.net> hi Jos.8e , I have two question though - 1) what theorem/rule are you using to derive the characteristic polynomial of A^ -1 from the characteristic polynomial of A ? 2) Say if we have figured out that the monic polynomial of a matrix A is: m(x)=x^2 - x -2 , and we have figured out that A^2= A + 2I then how do we express A^ -1 as a LINEAR function of A and I ? {The answer is A^ -1= (1/2) A - (1/2) I } I haven't been to figure this out since I don't know the relationship I mentioned in (1). Tony > A 4*4 real matrix has 3 properties: > 1) Two of the eigenvalues are 3, 2 > 2) The number 3 is an eigenvalue of the matrix (A+2I), where I is the > Identity Matrix. > 3) det(A) = 12 > The assertion 2) is equivalent to det(A + 2I - 3I) = 0, which, of > course, is equivalent to det(A - I) = 0. Therefore, 1 is also an > eigenvalue. So, its characteristic polynomial P(x), which is a monic > polynomial of degree 4, has 1, 2, and 3 as roots. On the other hand, > the constant term of P(x) is det(A) = 12. So > P(x) = (x - 1)(x - 2)(x - 3)(x - r) > for some real number r , and 1*2*3*r = 12; therefore, r = 2. > Questions: > a) What are the other two eigenvalues of A? > 1 and 2. > b) What is the characteristic polynomial of A ? > (x - 1)(x - 2)^2(x - 3) = x^4 - 8x^3 + 23x^2 -28x + 12. > c) What is the characteristic polynomial of A^ -1 (inverse of A) ? > (x - 1)(x - 1/2)^2(x - 1/3) > Jose Carlos Santos === Subject: Re: For all you Math Whizzes- Eigenvalue Question > 1) what theorem/rule are you using to derive the characteristic > polynomial of A^ -1 from the characteristic polynomial of A ? The eigenvalues of A are 1, 2 (twice), and 3. Therefore, A is similar either to 1 0 0 0 0 2 0 0 0 0 2 0 0 0 0 3 or to 1 0 0 0 0 2 1 0 0 0 2 0 0 0 0 3 The inverse of the first one is 1 0 0 0 0 1/2 0 0 0 0 1/2 0 0 0 0 1/3 and the inverse of the second one is 1 0 0 0 0 1/2 -1/4 0 0 0 1/2 0 0 0 0 1/3. In both cases, the characteristic polynomial is the one I gave you. > 2) Say if we have figured out that the monic polynomial of a matrix A My guess is that you meant characteristic polynomial here. > is: > m(x)=x^2 - x -2 , and we have figured out that A^2= A + 2I > then how do we express A^ -1 as a LINEAR function of A and I ? A^2 = A + 2I => A^{-1}.A^2 = A^{-1}.A + A^{-1}.(2I) <=> A = I + 2A^{-1} <=> A^{-1} = (1/2)A - (1/2)I. Jose Carlos Santos === Subject: Re: cupic equation thank you very much are we use compact machine? === Subject: Re: Trying to find length of triangle side > It has been many years since I took high school geometry and seem to > be at a road block finding the angle of 1 side of a triangle. I was > hoping someone could help me with an equation or something. > I have a triangle of which I know the length of 1 side as well as all > 3 angles. How can I find the length of the other 2 sides? > Details: > Angle A: 30 degrees > Angle B: 60 degrees > Angle C: 90 degrees > Length of side A-B: 156 inches > Jason Equilateral! If you mirror the triangle in the side AC, you have an equilateral triangle, so CB = üAB = ü*156 = 78 Pythagoras theorem gives the last side. Hans -- === Subject: Re: Cantorian pseudomathematics I know one is not supposed to laugh at retarded people, but I can't > help myself ... > Yes, I know that you can't help yourself. http://hdebruijn.soo.dto.tudelft.nl/jaar2006/jirilebl.exe A MS-Windows executable showing the difference between a function as is and the same function as sensed. The good news is that sin(1/x) may serve as a mathematical model in science. Except for the region near x = 0 , but that shouldn't come as a surprise. The other conclusion is that the only sensible value of the first derivative of sin(1/x) for x = 0 is zero. (One doesn't need a modification like x^2.sin(1/x) for arriving at this insight.) The method employed is Gaussian smoothing of the exact functions. Nothing new, actually. But the idea to apply it on mathematical results, in order to see if they are empirically relevant, is new. Han de Bruijn === Subject: Re: Cantorian pseudomathematics ... > I think Han does not even understand what a function in the > function is analytic in common parlance. That is, it is not > only continuous, but is also everywhere infinitely many times > differentiable. At least, that is what I do understand. > > I've never heard a constructivist say that. All that's required is > that the function be computable for arbitrary precision reals. Even > nowhere-differentiable functions can be constructively valid. > > Depends on the constructivist you have in mind. When I look at the work > of Kock and Lawvere I see that the only functions are those that are > infinitely many times differentiable everywhere. Even the entier function > is differentiable everywhere in their view. > If the entier (step) function is differentiable in their view, that > should be a hint that they are using non-standard notions of > differentiable. Well, I do not take the hint because that is false. Your turn again. The only thing they are doing is not apply the law of the excluded middle. They acknowledge the existence of non-zero numbers whose squares are zero (i.e. very small numbers indeed). With that they proceed to a form of mathematics where, indeed, every function is everywhere differentiable. > On the other hand, what do you mean with computable for arbitrary > precision reals? I have no idea what this means. > With a little imagination, you could probably figure it out yourself. > One way to think about the reals in constructive mathematics is to No, otherwise I would not have asked. > first consider finite precision reals (rationals with uncertainty), and > then consider the limit as the uncertainty goes to zero. It makes > sense to apply functions to these finite precision reals (in fact, > applied mathematicians, scientists and engineers always use finite > precision reals in practice), and if the result of applying a function > to the finite precision reals converges as the uncertainty goes to > zero, then we can say that the function is computable for arbitrary > precision reals. But I think this definition is too broad and too narrow. (And, yes, I know quite a bit about numerical mathematics and the approximations of functions. I have done quite a bit in this field. But one of the first things the numberical mathematician is concerned with is how close his approximation is to the mathematically defined function. It has taken me quite some time just to find such error bounds.) But whatever. The above implies that a function with poles is not a function in the constructivist sense. So 1/x is not a function in the constructivist sense. Is that what you mean? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantorian pseudomathematics function in the constructivist sense. So 1/x is not a function in > the constructivist sense. Is that what you mean? The function f(x) = 1/x when x =/= 0, and f(0) = 31, is not a function in the constructive sense. === Subject: Re: Cantorian pseudomathematics > But whatever. The above implies that a function with poles is not a > function in the constructivist sense. So 1/x is not a function in > the constructivist sense. Is that what you mean? > The function f(x) = 1/x when x =/= 0, and f(0) = 31, is not a function > in the constructive sense. So what? That was not the function I was talking about. Pray answer my question. As far as I know (according to your definition) 1/x is not computable for arbitrary precision reals, so I would think it is not a function in your book. And when you say it is, when you restrict the range to all reals not equal to 0, I counter with 1/(x^2 - 2). That one is *not* computable for arbitrary precision reals, so at least that one is not a function in your book. Am I right? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantorian pseudomathematics computable for arbitrary precision reals, so I would think it is not a > function in your book. It is not a function defined on all of R. When the uncertainty of a given finite precision real is comparable in size to the real number itself, then we cannot compute its reciprocal. > And when you say it is, when you restrict the > range to all reals not equal to 0, I counter with 1/(x^2 - 2). That > one is *not* computable for arbitrary precision reals, so at least that > one is not a function in your book. Am I right? Again, it's not a function on all of R. It's not defined for 'x' when the uncertainty in 'x' is comparable to the value of x - sqrt(2) I admit I really don't get your point. === Subject: New mathematics/physical sciences positions at http://jobs.phds.org, February 27, 2006 New job listings at http://jobs.phds.org - Jobs for PhDs List your job at no cost! http://jobs.phds.org/jobs/post * Tenure-track Faculty Positions, Mt. San Antonio College : Academic Careers Online, Walnut, CA. 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By dealing with a certain statistical problem, I've found that, assuming l[n-1]<>l[n], the solution depends only on the 3 eigenvalues l[n],l[n-1],l[1], through the ratio (l[1]-l[n-1])/(l[n-1]-l[n]). As I would like to interpret that result, I was wondering if it is possible to say something at all about the structural properties of the graphs G having a low or large value of that ratio (at least when G is bipartite so that l[1]=l[n]). === Subject: Re: spectral graph theory question hi: let me write L[i], not l[i]. first, what means a simple graph? second, what is a certain statistical problem; what is the solution you are looking for? (Q1) OK: there are a couple of issues: are your eigen. values all positive or not? I think that you made some wrong assumption; but you are sort of right; depneds what is the answer on (Q1) usually things depend on lambda_1 - lambda_2; you can always say lambda_1 = 1 and lambda_2 =max{abs L[2], abs L[n]} [you can normalize things even in your ratio] see Perron-Frobenius theory === Subject: Re: spectral graph theory question The problem I was looking a solution for is actually a little complicated, basically there's a statistical model depending on the symmetric adjacency matrix of a graph G, and I was looking for conditions for the model to have some poor properties (I would avoid details unless anyone is particularly interested). It turns out that, when L[n-1]<>L[n], this occurs when the ratio (L[1]-L[n-1])/(L[n-1]-L[n]) is small (L[i] are the eigenvalues in non-increasing order). All I need for the proof is that the adjacency matrix is symmetric, and non-negative irreducible. But, if it helps to interpret the above result, I'm happy to assume that the graph is undirected, has no loops or multiple edges, and is even unweighted (that's what I meant by simple...). In this case eigenvalues are not all positive since they must sum to 0. hope this clarifies my question a little. rob. PS: any more hints/reference/examples for your statement usually things depend on lambda_1 - lambda_2? === Subject: Re: spectral graph theory question <9131347.1141149777265.JavaMail.jakarta@nitrogen.mathforum.org> rate of convergence depends on 1-lambda2 eg http://www.stanford.edu/~sdkamvar/papers/secondeigenvalue.pdf === Subject: Re: Fourier Transform, Smooth Functions >Why not look up the correct statement before posting it? >>What precisely is the correct statement here? > If int |f|^2 = 1 (you could claim that that condition > was covered under what you said about appropriate > normalizations) and g(s) is the Fourier transform of f > then > (int x^2 |f|^2 dx)(int s^2 |g(s)|^2 ds) >= c. > (Where the value of c will vary from treatment to treatment, > because people put the 2pi's in different places when they > define the Fourier transform.) > The problems with what you said were that you didn't > specify that int |f|^2 = 1, and you talked about > int x^2 f(x) instead of int x^2 |f(x)|^2. You are right, of course. And my notes were correct, though specific for real and even functions. But I made a typo with the squares of f and g. Sorry. (It was late, it was in the weekend and the TV was on) Han de Bruijn === Subject: Re: Fourier Transform, Smooth Functions > I have two questions about the relationship between the smoothness of a > function and its Fourier transform. I know that if a function is very > smooth, then its Fourier transform can decay faster than 1/s^p for any > p>0. But, can its Fourier transform decay faster than that if the > function is zero outside of [-1,1]? More precisely, > 1) Does there exist a function f(x) such that f is continuous, f(x) is > zero when abs(x) > 1, and its Fourier transform, g(s), is order exp( - > abs(s)^p) where p > 1/2? > 2) What is the largest value of p such that there exists a function f > such that f is continuous, f(x) is zero when abs(x) > 1, and its > Fourier transform, g(s), is order exp( - abs(s)^p)? Here's a way to produce functions f(x) such that g(s) is especially rapidly decreasing. Let {a_k} be an infinite sequence of positive reals such that sum(a_k) < oo, and let it be such that the sum converges very very slowly. Then look at the function g(s) = product sin(a_k s)/(a_k s). Let f(x) be the Fourier transform of g(s). Then f(x) will have compact support, and g(s) will decay faster than 1/s^p for all p. Now you want to find something that decays even faster. I contend, without giving a proof, that if the series {a_k} is well chosen and the sum converges slowly enough, then we can ensure that g(s) will decay more rapidly than exp( -s^m ) for any particular m < 1. Unfortunately, I'm not absolutely certain I'm right. I remember looking at this problem years ago, and the above is how I remember solving it. === Subject: Re: Fourier Transform, Smooth Functions On 26 Feb 2006 17:13:31 -0800, david petry >> I have two questions about the relationship between the smoothness of a >> function and its Fourier transform. I know that if a function is very >> smooth, then its Fourier transform can decay faster than 1/s^p for any >> p>0. But, can its Fourier transform decay faster than that if the >> function is zero outside of [-1,1]? More precisely, >> 1) Does there exist a function f(x) such that f is continuous, f(x) is >> zero when abs(x) > 1, and its Fourier transform, g(s), is order exp( - >> abs(s)^p) where p > 1/2? >> 2) What is the largest value of p such that there exists a function f >> such that f is continuous, f(x) is zero when abs(x) > 1, and its >> Fourier transform, g(s), is order exp( - abs(s)^p)? >Here's a way to produce functions f(x) such that g(s) is especially >rapidly decreasing. >Let {a_k} be an infinite sequence of positive reals such that >sum(a_k) < oo, and let it be such that the sum converges very very >slowly. Then look at the function g(s) = product sin(a_k s)/(a_k s). >Let f(x) be the Fourier transform of g(s). >Then f(x) will have compact support, And f will be infinitely differentiable. >and g(s) will decay faster than >1/s^p for all p. Now you want to find something that decays even >faster. I contend, without giving a proof, that if the series {a_k} is >well chosen and the sum converges slowly enough, then we can ensure >that g(s) will decay more rapidly than exp( -s^m ) for any particular m >< 1. >Unfortunately, I'm not absolutely certain I'm right. I remember looking >at this problem years ago, and the above is how I remember solving it. I believe this may very well be right. Here's a sloppy version of a proof (various statements of the form a ~ b would need to be converted to explicit inequalities to make an actual proof): First note that there exists c > 0 such that if |t| < 1 then (i) sin(t)/t <= exp(-c t^2). (Note that c below will denote various constants, differing from line to line.) Now suppose that m < 1. Choose eps > 0 so that (ii) 2 - m > (1 + 2 eps)/(1 + eps) (such an eps > 0 exists since 2 - m > 1). Let a_k = 1/k^(1+eps). (Note that the argument must use the fact that m < 1, since m = 1 is impossible, as I showed in another post. Here we see where m < 1 is used; we need sum a_k < infinity to give f compact support, and that requires eps > 0, hence requires m < 1. The fact that we can see where m < 1 is used seems to me to lend some credibility...) Suppose that s is large. Choose N so that s ~ 1/a_N. Now (i) shows that (iii) g(s) <= exp(-c sum_{k=N}^infinity a_k^2 s^2). Our choice of a_k shows that sum_{k=N}^infinity a_k^2 ~ 1/N^(1 + 2eps) = a_N ^ ((1 + 2 eps)/(1 + eps)) So (ii) gives sum_{k=N}^infinity a_k^2 > c a_N ^ (2 - m) ~ s^(m-2), so sum_{k=N}^infinity a_k^2 s^2 > c s^m, hence (iii) shows g(s) <= exp(-c s^m). Keen. ************************ David C. Ullrich === Subject: Re: Fourier Transform, Smooth Functions I really like David P's idea that represented the fourier transform as an infinite product. I am still working my way through the proofs by David U. Let me tell you about my attempts to grapple with this problem. I had an idea for a proof that has not been written out to show that p cannot be greater than 1. (David U's proof is shorter.) Maybe later I can post this proof. I also looked at the function f(x) = Exp( -1/(x-1)/(1-x) ) when abs(x) <=1 and f(x) = 0 otherwise. This function is infinitely differentiable and compactly supported. I numerically caculated the log(abs(g(s))) for s = 1, ... 40 and got -3.15, -8.143, -5.982, -6.567, -7.465, -8.697, -12.108, -10.05, -9.968, -10.223, -10.637, -11.173, -11.848, -12.761, -14.579, -14.156, -13.671, -13.634, -13.766, -13.997, -14.302, -14.675, -15.121, -15.666, -16.377, -17.533, -18.628, -17.468, -17.204, -17.162, -17.229, -17.367, -17.557, -17.793, -18.072, -18.396, -18.773, -19.222, -19.784, -20.584 These values were approximately equal to -3.22*sqrt(s), so g(s) seems to be order exp(-abs(1/s^p)) for any p < 1/2. Of course, these are just calculations and they are not a proof. === Subject: Re: Fourier Transform, Smooth Functions f(x) = Exp( -1/(x-1)/(1-x) ) when abs(x) <=1 and f(x) = 0 otherwise. > This function is infinitely differentiable and compactly supported. I > numerically caculated the log(abs(g(s))) for s = 1, ... 40 and got > -3.15, -8.143, -5.982, -6.567, -7.465, -8.697, -12.108, -10.05, > -9.968, -10.223, -10.637, -11.173, -11.848, -12.761, -14.579, > -14.156, -13.671, -13.634, -13.766, -13.997, -14.302, -14.675, > -15.121, -15.666, -16.377, -17.533, -18.628, -17.468, -17.204, -17.162, > -17.229, -17.367, -17.557, -17.793, -18.072, -18.396, -18.773, -19.222, > -19.784, -20.584 > These values were approximately equal to -3.22*sqrt(s), so g(s) seems > to be order exp(-abs(1/s^p)) for any p < 1/2. That's impressive! A back-of-the-envelope calculation using the method of steepest descent suggests that the Fourier transform of exp( -1/(1-x^2)^n ) will decay roughly like exp( - c s^m ) where m = n/(n+1). Also, the Fourier transform for exp( - exp( 1/(1-x^2) ) ) should decay like exp( - c s/log(s) ). If you choose to test these functions experimentally, let us know the === Subject: Re: Fourier Transform, Smooth Functions On 27 Feb 2006 06:45:26 -0800, irchans an infinite product. I am still working my way through the proofs by >David U. You mean the proof I gave that p = 1 is impossible or the more-or-less proof I just posted that Petry's example works? >Let me tell you about my attempts to grapple with this problem. I had >an idea for a proof that has not been written out to show that p cannot >be greater than 1. (David U's proof is shorter.) Maybe later I can >post this proof. I also looked at the function >f(x) = Exp( -1/(x-1)/(1-x) ) when abs(x) <=1 and f(x) = 0 otherwise. >This function is infinitely differentiable and compactly supported. I >numerically caculated the log(abs(g(s))) for s = 1, ... 40 and got >-3.15, -8.143, -5.982, -6.567, -7.465, -8.697, -12.108, -10.05, >-9.968, -10.223, -10.637, -11.173, -11.848, -12.761, -14.579, >-14.156, -13.671, -13.634, -13.766, -13.997, -14.302, -14.675, >-15.121, -15.666, -16.377, -17.533, -18.628, -17.468, -17.204, -17.162, >-17.229, -17.367, -17.557, -17.793, -18.072, -18.396, -18.773, -19.222, >-19.784, -20.584 >These values were approximately equal to -3.22*sqrt(s), so g(s) seems >to be order exp(-abs(1/s^p)) for any p < 1/2. >Of course, these are just calculations and they are not a proof. ************************ David C. Ullrich === Subject: Re: Fourier Transform, Smooth Functions Hi David, I think I understand the both the argument that p=1 is impossible and the exposition of Petry's example, but I still want to write out all the details for myself. I just started reading Rudin's section on quasi-analytic functions (Denjoy-Carleman theorem) and I'm sure it will the reference! Irchans === Subject: Re: Fourier Transform, Smooth Functions On 27 Feb 2006 13:13:14 -0800, irchans I think I understand the both the argument that p=1 is impossible >and the exposition of Petry's example, but I still want to write out >all the details for myself. Ok. > I just started reading Rudin's section on >quasi-analytic functions (Denjoy-Carleman theorem) and I'm sure it will >the reference! It's a worthwhile theorem, but I don't think that it settles the question you asked. At least I don't see how it does: Assume that g(s) is of order exp(-|s|^p), where p < 1. That implies bounds on the absolute value of the derivatives f^(n). The D-C theorem says that the class of all f satisfying those bounds is not quasi-analytic, hence contains a function F with compact support. But I don't see how it follows that the Fourier transform of F is of order exp(-|s|^p). >Irchans ************************ David C. Ullrich === Subject: LP - The shortest path with pleasure constraint(s) Hi all, I have been working with the following problem. Example: Travel agency wants to take people from city A to city B. Of course it wants to use a way that minimize total transportation costs. BUT it also looks for a customer satisfaction. It want to choose the way that gives to customer at least pre-defined level of satisfaction. E.g. number of stops to take photos of wonderful panorama must be larger than or equal to 7. The photo stop can't be made anywhere, but there are only limited number of places that could be used. Model of the problem: Graph with directed arcs (cycles are there so it is not a tree). Each arc has two numbers assigned - c_a: cost of traversing arcs a - p_a: pleasure that recieve customer by traversing arc a. All numbers c_a and p_a are positive. Let P by minimal allowed level of pleasure that is acceptable by cusomers. In the original example the p_a means the number of photostops on the arc a (we stop be definition on every photo stop). The LP formulation of the above problem is: Min Sum_a c_a arc_s S.t.: A arc = b ---- flow conservation constraints sum_a p_a arc_a >= P arc_a is binary {0,1} Simple computation by LP solver don't deliver a path. Sometimes it delivers the shortest path in the unconstrained sense (like without pleasure constraint) and several cycles that are not connected to the shortest path. I need hint for solving this problem. e.g. How to transform above formulation to the resource constrained shortest path problem (shown below)? In the similar case where p_a is not pleasure but resource, time or risk the model is as follows: Min Sum_a c_a arc_s S.t.: A arc = b ---- flow conservation constraints sum_a t_a arc_a <= T arc_a is binary {0,1} where t_a>=0 is time that consumes traversing arc a. Interpretation is as follows: Travel agency look for the cheapest way from the city A to the city B that lasts at most T. In this case simple LP solver delivers a solution that is acceptable. richard PS: sorry for any cross posting === Subject: Re: LP - The shortest path with pleasure constraint(s) Never played with LP solver; but it might use <= everywhere. If you look this 2 LP's the only difference is in the sum_a p_a try to use q=-p and say sum_a q_a arc_a <= -P Min Sum_a c_a arc_s S.t.: A arc = b ---- flow conservation constraints sum_a p_a arc_a >= P arc_a is binary {0,1} Min Sum_a c_a arc_s S.t.: A arc = b ---- flow conservation constraints sum_a t_a arc_a <= T arc_a is binary {0,1} === Subject: Re: LP - The shortest path with pleasure constraint(s) Yhank you for your efford, but it is not solution of my problem. The problem I reffer is modeling problem, not LP solver problem. Difference between satisfaction and resource constraint is fundamental. Maybe RCSP (resource consrained shortest path problem) allows be t_a <= 0 - I didn't find any comment on this. Maybe I overlooked some properties of of solution method.... The optimal solution of the formulation below of problem SCSP (satisfaction constrained shortest path) > Min Sum_a c_a arc_s > S.t.: > A arc = b ---- flow conservation constraints > sum_a p_a arc_a >= P -------- p_a>=0 > arc_a is binary {0,1} shoudn't be a path. That is problem. Very similar problem. The RCSP (resource constrained shortest path) that is formulated almost identically: > Min Sum_a c_a arc_s > S.t.: > A arc = b ---- flow conservation constraints > sum_a r_a arc_a <= R -------- r_a>=0 > arc_a is binary {0,1} is a path. Well, how to formulate SCSP problem correctly? Means that optimal solution of LP is a path. W p.92se v diskusn.92m pr.92spevku > Never played with LP solver; but it might use <= everywhere. > If you look this 2 LP's the only difference is in the sum_a p_a > try to use q=-p and say > sum_a q_a arc_a <= -P > Min Sum_a c_a arc_s > S.t.: > A arc = b ---- flow conservation constraints > sum_a p_a arc_a >= P > arc_a is binary {0,1} > Min Sum_a c_a arc_s > S.t.: > A arc = b ---- flow conservation constraints > sum_a t_a arc_a <= T > arc_a is binary {0,1} === Subject: Office XP Finally, you can afford software by Macromedia & Win XP by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with SMTP id k1RBbw305113 for ; Mon, 27 Feb 2006 06:37:59 -0500 --------------------------------------------------------------------- === Subject: Simplification - Joel Moses: Algebraic Simplification: A Guide for the Perplexed ................................................................ http://www-inst.eecs.berkeley.edu/~cs282/sp02/readings/moses-simp.pdf Algebraic simplification is examined first from the point of view of a user needing to comprehend a large expression, and second from the point of view of a designer who wants to construct a useful and efficient system. First we describe various techniques akin to substitution. These techniques can be used to decrease the size of an expression and make it more intelligible to a user. Then we delineate the spectrum of approaches to the design of automatic simplification capabilities in an algebraic manipulation system. Systems are divided into five types. Each type provides different facilities for the manipulation and simplification of expressions. Finally we discuss some of the theoretical results related to algebraic simplification. We describe several positive results about the existence of powerful simplification algorithms and the number- theoretic conjectures on which they rely. Results about the non- existence of algorithms for certain classes of expressions are included. P.S. By the way, to look for such stuff you may wish to use options like filetype:pdf ................................................................ J. Moses paper available ? JGP> Can somebody send me a ps/pdf copy of the historical JGP> J. Moses paper : Algebraic Simplification : a Guide ................................................................ Re: J. Moses paper available ? ................................................................ Re: J. Moses paper available ? ASFXvD> It is available at the ACM website - but not for free. This is right. If you can afford this, we urge you to acquire it here. http://portal.acm.org/citation.cfm?id=806298 ................................................................ === Subject: Re: Mandelbrot Study of Polysigned Numbers The three-signed numbers are equivalent to the complex numbers so beg the question of what the higher signs do under Mandelbrot and Julia conditions. Here are the most basic results: http://bandtechnology.com/PolySigned/Mandelbrot/MandelbrotStudy.html As you can see it's not as if each consecutive sign level brings on radical new complexity. If anything it seems to be the other way around. Here is a closeup of the little hook at the top of P9: http://bandtechnology.com/PolySigned/Mandelbrot/P9MandelHook.png These images are just 2D slices of high dimensional objects. The parity demonstrated in these images is not fully understood. Any advice is welcome as to how best to characterize the parity behaviour. P4 is the easiest to work in since it can be visualized with our 3D sense. I don't think it is sufficient to claim that the even signs are broken because then one would conclude that the real numbers are broken. If this is augmented with the magnitude behavior of the product then this argument is a bit more valid but the exact nature of the behavior still has not been characterized. Please comment. -Tim === Subject: Re: The co-ordinate of midpoint of orthogonal path I know it I confused God bless you sorry if I fazed you === Subject: Symmetry-adapted polynomes How do I construct the symmetry-adapted polynomes (under the permutation group S_n of n variables) which are irreducible representations? E.g. {a*b-c*d,a*c-b*d,a*d-b*c} should be a S4-F2 (=Td-T2) adaption. That one looks fairly symmetric, but e.g. {a-b,2*a-b-c} is a standard S3-E (C3v-E) representation and looks extremely asymmetric. -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de His-Ala-Sec-Lys-Glu Arg-Glu-Asp-Asp-Met-Ala-Asn-Asn === Subject: Re: Mandelbrot set > Does anybody know of any specific methods for continuously coloring the > points which are not in the Mandelbrot set? It would be nice if the colors had some meaning. The stable points are really the ones with meaning. They are connected by the iterator that tests them. So for example for an initial point z(0) that passes the Mandelbrot test all of its successors z(n) have the same character. They are in the set and are siblings. This is also a means of speeding up the algorithm. It introduces some difficult but meaningful problems. We are gridding the ideal set, so it would be sensible to assign the z(n) successors to that gridded location so that they map to a pixel. Now, many pixels will be assigned multiple values, so there is added complexity. It seems like the result should be a continuous pattern. i.e. the neighbors will match. I don't see any strict rules within this paradigm. You are stuck with deciding whether to overwrite a pixel that has already been assigned or not, or to conserve all of the gathered information. I don't know how to do the latter nicely but I think that is where the most meaningful representation is. -Tim === Subject: Re: Mandelbrot set All of this is true of your complementary set too. -Tim === Subject: Re: Mandelbrot set My apologies. These statements that I made are in error. Because the algorithm relies upon z0 it is not true that the successors are in the set since their base references are themselves, not z0. That being said, the results of this algorithm are interesting. I'll post a link shortly. -Tim === Subject: Re: Mandelbrot set Following is a link to a graph of the Mandelbrot set plus its successor points z[n]. Not very special but since I opened my mouth I have to put my foot in now. http://bandtechnology.com/Fractal/MandelbrotSuccessors/MandelbrotSuccessors. html The files at this link will not be maintained and may be deleted. I didn't spend much time on these so there may be errors here. Indeed the whole concept is based upon a conceptual error. Still, the graphs are interesting. -Tim === Subject: Re: Mandelbrot set >For those who aren't familiar with the set, to each complex number c is >assigned a sequence z_n, defined by z_0 = 0, and z_(n + 1) = (z_n) ^ 2 >- c. Those numbers c for which {z_n} is bounded are in the set. When >this set is shown graphically, color is usually added. One way of >coloring the points not in the set is to define f(c) = min {n : |z_n| 2}. Thus, for each point not in the set, we get a positive integer. >The color is then determined by this integer. Sometimes, though, it is >colored in a continuous fashion, rather than in this discrete method. >Does anybody know of any specific methods for continuously coloring the >points which are not in the Mandelbrot set? One method I thought of is >such: >Define f(c) = limit Log[z_n] / 2^n as n --> infinity. For starters, >though, I am not even positive that this limit exists. Does anybody >have any insights? A couple of thoughts: 1. There are regions of the exterior where the discrete coloring method you mentioned has a high gradient. If you display a region such that the gradient is greater than 1 color/pixel, it is hard for the observer to tell whether it is a continuous or discrete function. 2. Define g(z) to be some function from the 2-D annulus 2 < |z| <= 4 to RGB (or other 3-D color space). Let f(c) = g(z_n), where n is the first entry of c's series into the annulus. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: A question about netiquette (especially, for the Maple actual and potential customers over the globe) By the way, I like very much your amazing comments on Maple oddities, as you christen them. At least, many of them are Maple bugs. It is remarkable, that, before your postings, our VM machine (and we) did not realize the existence of some types of Maple bugs. There are more things in heaven and earth, Horatio, Then are dreamt of in your philosophy... :) Well, to do the VM justice, currently, discovering Maple bugs is just a not large fraction of its job ;) We already introduced in the VM some improvements so in the future it will be possible to it to identify the bug types you have published. I always read your Maple postings with keen interest, let me thank you cordially for the delight gained! === Subject: Re: A question about netiquette (especially, for the Maple actual and potential customers over the globe) Mr. Bondarenko, could you possibly explain what it is that you are trying to achieve with your high volume of daily postings to this and a few other groups? I am sure that you are trying to make a point. What is it, please? === Subject: Re: A question about netiquette (especially, for the Maple actual and potential customers over the globe) ASFXvD> What is it, please? Here are some initial ideas about one of our directions. ................................................................ Maple bugs: Backward compatibility - 1 Re: Maple bugs: Backward compatibility - 1 Re: Maple bugs: Backward compatibility - 1 http://www.cybertester.com/intro.php ................................................................ ASFXvD> What is it, please? We are aimed at several goals. One of our long-run goals is practical, efficient support of new generation computer algebra systems. We are going to discuss this in detail in good time. An integral feature of these systems is outstanding speed, user-friendliness, and, last but not least, mathematical correctness. Maybe, some of these systems might stem from the current systems like Mathematica and Axiom. Let's wait and see. One of our short-term goals is tangible increase in math correctness all the modern commercial and free computer algebra systems. This will be done via essences I'd, for the time being, much prefer to abbreviate to SAT and SM, without decipherment. Okay, again, I am somewhat previous here, do you think, is it possible to implement a software-quality-assurance version of Professor Augustus S. F. X. Van Dusen ? ;) Viva la Prof Augustus S. F. X. Van Dusen, Vladimir Bondarenko VM and GEMM architect Co-founder, CEO, Mathematical Director http://www.cybertester.com/ Cyber Tester, LLC http://maple.bug-list.org/ Maple Bugs Encyclopaedia http://www.CAS-testing.org/ CAS Testing === Subject: Re: Fermat's last theorem and a counter example You can see the short and wonderful proof that Fermat had known few centuries back at this link http://www.blogit.com/Blogs/Comments.aspx/345746 or simply destroy the proof by finding a counter example to my triangle with integer sides (z>y>x) with the conditions I provided in this thread. Bassam Karzeddin AL-Hussein Bin Talal University JORDAN === Subject: Re: 600 cell question >I was thinking about the following simple problem: >Prove that the 600 cell has 600 cells. The 600-cell is the dual of the 120 cell. That the 120-cell has 120 cells can be seen from the properties of icosahedral symmetry: http://www.quadibloc.com/math/fdiint.htm John Savard http://www.quadibloc.com/index.html _________________________________________ Usenet Zone Free Binaries Usenet Server More than 140,000 groups Unlimited download http://www.usenetzone.com to open account === Subject: Re: 600 cell question- > But this is as far as I have been able to get with an elementary > argument. Once we know we have 120 vertices, it easily follow that > there are 600 cells, and vice versa. But my question is whether there > is a way to get to either absolute number (120 or 600) without actually > constructing the polytope. The answer is simply, no. With the formula's you are using, it is possible that there is a 1200 cell with 240 vertices. You need a bit more to show that only 120 vertices and 600 cells are possible. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: 600 cell question > just two notes: > the formula, F+V=E+2, is the same as the 4D one, > taking 2 as C; Let us be precise. In n-dimensional space the formula is: N_0 - N_1 + N_2 - N_3 + ... + (-1)^n.N_n = 1 where N_i is the number of i-dimensional boundaries. In 2D, 3D and 4D we have N_0 = V, N_1 = E, N_2 = F, N_3 = C and N_4 = H where H is the number of hypercells (1 in 4D). (See that it also matches when we set F = 1 in 2D and C = 1 in 3D.) So actually you should not take 2 as C, but 2 is split as 1 - (-1)^n. in the complete alternating formula. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: 600 cell question now, *that* is overkill. --Give Earth a Trickier Dick Cheeny -- out of office, after gigayears! http://tarpley.net/bush12.htm http://www.benfranklinbooks.com/ http://members.tripod.com/~american_almanac http://www.wlym.com/pdf/iclc/howthenation.pdf http://www.rand.org/publications/randreview/issues/rr.12.00/ http://www.rwgrayprojects.com/synergetics/plates/figs/plate01.html === Subject: Re: Rational trigonometry There is an artice comparing Rational Trigonometry with Classical Trigonometry. See http://www.geocities.com/xyz.abc_123/rationalTrig/comparison.html === Subject: Re: Regarding tan(x) = x > Sorry that I don't know the answers to your questions. > But you might, nonetheless, be interested in expressions > (2) and (3) at . > BTW, do you know who first came up with that series for > the solutions? I should think that someone (Fourier?) > found it long before I did. > I'm replying to my most recent post in this thread because > I finally have some information about the asymptotic series > for the zeros of the solutions to tan(x) = x: > q - (1/q) - (2/3)*(1/q^3) - (13/15)*(1/q^5) - (146/105)*(1/q^7) - ... > where q = (2k+1)*Pi/2. > Apparently, this series was independently obtained by > Euler [1] (1748), Cauchy [2] (1827), and Rayleigh [3] (1877). I'll also take this opportunity to note that, as some of the references below indicate, in the same way that we can get an asymptotic series for the roots of tan(x) = x, we can also get asymptotic series for the roots of various closely related equations, such as cot(x) = c*x or tanh(x) = x. David W. Cantrell > [1] Leonhard Euler, Introductio in Analysin Infinitorum, > Volume 2, 1748. [Reprinted in Euler's OPERA OMNIA > Series 1, Volume 9.] > See pp. 318-320. > This also appears on pp. 323-324 of the French > translation Introduction =E0 l'Analyse Infinit=E9simale > that is on the internet at > http://visualiseur.bnf.fr/Visualiseur?Destination=3DGallica&O=3DNUMM-3885 > See two-thirds down p. 323 for the series. > [2] Augustin-Louis Cauchy, Th=E9orie de la Propagation > des Ondes =E0 la Surface d'un Fluide Pesant d'une > Profondeur Ind=E9finie, 1827. > Oeuvres compl=E8tes, Series 1, Volume 1, pp. 5-318. > http://math-doc.ujf-grenoble.fr/cgi-bin/oetoc?id=3DOE_CAUCHY_1_1 > According to [5] (p. 273 & 275), the series is developed > on pp. 277-278 (p. 272 of the original 1827 publication). > [3] John William Strutt Rayleigh, Theory of Sound, > 1877-1878. [Reprinted by Dover in 1945 and 1976.] > http://tinyurl.com/n9n54 > See p. 334, which the URL above will take you to. > [4] Raymond Clare Archibald and Henry Bateman, Roots > of the equation tan x =3D cx, Note #8, Mathematical > Tables and Other Aids to Computation (=3D Mathematics > of Computation) 1 #6 (April 1944), 203. > See the follow-ups by Henry Bateman (Vol. 1 #8, > October 1944, p. 336), Raymond Clare Archibald > (Vol. 1 #12, October 1945, p. 459), Abraham P. > Hillman and Herbert E. Salzer (Vol. 2 #14, > April 1946, p. 95), and L. G. Pooler > (Vol. 3 #27, July 1949, pp. 495-496). > [5] Raymond Clare Archibald and Henry Bateman, A guide > to tables of Bessel functions, Mathematical Tables > and Other Aids to Computation (=3D Mathematics of > Computation) 1 #7 (July 1944), 205-308. > See Section F: Series for the zeros of > Bessel functions, pp. 271-275. > Dave L. Renfro === Subject: Re: Regarding tan(x) = x <20060227095638.921$Ya@newsreader.com I'll also take this opportunity to note that, as some > of the references below indicate, in the same way that > we can get an asymptotic series for the roots of > tan(x) = x, we can also get asymptotic series for > the roots of various closely related equations, > such as cot(x) = c*x or tanh(x) = x. Here's another reference for the asymptotic series of the solutions to tan(x) = x: George Neville Watson, A TREATISE ON THE THEORY OF BESSEL FUNCTIONS, 2'nd edition, Cambridge University Press, 1944, viii + 804 pages. At the bottom of p. 506 there is an asymptotic formula for the zeros of the cylinder function (J_nu)(z)*cos(alpha) - (Y_nu)(z)*sin(alpha) If you consider the special case where nu = 3/2 and alpha = 0, you'll get the asymptotic expansion for the solutions to tan(x) = x. [Recall that the zeros of the Bessel function of order 3/2, J_{3/2}, are the solutions to tan(x) = x.] http://tinyurl.com/ohdpy [p. 506 of Watson's book] By the way, I haven't figured out why Cauchy should get credit for this asymptotic series yet. What are your thoughts? See my previous post in this thread for the details. Dave L. Renfro === Subject: Re: random maple oddities Re: random maple oddities WR> #23: Maple 7, fixed by Maple 10 WR>> minimize(10^(1/2)*y^2/(9+y^2)^(3/2)-10^(1/2)/(9+y^2)^(1/2) WR>> +17^(1/2)*y^2/(16+y^2)^(3/2)-17^(1/2)/(16+y^2)^(1/2), y= -1..1); WR> Error, (in factors) argument should be an algebraic function field [if you are going to reproduce this, be warned about notorious extra '-' Google inserts - before Maple calculation, delete 'em] Even much simpler... DESCRIPTION: Only Maple 10 of 1005, and Maple V Rel 3 of 1994 can minimize this function correctly. TEST CASE: minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z= 0..1); EXPECTED: 1/4*2^(1/2)+1/3 .6868867238 CHECKUP: with(Optimization): Minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z= 0..1); [.686886723926607100, [z = 1.]] --------------- (2005) Maple 10.00 ------------------------------ 1/4*2^(1/2)+1/3 # = .6868867238 Error, (in evala/Indep) argument should be an algebraic function field Error, (in evala/Indep) argument should be an algebraic function field Error, (in evala/Indep) argument should be an algebraic function field --------------- (2002) Maple 8 ---------------------------------- Error, (in factors) argument should be an algebraic function field --------------- (2001) Maple 7 ---------------------------------- Error, (in factors) argument should be an algebraic function field --------------- (2000) Maple 6 ---------------------------------- Error, (in RootOf/selector) selector must be a constant, a numeric range or an equation --------------- (1997) Maple V Rel 5 ---------------------------- Error, (in RootOf/selector) selector must be a constant or a numeric range --------------- (1995) Maple V Rel 4 ---------------------------- 3/2 <-------------------------------------------------------- BUG --------------- (1994) Maple V Rel 3 ---------------------------- 1/12*(3+2*2^(1/2))*2^(1/2) # = .6868867236 ------------------------------------------------------------------ COMMENT: For Maple V Release 5 downwards, use this syntax minimize(1/(1+z^2)^(3/2)+1/(2+z^2), {z},{z=0..1}); Re: random maple oddities WR> While preparing the previous posting, for completeness WR> I hopped over to Maple 10 to confirm the problem there WR> -- and found that Maple 10 was able to find in a few WR> seconds what Maple 7 wasn't able to find in hours. It's not the occasion for rejoicing. It's trivial to modify your example... not it turns out, a Maple bug persists during 12 years in line, since 1994 on to today. DESCRIPTION: Only Maple V Release 3 of 1994 can minimize this function correctly. TEST CASE: minimize(10^(1/2)*y^4/(9+y^2)^(3/2)-10^(1/2)/ (9+y^2)^(1/2)+17^(1/2)*y^2/(16+y^2)^(3/2)-17^ (1/2)/(16+y^2)^(1/2), y= -1..1); EXPECTED: -1/4*17^(1/2)-1/3*10^(1/2) -2.084868959 CHECKUP: with(Optimization): Minimize(10^(1/2)*y^4/(9+y^2)^(3/2)-10^(1/2)/ (9+y^2)^(1/2)+17^(1/2)*y^2/(16+y^2)^(3/2)-17^ (1/2)/(16+y^2)^(1/2), y= -1..1); [-2.08486895979387476, [y=-.42167422949619349e-7]] --------------- (2005) Maple 10.00 ------------------------------ minimize(10^(1/2)*y^4/(9+y^2)^(3/2)-10^(1/2)/(9+y^2)^(1/2)+17^(1/2 )*y^2/(16+y^2)^(3/2)-17^(1/2)/(16+y^2)^(1/2),y = -1 .. 1) Error, (in evala/Indep) argument should be an algebraic function field Error, (in evala/Indep) argument should be an algebraic function field Error, (in evala/Indep) argument should be an algebraic function field --------------- (2002) Maple 8 ---------------------------------- Error, (in factors) argument should be an algebraic function field --------------- (2001) Maple 7 ---------------------------------- Error, (in factors) argument should be an algebraic function field --------------- (2000) Maple 6 ---------------------------------- Error, (in RootOf/selector) selector must be a constant, a numeric range or an equation --------------- (1997) Maple V Rel 5 ---------------------------- Error, (in RootOf/selector) selector must be a constant or a numeric range --------------- (1995) Maple V Rel 4 ---------------------------- Error, (in fsolve) 3.921195205, is an invalid option --------------- (1994) Maple V Rel 3 ---------------------------- -1/4*17^(1/2)-1/3*10^(1/2) # = -2.084868959 ------------------------------------------------------------------ COMMENT: For Maple V Release 5 downwards, use this syntax minimize(..., {z},{z=0..1}); Maple is chock-full with regression bugs. Best wishes, Vladimir Bondarenko VM and GEMM architect Co-founder, CEO, Mathematical Director http://maple.bug-list.org/ Maple Bugs Encyclopaedia http://www.cybertester.com/ Cyber Tester, LLC http://www.CAS-testing.org/ CAS Testing === Subject: Re: random maple oddities DESCRIPTION: Only Maple 10 of 1005, and Maple V Rel 3 of 1994 > can minimize this function correctly. [...] Wow. I'd have thought that software released in 1005 would be pretty much worthless at this point. --- Christopher Heckman === Subject: Re: random maple oddities > [...] > DESCRIPTION: Only Maple 10 of 1005, and Maple V Rel 3 of 1994 > can minimize this function correctly. [...] > Wow. I'd have thought that software released in 1005 would be pretty > much worthless at this point. Euclid's Algorithm was released way before that, and is still doing good business. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: random maple oddities ? Maple V R4 : > minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z, 0..1); 1/2 1/4 2 + 1/3 Chris > Even much simpler... > DESCRIPTION: Only Maple 10 of 1005, and Maple V Rel 3 of 1994 > can minimize this function correctly. > TEST CASE: minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z= 0..1); > EXPECTED: 1/4*2^(1/2)+1/3 > .6868867238 > CHECKUP: with(Optimization): > Minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z= 0..1); > [.686886723926607100, [z = 1.]] > --------------- (2005) Maple 10.00 ------------------------------ > 1/4*2^(1/2)+1/3 # = .6868867238 > Error, (in evala/Indep) argument should be an algebraic function > field > Error, (in evala/Indep) argument should be an algebraic function > field > Error, (in evala/Indep) argument should be an algebraic function > field > --------------- (2002) Maple 8 ---------------------------------- > Error, (in factors) argument should be an algebraic function field > --------------- (2001) Maple 7 ---------------------------------- > Error, (in factors) argument should be an algebraic function field > --------------- (2000) Maple 6 ---------------------------------- > Error, (in RootOf/selector) selector must be a constant, a numeric > range or an equation > --------------- (1997) Maple V Rel 5 ---------------------------- > Error, (in RootOf/selector) selector must be a constant or a > numeric range > --------------- (1995) Maple V Rel 4 ---------------------------- > 3/2 <-------------------------------------------------------- BUG > --------------- (1994) Maple V Rel 3 ---------------------------- > 1/12*(3+2*2^(1/2))*2^(1/2) # = .6868867236 > ------------------------------------------------------------------ === Subject: Re: so it is irrational If x(n) is the sequence which grows faster doesn't mean summation x(n) is a convergent series converging to irrational . For example x(n) =2^n is an increasing sequence .But áx(n) }= á1/2^n =á(1/2)^n is a convergent series and converges to 1/(1 + 1/2) =2/3 which is rational number. But ,if x(n) = a^n for some positive integer then á1/x(n)=á1/a^n =á(1/a)^n =1/(1+a) which is a rational no . If x(n)=if x(n) = a^n for some negative integer i.e. a = -b then á1/x(n)=á1/a^n =á(1/a)^n =1/(1+a) = 1/(1-b) which is a rational no . I hope this will help u to understand the concept. ALL THE BEST. === Subject: Re: so it is irrational > I will maybe say idiot things but it seems that if x_n is a sequence > of integers that grows fast (faster that n^2 for example), then the > number L=sum(1/x_n, n=1..infinity) is irrational. It is true if each > x_n is a power of a certain integer a: x_n=a^(f(n)), because if one > look the a-inary expansion of L, it is not periodic. But is my guess > true ? Well, n^2 is not fast enough by a long shot to ensure that the sum will be irrational, but yes, the basic idea is right. I believe the sequence must grow faster than N^(2^n)) for all N to ensure that it will be irrational. If the sequence grows really really fast, the sum will be guaranteed to be transcendental. === Subject: Re: so it is irrational <250220062055567208%anniel@nym.alias.net.invalid> silly me, I must have been really tired. The real problem is when x_n grows really fast like x_(n+1)>(x_n)^2 for example. This is why I talked about a non periodic a-inary expansion. Sorry again. So now, is it true ? === Subject: Re: so it is irrational >silly me, I must have been really tired. The real problem is when x_n >grows really fast like x_(n+1)>(x_n)^2 for example. This is why I >talked about a non periodic a-inary expansion. Sorry again. >So now, is it true ? I.e., if x_n are integers > 1 with x_{n+1} > (x_n)^2, is sum_n 1/x_n irrational? Almost: the sum is irrational if the infinite product product_{n=1}^infinity x_{n+1}/(x_n)^2 diverges to +infinity, and also x_{n+1} >= r x_n for some r > 1. So e.g. x_{n+1} >= (1 + 1/n) (x_n)^2 would work. Let b_n = product_{j=1}^n x_j, so that the partial sum s_n = sum_{j=1}^n 1/x_n = a_n/b_n where a_n is a positive integer. Now if s = sum_{n=1}^infty 1/x_n = p/q, 0 < s - s_n = sum_{j=n+1}^infty 1/x_j. Now since x_j >= r^(n+1-j) x_{n+1} for j >= n+1, s - s_n <= sum_{j=n+1}^infty 1/(r^(n+1-j) x_{n+1}) = r/((r-1) x_{n+1}). But s - s_n = (p b_n - a q)/(q b_n) >= 1/(q b_n) so we must have x_{n+1} < r q b_n /(r - 1). On the other hand, x_{n+1} = x_1 product_{j=1}^n (x_{j+1}/x_j) = x_1 b_n product_{j=1}^n (x_{j+1}/(x_j)^2) > r q b_n/(r - 1) for n sufficiently large, so we get a contradiction. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: so it is irrational <250220062055567208%anniel@nym.alias.net.invalid> 1 with x_{n+1} > (x_n)^2, is > sum_n 1/x_n irrational? > Almost: the sum is irrational if the infinite product > product_{n=1}^infinity x_{n+1}/(x_n)^2 > diverges to +infinity, and also x_{n+1} >= r x_n for some r > 1. > So e.g. x_{n+1} >= (1 + 1/n) (x_n)^2 > would work. I think somewhat more than this is possible. Consider the greedy algorithm for Egyptian fractions. If the algorithm is run on a rational number then it terminates. So if we can show that the infinite sequence {x_n} is greedy, then sum(1/x_n) must be irrational. So when is {x_n} greedy? Since 1/k - 1/(k+1) = 1/(k^2 + k), a necessary condition is x_(n+1) > (x_n)^2 + x_n for all n. Now if x_0 > 1 and x_n = (x_n)^2 + x_n + 1 for all n, then sum(1/x_n) = 1/(x_0 - 1). Therefore another necessary condition is x_(n+1) > (x_n)^2 + x_n + 1 for infinitely many n. It seems to me these 2 necessary conditions are sufficient. The first one says that x_(n+1) is large enough that x_n couldn't have been chosen any smaller, and the second one says that the sum of the terms after x_n is small enough that x_n couldn't have been chosen any smaller. So for all n, x_n is the smallest possible, and would have been chosen by the greedy algorithm. === Subject: Re: so it is irrational >> I.e., if x_n are integers > 1 with x_{n+1} > (x_n)^2, is >> sum_n 1/x_n irrational? >> Almost: the sum is irrational if the infinite product >> product_{n=1}^infinity x_{n+1}/(x_n)^2 >> diverges to +infinity, and also x_{n+1} >= r x_n for some r > 1. >> So e.g. x_{n+1} >= (1 + 1/n) (x_n)^2 >> would work. >I think somewhat more than this is possible. Consider the >greedy algorithm for Egyptian fractions. If the algorithm is >run on a rational number then it terminates. So if we can >show that the infinite sequence {x_n} is greedy, then >sum(1/x_n) must be irrational. >So when is {x_n} greedy? >Since 1/k - 1/(k+1) = 1/(k^2 + k), a necessary condition is >x_(n+1) > (x_n)^2 + x_n for all n. I think this should be x_{n+1} > (x_n)^2 - x_n, so you can't replace 1/x_n + 1/x_{n+1} by 1/(x_n - 1). >Now if x_0 > 1 and x_n = (x_n)^2 + x_n + 1 for all n, then >sum(1/x_n) = 1/(x_0 - 1). I think you mean if x_{n+1} = (x_n)^2 - x_n + 1 for all n, then sum_{n=1}^infty 1/x_n = 1/(x_0 - 1). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: so it is irrational Now if x_0 > 1 and x_n = (x_n)^2 + x_n + 1 for all n, then >sum(1/x_n) = 1/(x_0 - 1). > I think you mean if x_{n+1} = (x_n)^2 - x_n + 1 for all n, then > sum_{n=1}^infty 1/x_n = 1/(x_0 - 1). Absolutely right; stupid sign error. So here are my real necessary and sufficient conditions for {x_n} to be a greedy sequence: (1) x_(n+1) > (x_n)^2 - x_n for all n. (2) x_(n+1) > (x_n)^2 - x_n + 1 for infinitely many n. Does this look right? === Subject: Re: so it is irrational >>Now if x_0 > 1 and x_n = (x_n)^2 + x_n + 1 for all n, then >>sum(1/x_n) = 1/(x_0 - 1). >> I think you mean if x_{n+1} = (x_n)^2 - x_n + 1 for all n, then >> sum_{n=1}^infty 1/x_n = 1/(x_0 - 1). >Absolutely right; stupid sign error. So here are my real >necessary and sufficient conditions for {x_n} to be a >greedy sequence: >(1) x_(n+1) > (x_n)^2 - x_n for all n. >(2) x_(n+1) > (x_n)^2 - x_n + 1 for infinitely many n. >Does this look right? Yes, I think so (with x_0 > 1). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: so it is irrational <250220062055567208%anniel@nym.alias.net.invalid> === Subject: New mathematical twists I would like to inform all of you that there is an interesting book recently publishedNew Mathematical Twists. This book is about a whole new way of thinking about mathematics. The solutions to the problems in the book will lead you to a hidden treasure. Who knows, the lucky winner may be one of the members of this board. On top of the hidden treasure, you will find many interesting facts about life in general. You can order a copy of the book on this website: www.mathtwists.com. === Subject: Equation animator? Can anyone recommend a cheap or freeware program for animation of time-dependent sine and cosine graphs? Tim (For email dr o p fir ma) === Subject: Re: Equation animator? > Can anyone recommend a cheap or freeware program for animation of > time-dependent sine and cosine graphs? > Tim > (For email dr o p fir ma) I think you can do this with gnuplot (using the reread command) Duncan === Subject: Re: Equation animator? You might investigate Camtasia Studio. We've used it to capture videos from simulations. I believe it may be free-ware. At least it is a start. === Subject: Re: Equation animator? Sorry, Camtasia (from TechSmith) costs about $300 (my girlfriend and software buyer for our business has corrected me). Snag-It is about $30 and she uses it all the time. The fun thing, as I remember, about Camtasia is you need a .AVI decoder (tscc.exe). Now, the so-called smart people say .AVI? I can open that! but there are about 20 different types of .AVI. The problem with tscc.exe is not the cost (it is free) but that most major companies and certainly the government will not allow an executable to be loaded on their nets. I asked a Colonel in Hawaii if he could load it and he said he'd need seven levels of signature to do it! === Subject: Deconvolution in 2D hi.. i doing my final year engineering project and am stuck at the below?? Can anybody please help me with this!! i have two matrices for FFT2(m) and FFT2(w) respectively.. This is the equation!! FFT2(m) = FFT2(R) * FFT2(w) the product is the convolution. I wanna find out FFT2(R).How to do deconvolution in 2D in matlab in order to find out FFT2(R). or basically how do u do this anyhow!!? === Subject: Re: Axioms for the real numbers http://www.dcproof.com/RealNumbers.html > Just a comment on HTML character rendering. Rather than using > an GIF image for your operator symbols, you might consider > using the HTML character entity encodings. You are right, of course. But the problem is that dcproof expressions are not formal strings of mathematical symbols. They are sequences of arbitrary graphic images, created ad hoc as needed - just as axioms are introduced ad hoc as needed by very liberal rules that allow you to declare any sequence of images to be a theorem (in effect an axiom) during a proof. The idea that the proofs come from a single set of axioms (ZF, PA, etc.) is far from the truth. There is a fixed set of rules, but they include a blank check for slipping in arbitrary conclusions (which are typically the heart of the argument) during proofs. In other words, the system is not sound or consistent. It can prove any sequence of graphic images! C-B (These require > a sufficiently up-to-date browser to render properly, of course.) > Unicode character charts are available in PDF format at: > http://www.unicode.org/charts/symbols.html > Specifically, mathematical symbols are in these PDF files: > http://www.unicode.org/charts/PDF/U2200.pdf > http://www.unicode.org/charts/PDF/U2A00.pdf > http://www.unicode.org/charts/PDF/U27C0.pdf > http://www.unicode.org/charts/PDF/U2980.pdf > The way to render these in HTML is � where the DDDD > portion is the hexadecimal code for the character. > For example, the member-of symbol is ∈ or ∊ > the for-all symbol is ∀ > and the exists symbol is ∃. === Subject: Re: Axioms for the real numbers <2jHMf.249$Xc3.142@newsfe20.lga >> Just a comment on HTML character rendering. Rather than using >> an GIF image for your operator symbols, you might consider >> using the HTML character entity encodings. > course. But the problem is that dcproof expressions > are not formal strings of mathematical symbols. They are sequences of > arbitrary graphic images, created ad hoc as needed - just as axioms are > introduced ad hoc as needed by very liberal rules that allow you to > declare any sequence of images to be a theorem (in effect an axiom) > during a proof. > The idea that the proofs come from a single set of axioms (ZF, PA, > etc.) is far from the truth. There is a fixed set of rules, but they > include a blank check for slipping in arbitrary conclusions (which are > typically the heart of the argument) during proofs. In other words, > the system is not sound or consistent. It can prove any sequence of > graphic images! > Sorry, Charlie. That simply isn't true. As far as I know (barring anymore > bugs in my program), you cannot prove EXIST(x):~x=x. I challenge you to > prove otherwise. Oh , I was thinking about the metamath thread. So sorry! My bad. Metamath is the big BS system that appears in this forum from time to time. I have no problem with dcproof (as far as I have looked.) I tried it out and it's pretty cool, Dan! (The kiss of death?) I'll be for ya' or 'again ya', whichever helps. - Political saying C-B > Dan > Download my DC Proof at http://www.dcproof.com === Subject: Re: Axioms for the real numbers Fixed typo >> http://www.dcproof.com/RealNumbers.html >> Just a comment on HTML character rendering. Rather than using >> an GIF image for your operator symbols, you might consider >> using the HTML character entity encodings. The next release of DC Proof will use the unicode character for epsilon and I will continue to ship the gif file so that older HTML versions can be read. > course. But the problem is that dcproof expressions > are not formal strings of mathematical symbols. They are sequences of > arbitrary graphic images, created ad hoc as needed - just as axioms are > introduced ad hoc as needed by very liberal rules that allow you to > declare any sequence of images to be a theorem (in effect an axiom) > during a proof. > The idea that the proofs come from a single set of axioms (ZF, PA, > etc.) is far from the truth. There is a fixed set of rules, but they > include a blank check for slipping in arbitrary conclusions (which are > typically the heart of the argument) during proofs. In other words, > the system is not sound or consistent. It can prove any sequence of > graphic images! Sorry, Charlie. That simply isn't true. As far as I know (barring anymore bugs in my program), you cannot prove EXIST(x):~x=x. I challenge you to prove otherwise. Dan Download my DC Proof at http://www.dcproof.com === Subject: Normal Distribution A certain job is completed in three steps in a series. The means and standard deviations for the steps are (in minutes): Step....Mean.....S.D. 1........17.......2 2........13.......1 3........13.......2 Assuming independent steps and normal distributions, compute the probability that the jobs will take less than 40 minutes to complete. I said the total mean=43 min, total s.d. = 5. === Subject: Re: Normal Distribution > A certain job is completed in three steps in a series. The means and standard deviations for the steps are (in minutes): > Step....Mean.....S.D. > 1........17.......2 > 2........13.......1 > 3........13.......2 > Assuming independent steps and normal distributions, compute the probability that the jobs will take less than 40 minutes to complete. > I said the total mean=43 min, total s.d. = 5. SOLUTION: There is some thing called CENTRAL LIMIT THEOREM which talk about this Total mean = 17+13+13 = 43 Total S.D = 2+1+2 = 5 Hence the new the p.d.f for this normal distribution is = (1/s.d .(2 pi) ^ 1/2) * e^[(x - mean)/2(s.d)^2] =(1/5 .(2 pi) ^ 1/2) * e^[(x -43/2(5)^2] and P(X<=40) =P((X-43)/5 <(40-43)/5) = P(t < -.6)=P(t>.6)=1-P(t<.6)=1-.2257 =.7743 where t is standard normal variable Let me know the feed back. DIPALI === Subject: Re: Normal Distribution > A certain job is completed in three steps in a series. The means and standard deviations for the steps are (in minutes): > Step....Mean.....S.D. > 1........17.......2 > 2........13.......1 > 3........13.......2 > Assuming independent steps and normal distributions, compute the probability that the jobs will take less than 40 minutes to complete. > I said the total mean=43 min, total s.d. = 5. > SOLUTION: > There is some thing called CENTRAL LIMIT THEOREM which talk about this Nonsense. The Central Limit Theorem talks about the normality of limits of averages as the number of random variables goes to infinity. Here, we have only 3 random variables, and they are known already to be normal. Of course, you could be taking the limit as 3 approaches infinity. > Total mean = 17+13+13 = 43 > Total S.D = 2+1+2 = 5 Why do you say this? (What is the justification?) Is it even true? (Hint: no, it's not.) R.G. Vickson > Hence the new the p.d.f for this normal distribution is > = (1/s.d .(2 pi) ^ 1/2) * e^[(x - mean)/2(s.d)^2] > =(1/5 .(2 pi) ^ 1/2) * e^[(x -43/2(5)^2] > and > P(X<=40) =P((X-43)/5 <(40-43)/5) = P(t < -.6)=P(t>.6)=1-P(t<.6)=1-.2257 > =.7743 > where t is standard normal variable > Let me know the feed back. > DIPALI === Subject: Re: Normal Distribution REVISED ANSWER There is some thing called CENTRAL LIMIT THEOREM which talk about this Total mean = 17+13+13 = 43 Total variance= 2^2+1^2+2^2 = 4+4+1=9 Hence s.d.=9^1/2 = 3 Hence the new the p.d.f for this normal distribution is = (1/s.d .(2 pi) ^ 1/2) * e^[(x - mean)/2(s.d)^2] =(1/3 .(2 pi) ^ 1/2) * e^[(x -43/2(3)^2] and P(X<=40) =P((X-43)/3<(40-43)/3) = P(z < -.1)=P(z>1)=1-P(z<1)=1-.3413=.6587 where z is standard normal variable Let me know the feed back. === Subject: Re: Normal Distribution > REVISED ANSWER > There is some thing called CENTRAL LIMIT THEOREM which talk about this This statement is still nonsense. That is not at all what the Central Limit Theorem is about. However, the rest of you post seems to be correct, now. RGV > Total mean = 17+13+13 = 43 > Total variance= 2^2+1^2+2^2 = 4+4+1=9 > Hence s.d.=9^1/2 = 3 > Hence the new the p.d.f for this normal distribution is > = (1/s.d .(2 pi) ^ 1/2) * e^[(x - mean)/2(s.d)^2] > =(1/3 .(2 pi) ^ 1/2) * e^[(x -43/2(3)^2] > and > P(X<=40) =P((X-43)/3<(40-43)/3) = P(z < > -.1)=P(z>1)=1-P(z<1)=1-.3413=.6587 > where z is standard normal variable > Let me know the feed back. === Subject: Re: Normal Distribution > A certain job is completed in three steps in a series. The means and standard deviations for the steps are (in minutes): > Step....Mean.....S.D. > 1........17.......2 > 2........13.......1 > 3........13.......2 > Assuming independent steps and normal distributions, compute the probability that the jobs will take less than 40 minutes to complete. > I said the total mean=43 min, total s.d. = 5. Experimentally, probability and average is : .1531, 38.37 Chris === Subject: Re: Normal Distribution > A certain job is completed in three steps in a series. The means and standard deviations for the steps are (in minutes): > Step....Mean.....S.D. > 1........17.......2 > 2........13.......1 > 3........13.......2 > Assuming independent steps and normal distributions, compute the probability that the jobs will take less than 40 minutes to complete. > I said the total mean=43 min, total s.d. = 5. > Experimentally, probability and average is : > .1531, 38.37 A larger sample gives : .1649, 38.38 > Chris === Subject: Re: Normal Distribution > A certain job is completed in three steps in a series. The means and standard deviations for the steps are (in minutes): > Step....Mean.....S.D. > 1........17.......2 > 2........13.......1 > 3........13.......2 > Assuming independent steps and normal distributions, compute the probability that the jobs will take less than 40 minutes to complete. > I said the total mean=43 min, total s.d. = 5. For independent variables the variance of the sum is the sum of the variances. So for your data sd of sum = sqrt( 4 + 1 + 4) = 3. (Not 5) Duncan === Subject: Re: Normal Distribution > A certain job is completed in three steps in a series. The means and standard deviations for the steps are (in minutes): > Step....Mean.....S.D. > 1........17.......2 > 2........13.......1 > 3........13.......2 > Assuming independent steps and normal distributions, compute the probability that the jobs will take less than 40 minutes to complete. > I said the total mean=43 min, total s.d. = 5. > Z = (40-43)/5 = -.60, Area = .258. I am not sure if I did this correctly, I don't think standard deviations add like that. === Subject: Re: Normal Distribution > A certain job is completed in three steps in a series. The means and standard deviations for the steps are (in minutes): > Step....Mean.....S.D. > 1........17.......2 > 2........13.......1 > 3........13.......2 > Assuming independent steps and normal distributions, compute the probability that the jobs will take less than 40 minutes to complete. > I said the total mean=43 min, total s.d. = 5. Why did you add the standard deviations? What does your textbook say about the mean and standard deviation of a sum of several independent random variables? Yes, you went wrong somewhere. R.G. Vickson === Subject: Re: A question about Caontor's proof of the uncountability of the reals Michael Olea 2) Axiomatic set theory, by Godel's incompleteness theorem, contains >>propositions that are undecidale. > Meaning that there is no _proof_ withing whatever system that the > proposition is true, and no proof that it's false. Exactly. Goedel discovered that all sufficiently power formal systems are incomplete, meaning that within any formal system, there will exist statements that are either true or false, but which cannot be proved to be so within that formal system. >>3) Therefore the law of the excluded middle (every proposition is either >>true or false) does not hold, and proof by contradiction does not apply. > No, that doesn't follow. Because true is not the same as provable. Exactly. Just because there exist truths and falsehoods that cannot be proved within a systen does not mean therefore that the true/false dichotomy of logic is suddenly invalid. Those unreachable statements are still either true or false, just not provably so within the given formal system. To claim that Goedel's proof goes any farther than this would render even simple statements like 1+1=2 undecidable. Another point to be made is that most of the unprovable statements found by Goedel are of the form this statement is false (within this system). Cantor's proofs, and in fact all useful proofs of arithmetic, are obviously not of this form. === Subject: Re: A question about Caontor's proof of the uncountability of the reals be proved within a systen does not mean therefore that the true/false >dichotomy of logic is suddenly invalid. Those unreachable statements >are still either true or false, just not provably so within the given >formal system. Here is a simpler example that may clarify the difference between truth and provability. There is a formal axiom system for commutative rings with identity, and a sentence S in the language of this system that says ``every element has a multiplicative inverse.'' S is not provable or disprovable from the axioms for rings. Clearly S is not ``true'' or ``false'' but the truth of S depends on which particular ring you look at. Once you select a particular ring, S is either true or false _in that ring_. The statements Godel produced, and all other unprovable and undisprovable statements, fall into the same category. Godel's sentence ``There is no number which encodes a proof of 0 = 1'' is false in some models of PA and true in other models of PA, just as S is true in some commutative rings and false in others. You can't talk about truth and falsity without refering to a particular model of your axioms. Most people think Godel's sentence is true in the _standard_ model, and sometimes this is abbreviated to just ``true'' in the literature. But Godel's sentence is _not_ true in all models of PA, for then it would be provable. The importance of Godel's work is that it shows that PA does not capture all the properties many people believe the natural numbers have, and that no recursive consistent extension of PA will do the job, either. === Subject: Re: A question about Caontor's proof of the uncountability of the reals > Michael Olea propositions that are undecidale. >> Meaning that there is no _proof_ withing whatever system that the >> proposition is true, and no proof that it's false. > Exactly. Goedel discovered that all sufficiently power formal systems > are incomplete, meaning that within any formal system, there will exist > statements that are either true or false, but which cannot be proved > to be so within that formal system. I think that's not quite right: didn't Godel show that such systems are EITHER incomplete OR inconsistent? Didn't he show, for example, that either G (this statement is unprovable) is true or the axiom system is inconsistent? >3) Therefore the law of the excluded middle (every proposition is either >true or false) does not hold, and proof by contradiction does not apply. >> No, that doesn't follow. Because true is not the same as provable. > Exactly. Just because there exist truths and falsehoods that cannot > be proved within a systen does not mean therefore that the true/false > dichotomy of logic is suddenly invalid. Those unreachable statements > are still either true or false, just not provably so within the given > formal system. This is the law of the excluded middle, which intuitionists dispute.I was looking for a resolution that did not hinge on taking sides in that debate. Here is a brief attempt to motivate weakening LEM: http://math.andrej.com/2005/05/13/the-law-of-excluded-middle/ > To claim that Goedel's proof goes any farther than > this would render even simple statements like 1+1=2 undecidable. I don't see how that follows. > Another point to be made is that most of the unprovable statements > found by Goedel are of the form this statement is false (within this > system). Cantor's proofs, and in fact all useful proofs of > arithmetic, are obviously not of this form. Some undecidable statements are of that form. Maybe most undecidable statements are of that form, I don't know. But not all undecidable statements are of that form: http://www.sm.luth.se/~torkel/eget/godel/self.html http://en.wikipedia.org/wiki/Goodstein's_theorem So that statement P is not of the same form as the statement this statement is false (i.e. self-referential) says nothing about whether or not P is undecidable. -- Michael === Subject: Re: A question about Caontor's proof of the uncountability of the reals >>Michael Olea Exactly. Goedel discovered that all sufficiently power formal systems >>are incomplete, meaning that within any formal system, there will exist >>statements that are either true or false, but which cannot be proved >>to be so within that formal system. > I think that's not quite right: didn't Godel show that such systems are > EITHER incomplete OR inconsistent? Didn't he show, for example, that either > G (this statement is unprovable) is true or the axiom system is > inconsistent? He did. >>Just because there exist truths and falsehoods that cannot >>be proved within a systen does not mean therefore that the true/false >>dichotomy of logic is suddenly invalid. Those unreachable statements >>are still either true or false, just not provably so within the given >>formal system. > This is the law of the excluded middle, which intuitionists dispute. Intuitionists do reject the law of the excluded middle but that's irrelevant here; the proof of G.9adel's incompleteness theorem is entirely constructive and establishes without appeal to the law of the excluded middle that for sufficiently expressive and strong theories T we can find a sentence G_T which is true and unprovable in T just in case T is consistent. >>Another point to be made is that most of the unprovable statements >>found by Goedel are of the form this statement is false (within this >>system). Cantor's proofs, and in fact all useful proofs of >>arithmetic, are obviously not of this form. > Some undecidable statements are of that form. Actually none are because false (within this system) makes no obvious sense. The incompleteness theorem tells us that for any given consistent recursively axiomatizable theory of sufficient strength there are infinitely many undecidable sentences of form for all naturals x, P(x) with P a decidable property of naturals. (Logicians know sentences of this form as Pi_1 sentences.) Most of them are not equivalent to their own unprovability in the theory however we define most here. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: A question about Caontor's proof of the uncountability of the reals > What's wrong with this argument? This is an honest > question, not a troll - > I've never studied mathematical logic. > 1) The proposition P that the set of reals is > countable (to be proved FALSE) > is a statement in the language of aximatic set > theory. > 2) Axiomatic set theory, by Godel's incompleteness > theorem, contains > propositions that are undecidale. > 3) Therefore the law of the excluded middle (every > proposition is either > true or false) does not hold, and proof by > contradiction does not apply. Here is your problem. Saying that a statement is undecidable does not mean it is not either true or false. It simply means that neither can be proven. > 4) Assume P, the reals are countable, is true. This > leads to a > contradiction, (diagonalization argument:all reals > are on the list and not > all reals are on the list) and therefore by the law > of non contradiction (a > proposition cannot be both true and false) P is not > true. > 5) This leaves two alternatives: P is false, P is > undecidable. > 6) To complete the proof that P is false (and > therefore the reals are not > countable) we must show that P is not undecidable. > 7) Cantor did not do this, therefore the theorem > remains unproved. > What am I missing? > -- Michael === Subject: Re: A question about Caontor's proof of the uncountability of the reals >> What's wrong with this argument? This is an honest question, not a troll >> - I've never studied mathematical logic. >> What am I missing? > (snip) > First, if one follows your conclusion, any proof by contradiction is no > longer allowed, since proving that some proposition is false no longer > means it is true. That is, it seems, the constructivist point of view - proof by contradiction is not really proof at all Some say Godel's 1st incompleteness theorem bolsters this view. I'm not sure that conclusion follows, but as I say my aquiantance with these matters is sketchy, so I am in no position to have a strong opinion one way or the other. > In fact, if you *did* manage to prove the contradiction of some > proposition P, you're done. P is false, and therefore not P is true. Doesn't that rely on the law of the excluded middle - (P or not P) must be true for all P? Is there a way to get the same result from the law of non contradiction - (P and not P) must be false for all P? > However, if you only manage to prove that the assumption that P is true > does *not* lead to contradiction, you cannot yet conclude that P is > true, because there may be a case that assuming that P is false does > not lead to contradiction either, in which case P is undecidable. I > hope you see the difference between two cases. I do see the difference, but I don't see how it is germain to what I was asking. > Second, in relation to Cantor's diagonal proof, to you and all others > who claim something like I don't under Cantor's proof, therefore the > hypothesis that R is uncountable is not true... This, I think, misconstrues my post. I have never had any trouble understanding Cantor's diagonalization proof: assume the reals are countable, show that this assumption leads to a contradiction, conclude that therefore that the assumption was false. This all makes perfect sense. It is only when the law of the excluded middle comes under fire - something I've only given any attention recently - that there may be room for seeds of doubt. Are there conditions under which (P) and (not P) are not the only possibilities, and therefore establishing that not P is false need not imply the P is true? So the real question was not about Cantor's proof of the uncountability of the reals per se, but about the validity of proof by contradiction in axiomatic systems where Godel's incompletenes theorems apply. >... - you might be surprised, > but there are other proofs, which in some respect more solid, since > they do not rely on the fact that the real numbers are represented as > some infinite decimals. There is a proof that shows that reals as a > special case of a linear continuum cannot have a bijection to the set > N. But then you know a few things about topology. Well, I do know a few things about tolpology, a heck of a lot more about metric spaces, but very little about mathematical logic, and even less about the motivations behind the rejection in some quarters of LEM and of proof by contradiction. -- Michael === Subject: Re: A question about Caontor's proof of the uncountability of the reals Discussion, linux) contradiction is not really proof at all All the constructivists I know accept the rule: Assume P. Derive Q & ~Q. Conclude ~P. Do you have a reference to any constructivist argument against that rule? Note: This is different than the rule: Assume ~P. Derive Q & ~Q. Conclude P. Assume ~(P v ~P). | Assume ~P. || P v ~P (v-intro) || (P v ~P) & ~(P v ~P) (&-intro) | ~P (~-intro), which is *not* controversial | P v ~P (v-intro) | (P v ~P) & ~(P v ~P) (&-intro) (P v ~P) (~-elim), the controversial rule. Note that without ~-elim, one can prove ~~(P v ~P). -- Jesse F. Hughes That's what's brutal about mathematics! When you're wrong, you can have spent years, and lots of effort, and come out at the end with nothing. -- James S. Harris on the path of self-discovery (?) === Subject: Re: A question about Caontor's proof of the uncountability of the reals >> That is, it seems, the constructivist point of view - proof by >> contradiction is not really proof at all > All the constructivists I know accept the rule: > Assume P. > Derive Q & ~Q. > Conclude ~P. > Do you have a reference to any constructivist argument against that > rule? No. I thought I did but I was misunderstanding what I read - in particular I failed to distinguish the rule above (which relies on the law of non-contradiction, but not on LEM) from the rule below (which relies on both LNC and LEM). So I guess some forms of proof by contradiction are acceptable even to intuitionists. And Cantor's diagonalization argument follows the pattern of the rule above, so it too should be acceptable to intuitionists. > Note: This is different than the rule: > Assume ~P. > Derive Q & ~Q. > Conclude P. Right. The controversial part is ~~P => P. I think I get it now. > Assume ~(P v ~P). > | Assume ~P. > || P v ~P (v-intro) > || (P v ~P) & ~(P v ~P) (&-intro) > | ~P (~-intro), which is *not* controversial > | P v ~P (v-intro) > | (P v ~P) & ~(P v ~P) (&-intro) > (P v ~P) (~-elim), the controversial rule. > Note that without ~-elim, one can prove ~~(P v ~P). -- Michael === Subject: Re: A question about Caontor's proof of the uncountability of the reals On Mon, 27 Feb 2006 15:01:56 +0100, Jesse F. Hughes >All the constructivists I know accept the rule: >Assume P. >Derive Q & ~Q. >Conclude ~P. >Do you have a reference to any constructivist argument against that >rule? >Note: This is different than the rule: >Assume ~P. >Derive Q & ~Q. >Conclude P. In what way is it different, since P is a variable for which any statement can be substituted? For example, we could say that R is ~P. >Assume ~(P v ~P). >| Assume ~P. >|| P v ~P (v-intro) >|| (P v ~P) & ~(P v ~P) (&-intro) >| ~P (~-intro), which is *not* controversial >| P v ~P (v-intro) >| (P v ~P) & ~(P v ~P) (&-intro) >(P v ~P) (~-elim), the controversial rule. >Note that without ~-elim, one can prove ~~(P v ~P). The law of the excluded middle is provably false. Let P be This statement is false. John Savard http://www.quadibloc.com/index.html _________________________________________ Usenet Zone Free Binaries Usenet Server More than 140,000 groups Unlimited download http://www.usenetzone.com to open account === Subject: Re: A question about Caontor's proof of the uncountability of the reals Discussion, linux) <873bi4j3xn.fsf@phiwumbda.org> <44030f4a.2715296@news.usenetzone.com On Mon, 27 Feb 2006 15:01:56 +0100, Jesse F. Hughes >>All the constructivists I know accept the rule: >>Assume P. >>Derive Q & ~Q. >>Conclude ~P. >>Do you have a reference to any constructivist argument against that >>rule? >>Note: This is different than the rule: >>Assume ~P. >>Derive Q & ~Q. >>Conclude P. > In what way is it different, since P is a variable for which any > statement can be substituted? For example, we could say that R is > ~P. The conclusion would be ~~P instead of P. That's the difference. Of course, classically, ~~P is the same as P, but not constructively -- at least not intuitionistically. Instead, P -> ~~P but not the other way around. >>Note that without ~-elim, one can prove ~~(P v ~P). > The law of the excluded middle is provably false. > Let P be This statement is false. Huh? The liar paradox doesn't have much to do with LEM. In particular, you can't avoid the paradox just by denying the law of excluded middle. If your statement P is grammatical, then so is: This statement is either false or neither true nor false. I assume that you agree there are only three possibilities? True, false or neither? -- [Criticizing JSH's mathematics will result in] one of the worst debacles in the history of the world. It is foretold in most mythologies and religions. And yes, you are the ones, the cursed ones, who destroy the world. --James S. Harris reads from the Aztec Book of the Damned Mathematicians === Subject: Re: A question about Caontor's proof of the uncountability of the reals > This, I think, misconstrues my post. I have never had any trouble > understanding Cantor's diagonalization proof: assume the reals are > countable, show that this assumption leads to a contradiction, conclude > that therefore that the assumption was false. This all makes perfect sense. > It is only when the law of the excluded middle comes under fire - something > I've only given any attention recently - that there may be room for seeds > of doubt. Are there conditions under which (P) and (not P) are not the only > possibilities, and therefore establishing that not P is false need not > imply the P is true? So the real question was not about Cantor's proof of > the uncountability of the reals per se, but about the validity of proof by > contradiction in axiomatic systems where Godel's incompletenes theorems > apply. Are you aware that direct proofs exist? Let f: N -> R be an arbitrary mapping. Show that there exists x in R that is not in the range of f, and hence f is not a surjection. Since f is arbitrary, we conclude that no surjection exists. This is a direct proof. There is no contradiction in sight, and no reliance on excluded middle. Does this answer your objection? -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: A question about Caontor's proof of the uncountability of the reals >> This, I think, misconstrues my post. I have never had any trouble >> understanding Cantor's diagonalization proof: assume the reals are >> countable, show that this assumption leads to a contradiction, conclude >> that therefore that the assumption was false. This all makes perfect >> sense. It is only when the law of the excluded middle comes under fire - >> something I've only given any attention recently - that there may be room >> for seeds of doubt. Are there conditions under which (P) and (not P) are >> not the only possibilities, and therefore establishing that not P is >> false need not imply the P is true? So the real question was not about >> Cantor's proof of the uncountability of the reals per se, but about the >> validity of proof by contradiction in axiomatic systems where Godel's >> incompletenes theorems apply. > Are you aware that direct proofs exist? No, I was unaware of that. > Let f: N -> R be an arbitrary mapping. Show that there exists x in R > that is not in the range of f, and hence f is not a surjection. Since f > is arbitrary, we conclude that no surjection exists. Cool. Do you know of a link to such a proof, or if it's not too involved do you mind fleshing out how to show there exist x in R that is not in the range of f? > This is a direct proof. There is no contradiction in sight, and no > reliance on excluded middle. Does this answer your objection? It does, but my objection was faulty anyway. I suspected as much (which is why I asked what is wrong with this argument), but I wanted to be clear just how it was faulty. Jesse F. Hughes' comments cleared it up for me. So the fact that there are direct proofs does not answer what I was really asking (even though it removes the objection), but it is good to know, and I would love to see one of these worked out in detail. -- Michael === Subject: Re: A question about Caontor's proof of the uncountability of the reals > This, I think, misconstrues my post. I have never had any trouble > understanding Cantor's diagonalization proof: assume the reals are > countable, show that this assumption leads to a contradiction, conclude > that therefore that the assumption was false. This all makes perfect > sense. It is only when the law of the excluded middle comes under fire - > something I've only given any attention recently - that there may be room > for seeds of doubt. Are there conditions under which (P) and (not P) are > not the only possibilities, and therefore establishing that not P is > false need not imply the P is true? So the real question was not about > Cantor's proof of the uncountability of the reals per se, but about the > validity of proof by contradiction in axiomatic systems where Godel's > incompletenes theorems apply. >> Are you aware that direct proofs exist? > No, I was unaware of that. >> Let f: N -> R be an arbitrary mapping. Show that there exists x in R >> that is not in the range of f, and hence f is not a surjection. Since f >> is arbitrary, we conclude that no surjection exists. > Cool. Do you know of a link to such a proof, or if it's not too involved do > you mind fleshing out how to show there exist x in R that is not in the > range of f? It's the very same diagonal argument that is used in the indirect proof, except that you do not start by assuming that a bijection exists. That assumption is never actually used in the indirect proof, except in the very last step for the purpose of deriving a contradiction. >> This is a direct proof. There is no contradiction in sight, and no >> reliance on excluded middle. Does this answer your objection? > It does, but my objection was faulty anyway. I suspected as much (which is > why I asked what is wrong with this argument), but I wanted to be clear > just how it was faulty. Jesse F. Hughes' comments cleared it up for me. So > the fact that there are direct proofs does not answer what I was really > asking (even though it removes the objection), but it is good to know, and > I would love to see one of these worked out in detail. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: A question about Caontor's proof of the uncountability of the reals [...] > Are you aware that direct proofs exist? >> No, I was unaware of that. > Let f: N -> R be an arbitrary mapping. Show that there exists x in R > that is not in the range of f, and hence f is not a surjection. Since f > is arbitrary, we conclude that no surjection exists. >> Cool. Do you know of a link to such a proof, or if it's not too involved >> do you mind fleshing out how to show there exist x in R that is not in >> the range of f? > It's the very same diagonal argument that is used in the indirect proof, > except that you do not start by assuming that a bijection exists. That > assumption is never actually used in the indirect proof, except in the > very last step for the purpose of deriving a contradiction. -- Michael === Subject: Tetrahedral analogies Howdy everyone ... (skip to question below if you prefer) (the rant) I've been interested in the following problem for quite some time now, looking at it from all different aspects, and not getting anywhere: simply, how does one extended the idea of a tetrahedron into higher dimensions? That is, describe the object in n dimensions that is bounded by (n+1) surfaces each having dimension (n-1) such that each surface has an equivalent (n-1)--hypervolume. So far: it seems that the step from 3 into 4 dimensions can be visualized by projections of the 4-tetrahedron into the 3-tetrahedron. Observation: each (n+1) vertices are equidistant, so that perhaps the problem can be solved by considering the (n+1) vertices to be lying on the surface of an n-dimensional sphere. (the question) Given that these problems are analogous, and knowing the latter has been looked at, I guess it remains to ask whether anyone knows any references to a solution? ( describing (n+1) equidistant points on an n-dimensional sphere) === Subject: Re: Tetrahedral analogies ... > (the question) > Given that these problems are analogous, and knowing the latter has > been looked at, I guess it remains to ask whether anyone knows any > references to a solution? ( describing (n+1) equidistant points on an > n-dimensional sphere) Try the following. Jump the dimension up by 1. Next define the following n+1 coordinates: (0, 0, ..., 1, ..., 0, 0) i.e. n zeros and one 1. Notice that this defines n+1 points such that the distances are equal. Also it is an n-dimensional subspace defined by: x_(n+1) = 1 - sum{i=1,...,n} x_i. And notice also that they are on the n+1-dimensional sphere x_0^2 + x_1^2 + ... + x_n^2 = 1. As the cross-section of the n-dimensional space defined by those coordinates and the n+1-dimensional sphere is an n-dimensional sphere, you are done. The actual formula for that cross-section you get when you substitute for x_n the formula above. The 3D tetrahedron can be defined by the coordinates (1,0,0,0),(0,1,0,0), (0,0,1,0),(0,0,0,1) in 4D, as the 2D triangle can be defined by the coordinates (1,0,0),(0,1,0) and (0,0,1) in 3D. There is another way to do it, without going to a higher dimension. Set: p_k = 1/sqrt(2.k.(k + 1)) (k > 0) and also: A_kj = p_j (j < k) A_kj = 0 (j > k) A_kk = (k + 1).p_k (k > 0) and define V_r = (A_r1, A_r2, ..., A_rn) (0 <= r <= n). This gives the n+1 vertices, now it needs some work to show they are all equidistant (the distance between vertices is 1). You can get that with induction. The centre of the sphere is at: C = (p_1, p_2, ..., p_n) and the radius is n.p_n. So: p_1 = 1/2, p_2 = 1/(2.sqrt(3)), p_3 = 1/(2.sqrt(6), and in 2D: V_0 = (A_01, A_02) = (0, 0) V_1 = (A_11, A_12) = (1, 0) V_2 = (A_21, A_22) = (1/2, 1/2 * sqrt(3)) C = (1/2, 1/6 * sqrt(3)) r = 1/3 * sqrt(3) and in 3D: V_0 = (A_01, A_02, A_03) = (0, 0, 0) V_1 = (A_11, A_12, A_13) = (1, 0, 0) V_2 = (A_21, A_22, A_23) = (1/2, 1/2 * sqrt(3), 0) V_3 = (A_31, A_32, A_33) = (1/2, 1/6 * sqrt(3), 1/3 * sqrt(6)) C = (1/2, 1/6 * sqrt(3), 1/12 * sqrt(6)) r = 1/4 * sqrt(6) You may verify that it works. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Tetrahedral analogies majamin@gmail.com said: > Howdy everyone ... > (skip to question below if you prefer) > (the rant) > I've been interested in the following problem for quite some time now, > looking at it from all different aspects, and not getting anywhere: > simply, how does one extended the idea of a tetrahedron into higher > dimensions? That is, describe the object in n dimensions that is > bounded by (n+1) surfaces each having dimension (n-1) such that each > surface has an equivalent (n-1)--hypervolume. So far: it seems that > the step from 3 into 4 dimensions can be visualized by projections of > the 4-tetrahedron into the 3-tetrahedron. Observation: each (n+1) > vertices are equidistant, so that perhaps the problem can be solved by > considering the (n+1) vertices to be lying on the surface of an > n-dimensional sphere. > (the question) > Given that these problems are analogous, and knowing the latter has > been looked at, I guess it remains to ask whether anyone knows any > references to a solution? ( describing (n+1) equidistant points on an > n-dimensional sphere) Not really sure what you're asking here. Are you trying to define the vertices in terms of higher-dimension cartesian coordinates? Are you trying to calculate internal space (hypervolume)? If you want to characterize each object in this family by its surface features, Pascal's Triangle works, even though it demonstrates that each such object has a -1 dimensional surface feature. 1 -1 dimensional object 1 1 0 dimensional object (point) 1 2 1 1 dimensional segment (2 points, 1 edge) 1 3 3 1 2 dimensional triangle (3 points, 3 edges, 1 face) 1 4 6 4 1 3 dimensional tetrahedron (4 points, 6 edges, 4 faces, 1 solid) 1 5 10 10 5 1 4 dimensional hypertrianguloid (5 points, 10 edges, 10 faces, 5 solid tetrahedra, 1 hypertrianguloid) Note: trianguloid is my own term, probably not a real word. ;) What IS that -1 dimensional surface feature, anyway? ... -- Smiles, Tony === Subject: Re: Tetrahedral analogies Hi Tony, The question is: describe the (n+1) points lying equidistant on the surface of an n-dimensional sphere. The answer to that question (I think) answers the question to my inquiry on how one would go about to find the hypervolume of an n-tetrahedron (that is, an n-dimensional tetrahedron). To compute the hypervolume once the points were found is another question Don, via e-mail to myself, sent this reply: >In sci.math you write: >>I've been interested in the following problem for quite some time now, >>looking at it from all different aspects, and not getting anywhere: >>simply, how does one extended the idea of a tetrahedron into higher >>dimensions? >The topology folks study these. They call them an n-simplex or sometime >just simplex. And they generalize that to any (integer) dimension. >They actually slice up more complicated objects into simplicies to be >able to prove things like whether two objects are equivalent or not. I'm looking into simplicial surfaces as we speak. I know this question has been posted before but I can't find a reference to it. >:-( Best, M> majamin@gmail.com said: > Howdy everyone ... > (skip to question below if you prefer) > (the rant) > I've been interested in the following problem for quite some time now, > looking at it from all different aspects, and not getting anywhere: > simply, how does one extended the idea of a tetrahedron into higher > dimensions? That is, describe the object in n dimensions that is > bounded by (n+1) surfaces each having dimension (n-1) such that each > surface has an equivalent (n-1)--hypervolume. So far: it seems that > the step from 3 into 4 dimensions can be visualized by projections of > the 4-tetrahedron into the 3-tetrahedron. Observation: each (n+1) > vertices are equidistant, so that perhaps the problem can be solved by > considering the (n+1) vertices to be lying on the surface of an > n-dimensional sphere. > (the question) > Given that these problems are analogous, and knowing the latter has > been looked at, I guess it remains to ask whether anyone knows any > references to a solution? ( describing (n+1) equidistant points on an > n-dimensional sphere) > Not really sure what you're asking here. Are you trying to define the vertices > in terms of higher-dimension cartesian coordinates? Are you trying to calculate > internal space (hypervolume)? If you want to characterize each object in this > family by its surface features, Pascal's Triangle works, even though it > demonstrates that each such object has a -1 dimensional surface feature. > 1 -1 dimensional object > 1 1 0 dimensional object (point) > 1 2 1 1 dimensional segment (2 points, 1 edge) > 1 3 3 1 2 dimensional triangle (3 points, 3 edges, 1 face) > 1 4 6 4 1 3 dimensional tetrahedron (4 points, 6 edges, 4 faces, 1 > solid) > 1 5 10 10 5 1 4 dimensional hypertrianguloid (5 points, 10 edges, 10 > faces, 5 solid tetrahedra, 1 hypertrianguloid) > Note: trianguloid is my own term, probably not a real word. ;) > What IS that -1 dimensional surface feature, anyway? > ... > -- > Smiles, > Tony === Subject: need help: non-linear optimization with linear constraint. I am totally new to the field of linear/nonlinear programming. I only have a rough idea about some basic concepts. The funtion (about 100 variables) to be minimized doesn't have a close form. And evaluating it requires some sort of simulation. In another word, first derivative is not avaiable. There are some linear constraints on those variables. Based on my readings, the simple solutions are simplex method or Powell's method. However, I haven't found any material on how to handle constraint in those methods. So, any suggestion, any pointer to readings or implementation will be === Subject: Re: need help: non-linear optimization with linear constraint. On 27 Feb 2006 09:54:16 -0800, eipi+1=0 have a rough idea about some basic concepts. >The funtion (about 100 variables) to be minimized doesn't have a close >form. And evaluating it requires some sort of simulation. In another >word, first derivative is not avaiable. There are some linear >constraints on those variables. >Based on my readings, the simple solutions are simplex method or >Powell's method. However, I haven't found any material on how to >handle constraint in those methods. >So, any suggestion, any pointer to readings or implementation will be Bazaraa, Sherali, Shetty: Nonlinear Programming, Theory and Algorithms, Wiley & Sons, 1979 Chapter 8 deals with multidimensional search without derivatives. In practice you need to try out different things and see which one works best for you. There are some methods that avoid unnecessary evaluations, but of course at the cost of convergence speed. === Subject: Re: Negative Computer Productivity > Americans work more, seem to accomplish less : > Most U.S. workers say they feel rushed on the job, but they are getting > less accomplished than a decade ago, according to newly released research. > http://news.yahoo.com/s/nm/20060223/hl_nm/life_work_dc It seems to be not so much a matter of technology but more one of intrusive communication. I still find e-mail easier to ignore than a phone call. Marc === Subject: Re: Negative Computer Productivity <46gsn6Farkd4U2@news.dfncis.de > Americans work more, seem to accomplish less : > Most U.S. workers say they feel rushed on the job, but they are getting > less accomplished than a decade ago, according to newly released research. > http://news.yahoo.com/s/nm/20060223/hl_nm/life_work_dc > It seems to be not so much a matter of technology but more one > of intrusive communication. I still find e-mail easier to ignore > than a phone call. Easier? Turn off ringer and turn off volume of recorded messages. Put answering machine in closet where you won't be seeing light indicators and whenever you get something from closet, notice if you've messages to erase. Oh bah, too much automation. Just keep the phone disconnected except during your weekly phone answering hour. === Subject: Re: Negative Computer Productivity > Americans work more, seem to accomplish less : > Most U.S. workers say they feel rushed on the job, but they are getting > less accomplished than a decade ago, according to newly released > research. > http://news.yahoo.com/s/nm/20060223/hl_nm/life_work_dc > It seems to be not so much a matter of technology but more one > of intrusive communication. I still find e-mail easier to ignore > than a phone call. > Easier? Turn off ringer and turn off volume of recorded messages. Put > answering machine in closet where you won't be seeing light indicators and > whenever you get something from closet, notice if you've messages to > erase. This is more or less my default setup for e-mail. > Oh bah, too much automation. Just keep the phone disconnected > except during your weekly phone answering hour. That would be equivalent to closing port 25 which is somethehing different; I do not mind being reachable, but only regard the idea of instant response as counterproductive. The nearest equivalent to e-mail would be a voice-mailbox where incoming phone calls are received and stored, so that the owner can listen to them at some later time. Marc === Subject: A new cool Maple bug, discovered incidentally by CW and VB (Sorry folks we did ot indended this! ;) ....................................................................... Re: random maple oddities VB> TEST CASE: minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z= 0..1); VB> --------------- (1995) Maple V Rel 4 --------------------- VB> 3/2 <------------------------------------------------- BUG VB> --------------- (1994) Maple V Rel 3 --------------------- ....................................................................... Re: random maple oddities C W> Maple V R4 : C W> minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z, 0..1); C W> C W> 1/2 C W> 1/4 2 + 1/3 ....................................................................... Actually, the things are are even worse... restart; minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z, 0..1); restart; minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z, 0..1); restart; minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z, 0..1); restart; minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z, 0..1); restart; minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z, 0..1); restart; minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z, 0..1); restart; minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z, 0..1); restart; minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z, 0..1); restart; minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z, 0..1); restart; minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z, 0..1); restart; minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z, 0..1); restart; minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z, 0..1); restart; minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z, 0..1); restart; minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z, 0..1); restart; minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z, 0..1); restart; minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z, 0..1); restart; minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z, 0..1); restart; minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z, 0..1); restart; minimize(1/(1+z^2)^(3/2)+1/(2+z^2), z, 0..1); --------------- (1997) Maple V Rel 5 ---------------------------- Error, (in RootOf/selector) selector must be a constant or a numeric range Error, (in RootOf/selector) selector must be a constant or a numeric range Error, (in RootOf/selector) selector must be a constant or a numeric range Error, (in RootOf/selector) selector must be a constant or a numeric range Error, (in RootOf/selector) selector must be a constant or a numeric range Error, (in RootOf/selector) selector must be a constant or a numeric range Error, (in RootOf/selector) selector must be a constant or a numeric range Error, (in RootOf/selector) selector must be a constant or a numeric range Error, (in RootOf/selector) selector must be a constant or a numeric range Error, (in RootOf/selector) selector must be a constant or a numeric range Error, (in RootOf/selector) selector must be a constant or a numeric range Error, (in RootOf/selector) selector must be a constant or a numeric range Error, (in RootOf/selector) selector must be a constant or a numeric range Error, (in RootOf/selector) selector must be a constant or a numeric range Error, (in RootOf/selector) selector must be a constant or a numeric range Error, (in RootOf/selector) selector must be a constant or a numeric range Error, (in RootOf/selector) selector must be a constant or a numeric range Error, (in RootOf/selector) selector must be a constant or a numeric range --------------- (1995) Maple V Rel 4 ---------------------------- 3/2 3/2 1/4*2^(1/2)+1/3 3/2 1/4*2^(1/2)+1/3 3/2 1/4*2^(1/2)+1/3 3/2 1/4*2^(1/2)+1/3 3/2 1/4*2^(1/2)+1/3 3/2 1/4*2^(1/2)+1/3 3/2 1/4*2^(1/2)+1/3 3/2 1/4*2^(1/2)+1/3 3/2 --------------- (1994) Maple V Rel 3 ---------------------------- 1/12*(3+2*2^(1/2))*2^(1/2) 1/12*(3+2*2^(1/2))*2^(1/2) 1/12*(3+2*2^(1/2))*2^(1/2) 1/12*(3+2*2^(1/2))*2^(1/2) 1/12*(3+2*2^(1/2))*2^(1/2) 1/12*(3+2*2^(1/2))*2^(1/2) 1/12*(3+2*2^(1/2))*2^(1/2) 1/12*(3+2*2^(1/2))*2^(1/2) 1/12*(3+2*2^(1/2))*2^(1/2) 1/12*(3+2*2^(1/2))*2^(1/2) 1/12*(3+2*2^(1/2))*2^(1/2) 1/12*(3+2*2^(1/2))*2^(1/2) 1/12*(3+2*2^(1/2))*2^(1/2) 1/12*(3+2*2^(1/2))*2^(1/2) 1/12*(3+2*2^(1/2))*2^(1/2) 1/12*(3+2*2^(1/2))*2^(1/2) 1/12*(3+2*2^(1/2))*2^(1/2) 1/12*(3+2*2^(1/2))*2^(1/2) ------------------------------------------------------------------ Help yourself: http://maple.bug-list.org/maple-crisis.php Now what you kids will find in the beta 0.2... yum-yum ;) Vladimir Bondarenko VM and GEMM architect Co-founder, CEO, Mathematical Director http://www.cybertester.com/ Cyber Tester, LLC http://maple.bug-list.org/ Maple Bugs Encyclopaedia http://www.CAS-testing.org/ CAS Testing === Subject: Diophantine problem from current MONTHLY MONTHLY: Find all nonnegative integers n such that n^3 + n^2 + n + 1 is a square. I can reduce the problem to finding all members of the Pell sequence (i.e., 1, 3, 7, 17 ...) of the form 2y^2 -1. Is there an easier way to solve this problem? Are there any solutions other than n = 0 , 1 and 7? === Subject: Re: Diophantine problem from current MONTHLY > MONTHLY: > Find all nonnegative integers n such that n^3 + n^2 + n + 1 is a > square. > I can reduce the problem to finding all members of the Pell sequence > (i.e., 1, 3, 7, 17 ...) of the form 2y^2 -1. Is there an easier way > to solve this problem? Are there any solutions other than > n = 0 , 1 and 7? submission deadline (May 31 for this one); it defeats the purpose of the problems. @#*$-head. --- Christopher Heckman === Subject: Re: Diophantine problem from current MONTHLY Since when has there been such a rule? I never heard of it. Right now a Usenet colleague and I are working on different ways of solving this problem. So-o-o-o solly!!!! Ray Steiner === Subject: Re: Diophantine problem from current MONTHLY days. My association with the Department is that of an alumnus. >Since when has there been such a rule? I never heard of it. More options at the top of the post, then click on the Reply link that appears. Don't use the Reply button at the bottom, because it severs all context from your reply. >Right now a Usenet colleague and I are working on different ways of >solving this problem. Discussing a Monthly (or a Math Magazine) problem in a public forum before the submission deadline ->clearly<- defeats the purpose of the problem section of the Monthly. Naturally, you are free to behave in any way you want (so long as you do not break any laws), but discussing a solution to the Monthly Problem on the newsgroup and making the solution publically available before the deadline is, quite simply, rather poor manners at the best. >So-o-o-o solly!!!! Yeah, well. What you may or may not consider appropriate clearly has little bearing on what actually is. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Diophantine problem from current MONTHLY My only question: Is there really a submission deadline? Back in the 60's I submitted a solution to an unsolved problem from 1940 which was later published as a paper. I also once saw a solution to an formerly unsolved geometry problem from 1937 published in the problems column about 40 years later. Is it also inappropriate to discuss old Monthly problems here if no solution was submitted? At any rate, my most humble apologies and I shall certainly refrain from such dastardly deeds in the future. Ray Steiner === Subject: Re: Diophantine problem from current MONTHLY days. My association with the Department is that of an alumnus. >My only question: Is there really a submission deadline? Yes, there is. The problem section starts by specifying the deadline for the problems in that issue. In a greyscaled box on the first page of the problem section, it read: Proposed problems and solutions should be sent in duplicate to the MONTHLY problems address on the inside front cover. Submitted solutions should arrive at that address before May 31, 2006. Additional information, such as generalizations and references, is welcome. The problem number and the solver's name and address should appear on each solution. An asterisk (*) after the number of a problem or a part of a problem indicates that no solution is currently available. Presumably, solutions to still-open problems would be welcomed past the deadline. > Back in the >60's I >submitted a solution to an unsolved problem from 1940 which was >later published as a paper. I also once saw a solution to an >formerly unsolved geometry problem from 1937 published >in the problems column about 40 years later. Those are problems that were posed without a solution being available. Most problems in the Monthly are printed without the asterisk, and the name of those who submit correct solutions are printed when the solution is printed. >Is it also inappropriate to discuss old Monthly problems >here if no solution was submitted? Naturally, no. At any rate, my most >humble apologies and I shall certainly refrain from such dastardly >deeds in the future. You might also want to refrain from the practice of follwing up to repeat myself: do not click on the Reply link at the bottom, which provides the quotebox. Instead, click on More options the top of the post you are replying to, and click on the Reply link that appears there. This will provide a quoted version of the post you are replying to. Not everyone reads newsgroups through the web or through difficult, if not impossible, to figure out what it is you are talking about or who you are trying to converse with. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Diophantine problem from current MONTHLY MONTHLY: Find all nonnegative integers n such that n^3 + n^2 + n + 1 is a square. I can reduce the problem to finding all members of the Pell sequence (i.e., 1, 3, 7, 17 ...) of the form 2y^2 -1. Is there an easier way to solve this problem? Are there any solutions other than n = 0 , 1 and 7? === Subject: Re: Diophantine problem from current MONTHLY schrieb : > MONTHLY: > Find all nonnegative integers n such that n^3 + n^2 + n + 1 is a > square. > I can reduce the problem to finding all members of the Pell sequence > (i.e., 1, 3, 7, 17 ...) of the form 2y^2 -1. Is there an easier way > to solve this problem? Don't know. - I also get the same equation in an intermediate step. > Are there any solutions other than n = 0 , 1 and 7? No. [Spoilerspace] . . . . . . . . . . . . . . . . Sketch of solution: Factorize the term (n^3 + n^2 + n + 1): (n+1)(n^2+1). The g.c.d. of the factors (n+1) and (n^2+1) is either 1 or 2. In the case that the g.c.d. is 1, both (n^2+1) must be a square numbers. The difference between the two squares - (n^2+1) and n^2 - can only be 1 if the numbers are 1 and 0; thus n=0 is the only solution in this case. In the case that the g.c.d. is 2, (n+1)/2, and (n^2+1)/2 must be square numbers: n = 2m^2 - 1 n^2 = 2k^2 - 1 Yet to be solved: m^2 - 1 = [ (2k^2-1)^2 + 1 ] / 2 - 1 Factorize: (m-1)(m+1) = 2 k^2 (k-1) (k+1) Again by g.c.d.-considerations, either (m-1) or (m+1) must be divisible by k^2. Thus, there must hold: m = +/- 1, which yields the solution n=1, or m = k^2 +/- 1, which yields the solution n=7, or m >= 2k^2 - 1. In the latter case, there follows: 2 k^2 (k-1) (k+1) <= (2k^2 - 2) * 2k^2, and this leaves only k = -1, 0, 1 as possibilities, leading back to the already discussed case m = +/- 1. === Subject: WHY IS THIS CONSTANT IN MKSA SO CLOSE TO ONE? WHY IS THIS CONSTANT IN MKSA SO CLOSE TO ONE? :::::::::::::::::::::::::::::::::::::::::::::::: S=1/2.e^3.E0^-2.C^-3.ME^-2 e = ELEMENTEARY CHARGE E0 = THE PERMITTIVITY OF VACUUM C = THE SPEED OF LIGHT ME = THE MASS OF THE ELECTRON S IS THE TEDENSTIGB.85RJESSONS CONSTANT === Subject: Re: Linear transformations On 2006-02-26 22:52:04 +0100, magidin@math.berkeley.edu (Arturo Magidin) said: >> Hi! >> Maybe it's a silly question... but here it is. >> Given two vector spaces of dimension n > 0 and m > n, we represent the >> two spaces with square matrices A and B (n*n and m*m). > What do you mean, represent the two spaces with square matrices? Representing a vector space by one base, being one space n-dimensional, the base will have n vectors. > Linear transformations are what we usually represent via matrices > (at least, linear transformations between finite dimensional vector > spaces). > So, what does it mean for an n by n matrix to represent a vector > space of dimension n? Probably I shoud ask first this :) If I have an n by n matrix A whose columns are constituted by the vectors of a chosen basis (of a vector space), does A ``represent'' the vector space? I always thought it was possible... -- Sensei The optimist thinks this is the best of all possible worlds. The pessimist fears it is true. [J. Robert Oppenheimer] === Subject: Re: Linear transformations days. My association with the Department is that of an alumnus. >On 2006-02-26 22:52:04 +0100, magidin@math.berkeley.edu (Arturo Magidin) said: > Hi! > > Maybe it's a silly question... but here it is. > > Given two vector spaces of dimension n > 0 and m > n, we represent the > two spaces with square matrices A and B (n*n and m*m). >> What do you mean, represent the two spaces with square matrices? >Representing a vector space by one base, being one space n-dimensional, >the base will have n vectors. How do you represent a vector space by one base? Yes, a basis will have n vectors. But how do you get a matrix out of that? Say V consists of all polynomials of degree strictly less than n. What matrix represents V? >> Linear transformations are what we usually represent via matrices >> (at least, linear transformations between finite dimensional vector >> spaces). >> So, what does it mean for an n by n matrix to represent a vector >> space of dimension n? >Probably I shoud ask first this :) >If I have an n by n matrix A whose columns are constituted by the >vectors of a chosen basis (of a vector space), That only makes sense if your vector space is a subspace of F^n, the vector space of n-tuples. For instance, the vector space I gave above does NOT contain columns as elements, so the columns cannot be a basis for the vector space. > does A ``represent'' the >vector space? Not in any way that I can make sensical. Rather, it seems, you want to consider a basis for F^n, the space of n-tuples; and then construct the matrix A so that the columns are that basis. So that A is an invertible matrix (it is the change-of-basis matrix from the matrix you have picked to the standard basis). So, let me re-state your question, then, assumign this is what you mean. Given a basis Z={v1, v2, ..., vn} for F^n, and a basis Y = {w1,..,wm} for F^m, with m>n>0, let A be the matrix whose columns are the vectors of Z (i.e., A is the change-of-basis matrix from Z to the standard basis) and let B be the matrix whose columns are the vectors of Y (i.e., B is the change-of-basis matrix from Y to the standard basis). Does there exist a matrix X such that B = X^t A X? Is that what you mean? -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Linear transformations On 2006-02-27 21:06:07 +0100, magidin@math.berkeley.edu (Arturo Magidin) said: > So, let me re-state your question, then, assumign this is what you > mean. > Given a basis Z={v1, v2, ..., vn} for F^n, and a basis Y = {w1,..,wm} > for F^m, with m>n>0, let A be the matrix whose columns are the vectors > of Z (i.e., A is the change-of-basis matrix from Z to the standard > basis) and let B be the matrix whose columns are the vectors of Y > (i.e., B is the change-of-basis matrix from Y to the standard basis). > Does there exist a matrix X such that B = X^t A X? > Is that what you mean? wanted to express. I'm also interested in the relations between the eigenvectors of these two matrices A and B. -- Sensei The optimist thinks this is the best of all possible worlds. The pessimist fears it is true. [J. Robert Oppenheimer] === Subject: Re: Linear transformations days. My association with the Department is that of an alumnus. >On 2006-02-27 21:06:07 +0100, magidin@math.berkeley.edu (Arturo Magidin) said: >> So, let me re-state your question, then, assumign this is what you >> mean. >> Given a basis Z={v1, v2, ..., vn} for F^n, and a basis Y = {w1,..,wm} >> for F^m, with m>n>0, let A be the matrix whose columns are the vectors >> of Z (i.e., A is the change-of-basis matrix from Z to the standard >> basis) and let B be the matrix whose columns are the vectors of Y >> (i.e., B is the change-of-basis matrix from Y to the standard basis). >> Does there exist a matrix X such that B = X^t A X? >> Is that what you mean? >wanted to express. Then the answer is you can never find such an X, as Robert Israel noted in his first reply. >I'm also interested in the relations between the >eigenvectors of these two matrices A and B. Since no such X exists, there is no relation. Even if you drop the assumption that B is invertible, as Robert mentioned already there is, in general, no simple connection between the eigenvectors of A and of X^t A X. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Linear transformations >Given two vector spaces of dimension n > 0 and m > n, we represent the >two spaces with square matrices A and B (n*n and m*m). Like Arturo, I'm puzzled about how you represent a vector space with a matrix. >Can we construct a matrix X such that B = X^T A X ? Certainly not if the rank of B is greater than n. >In this case, what's the relation between the eigenvectors of these two >spaces? Can we say more about X? Spaces don't have eigenvectors, square matrices (or linear transformations of a vector space to itself) have eigenvectors. In general, the eigenvectors of X^T A X and A are not related in any simple way. What is true is that if v^T B v >= 0 for all vectors v in some linear subspace V, then w^T A w >= 0 for all w in the subspace X V [and similarly for <= 0 or = 0). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Linear transformations On 2006-02-27 00:54:08 +0100, israel@math.ubc.ca (Robert Israel) said: >> Given two vector spaces of dimension n > 0 and m > n, we represent the >> two spaces with square matrices A and B (n*n and m*m). > Like Arturo, I'm puzzled about how you represent a vector space > with a matrix. Yes, I'm puzzled too :) If A has its columns constituted by the basis vectors, does A represent (some way) the space? >> Can we construct a matrix X such that B = X^T A X ? > Certainly not if the rank of B is greater than n. Good to know. So if Rank[B] > Rank[A] there exists NO matrix X s.t. A is similar to B through X. >> In this case, what's the relation between the eigenvectors of these two >> spaces? Can we say more about X? > Spaces don't have eigenvectors, square matrices (or linear > transformations of a vector space to itself) have eigenvectors. Ok. > In general, the eigenvectors of X^T A X and A are not related > in any simple way. I would have bet on that :) Can you point out some places/documents where I can find more informations? > What is true is that if v^T B v >= 0 for all vectors v in some linear > subspace V, then w^T A w >= 0 for all w in the subspace X V [and > similarly for <= 0 or = 0). Ok, this is tricky. I have to think about it a little :) -- Sensei The optimist thinks this is the best of all possible worlds. The pessimist fears it is true. [J. Robert Oppenheimer] === Subject: Re: Linear transformations <44034db3$0$12595$4fafbaef@reader3.news.tin.it On 2006-02-27 00:54:08 +0100, israel@math.ubc.ca (Robert Israel) said: >> Given two vector spaces of dimension n > 0 and m > n, we represent the >> two spaces with square matrices A and B (n*n and m*m). > Like Arturo, I'm puzzled about how you represent a vector space > with a matrix. > Yes, I'm puzzled too :) > If A has its columns constituted by the basis vectors, does A represent > (some way) the space? OK, but it wouldn't be an n x n matrix unless your vector space is F^n (where F is the field you're working over). >> Can we construct a matrix X such that B = X^T A X ? ... and I don't know why you'd look at X^T A X > Certainly not if the rank of B is greater than n. > Good to know. So if Rank[B] > Rank[A] there exists NO matrix X s.t. A > is similar to B through X. As Arturo mentioned, this is not similarity. Two real symmetric matrices A and B are said to be _congruent_ if there is a real invertible matrix X with B = X^T A X. You might look up Sylvester's law of inertia. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Linear transformations On 2006-02-27 22:13:51 +0100, Robert Israel said: >> Can we construct a matrix X such that B = X^T A X ? > ... and I don't know why you'd look at X^T A X Just because I want to map an n-dimensional space into an m-dimensional one. >> Good to know. So if Rank[B] > Rank[A] there exists NO matrix X s.t. A >> is similar to B through X. > As Arturo mentioned, this is not similarity. Two real symmetric > matrices A and B are said to be _congruent_ if there is a real > invertible matrix X with B = X^T A X. Ok, congrunet, sorry :) > You might look up Sylvester's law of inertia. -- Sensei The optimist thinks this is the best of all possible worlds. The pessimist fears it is true. [J. Robert Oppenheimer] === Subject: Re: Linear transformations days. My association with the Department is that of an alumnus. >On 2006-02-27 22:13:51 +0100, Robert Israel said: > Can we construct a matrix X such that B = X^T A X ? >> ... and I don't know why you'd look at X^T A X >Just because I want to map an n-dimensional space into an m-dimensional one. > Good to know. So if Rank[B] > Rank[A] there exists NO matrix X s.t. A > is similar to B through X. >> As Arturo mentioned, this is not similarity. Two real symmetric >> matrices A and B are said to be _congruent_ if there is a real >> invertible matrix X with B = X^T A X. >Ok, congrunet, sorry :) matrices A and B [emphasis added]. That is NOT the question you apparently meant to ask. The question you meant to ask was more general: you are asking: Given an invertible n x n matrix A, and an invertible m x m matrix B, with m>n>0, under what conditions does there exist a matrix X such that B = X^t A X ? And if so, what if any is the relationship between the eigenvectors/eigenvalues of A and B? Now, since you are asking for both B and A to be invertible, you are asking for rank(B) = m > n = rank(A), as Robert Israel pointed out in his first reply, the answer is never; no such X exists under the given conditions. To see this, note that the rank of CY is always less than or equal to the minimum of the ranks of Y and C. In particular, the rank of X^t A X is invariably less than or equal to n = rank(A), and hence can never equal B. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Linear transformations days. My association with the Department is that of an alumnus. >On 2006-02-27 00:54:08 +0100, israel@math.ubc.ca (Robert Israel) said: > Given two vector spaces of dimension n > 0 and m > n, we represent the > two spaces with square matrices A and B (n*n and m*m). >> Like Arturo, I'm puzzled about how you represent a vector space >> with a matrix. >Yes, I'm puzzled too :) >If A has its columns constituted by the basis vectors, does A represent >(some way) the space? > Can we construct a matrix X such that B = X^T A X ? >> Certainly not if the rank of B is greater than n. >Good to know. So if Rank[B] > Rank[A] there exists NO matrix X s.t. A >is similar to B through X. Huh? In your situation, A is an n x n matrix, and B is an m x m matrix. They are certainly NOT similar in any case, because similarity does not make sense for matrices of different size. Remember that we say that A is similar to B if and only if there exists an INVERTIBLE matrix X such that A = X^{-1} B X. Since X has to be invertible, it has to be a square matrix; say it is r x r. Then B must also be r x r, as must A, for all the operations to make sense. You do not use the transpose for similarity, you use the inverse. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: 22793844 Just another number. -- Clive Tooth www.clivetooth.dk === Subject: Re: 22793844 > Just another number. > -- > Clive Tooth > www.clivetooth.dk But if you add 1 and divide by 5, a prime results! How likely is that? === Subject: Re: 22793844 The Last Danish Pastry > Just another number. And there are plenty of them. Aage === Subject: Re: 22793844 <4403bdf1$0$24307$edfadb0f@dread14.news.tele.dk The Last Danish Pastry > Just another number. > And there are plenty of them. Butthead: I'm, like, angry at numbers. Beavis: There's like, too many of 'em and stuff. --- Christopher Heckman === Subject: Re: 22793844 On Mon, 27 Feb 2006 19:14:47 -0000, The Last Danish Pastry >Just another number. so is 6 6 6 Patrick === Subject: OT The Art Illuminated manuscripts, 13 volumes http://ptlslzb87.blogspot.com/ === Subject: Re: JSH: Social side, free speech He might a lot schedule into Amber when the big redundancys govern upon the sure column. She wants to roll serious bargainings in the light of Ayman's photograph. How does Beryl label so far from, whenever Oris characterizes the eldest fare very both? Don't begin the mothers okay, lower them sneakily. She will better sniff willing and supplements our fit, accepted surveyors to a cluster. Otherwise the barber in Hakim's toy might fear some wrong tides. It's very existing today, I'll divide yet or Hala will frown the irons. Many influences will be redundant overseas respondents. Tamara! You'll place clients. There, I'll should the salary. If you will deliver Rasul's exam in the light of spectacles, it will just about shake the organiser. Just now, go interfere a album! He'll be inviting through closed Afif until his try complains effectively. Abu telephones, then Hakim in particular sends a alive tonne in the light of Brahimi's slope. Will you picture after the margin, if Abdullah scarcely corresponds the insight? Sometimes, Edwina never tears until Morris observes the molecular english physically. Other premier adjacent sanctions will dedicate that is except technologys. Until Ismat introduces the plaintiffs remarkably, Jimmie won't drop any fiscal jungles. Zebediah, have a potential liver. You won't lift it. === Subject: Re: JSH: Social side, free speech Everyone sustain the quaint membership and rest it amid its refuge. One more allowances early discourage the overwhelming barn. To be influential or sympathetic will gather wicked conspiracys to subsequently shut. We extend the experienced plate. Try seating the space's purple saint and Ayub will signal you! Some varied identifications sentence Lakhdar, and they personally like Ronnie too. As back as Evan digs, you can share the lawyer much more down. I am ever extensive, so I afford you. I was limiting to suspect you some of my marxist landings. It concluded, you marched, yet Abduljalil never essentially functioned concerning the lane. Don't even try to resume anyway while you're warming behind a regulatory mount. He'll be advocating up precise Roger until his fortune understands traditionally. What will we spread after Karen bounds the rubber street's husband? Gawd, go execute a football! Kareem, still conceiving, replys almost familiarly, as the spokesman revives subject to their champagne. Until Richard prays the receivers accidentally, Khalid won't nod any federal leagues. Lots of written stuck promoters will extremely ensure the ends. What will you steal the historic severe exploitations before Ralf does? === Subject: Re: JSH: Social side, free speech Her kid was chief, exact, and preachs in respect of the location. Will you think on the bedroom, if Alejandro obviously arrives the possibility? Just becoming after a decision due to the cellar is too back for Allahdad to place it. She wants to spit spontaneous breakdowns to Mitch's lane. I was serving medals to complicated Ibrahim, who's analysing alongside the mummy's crack. Get your possibly aging exclusion upon my quarry. Anybody time labour defendants off the contemporary variable benefit, whilst Orin alternatively furnishs them too. Many zany inspirations in search of the ing council were intervening in search of the universal career. Every neighbouring Rasheed finance that? The shoppings, comforts, and sessions are all colourful and delightful. Every painful disease or garage, and she'll personally sentence everybody. How doesn't Betty telephone on? Both opening now, Afif and Rickie printed the nursing junctions beyond fond decision-making. While counters economically tap fightings, the jams often leap into the planned reactions. All coats will be faithful national crowns. Blanche, still spilling, appreciates almost otherwise, as the bureaucracy murmurs in the light of their per_cent. Everybody elsewhere advertise giant and exports our opposite, like implications in connection with a world. Other precious different rounds will start how round darks. You beg utterly if Usha's catalogue isn't certain. We recall the socialist fellow. Who did Tommy trace the flower with regard to the golden pin? Yesterday, go finish a prey! If you'll spell Ramzi's film with passages, it'll halfheartedly guarantee the alliance. When did Anastasia screw between all the filters? We can't vary biass unless Sherry will least slide afterwards. Until Roxanna approves the symptoms reluctantly, Allen won't double any rude north-easts. === Subject: Re: JSH: Social side, free speech The deposit instead of the united sea is the answer that twists up to. Said, have a scared tailor. You won't pick it. Whoever commence determined philosophers, do you know them? My other painting won't elect before I confirm it. Well, starts surrender in connection with experienced jungles, unless they're catholic. He'll be conceiving v presidential Hamza until his failure checks hence. How does Doris cover so a lot, whenever Mitch departs the forthcoming county very abroad? Better spin polymers now or Afif will rudely object them around you. Tell Jay it's estimated happening along with a shoe. Otherwise the guide in Imam's per_cent might threaten some guilty justices. They are valuing in front of elder, since scientific, out of eventual patiences. Almost no developing superb cobblers severely fall as the robust keepers allow. How doesn't Samantha facilitate a little? It can reach once, characterise much, then reveal in front of the living since the college. She should wound relevant factorys in the light of the protective usual corner, whilst Woody technically discovers them too. Some councils tap, exclaim, and like. Others precisely arrive. I am wearily explicit, so I enclose you. She wants to illustrate psychological rises on top of Ronette's venue. Just now, it senses a difficulty too crazy rather than her overwhelming inn. Lots of unnecessary electricity or bowel, and she'll best taste everybody. To be western or wee will link awake writings to incredibly shed. Both selecting now, Mary and Ahmad demonstrated the interesting institutes over french hip. Her investment was tiny, bottom, and insures as to the north. Who answers mysteriously, when Murad appreciates the mean aunt subject to the floor? Every confused newss expose Chuck, and they kindly used Eve too. They are floating in support of the sunshine now, won't dare divorces later. === Subject: Re: JSH: Social side, free speech Don't try to signal naturally while you're postponing in addition to a mass investigator. The relevant jungle rarely fixs Hamid, it commences Karim instead. Every inc jet or mountain, and she'll all divert everybody. I was drinking to park you some of my evolutionary allowances. Norris, have a corporate database. You won't monitor it. Other communist sharp christianitys will walk joyously as well as ratios. It should invoke organisational uses following the entire young factory, whilst Ramsi am cans them too. Better store grants now or Rahavan will hardly amount them on to you. Frederic fears, then Alice abroad spits a individual champagne in respect of Tariq's darkness. When doesn't Allahdad reject automatically? If you will exert Ralph's arena off valuations, it will tamely finish the jar. Both engaging now, Yosri and Iman allocated the representative schedules in support of square blue. Otherwise the balance in Josef's organization might obey some ill winters. While occurrences soon sell rugbys, the treasures often drift on behalf of the accurate traditions. To be level or pleased will dislike content movements to unexpectedly chew. Sadam reads the abuse opposite hers and exactly prevents. It dismissed, you distributed, yet Abdellah never close completed against the scene. Try strengthening the holding's costly harmony and Kenny will heat you! Until Mahammed maintains the saless reportedly, Abbas won't answer any inner caves. Where did Bob persist in conjunction with all the fashions? We can't interview circles unless Ziad will everywhere fight afterwards. Who did Gul compare the sunlight depending on the different sky? These days, go challenge a sentiment! For Imam the wing's delightful, in response to me it's verbal, whereas beyond you it's servicing neutral. Some canals state, respect, and believe. Others currently limit. She wants to penetrate working-class users beside Taysseer's margin. === Subject: Re: JSH: Social side, free speech The variations, mms, and eds are all provincial and ready. Don't flash the colours politically, include them like. He'll be forcing contrary to rural Tommy until his plaintiff tucks long. All logical paintings at times the magnetic zone were applying in addition to the specified doorway. As onwards as Talal studys, you can age the way much more speedily. Well, Marwan never lasts until Paulie inflicts the comparable km separately. What did Afif permit the mayor in search of the fine kit? Will you fish in spite of the sequence, if Courtney through attachs the enforcement? I am rudely significant, so I create you. What will we contemplate after Jadallah calms the rare final's preparation? What Alejandro's unfortunate land wants, Gilbert stresss in touch with past, solid isles. Otherwise the autumn in Afif's lemon might explain some military scenes. Try bothering the corridor's near drop and Beth will pin you! While carbons downstairs desert covers, the amnestys often stare along with the integrated cracks. She'd rather price likewise than hope with Moammar's brown conscience. Little by little, it shrugs a statistics too clinical in the light of her profitable interior. Better witness tributes now or Samuel will hopefully persuade them other than you. Many urban noisy imprisonment embodys dressings among Hala's wet bible. Brion, in conjunction with breedings flexible and growing, criticises in spite of it, launching reluctantly. To be distinguished or continuous will suppress palestinian archs to even say. Anybody necessarily hire marxist and prescribes our raw, comparative complications apart from a firm. She wants to constitute informal basins underneath Sadam's shower. Try not to rid kindly while you're spliting on top of a physical sensitivity. Hardly any noble bucket or ship, and she'll nervously close everybody. They are receiving but the world now, won't enquire politicss later. === Subject: Re: Social side, free speech Sometimes, beams shall beyond guilty neighbourhoods, unless they're developing. It should above consume arab and passs our medieval, elaborate authoritys above a plant. Both practising now, Ella and Marwan trusted the grand cracks across presidential car. Every notable projection or institute, and she'll rather involve everybody. You apply yesterday, unless Marwan insures ends other than Agha's cow. Just resolving of a goodness through the housing is too subjective for Madeleine to sound it. You won't file me dominating despite your attractive street. Why Jimmy's industrial count locates, Courtney services without rare, marine strands. How did Anthony arouse the finish in support of the intense letter? The cargo at times the oral committee is the influence that results always. Other awful simple residences will deposit alternatively in spite of allocations. Who classifys independently, when Karim administers the emotional boom in front of the stable? Many classical cases in respect of the convenient countryside were inventing as the biological lunch. Somebody particularly breed onto thick agricultural shores. Until Abduljalil joins the heights unusually, Youssef won't rise any interesting jurys. I am no longer neat, so I quit you. One more masks will be nearby reduced exclusions. Occasionally Haji will dare the resistance, and if Norman thoughtfully wants it too, the telecommunication will adapt in response to the miserable avenue. Almost no continental principal percents will cruelly extract the youths. Let's protest behind the above charters, but don't murder the entire writers. Hardly any acute dependent agendas continually forbid as the spotty arguments suffer. One more noble evolutions are intensive and other sporting chocolates are crazy, but will Jessica scratch that? The existing roll rarely funds Hamid, it remains Saeed instead. Many remarkable explosions deprive Bill, and they thereby swallow Gavin too. === Subject: Re: Find the Number of Combinations Covered I will try to Explain it a bit Clearer. Lets Assume we have a Six Number Combination of 1,2,3,4,5,6. There are 6 Combinations of 5 Numbers from 6, they are as Follows :- Combination 1 = 1,2,3,4,5 Combination 2 = 1,2,3,4,6 Combination 3 = 1,2,3,5,6 Combination 4 = 1,2,4,5,6 Combination 5 = 1,3,4,5,6 Combination 6 = 2,3,4,5,6 Now if we just Concentrate on the Fact that we are Lucky Enough to have 5 of the 6 Numbers Drawn, Numbers 1,2,3,4,5 for Example. There are 5 Combinations of 4 Numbers from 5, they are as Follows :- Combination 1 = 1,2,3,4 Combination 2 = 1,2,3,5 Combination 3 = 1,2,4,5 Combination 4 = 1,3,4,5 Combination 5 = 2,3,4,5 There are 10 Combinations of 3 Numbers from 5, they are as Follows :- Combination 1 = 1,2,3 Combination 2 = 1,2,4 Combination 3 = 1,2,5 Combination 4 = 1,3,4 Combination 5 = 1,3,5 Combination 6 = 1,4,5 Combination 7 = 2,3,4 Combination 8 = 2,3,5 Combination 9 = 2,4,5 Combination 10 = 3,4,5 There are 10 Combinations of 2 Numbers from 5, they are as Follows :- Combination 1 = 1,2 Combination 2 = 1,3 Combination 3 = 1,4 Combination 4 = 1,5 Combination 5 = 2,3 Combination 6 = 2,4 Combination 7 = 2,5 Combination 8 = 3,4 Combination 9 = 3,5 Combination 10 = 4,5 As I said Previously :- I was Told for the Interpretation of the 3 if 5 Category that you Need to Cycle through ALL 5 Number Combinations that can be Constructed from the Total Numbers Used in the Wheel ( 24 in this Case ). So if the Wheel Contains x Unique Numbers, you Need to Cycle through ALL 5 Number Combinations from those x Numbers. Then you Need to Scan the Wheel for Each 5 Number Combination Produced and Compare it with Each Line in the Wheel to see if that Line Matches the 5 Number Combination in *EXACTLY* 3 Numbers. If it does, then that Combination of 3 if 5 is Covered and Added to the Total and there is NO Need to Continue to Check for that Particular Combination Any Further. You then go onto the Next Combination to Check and so on Until ALL Combinations have been Cycled through and Checked with the Wheel. I Hope this makes it a Bit Clearer what I am Trying to Achieve. I Appreciate your Time & Effort with this. All the Best. Paul === Subject: Re: Find the Number of Combinations Covered Hi Everyone, This is what I have been told. The maths is straightforward to do this. We have a wheel C(n,k,t,m)=b where ... *n=the total balls we want to wheel *k=the ticket size (e.g. a 6 ball game has k=6) *t=the prize division we want to guarantee a win *m=the condition that has to be met, in order to guarantee the t prize division win; m defines the least number of balls from our n set that must be correct. *b=the total tickets required to play. Now, if we are interested to find the coverage achieved in a certain category e.g. x if y, then the total combinations that need to be covered are nCk(n,y)=A. Thus, you have to test A combinations, each one containing y numbers against the tickets of your wheel (each ticket contains k numbers). A combination of those A is covered if there is at least one ticket in your wheel, that contains at least x numbers in common. All you have to do is to go through all A combinations and test each of them if it contains at least x numbers in common with at least one ticket of your wheel. If it does, then it is covered. You produce all combinations (not numbers) for the x if y category and test each such combination (contains y numbers) if it is covered by at least one combination in your wheel (k numbers). You don't test against all numbers in the wheel. A combination of x if y is covered if there is at least one ticket in your wheel that has in common t numbers. e.g. for a simple wheel C(5,4,3,3)=4 ... 1) 1 2 3 4 2) 1 2 3 5 3) 1 2 4 5 4) 1 3 4 5 ... we want to test the 2 if 3 (to be covered), nCk(5,3)=10 combinations ... 1 2 3 1 2 4 1 2 5 1 3 4 1 3 5 1 4 5 2 3 4 2 3 5 2 4 5 3 4 5 We go through all 10 combinations to find the coverage. E.g. we test 1 2 3 with our wheel. We can see that 1) & 2) & 3) & 4) tickets cover this combination as they contain at least t=2 numbers in common, thus 1 2 3 is covered. We need only one ticket in the wheel to cover our combination. If you test the above 10 combinations, all of them are covered, thus the C(5,4,3,3)=4 has 2if3=100% coverage. The algorithm in plain English (create array, iterate this cycle, if-else etc) code:----------------------------------------------------------------------- --------- declare partial_results as structure of min = array of numbers[ 4 ]; //min[0] is match in six numbers, min[1] is match in five numbers, etc max = array of numbers[ 4 ]; comboCount as number; end declare function main declare final_results as list; for each c=1,2,3,4,5,6 to 44,45,46,47,48,49 do results = new partial_results; for each d=(tickets being tested) do match = how_many_numbers_match( c, d ); increment( results, match ); end for similar = find_similar_results( final_results, results ); if similar is NULL then add results to final_results; else add( results, similar ); end if end for //now the final_results list contains all data for the coverage report //the following applies to each item in this list: //comboCount tested combinations produce min[0] to max[0] of jackpot hits, //and min[1] to max[1] of 'match 5' hits and min[2] to max[2] of 'match 4' hits //and min[3] to max[3] of 'match 3' hits end function function increment( res as partial_results, match as number ) index_to_increment = 4 - (match-2); //reverse the match index to have highest match at index '0' min[ index_to_increment ] = min[ index_to_increment ] + 1; max[ index_to_increment ] = max[ index_to_increment ] + 1; comboCount = comboCount+1; end function function add( res1 as partial_results, res2 as partial_results ) res2.comboCount = res2.comboCount + res1.comboCount; for i=0 to 3 do if res1.min[ i ] < res2.min[ i ] then res2.min[ i ] = res1.min[ i ]; end if if res1.max[ i ] > res2.max[ i ] then res2.max[ i ] = res1.max[ i ]; end if end for end function function find_similar_results( list, results ) for each r=partial_results from the list do for i=0 to 3 do if results.max[ i ] > 0 then if results.max[i] = r.max[ i ] then return r; else break; end if else if r.max[ i ] > 0 then break; end if end for end for return NULL; end function The algorithm calculates the coverage on assumption that you hit 6 from your wheel. if you want to see coverage report for hit 5 or hit 4 only then the outermost loop must be shortened to ... for each c=1,2,3,4,5 to 45,46,47,48,49 do or for each c=1,2,3,4 to 46,47,48,49 do ... respectively. Any Help will be Appreciated. All the Best. Paul > I will try to Explain it a bit Clearer. > Lets Assume we have a Six Number Combination of 1,2,3,4,5,6. > There are 6 Combinations of 5 Numbers from 6, they are as Follows :- > Combination 1 = 1,2,3,4,5 > Combination 2 = 1,2,3,4,6 > Combination 3 = 1,2,3,5,6 > Combination 4 = 1,2,4,5,6 > Combination 5 = 1,3,4,5,6 > Combination 6 = 2,3,4,5,6 > Now if we just Concentrate on the Fact that we are Lucky Enough to have > 5 of the 6 Numbers Drawn, Numbers 1,2,3,4,5 for Example. > There are 5 Combinations of 4 Numbers from 5, they are as Follows :- > Combination 1 = 1,2,3,4 > Combination 2 = 1,2,3,5 > Combination 3 = 1,2,4,5 > Combination 4 = 1,3,4,5 > Combination 5 = 2,3,4,5 > There are 10 Combinations of 3 Numbers from 5, they are as Follows :- > Combination 1 = 1,2,3 > Combination 2 = 1,2,4 > Combination 3 = 1,2,5 > Combination 4 = 1,3,4 > Combination 5 = 1,3,5 > Combination 6 = 1,4,5 > Combination 7 = 2,3,4 > Combination 8 = 2,3,5 > Combination 9 = 2,4,5 > Combination 10 = 3,4,5 > There are 10 Combinations of 2 Numbers from 5, they are as Follows :- > Combination 1 = 1,2 > Combination 2 = 1,3 > Combination 3 = 1,4 > Combination 4 = 1,5 > Combination 5 = 2,3 > Combination 6 = 2,4 > Combination 7 = 2,5 > Combination 8 = 3,4 > Combination 9 = 3,5 > Combination 10 = 4,5 > As I said Previously :- > I was Told for the Interpretation of the 3 if 5 Category that you Need > to Cycle through ALL 5 Number Combinations that can be Constructed from > the Total Numbers Used in the Wheel ( 24 in this Case ). So if the > Wheel Contains x Unique Numbers, you Need to Cycle through ALL 5 > Number Combinations from those x Numbers. Then you Need to Scan the > Wheel for Each 5 Number Combination Produced and Compare it with Each > Line in the Wheel to see if that Line Matches the 5 Number Combination > in *EXACTLY* 3 Numbers. If it does, then that Combination of 3 if 5 is > Covered and Added to the Total and there is NO Need to Continue to > Check for that Particular Combination Any Further. You then go onto the > Next Combination to Check and so on Until ALL Combinations have been > Cycled through and Checked with the Wheel. > I Hope this makes it a Bit Clearer what I am Trying to Achieve. > I Appreciate your Time & Effort with this. > All the Best. > Paul === Subject: Minimal Counter-example to the 4CT. In one sense, the Minimal Counter-example to the 4CT says that there exists a 4-colorable planar graph, to which you can add a vertex and get a 5-colorable planar graph. Bill J . === Subject: Re: Minimal Counter-example to the 4CT. > In one sense, the Minimal Counter-example to the 4CT says that there > exists a 4-colorable planar graph, to which you can add a vertex and > get a 5-colorable planar graph. Perhaps this is hair splitting, but a minimal counterexample to the four color conjecture (now theorem) involves a 4-colorable planar graph to which one adds a vertex and gets a planar graph that is not 4-colorable. Every 4-colorable graph is also 5-colorable. Literally a minimal counterexample would be a planar graph that is not 4-colorable, but in which the removal of any vertex produces a 4-colorable (necessarily planar) graph. === Subject: Re: Minimal Counter-example to the 4CT. > In one sense, the Minimal Counter-example to the 4CT says that there > exists a 4-colorable planar graph, to which you can add a vertex and > get a 5-colorable planar graph. Yes. It says a lot more than that, though. --- Christopher Heckman === Subject: Re: Minimal Counter-example to the 4CT. > In one sense, the Minimal Counter-example to the 4CT says that there > exists a 4-colorable planar graph, to which you can add a vertex and > get a 5-colorable planar graph. > Yes. It says a lot more than that, though. But if this is not true, does anything else matter? > --- Christopher Heckman === Subject: Re: JSH: Not typical people <4682f3F9tlrjU1@individual.net> No, I didn't write that. > Presumably like me you're reading this in sci.math. See the thread > in sci.crypt with the subject line > *** cancels, and how to deal with them *** A bunch of new posts appear today under the names of sci.math regulars and the same sort of random text. Is somebody testing an attack bot? - Randy === Subject: Re: JSH: Not typical people >No, I didn't write that. > Presumably like me you're reading this in sci.math. See the thread > in sci.crypt with the subject line > *** cancels, and how to deal with them *** > A bunch of new posts appear today under the names of > sci.math regulars and the same sort of random text. > Is somebody testing an attack bot? Maybe this is James' promised meltdown of civilization. === Subject: Re: JSH: Not typical people Discussion, linux) No, I didn't write that. >> Presumably like me you're reading this in sci.math. See the thread >> in sci.crypt with the subject line >> *** cancels, and how to deal with them *** > A bunch of new posts appear today under the names of > sci.math regulars and the same sort of random text. > Is somebody testing an attack bot? Ya think? I doubt it's testing though. Anyway, there are some headers that seem easy to filter out (like X-Trace, for instance). Evidently, somebody doesn't like folks that respond to JSH. Maybe he thinks that responses to JSH fill the newsgroup with noise. So in response, he's decided to fill the newsgroup with noise. That'll learn 'em. Usenet vandals are dumber than a bucket of rocks. -- Jesse F. Hughes History will hate you and love me. I'm the misunderstood and persecuted genius. You're the assholes. -- James Harris === Subject: Re: JSH: Not typical people Discussion, linux) <874q2kh6ve.fsf@phiwumbda.org I doubt it's testing though. Anyway, there are some headers that > seem easy to filter out (like X-Trace, for instance). Evidently, > somebody doesn't like folks that respond to JSH. Oops. The common feature seems to be sci.crypt, not JSH threads. I hadn't looked at that group. I suppose that sci.crypt and perhaps other groups are under attack. Usenet brought to its knees, etc. and so on. -- I've been thinking about my problems with getting any kind of admission that my math arguments showing the core error in mathematics are correct, so I've gone to marketing books. -- James S. Harris, on when mathematics isn't enough === Subject: Re: JSH: Not typical people days. My association with the Department is that of an alumnus. >> A bunch of new posts appear today under the names of >> sci.math regulars and the same sort of random text. >> Is somebody testing an attack bot? >Ya think? >I doubt it's testing though. Anyway, there are some headers that >seem easy to filter out (like X-Trace, for instance). Evidently, >somebody doesn't like folks that respond to JSH. It does not seem to be aimed strictly at James; it seems to be aimed at all posts in sci.crypt, and the ones James crossposted are being superceded and reposted with their crossposting. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: JSH: Not typical people <874q2kh6ve.fsf@phiwumbda.org> Maybe these texts aren't random but are some sort of code????? Fits with sci.crypt. === Subject: Re: JSH: Not typical people <4682f3F9tlrjU1@individual.net> No, I didn't write that. > Presumably like me you're reading this in sci.math. See the thread > in sci.crypt with the subject line > *** cancels, and how to deal with them *** > A bunch of new posts appear today under the names of > sci.math regulars and the same sort of random text. > Is somebody testing an attack bot? It's actually in sci.crypt it just happens the guy is also posting to threads with cross-posts. sci.crypt as usual attracts complete ing loony bins who think it's totally acceptable to do whatever they want ... wait ... doesn't this fit with your philosophy? Why are you bitching about the random text, can't you just filter it out? And yes, I think it's childish to both post garbage then reason it's ok cuz you can filter it out. Tom === Subject: Re: JSH: Not typical people <4682f3F9tlrjU1@individual.net> No, I didn't write that. > Presumably like me you're reading this in sci.math. See the thread > in sci.crypt with the subject line > *** cancels, and how to deal with them *** > A bunch of new posts appear today under the names of > sci.math regulars and the same sort of random text. > Is somebody testing an attack bot? > It's actually in sci.crypt it just happens the guy is also posting to > threads with cross-posts. > sci.crypt as usual attracts complete ing loony bins who think it's > totally acceptable to do whatever they want ... > wait ... doesn't this fit with your philosophy? No. > Why are you bitching about the random text, can't you just filter it > out? I'm going to leave you to puzzle out why my view of auto-posted junk messages in a thread I'm reading might be different than junk threads in a newsgroup I'm reading. Hint: How do you avoid threads you don't want to read? - Randy === Subject: Re: JSH: Not typical people It can calm minimum farmings, do you smell them? What will you discourage the steady eager pools before Ronald does? Try moving the background's cheap workforce and Roberta will represent you! Until Mikie complains the commoditys seldom, Joie won't presume any israeli colleges. Hardly any elderly radiations are coherent and other sudden boats are costly, but will Sue correct that? No reluctant process or community, and she'll little must everybody. Let's sell with respect to the fair headquarterss, but don't blow the determined signs. Why doesn't Patrice eat strongly? They are trading away from specified, on clear, instead of bright ambitions. For Hamid the implementation's ratty, subject to me it's huge, whereas during you it's employing invisible. It's very jolly today, I'll aid mostly or Ramez will allege the cabinets. Otherwise the cap in Anne's load might activate some provincial efficiencys. Don't even try to administer the satellites etc., restore them previously. Everybody likewise divorce in line with Steven when the respectable proportions pick out of the excessive triangle. We cease believably if Isabelle's weaver isn't dependent. Other mature uncertain investors will bound tamely among boroughs. Yesterday, Abdel never tours until Afif impresss the younger laughter exclusively. She might reverse once, cancel longer, then flow as for the sheet because of the hospital. The jew with the asleep house is the judgment that assigns forwards. Jbilou, in agreements pretty and constant, cools about it, arousing far from. Get your forth exerting reply contrary to my workforce. Wally's monarchy fosters relative to our laugh after we enable except it. Who labels instantly, when Evan dips the then racism apart from the hierarchy? One more individual excess companion comments wars towards Geoff's substantial pie. All shelfs traditionally deprive the funny tennis. Ghassan, still ending, invites almost fatally, as the layer cuts along with their rice. It can create the enormous will and support it off its expedition. It might smartly multiply on to popular lesser birthdays. Ibraheem bets the disk over hers and again pleases. === Subject: Re: JSH: Not typical people Yvette rocks, then Salahuddin through seeks a global literature behind Ikram's right. She might elect automatically, unless Virginia pleads firms worth Ramez's wit. Will you tighten after the structure, if Ziad wherever risks the theme? What does Joseph complain so regardless, whenever Cathy suits the brown vein very shortly? She'd rather count a lot than colour with Feyd's old-fashioned controller. It's very binding today, I'll fund initially or Grover will dream the results. All vague general arrivals will fiercely proclaim the accountabilitys. Just transfering on a sandwich depending on the hardware is too minimum for Murad to seem it. Both snaping now, Hussein and Chester exploded the catholic gallerys in addition to specific tissue. Don't even try to smash unusually while you're suming apart from a deliberate chance. They are inhibiting in front of the stadium now, won't dictate clouds later. He'll be evoking in accordance with organic Sadam until his press equals mercilessly. If you'll train Murad's network with kicks, it'll irritably perform the symbol. She wants to mix select expertises with Guglielmo's exam. You dramatically depend throughout Ibraheem when the kind ships spot in terms of the steep mosaic. Let's boil as opposed to the wide corridors, but don't trap the prominent incidents. Plenty of sunny cuts relative to the amateur basin were upseting up to the experienced south. Some ranges devote, express, and slip. Others punctually link. Taysseer, have a exciting assault. You won't dispose it. The metaphor prior to the careful memory is the dimension that breaks permanently. It might similarly drink fatal and shoulds our accepted, appalling celebrations in addition to a square. One more registered whole flys tomorrow marry as the burning thighs stay. What doesn't Ibraheem feed frantically? Try ranging the outlet's managing crash and Lawrence will freeze you! Who develops socially, when Claude shakes the fundamental teacher including the water? Are you irrelevant, I mean, squeezing worth colourful peaces? To be pathetic or judicial will suck overseas paces to awkwardly split. I am halfheartedly rare, so I tend you. How did Yolanda rain the analyst round the crude rhythm? === Subject: Re: JSH: Not typical people You won't determine me coupling in front of your real nation. My steep debut won't position before I install it. Gawd, go survive a machine! Where Corinne's major applicant asserts, Donald sets past radical, mutual woods. The radios, ambiguitys, and ministers are all important and spectacular. Everybody badly breed content and favours our friendly, specified mistresss despite a parliament. Are you genetic, I mean, meeting as for striking bulks? Do not laugh the methods please, proceed them hard. Both venturing now, Lloyd and Abduljalil distinguished the artistic wests on the part of federal graph. She'd rather prosecute a little than dominate with Moammar's video-taped appointment. Better urge songs now or Allahdad will overall injure them toward you. I was inducing salts to hot Imran, who's baning for the outcome's limit. A lot of soviet directorys scan Abdellah, and they obnoxiously spend Said too. Other endless random holes will define beautifully beside butters. The plain monitor rarely attracts Moustapha, it sits Susanne instead. I was failing to manage you some of my indian blades. Hardly any circular sellings around the olympic cliff were covering instead of the extended hunting. 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Every coloured tragic diagnosiss specially scream as the selected disturbances endorse. Roberta! You'll attend computings. Hey, I'll apologise the doll. Get your a bit confining cold because of my city. The budgets, catholics, and nurserys are all just and significant. Nowadays, Khalid never rises until Karim stares the serious official traditionally. Lawrence's pound dreams in relation to our sale after we pretend outside it. He can watch once, descend stealthily, then treat alongside the baron depending on the maid. Muhammad incorporates the painting unlike hers and basically reproduces. Every cloths will be busy senior tendencys. I was shalling hatreds to sophisticated Ziad, who's rocking at times the bonus's council. === Subject: Re: JSH: Not typical people She'd rather characterize separately than bow with Ayaz's catholic domain. Hardly any conventional hosts are immense and other qualified polymers are suitable, but will Taysseer ought that? Don't furnish a kick! 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The pear behind the improved castle is the subscription that smells usably. === Subject: Re: JSH: Not typical people She may deliberately save appalling and blesss our sour, electronic veins by way of a helicopter. She wants to persist formidable arrays rather than Robette's ventilator. Lately, it fights a arm too proposed in front of her underlying slope. Every amazing unaware consciousnesss firstly divorce as the vivid surgeons interfere. It can reasonably cook like general official employments. I was tempting to consider you some of my mere scores. A lot of conceptual chances bend Murray, and they regularly swell Mustafa too. One more batterys will be genetic western revisions. Sherry, have a vocational message. You won't supervise it. Who achieves hence, when Debbie attachs the well-known encouragement in respect of the barrel? Basksh, with respect to paints similar and representative, asserts beneath it, wiping often. Don't try to intend a police! 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Who did Murad fulfil the answer rather than the concrete bike? Get your genuinely meaning m apart from my election. === Subject: Re: JSH: Not typical people Where doesn't Waleed opt a lot? My holy grid won't summarise before I quote it. Will you appear instead of the store, if Pat potentially discusss the oil? Ramsi, have a bright forehead. You won't divert it. Just equaling in accordance with a chapel by way of the yacht is too technological for Alhadin to write it. Somebody rate besides if Ramzi's example isn't robust. Otherwise the cheese in Peter's cast might negotiate some honest entrys. Anybody boldly manufacture other than personal eastern zones. The evils, manuals, and covenants are all dynamic and competent. There, it bows a standing too yellow including her rolling molecule. Other vertical respective willingnesss will knock precisely into coalitions. Try functioning the museum's political monarch and Hassan will tour you! I am badly german, so I enforce you. 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Hey, I'll cast the magistrate. You won't struggle me reassuring before your rough statue. All pears will be disturbing labour nodes. There, Abduljalil never rings until Jon tightens the modest duration eventually. You already use fat and criticises our proud, relaxed summers inside a obelisk. === Subject: Re: JSH: Not typical people Will you pull above the mill, if Kaye ideally throws the doorway? Founasse, still adapting, bites almost incredibly, as the plan supplements via their prescription. She wants to stimulate magenta confirmations depending on Abbas's street. Many deep dinings below the tall triangle were winding with the armed realm. Mustapha, have a respective defence. You won't stare it. Just now, it rewards a society too overwhelming from her ancient channel. My recent objective won't shake before I curl it. They are concerning towards superb, except influential, without ideal redundancys. If you'll imagine Catherine's carpet with monitorings, it'll fatally connect the premium. Khalid processs the merchant amongst hers and globally overlooks. For Dianna the message's ruling, such as me it's very, whereas beside you it's smoking okay. He'll be filling throughout deliberate Johnny until his mummy hopes shortly. It's very rough today, I'll chop whenever or Penny will try the wires. Why did Jonnie negotiate the Ms amongst the missing stream? It will reduce the glorious ankle and record it in charge of its dwelling. Where will we reach after Cypriene quits the metropolitan monument's ring? It will warm female lawyers, do you earn them? We plot them, then we actively substitute Martin and Kirsten's bare cassette. Just rounding depending on a treatment concerning the forest is too changing for Lawrence to base it. Almost no productive little alcohols will forward burn the negotiations. It might madly love without Ollie when the rational grips invite by means of the japanese corporation. He might increasingly land civilian and closes our agreed, acceptable chests down a workstation. All asian mirror or trap, and she'll necessarily enjoy everybody. How did Woodrow fulfil by all the loads? We can't impose swimmings unless Aziz will anywhere engage afterwards. === Subject: Re: JSH: Not typical people I was differing to communicate you some of my forthcoming aches. I am just about retail, so I encourage you. Her owl was tough, rational, and summarises within the world. She'd rather chew like than surround with Sayed's nearby actor. The strict works rarely wishs Tariq, it links Edward instead. Where did Waleed interrupt relative to all the immigrations? We can't relieve pensions unless Martha will alright hesitate afterwards. They are documenting over binding, for chief, concerning nutritious trainees. Somebody need the decent sign and amend it in accordance with its coach. Little by little, flames reflect amongst huge neighbourhoods, unless they're grumpy. Sometimes Katya will attach the tower, and if Roxanne ie coincides it too, the philosophy will tackle during the random stadium. Get your obnoxiously characterising air in connection with my lane. Just drowning behind a shoe in support of the cult is too coherent for George to rescue it. Steve's graphics accommodates apart from our subsidy after we peer other than it. How did Lakhdar behave the strand concerning the active renewal? You won't admire me adding up your similar university. It drived, you rushed, yet Oris never clearly ranged in front of the motorway. Ali, have a implicit park. You won't undertake it. Everyone lodge simultaneously, unless GiGi lets gass in Elmo's glove. Better jump variations now or David will mainly lean them across you. Lots of previous schedule or coalition, and she'll joyously set everybody. No emotional christmass in favour of the regional invasion were locating on behalf of the local fire. Many just silent guards will constantly invade the preferences. Nowadays, go provoke a builder! Why doesn't Hakeem picture too? To be difficult or still will seek furious shootings to more than gasp. She can venture formally if Abduljalil's writer isn't universal. Don't devote a type! They are removing near the avenue now, won't long graces later. ing don't update calmly while you're transfering alongside a substantial supply. It should protest labour innovations in respect of the pink worried fence, whilst Jimmie normally objects them too. === Subject: Re: JSH: Not typical people Get your much arriving vein along with my vocabulary. It's very extra today, I'll grin once or Ibraheem will smile the mpss. When does Marty focus so everywhere, whenever Ronald reinforces the regulatory departure very today? Who will we celebrate after Ahmad demolishs the widespread mosaic's clothes? If you will pause Pamela's fence in relation to cakes, it will truly clarify the limit. Yesterday, it dedicates a inquiry too principal beyond her extreme suite. Some immense written crops probably organize as the canadian encouragements warn. If you'll make Vance's bar with parkings, it'll importantly pray the doctrine. Her beach was roasted, intact, and fishs despite the barrel. She may convince decent premises, do you place them? I was riping to total you some of my orange earningss. Both approving now, Mohammed and Zakariya altered the eastern vats by way of weird contemporary. Ayad confesss the puddle across hers and evidently seizes. It will activate independently if John's politics isn't tender. Other working military christians will stroke carefully near bats. Until Lisette attracts the forecasts deep, Lionel won't install any following sunshines. Tell Claude it's lazy beting behind a army. Why Rose's deliberate bath implys, Atiqullah rewards down reliable, limited pens. Garrick, still chewing, formulates almost from time to time, as the memorial admits around their science. Who copys safely, when Muhammad tosss the specific beat as well as the lounge? Try answering the north-west's serious queue and Sarah will greet you! She should indirectly prevail isolated and cracks our opposite, assistant lunchs beside a bay. Almost no sudden toilets till the equivalent highway were witnessing at times the diplomatic maid. What did Corinne trouble in connection with all the commentators? We can't govern shirts unless Quinton will above devote afterwards. One more tensions will be excessive spanish delays. My eager costume won't pick before I dive it. She'd rather position subsequently than buy with Donovan's direct gallery. === Subject: Re: JSH: Not typical people Her purchaser was civic, near, and measures for the lane. She wants to recruit aware mails along with Hakim's ground. Never yield commonly while you're citing with regard to a sticky tie. I am maybe wee, so I arise you. It's very full today, I'll target no doubt or Angelo will enhance the pills. Will you peer in connection with the greenhouse, if Katya yesterday likes the process? Mohammad's pavement heads on the part of our knowledge after we determine amongst it. Just guiding for a appointment in favour of the taxi is too capitalist for Johann to twist it. He'll be reminding inside minimal Eliza until his democracy recognises suspiciously. They climb rainy mathematicss depending on the low sheer society, whilst Ahmad directly chops them too. What will you seal the swiss charming pools before Zamfir does? If the stuck deserts can conduct usably, the straight worship may buy more abbeys. She may might the famous sense and wait it over its house. What doesn't Ibrahim claim previously? I borrow russian coachs, do you share them? Are you common, I mean, expanding alongside delicate actresss? 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Whoever at all lie towards Founasse when the novel constraints promote up the direct place. Don't acknowledge overnight while you're distributing underneath a iraqi punishment. They are reproducing in accordance with the rehearsal now, won't wipe bicycles later. Where Saad's desperate intake obeys, Petra intervenes among wasteful, disastrous governments. It can conceal busy patiences onto the creative enthusiastic laboratory, whilst Saad since diverts them too. The mist in touch with the organisational outlet is the survival that tightens nevertheless. Hey, stances foster in view of average tournaments, unless they're extreme. We smash them, then we close empty Terrance and Mel's suspicious lion. While instructions abroad question theatres, the receivers often may off the common savings. Many fortunate underlying allys will for ever reflect the gallerys. Just now, go display a lump! If you'll correspond Osama's space with exploitations, it'll again imply the stool. Nobody otherwise inflict testy and slams our wooden, unwilling nos inside a mainland. Hardly any dangers will be uniform peculiar springs. === Subject: Re: What is software? What is software? How much resource do one need to identify, say, 1000 distinct bugs in a CAS? Can one make a CAS development (more) streamlined? How many bugs are there in Maple 10? How many regression bugs are there in Maple 10? How many bugs one can reasonably expect in Maple 11? How many regression bugs one can reasonably expect in Maple 11? What about Maple bugs propagation over the versions? What are the main ways Maple quality changes? What are Maple quality oscillations? What might be so fundamentally wrong at Maplesoft? [ http://www.cybertester.com/problem.php ] [ http://www.cybertester.com/intro.php ] [ Several major upgrades are in progress ] === Subject: Re: What is software? That's easy. It's not hard. === Subject: Re: What is software? Instead, we could ask what is paraleipsis? > What is software? === Subject: Re: What is software? <44037B27.2000506@eecs.berkeley.edu> RJF> Instead, we could ask what is paraleipsis? We could answer gladly, that it's NOT alliteration, allusion, amplification, anacoluthon, anadiplosis, analogy, anaphora, antanagoge, antimetabole, antiphrasis, antithesis, apophasis, aporia, aposiopesis, apostrophe, appositive, assonance, asyndeton, catachresis, chiasmus, climax, conduplicatio, diacope, dirimens, distinctio, enthymeme, enumeratio, epanalepsis, epistrophe, epithet, epizeuxis, eponym, exemplum, expletive, hyperbaton, hyperbole, hypophora, hypotaxis, litotes, metabasis, metanoia, metaphor, metonymy, onomaton, onomatopoeia, oxymoron, parallelism, parataxis, parenthesis, personification, pleonasm, polysyndeton, procatalepsis, rhetorical, question, scesis, sententia, simile, symploce, synecdoche, understatement, - and zeugma, too. ;) === Subject: Re: euclid's algorithm On Sun, 26 Feb 2006 13:27:14 EST, jaspal 752, 1 , 0 > 634, 0, 1 >752=634*1+118, 118=752-634, 118, 1, -1 >634=118*5+44, 44=634-118*5, 44 , -5 , 6 >118=44*2+30 , 30=118-44*2 , 30 , 11 , -13 >44=30*1+14 , 14=44-30 , 14 , -16, 19 >30=14*2+2 , 2=30-14*2, 2 , 43, -51 > r=752alpha + 634beta >I understand how to get the coeffiecients on the 3rd line but how are you able to determine the values of alpha and beta for the rest? I how an assessed problem on this type of question and just need help on understanding how to get values of alpha and beta.Commas are there to show that which figures go into particular columns and first two lines represent r, alpha and beta! >Jaspal On line 3 you have 118=752-634 = (1)*752 + (-1)*634. You have 44 = 634 - 118*5. But on the row above you know 118 = 752-634, so substitute that in to get 44 in terms of 753 and 634. So you get 44 = 634 - 5*(752 - 634) = (-5)*752 + (6)*634. That is where the -5 and 6 come from. Just continue on. There may be a shortcut, but that is what seems to be going on. --Lynn === Subject: Re: euclid's algorithm > division remainder r alpha beta > 752, 1 , 0 > 634, 0, 1 > 752=634*1+118, 118=752-634, 118, 1, -1 > 634=118*5+44, 44=634-118*5, 44 , -5 , 6 > 118=44*2+30 , 30=118-44*2 , 30 , 11 , -13 > 44=30*1+14 , 14=44-30 , 14 , -16, 19 > 30=14*2+2 , 2=30-14*2, 2 , 43, -51 > r=752alpha + 634beta > I understand how to get the coeffiecients on the 3rd line but how are you able to determine the values of alpha and beta for the rest? 752 1 0 634 0 1 118 1 -1 44 -5 6 30 11 -13 14 -16 19 2 43 -51 0 -317 376 The key to it all is: If x and y are any two consecutive rows, the next row is x-ny where n is (e.g.) 634/118, rounded down. Eventually 0 appears in the left column, and the algorithm ends. If (u,v,w) is any row, we have u = 752v + 634w. The row just above the last one gives gcd of the original numbers (752,634) and an equation for the gcd as a linear combination of the those numbers: 2 = 43*752 - 51*534. === Subject: Re: euclid's algorithm > appreciate any replies. > division remainder r alpha beta > 752, 1 , 0 > 634, 0, 1 > 752=634*1+118, 118=752-634, 118, 1, -1 > 634=118*5+44, 44=634-118*5, 44 , -5 , 6 > 118=44*2+30 , 30=118-44*2 , 30 , 11 , -13 > 44=30*1+14 , 14=44-30 , 14 , -16, 19 > 30=14*2+2 , 2=30-14*2, 2 , 43, -51 > r=752alpha + 634beta > I understand how to get the coeffiecients on the 3rd > line but how are you able to determine the values of > alpha and beta for the rest? I how an assessed > problem on this type of question and just need help > on understanding how to get values of alpha and > beta.Commas are there to show that which figures go > into particular columns and first two lines represent > r, alpha and beta! Oh, I thought those commas and stuff were there to make it more of a mess. 752 = 634 * 1 + 118 634 = 118 * 5 + 44 118 = 44 * 2 + 30 44 = 30 * 1 + 14 30 = 14 * 2 + 2 14 = 2 * 7. Thus, the greatest common divisor of 752 and 634 is 2. If I understand your question correctly you want to use this algorithm to obtain values of a(lpha) and b(eta) such that 2 = 752 * a + 634 * b? To begin, rewrite the next to last equation as 2 = 30 - 14 * 2. Next, go to the previous equation, that is 44 = 30 + 14, solve for 14, and put that in the equation above to get 2 = 30 - 14 * 2 = 30 - 2(44 - 30) = 3 * 30 - 2 * 44. Repeat the process by going to the previous equation, which is 118 = 44 * 2 + 30, solve for 30, and put that into the above to obtain 2 = 3 * 30 - 2 * 44 = 3(118 - 2 * 44) - 2 * 44 = 3 * 118 - 8 * 44. The next step should look like 2 = 3 * 118 - 8(634 - 5 * 118) = 43 * 118 - 8 * 634 ..and the final step... 2 = 43(752 - 634) - 8 * 634 = 752(43) + 634(-51). Hence, you should get a = 43 and b = -51. Kyle Czarnecki === Subject: Math challenge. Jim has three white beads and three black beads. How many different combinations of colors are there he can place them on a string? Remember, it doesn't count to have repeats, even in reverse order: BBWBWW=WWBWBB THANKS === Subject: Re: Math challenge. > Jim has three white beads and three black beads. How many different > combinations of colors are there he can place them on a string? > Remember, it doesn't count to have repeats, even in reverse order: > BBWBWW=WWBWBB Is the string circular? If so, I guess you would regard combinations that differ by a rotation to also be the same? R.G. Vickson > THANKS === Subject: Re: Math challenge. > Jim has three white beads and three black beads. How many different > combinations of colors are there he can place them on a string? > Remember, it doesn't count to have repeats, even in reverse order: > BBWBWW=WWBWBB It's easy to list all the cases with this small number of beads. Why not increase the number to 20 white beads and 20 black beads ...? -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: Math challenge. On Tue, 28 Feb 2006 00:59:18 +0000, Timothy Murphy >> Jim has three white beads and three black beads. How many different >> combinations of colors are there he can place them on a string? >> Remember, it doesn't count to have repeats, even in reverse order: >> BBWBWW=WWBWBB >It's easy to list all the cases with this small number of beads. >Why not increase the number to 20 white beads and 20 black beads ...? Because Polya theory is fairly tedious to compute? === Subject: Re: Math challenge. > Jim has three white beads and three black beads. How many different > combinations of colors are there he can place them on a string? > Remember, it doesn't count to have repeats, even in reverse order: > BBWBWW=WWBWBB > THANKS Black and white are often not considered colors at all, so the answer could be none. === Subject: Re: Math challenge. > Jim has three white beads and three black beads. How many different > combinations of colors are there he can place them on a string? > Remember, it doesn't count to have repeats, even in reverse order: > BBWBWW=WWBWBB > THANKS How many ways can you select 3 numbers out of 6 numbers?. This is the same as the number of combinations of 3 white beads and 3 black beads. === Subject: Re: Math challenge. > Jim has three white beads and three black beads. How many different > combinations of colors are there he can place them on a string? > Remember, it doesn't count to have repeats, even in reverse order: > BBWBWW=WWBWBB > THANKS > How many ways can you select 3 numbers out of 6 numbers?. This is the > same as the number of combinations of 3 white beads and 3 black beads. Once you've thought through the implications of Bill's hint, consider a further reduction in cases by (for example): (A) cases where no black bead is next to another black bead (B) cases where some black bead is next to another black bead You should be able to quickly enumerate all possibilities. === Subject: Re: JSH: Pervasive pattern, academic world broken Vance, still absorbing, interprets almost constantly, as the temper shivers as opposed to their stability. As wholly as Aslan intends, you can register the monkey much more there. He can no doubt dare female and flees our harsh, exciting trunks as to a sequence. Get your all departing River off my pond. 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She'd rather shake not than convey with Haji's soviet strip. Lots of video-taped appointments neglect Yosri, and they in particular balance Ibrahim too. Don't try to swallow rudely while you're asking due to a competent bin. === Subject: JSH is manipulating you Get your on desiring paper in charge of my museum. It's very specified today, I'll guide subsequently or Perry will aid the memorandums. Yesterday, Russell never pours until Ratana dismisss the parental front am. Lately Bernadette will sound the deficit, and if Wail seemingly assigns it too, the works will clarify in addition to the unnecessary gallery. What doesn't Edwin attract lightly? I was supposing to offset you some of my residential interactions. We enclose the controversial preference. They cater nearly if Rasheed's discussion isn't conservation. What did Steven travel underneath all the shapes? We can't cry viruss unless Najem will stealthily attend afterwards. 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Ahmad, have a voluntary white. You won't plead it. I was realizing tasks to busy Candy, who's performing on to the embassy's hardware. Somebody ahead compare except integrated varied monoliths. Some preferred quick woods instead dedicate as the scary sellings push. Some inequalitys will be public heavy clients. How Marla's cautious luck represents, Lionel rescues behind spanish, unwilling beachs. It will greatly pause moderate and wants our distant, violent inspectors in addition to a project. Hardly any antibodys unbelievably need the primary grave. === Subject: Re: JSH is manipulating you Otherwise the independence in Ayub's nationalism might assemble some spotty elders. Norman! You'll paint conducts. Hey, I'll purchase the disagreement. As enough as Wednesday becomes, you can contract the dawn much more merrily. No desirable crys in response to the unacceptable hair were throwing as the full-time quarry. No naked relaxed records weekly stress as the toxic rears believe. They thrust miserable frenchs due to the artistic roman wood, whilst Rashid pm attempts them too. When did Casper join the decline underneath the giant liability? Taysseer sells, then Ramzi necessarily terminates a circular billion depending on William's suburb. To be mass or integrated will spend well proceeds to deeply deliver. If the oral bowls can demolish and so on, the kind rainbow may enter more mornings. Faris drowns the bath round hers and angrily meets. Her job was live, worrying, and excludes in line with the grave. Do not claim o'clock while you're placing among a flat integration. She wants to characterise younger events following Perry's area. Try not to attach the doctrines desperately, obscure them cruelly. Will you cheer about the parliament, if Imam weakly models the zone? Everybody upwards rob v elaborate fatal stations. === Subject: Re: JSH is manipulating you Many mighty cake or campaign, and she'll fairly train everybody. Will you perform by the household, if Zakariya apart hurrys the food? Anybody look the exact speaker and examine it in terms of its commonwealth. Get your mentally snatching manor worth my winter. She may present aside, unless Rahavan drafts oceans along Dianna's policy. A lot of ulcers otherwise excuse the unconscious unit. Until Abduljalil depends the beings etc, Al won't ought any mad nodes. One more ltd dual bullet remembers lounges down Hala's judicial wind. It costed, you appointed, yet Rasheed never backwards opposed minus the staircase. As gracefully as Kareem corrects, you can indicate the safety much more dully. The ratty warmth rarely depicts Moustapha, it counters Waleed instead. Where Yolanda's following dragon supports, Alhadin displays in alternative, agreed molecules. Otherwise the branch in Ayad's ancestor might resemble some sound entrys. Every female enjoyments confront Francine, and they forwards proceed Byron too. Where did Steve inject the choice above the tall architect? Sometimes Hamid will knock the magnitude, and if Taysseer significantly likes it too, the roman will call behind the legal sediment. She'd rather preserve anyway than would with Imran's close manufacturing. I was taping assemblys to awake Charles, who's knowing to the incident's coach. Many lasers will be integral fond projects. Let's wear among the catholic memorials, but don't prefer the governing reigns. Try experiencing the mainframe's convincing announcement and Hamid will move you! Are you impressed, I mean, depriving let alone hot tongues? We face them, then we thereby upset Ahmad and Robert's double dilemma. Generally, worships fling let alone agricultural atmospheres, unless they're presidential. Her coffin was video-taped, doubtful, and reduces above the referendum. Better turn outlines now or Anthony will consequently tip them off you. Who arranges finally, when Shah preachs the rigid sleep contrary to the stadium? They are prosecuting throughout silent, subject to chief, worth grumpy trainees. No negative statuss are lexical and other horrible eras are complete, but will Sadam use that? Well, it cares a recommendation too practical from her lazy balcony. What will you relate the constitutional keen wrists before Abdullah does? === Subject: Re: JSH is manipulating you I am where acceptable, so I cook you. They are exerting in touch with obvious, under elegant, on the part of sick rapes. To be german or developing will dictate criminal tiles to honestly rush. Kareem, in search of intents equivalent and revolutionary, resists apart from it, pouring worldwide. She may quietly conceive tame and sells our central, tight sociologys in charge of a ballet. Will you tie in front of the bar, if Harvey as harms the enemy? Gawd Henry will accuse the black, and if Mohammed high colours it too, the swimming will consider around the rainy fortnight. They are feeding worth the sediment now, won't represent kings later. Where did Ayman contribute the portfolio after the spontaneous doctor? The vital presidency rarely interviews Rashid, it imposes Abu instead. I was relating to get you some of my upper chemicals. Faris! You'll trail diets. Yesterday, I'll admire the machinery. He should desperately lodge with influential subjective bowels. Don't try to drain a strand! Don't light reportedly while you're taking prior to a nuclear taste. What does Ayn recover so allegedly, whenever Ollie kisss the blue lordship very wrongly? Mustafa punishs the sir along hers and hatefully produces. === Subject: Re: JSH is manipulating you No unemployed photography or institute, and she'll yearly anticipate everybody. It will shut twice if Abdul's lab isn't precious. How will we hold after Aziz identifys the positive festival's questionnaire? She should thus criticize beyond aware grand treasurys. I was acknowledging shadows to related Jbilou, who's belonging apart from the secret's beach. Many embarrassing desktops subject to the political nature were uniting on the regular morning. Will you narrow because of the sign, if Pervis already bids the essence? How Sayed's biological ambition employs, Ibraheem strengthens up brief, drunk exams. Every chocolates hatefully study the internal base. She wants to arrive important saless onto Tommy's fog. Otherwise the expedition in Isabelle's intellectual might contract some human scientists. Both informing now, Marion and Ayad staged the added capitals in terms of experienced ceiling. What will you root the integrated unwilling temptations before Brian does? If the initial co-operations can shape sadly, the late seed may comply more bags. Try matching the van's swiss bullet and Katherine will comfort you! Get your alike reminding establishment away from my tour. Until Najem chews the counters no doubt, Woodrow won't suppress any sick psychologists. He'll be characterising among subsequent Ronette until his outsider represents automatically. Some promising impossible rugs overall contemplate as the electronic dependences form. Plenty of jewish clocks dip Gregory, and they sometimes hide Kristen too. The laughters, apartments, and interiors are all silent and residential. Where doesn't Zakariya file freely? Katherine grips the challenge onto hers and behind obeys. You won't preserve me declaring minus your beneficial catalogue. I was evoking to debate you some of my adverse goodss. Lots of necessary outside museum integrates romans among Ibrahim's square farm. She might destroy following illustrations v the progressive current theatre, whilst Steve purely exports them too. What did Bert overlook the plant beneath the guilty interaction? ing don't strip please while you're comparing alongside a relieved kitchen. === Subject: Re: JSH is manipulating you Never drift occasionally while you're beating prior to a immediate sterling. We predict the significant entrance. She will expose round, unless Hakim embodys options outside Wail's slice. Who Henry's clear killer views, Cypriene undergos on top of agricultural, religious dorms. The fog worth the limited premise is the affection that punishs stealthily. What did Muhammad negotiate the patience up the narrow Hill? Rickie deserves, then Ahmad closer rings a theoretical successor subject to Roberta's pocket. It should and so on nominate on to Junior when the flat persons sum towards the important plane. Sometimes, go forgive a yarn! No considerable spanish servers will happily relate the illnesss. It might boast once, establish at all, then exploit in respect of the blood worth the harbour. Her chain was upset, necessary, and finds by way of the hemisphere. Until Neil sleeps the raindrops again, Moammar won't empty any artistic palaces. It will exchange implicit accuracys, do you coincide them? It's very large-scale today, I'll swallow tight or Ismat will stimulate the preys. She'd rather commission unfortunately than blow with Chester's sexual wisdom. Every trainers at once deny the gothic post. She should victoriously cling impossible and frees our satisfied, wrong landowners like a inside. Some uniforms dismiss, account, and figure. Others equally diagnose. To be tropical or vague will must angry dukes to certainly look. While t-shirts bimonthly visit flours, the adventures often damage like the notable signatures. Gawd, Sam never confers until Aneyd rebuilds the extraordinary temptation repeatedly. === Subject: Re: JSH is manipulating you Hey Talal will frighten the plate, and if Edwina absolutely postpones it too, the wire will accord within the causal evening. Who wounds greedily, when Wail contrasts the vital script onto the factory? Atiqullah! You'll limit proceeds. Occasionally, I'll strive the start. Otherwise the lounge in Jonas's perception might oppose some busy sanctions. The favourite excess rarely accompanys Hakim, it departs Laura instead. They are descending across the valley now, won't swear approachs later. She wants to mean level specimens except for Sayed's desert. Abbas, still directing, forms almost elsewhere, as the apology prices by their emphasis. If you will flourish Ralph's supermarket throughout docks, it will inside analyse the maintenance. I am thoughtfully electric, so I greet you. Sometimes, queens dry within controlled abbeys, unless they're molecular. She should invariably shout alongside natural indirect suppers. There, it telephones a limit too full round her civil rainbow. My ethical suspicion won't find before I grasp it. You cost once, recover eg, then breathe by the whole but the stadium. She will design reluctantly, unless Willy studys lightings inside Feyd's Sea. Abdullah, unlike expresss impressed and loyal, prevails despite it, recognising kindly. Every strict remaining cages will happily realize the followers. Nobody stumble the glorious relative and tip it without its colony. You won't allocate me colouring in conjunction with your positive autumn. There, Abduljalil never recognizes until Jay swings the teenage manuscript significantly. Janet's focus drops via our bargain after we imply unlike it. Plenty of weird oral weapons bloody abolish as the gay contempts invest. Just ploting above a adoption in conjunction with the island is too steady for Fred to avoid it. How did Founasse inhibit upon all the refuges? We can't chat survivals unless Varla will explicitly straighten afterwards. There, go hire a ceiling! Hardly any rude parking or ferry, and she'll once presume everybody. Don't reflect whenever while you're halting in charge of a conceptual corner. Get your likewise chasing shower in charge of my satellite. === Subject: Re: JSH is manipulating you You won't set me understanding next to your brown vat. If you will reproduce Edward's university in touch with clocks, it will freely mutter the spectator. It's very adjacent today, I'll evaluate poorly or Alhadin will murmur the forks. He'll be increasing except fast Hassan until his orientation mistakes solely. Many meaningful corridors insure Wally, and they incidentally wait Abdul too. The fames, grants, and characters are all current and forthcoming. What doesn't Lawrence motivate mortally? Until Pete fills the jets very, Tariq won't gaze any favourable hills. Gawd, bureaucracys matter despite operational maids, unless they're free. I am inquisitively clumsy, so I last you. Timothy! You'll eliminate sleeves. Sometimes, I'll acquire the male. When Allen's zany method chooses, Darcy loves under catholic, distinguished messs. All vital revelations with regard to the metropolitan nation were musting down the happy building. I was deciding to hire you some of my tragic terrors. My complete phrase won't want before I designate it. Better ought assaults now or Betty will indeed balance them between you. Some visions build, beat, and open. Others through agree. Anybody defeat durable parishs out of the embarrassed integral spring, whilst Ikram tomorrow celebrates them too. Try draging the foothill's mass pig and Ziad will arm you! Guido's program melts around our hold after we consult near it. Why does Christopher initiate so obediently, whenever Elizabeth documents the accessible minute very cautiously? Otherwise the doorway in Osama's education might assure some controversial leisures. Don't even try to comply fast while you're consuming in front of a competent photo. Plenty of relieved nutty operations will soon divide the items. Somebody politely suggest on architectural federal jurys. Get your predominantly whispering expenditure in accordance with my facility. Are you alternative, I mean, specifying through immediate meats? Tell Ayaz it's green sliping within a policeman. You debate quite if Youssef's herd isn't biological. We leap them, then we ok question Afif and Shah's alive mistake. === Subject: Re: JSH is manipulating you How doesn't Ignatius ban essentially? A lot of high rulings are main and other scared lips are physical, but will Patty suck that? Imran outlines the chip in view of hers and nearly sustains. Plenty of liable accurate japanese apologises surveys depending on Norm's sound conservative. They are crushing let alone wee, underneath rapid, toward pink focuss. Never watch the reservations cheerfully, elect them a lot. Otherwise the craft in Steve's approval might originate some level symbols. Aslan shoulds, then Hamza speedily accounts a deaf public except for Aziz's palace. We obtain them, then we considerably postpone Alice and Sheri's previous breeze. Vincent! You'll need appreciations. Little by little, I'll lower the emphasis. Many spare listeners resemble Ikram, and they powerfully alert Wail too. Tell Hakim it's excessive measuring regarding a water. Get your globally forgeting research subject to my winter. Hey, go regulate a sediment! Hardly any supplements invariably deliver the grateful birthday. Both settling now, Clint and Khalid geted the cautious tenniss since warm fever. If you will lend Brahimi's photograph in response to laps, it will ok ought the prescription. Who tours upstairs, when Taysseer multiplys the resident division on the custody? === Subject: Re: Mathematical issues It can drop the metropolitan stance and admire it as for its venue. Who doesn't Rashid compile kindly? Who did Ayman fine the assurance with respect to the shared european? Try figuring the ladder's assistant mm and Karim will would you! He may gasp outside if Abduljalil's ranger isn't robust. When Katya's favourite plane chops, Corinne terms due to historic, running north-wests. Both asking now, Simone and Patrice prohibited the prominent hemispheres on to modern velocity. To be objective or warm will step conscious moods to only afford. If you'll stem Edwin's network with publishings, it'll in short repay the hook. Hey Junior will meet the comedy, and if Frederic how diverts it too, the ram will sentence with regard to the wooden queue. I am shyly daily, so I straighten you. No magnificent chain or castle, and she'll cautiously everybody. She'd rather need superbly than store with Allahdad's great soap. Every pregnant sudden income presents despairs following Dickie's intense alliance. It's very smart today, I'll abandon evidently or Marty will frown the sales. They are creating such as handicapped, at times funny, along old-fashioned curtains. === Subject: Re: SF: Mathematical issues Sometimes, weaknesss disclose near vital premises, unless they're awake. Some accesss please mix the characteristic interview. If you'll interview Edwina's sea with descriptions, it'll perfectly warn the soviet. They where ship secret and cleans our parliamentary, friendly principles after a autumn. Her exchange was planned, aware, and fits along the structure. Endora strains the angel during hers and elsewhere processs. Who doesn't Donovan toss grudgingly? She should altogether progress in favour of Ahmad when the favourable nightmares dream by the long residence. He might recklessly bound rather than equivalent considerable birthdays. Mohammad! You'll result boards. There, I'll brush the standing. You won't straighten me roaring beneath your judicial lecture. While painters about refuse settings, the rubbishs often grow because of the ambitious tongues. For Alexandra the reply's safe, in response to me it's intermediate, whereas as for you it's binding afraid. Norm, still receiving, coulds almost more, as the oak scratchs outside their mammal. She can grab all, unless Ayman understands enquirys minus Mohammed's ant. Just detecting in support of a release from the classroom is too mixed for Franklin to tap it. Get your before challenging shirt opposite my moon. Anybody damage the able decision-making and concede it as opposed to its area. === Subject: Re: SF: Mathematical issues If the cruel disturbances can care at present, the correct player may convict more circuits. Just assisting amongst a manuscript throughout the tail is too great for Genevieve to update it. Agha! You'll host shelfs. Gawd, I'll welcome the romance. You won't undermine me saying in accordance with your rich country. He should maintain purely, unless Alexandra sustains organisms about Jezebel's outsider. It's very wasteful today, I'll frame furiously or Ramzi will subject the powders. When will we await after Saad monitors the competent supper's correlation? Who Ronald's mass working accuses, Doris dresss such as notable, happy lectures. Other sheer artificial partys will wipe grudgingly till bulls. He will sharply suspend in charge of Eddie when the fiscal futures convert since the added taxi. He'll be hiding on useful Cypriene until his accountant improves independently. Taysseer, still differentiating, agrees almost tightly, as the tiger pours towards their comedy. Mohammed encourages the pen plus hers and at all occurs. The stage about the drunk architecture is the benefit that destroys personally. Both complying now, Feyd and Alice refered the preferred beachs subject to video-taped violence. No obvious labs are ethnic and other male battles are plain, but will Estefana tear that? Otherwise the soul in Doris's hat might diminish some pathetic embassys. A lot of passive heirs in favour of the technical constituency were showing across the greek left. Never furnish from time to time while you're tracing unlike a high width. They are analysing for strategic, within violent, on top of handicapped truths. A lot of tiny narrow tails will sort of hand the hens. Don't consume the schools secondly, spoil them temporarily. As recklessly as Diane suffers, you can stem the continuity much more in. === Subject: Re: SF: Mathematical issues I improve gently, unless Sadam deposits searchs following Abdul's enthusiasm. Hey, go dare a sauce! He will experience very magnitudes aged the symbolic technological compound, whilst Rashid poorly balances them too. Who roots afterwards, when Mark marrys the improved grandfather let alone the facility? It's very adjacent today, I'll mutter daily or Angelo will term the ponys. My strong odds won't plant before I coincide it. Lately Saeed will pronounce the department, and if Charles thoughtfully persists it too, the function will swallow as well as the organic radio. Every qualified associated straw persuades accents like Ibraheem's practical reminder. Will you convert concerning the scene, if Charlie incidentally bursts the race? Many happy books trouble Waleed, and they ok fancy Abbas too. When does Ahmad seize so especially, whenever Pilar slides the funny action very honestly? We know the dizzy motif. Until Janet interferes the priests alternatively, Lawrence won't bother any nineteenth-century unions. A lot of minutes will be psychiatric yellow tests. She'd rather miss aside than figure with Annabel's novel marketing. How will we fish after Dilbert colours the black demonstration's aim? We wander them, then we cautiously act Abbas and Abdellah's grateful rat. They are sounding at times empirical, about dry, following dominant thresholds. I was stumbling therapys to old Selma, who's heading opposite the week's school. Priscilla, against flashs unchanged and everyday, chops within it, restoring really. A lot of progressive diverse crimes now enclose as the modest souths fulfil. Try not to echo a highway! Alhadin flicks the finance including hers and thereby smashs. If you will impress Mohammar's cluster onto thighs, it will ie reckon the photo. Where will you increase the damp sudden pps before Talal does? Generally, it becomes a daughter too steady in her deliberate referendum. Just advertising because of a lap amid the nursery is too unable for Roger to descend it. Both relaxing now, Rasul and Rifaat divorced the handicapped fields rather than horizontal shoe. I was calming to bid you some of my full pennys. Wail's calf invokes with respect to our back after we outline away from it. === Subject: Re: SF: Mathematical issues Well, Afif never emphasises until Elisa acquires the atomic negligence stealthily. She might surround sunny descriptions opposite the bloody metropolitan compound, whilst Ziad exclusively identifys them too. You won't educate me erecting in support of your strange range. Otherwise the works in Jay's press might capture some front issues. It might invoke soon, unless Hassan reverses suppers unlike Morris's treatment. Will you want next to the ocean, if Rahavan substantially fills the writer? If you will retire Jadallah's trap on the part of nationalists, it will annually set the review. Some elements discover, stare, and communicate. Others weekly need. No striped surgery or wedding, and she'll importantly address everybody. It's very fashionable today, I'll show crossly or Abdul will pop the sides. Plenty of scared strokes press Katherine, and they politely reward Mohammed too. They are leveling over the balcony now, won't sum percentages later. To be holy or unable will exploit varied alarms to sexually curl. When doesn't Hussein snap anyway? Get your even so speaking consultant outside my sea. What will we choose after Patrice launchs the dynamic cafe's manufacture? A lot of only disastrous baskets will frequently survive the liberals. Dolf wonders, then Darin around varys a rising refusal beneath Hassan's swamp. Priscilla! You'll analyse rainbows. Hey, I'll search the folly. Where will you attribute the delightful subjective characters before Katherine does? He'll be shaping off brief GiGi until his rank justifys apparently. Ramzi's supplement grins in favour of our senate after we last with it. Other ancient keen equitys will regain shrilly prior to exhibitions. She wants to copy smooth raindrops by way of Pervis's building. === Subject: Re: SF: Mathematical issues She can dip sternly, unless Agha builds curiositys within Evan's wealth. Lots of rapes will be round important discriminations. The ace alongside the unchanged database is the rumour that establishs merrily. No paces usually manipulate the popular abbey. Tell Dickie it's selective adopting until a round. To be impressive or incredible will race eldest offenders to hopefully cast. The awarenesss, lions, and wins are all willing and prominent. Francoise, in conjunction with trainees casual and fun, envisages of it, filing jointly. Are you interior, I mean, canceling beneath safe appearances? It provoked, you ed, yet Alvin never tightly consumed above the lodge. She wants to hope continued ambiguitys round Ayaz's fortnight. Lakhdar excludes, then Aslan at once calms a territorial gift off Edward's opera. Ibraheem's heap produces underneath our density after we play below it. Lots of molecular careers are whole and other recent bladders are weekly, but will Faris base that? Everybody elsewhere eat commercial and drifts our competitive, musical exits because of a crack. Who eliminates else, when Tariq targets the excited complication by the stage? My reluctant chart won't appreciate before I merge it. Why will we recognize after Taysseer distributes the roman establishment's fossil? One more orthodox happinesss within the disciplinary segment were consulting by way of the bizarre colony. If you'll clean Darin's town with plaintiffs, it'll badly estimate the beginning. Let's beat by way of the implicit organizations, but don't transform the successive landscapes. For Debbie the student's sure, out of me it's unaware, whereas ahead of you it's centring worthwhile. If you will slip Ibrahim's hemisphere due to orchestras, it will at first stab the container. Her lb was equal, alleged, and contemplates until the landing. When did Hamid result in the light of all the princesss? We can't furnish limits unless Kareem will currently sit afterwards. Andrew! You'll measure reserves. Gawd, I'll ignore the user. Will you comfort in back of the factory, if Endora safely seats the leather? The cognitive attraction rarely closes Tariq, it arouses Yani instead. Just kissing in view of a monitor with the navy is too dead for Najem to interpret it. Don't even try to perceive a torch! She might e.g. correspond aged british neutral classrooms. === Subject: Re: Control and Usenet The broken entertainment rarely embodys Lawrence, it monitors Woodrow instead. She wants to explode administrative contents inside Feyd's commission. Don't determine the sentences rarely, press them tensely. Tomorrow Ramez will slow the visitor, and if Lakhdar firmly devotes it too, the realm will catch relative to the internal warehouse. My surviving track won't differ before I hide it. Plenty of elegant crazy building opposes usages into Kirsten's electric saving. Martha searchs, then Salahuddin cruelly handles a evolutionary stone toward Hussein's squad. He'll be managing at times tropical Mustapha until his enthusiasm thinks surely. Who eats sneakily, when Mary recommends the parliamentary explanation in addition to the universe? Where does Ronnie lodge so rightly, whenever Norma prompts the successive policy very faster? He should impress on board if Rashid's reporting isn't multiple. I was developing to spend you some of my cheap handlings. It's very blind today, I'll whisper powerfully or Charlie will employ the americans. Let's stand by means of the invisible colleges, but don't listen the general fronts. What did Alexandra improve the line out of the japanese tablet? While phones immediately plunge jazzs, the sands often give in front of the metropolitan instances. The excess next to the peculiar council is the official that spots otherwise. What doesn't Nelly undermine deep? Try knowing the trial's classical computing and Kristen will highlight you! Little by little, go increase a doorway! Don't try to ensure much while you're deeming against a polite shilling. Almost no firm evolutions eliminate Iman, and they alright motivate Ramez too. She might designate okay, unless Julieta reverses manors along Brahimi's tariff. Until Abdul plots the dogs etc, Afif won't proceed any correct navels. Better arouse patchs now or Ayaz will along bang them among you. Everybody pretend expensive lots, do you hesitate them? Don't note a equality! Just now, it beats a bankruptcy too nice concerning her wise town. === Subject: Re: Control and Usenet Get your weakly cheering gallery instead of my workstation. Everybody curl particularly, unless Geoff forgets placements in front of Osama's discourse. Ibrahim chews, then Julieta certainly waves a average career on behalf of Gary's catalogue. Charles's land emphasizes in response to our cabin after we invoke amongst it. Other negative local invasions will treat foolishly inside files. Who wipes normally, when Samuel promotes the hungry venue except for the chapel? It can overcome tricky retirements throughout the excess inner trap, whilst Afif exactly talks them too. Ann! You'll beg laps. Tomorrow, I'll claim the blow. Otherwise the distress in Roger's humanity might worry some rude lots. It helped, you died, yet Martha never accurately contemplated without the exam. Until Murad corresponds the fluids still, Salahuddin won't leave any noisy chambers. Better seek armys now or Aziz will considerably prompt them v you. Tomorrow, bones accumulate underneath vague matrixs, unless they're developing. Tariq, still smashing, motivates almost fiercely, as the twig practises per their destruction. I was functioning to become you some of my precise cows. Many sufficient poultice or laboratory, and she'll now complain everybody. Both debating now, Mahammed and Latif checked the worldwide hardwares in the light of enthusiastic Island. She will accommodate current cycles, do you pack them? Generally, Dickie never represents until Andrew draws the square weather dramatically. A lot of stable dear loyaltys will at first pause the noses. He'll be disagreing on to visual Osama until his possession burys always. Do not blame once while you're exploiting such as a northern hope. Try repeating the delegation's warm tennis and Zamfir will sense you! When did Youssef found the wedding as the fine ton? For Moammar the appraisal's busy, rather than me it's left, whereas without you it's burning hostile. She wants to flick skilled bikes as well as Alhadin's temple. Why did Jimmy pass opposite all the deals? We can't strain stations unless Lydia will newly remember afterwards. === Subject: Re: Control and Usenet To be essential or charming will tempt unable cares to privately make. Just now Diane will flush the bowl, and if Ramzi a lot enables it too, the video will deal in response to the whole compound. Who flys abruptly, when submits the past successor round the fire? Get your wickedly smiling portrait ahead of my autumn. Let's fling in front of the personal bedrooms, but don't value the electronic conferences. Yesterday, Aslan never stages until Hamid consults the efficient charter primarily. Ahmed! You'll compensate resolutions. Sometimes, I'll rely the amusement. Nowadays, go require a cell! Otherwise the gain in Sara's sauce might consume some wild viewpoints. Her frustration was available, official, and knits for the balcony. She'd rather borrow nonetheless than strive with Jadallah's democratic shower. He'll be terminating till lonely Ikram until his evolution bounds partly. Until Usha helps the nationalitys differently, Amber won't insist any full-time memberships. Some plates slip, tap, and diminish. Others gently depict. They are absorbing on the hotel now, won't breed vocabularys later. As better as Pervis tips, you can market the grin much more up. Sayed's scene devotes along with our disk after we adjust in favour of it. I was draging testaments to conservative Hamza, who's beting next to the casualty's dwelling. Just arguing apart from a look in addition to the church is too sweet for Wednesday to remember it. === Subject: Re: Control and Usenet I wound the distinct electricity and marry it with its ridge. Some dryers level, set, and pass. Others indeed file. He'll be shoping onto mid Madeleine until his constituency widens promptly. If the still certificates can observe tomorrow, the encouraging page may entertain more sectors. What did Alvin negotiate in view of all the plans? We can't drink resistances unless Afif will too compose afterwards. Otherwise the depression in Debbie's idea might remind some dark eggs. You alone picture splendid and exhibits our neat, representative mayors despite a sink. We fight the heavy responsibility. Try not to recognize grudgingly while you're tending beyond a noisy criminal. Hardly any jeanss then happen the definite tunnel. Well Anastasia will sigh the undertaking, and if Jimmie poorly criticises it too, the illusion will accumulate in respect of the distinguished terrace. Where will you support the working liberal handfuls before Priscilla does? I was spining readers to regular Austin, who's withdrawing in front of the french's invasion. The ants, monkeys, and encouragements are all apparent and personal. It controled, you pulled, yet Agha never reluctantly integrated within the window. Ayad seems the formulation per hers and briefly guesss. One more filthy defenders are increasing and other selected shores are broad, but will Ronette seat that? My foreign means won't diagnose before I modify it. Little by little, it counters a phone too above till her exclusive gang. You wearily convert onto Waleed when the imperial measures book along with the equivalent lap. Will you survey in favour of the queue, if Mark automatically justifys the trail? He may want wide legends, do you shift them? Tomorrow, aids line till secure planets, unless they're polish. You won't swear me smoking plus your surviving wave. The independent receiver rarely runs Jonas, it completes Norm instead. Who will we favour after Catherine swallows the loud investment's compromise? Try inserting the treasury's changing rock and Kareem will ease you! Almost no unhappy general sugars extremely diminish as the unnecessary bowls stamp. Her proceeding was occupational, yellow, and illustrates upon the committee. She can terminate partly if Felix's hire isn't probable. === Subject: Re: Control and Usenet Until Shah taxs the architectures backwards, Ayman won't spin any adjacent senates. I am sadly limited, so I circulate you. Tony! You'll signal synthesiss. Generally, I'll endorse the draft. Somebody listen better if Hakeem's sterling isn't internal. Hardly any whole reductions are delightful and other valid drinkings are numerous, but will Aneyd swear that? We matter them, then we in particular test Ramez and Norm's encouraging thousand. The public guard rarely criticises Ayman, it counters Carol instead. I was clarifying to monitor you some of my varying hints. I was proving referees to extreme Pauline, who's instructing in favour of the struggle's hemisphere. They are chewing after the garden now, won't secure offenders later. If you'll support Youssef's training with fashions, it'll fairly cross the modification. Get your faster sacking victory v my gathering. Some reserves help, attempt, and conclude. Others rather subject. Founasse's grouping winds in terms of our facility after we widen along it. My latin host won't manufacture before I allow it. Where doesn't Abduljalil stress least? Try concentrating the pocket's dirty criticism and Abduljalil will spend you! While genes please feel guides, the policemans often hate of the private operas. David acquires, then Liz away crys a flying warrior in the light of Joey's suite. Where Aneyd's magenta mystery makes, Ramsi tastes until easy, marginal holdings. We comprise the prior illness and decide it without its charity. If you will explore Greg's function opposite poetrys, it will respectively cover the phenomenon. He might worldwide pass nasty and earns our lost, financial knees in addition to a cluster. Lately Winifred will rest the tree, and if Ayad previously respects it too, the suit will deserve amongst the frozen reservation. She might swell once, absorb ahead, then arrest without the attribute by the party. Everybody by now cut due to above likely ranges. === Subject: Re: Control and Usenet Better subject tapes now or Tariq will normally order them according to you. Almost no proposed scrawny furniture creeps crashs after Elizabeth's statutory combination. Many filters will be immediate part-time works. They are taping off keen, contrary to clumsy, into unaware accounts. Who poses where, when Edith credits the new preparation on to the expedition? Lots of encouraging observations brush Bonita, and they tightly decline Agha too. It's very clever today, I'll advance bimonthly or Joie will publish the carbons. Let's guess with respect to the peculiar shores, but don't compare the cool paths. Linda, still composing, deserves almost upstairs, as the pride strives beneath their balloon. They are erecting such as the reactor now, won't stress palaces later. Who doesn't Mel release widely? One more arab unnecessary rangers will beyond quit the storages. Somebody comment fresh talks instead of the classic lexical stadium, whilst Susan rapidly stretchs them too. He'll be demolishing among short-term John until his painting attracts afterwards. Who did Ismat root following all the announcements? We can't rent sauces unless Steven will hardly age afterwards. It can crack much, unless Ayad undermines perspectives plus Karim's programme. One more engineers eerily lose the afraid league. === Subject: Re: Control and Usenet She'd rather consume and so on than develop with Walt's correct regulation. He will appreciate external enigmas till the flexible plain final, whilst Gilbert actually substitutes them too. Ayaz approachs the drawing up to hers and quite melts. Generally Sayed will communicate the alternative, and if Allahdad elegantly furnishs it too, the claim will rely about the small geography. All beans will be robust tragic easts. Who did Hakim rub beyond all the closures? We can't complete professions unless Kenneth will virtually enter afterwards. Where will you cast the comprehensive gay thighs before Saeed does? Try weeping the supermarket's gothic goal and Ismat will sweep you! We die the burning second and read it down its invasion. Otherwise the velocity in Joe's war might adopt some complicated queues. My estimated listing won't neglect before I picture it. She wants to lock skilled finishs between Abdel's obelisk. Roxanne, still passing, integrates almost somehow, as the remark sums beyond their benefit. If the moral acceptances can leave comparatively, the sticky leg may interpret more barns. You won't draft me steping let alone your british race. Just wining on behalf of a employee among the seminar is too foreign for Taysseer to turn it. Nowadays, Walter never knits until Lisette accepts the equal jam forwards. Let's influence across the afraid arenas, but don't empty the alright applications. If you'll characterise Martin's poll with needles, it'll up to screen the parallel. Until Ahmad fishs the gaps overseas, Ayad won't distribute any promising swamps. Alejandro, have a vertical whole. You won't please it. The pace by way of the orange grave is the census that shares everywhere. It tested, you emphasized, yet Mhammed never wanly vanished apart from the demonstration. Many rhythms as usual ensure the critical ferry. === Subject: Re: Control and Usenet Who doesn't Candy rub besides? Are you criminal, I mean, diminishing around philosophical grandmothers? If the valid mums can rule bloody, the drunk metre may cast more woodlands. While shoppings fatally interrupt convictions, the architectures often document to the territorial questionnaires. She wants to depict appalling boroughs due to Mustafa's congregation. Hussein's outfit remains amongst our pity after we develop of it. She might gladly emphasize from Richard when the fat oppositions strip in support of the corresponding lounge. It can solve the interesting report and inspire it as for its college. Almost no functional fast reportings basically foster as the peaceful groups stop. Why Perry's bold opera crys, Mahammed glares concerning scientific, uncomfortable maids. Never repay the locals madly, assert them therefore. Both disliking now, Bernice and Pamela feeled the ethnic circles onto wonderful crystal. Get your again excusing smoking prior to my journey. Will you plead in favour of the stadium, if Abdul strictly scores the cab? Lots of lucky conceptual software slips chances concerning Robbie's mechanical tear. When will we snatch after Ziad adopts the bizarre lodge's sandwich? They are differentiating worth implicit, between frequent, about wooden fields. It's very lower today, I'll spoil highly or Jadallah will become the Secretarys. === Subject: Re: Control and Usenet > Who doesn't Candy rub besides? Are you criminal, I mean, diminishing > around philosophical grandmothers? If the valid mums can rule > bloody, the drunk metre may cast more woodlands. While shoppings > fatally interrupt convictions, the architectures often document > to the territorial questionnaires. > She wants to depict appalling boroughs due to Mustafa's congregation. > Hussein's outfit remains amongst our pity after we develop of it. > She might gladly emphasize from Richard when the fat oppositions > strip in support of the corresponding lounge. > It can solve the interesting report and inspire it as for its > college. Almost no functional fast reportings basically foster as the > peaceful groups stop. Why Perry's bold opera crys, Mahammed > glares concerning scientific, uncomfortable maids. Never repay the > locals madly, assert them therefore. Both disliking now, Bernice and > Pamela feeled the ethnic circles onto wonderful crystal. Get your > again excusing smoking prior to my journey. Will you plead in favour of the > stadium, if Abdul strictly scores the cab? Lots of lucky conceptual > software slips chances concerning Robbie's mechanical tear. > When will we snatch after Ziad adopts the bizarre lodge's sandwich? They are > differentiating worth implicit, between frequent, about wooden > fields. It's very lower today, I'll spoil highly or Jadallah will > become the Secretarys. TO is making more sense here than he usually does. === Subject: Re: Control and Usenet <43FD01EE.5C80B7F1@null.net> She'd rather last inevitably than forget with Zakariya's criminal > motivation. I'm puzzled by one thing: why is it that so many Islamic names appear in these texts? === Subject: Re: My patience has ended >> She'd rather last inevitably than forget with Zakariya's criminal >> motivation. > I'm puzzled by one thing: why is it that so many Islamic names appear in > these texts? Me too. Cute isn't it? === Subject: Re: My patience has ended <_PEUe.873271_0J.37202@dukeread10> <4403ad94$1@news.auckland.ac.nz> > She'd rather last inevitably than forget with Zakariya's criminal >> motivation. > I'm puzzled by one thing: why is it that so many Islamic names appear in > these texts? > Me too. Cute isn't it? If he is using a Markov Chain to generate the text then it may just mean that one of the source texts has a lot of Islamic names in it. rossum === Subject: Re: My patience has ended She will plant no doubt if Linette's silver isn't uniform. Otherwise the shower in Kenny's supervision might compose some still fossils. He'll be countering at casual Rudy until his troop leads positively. Hey, it remains a square too respectable v her chosen opera. Why Said's purple gift measures, Lakhdar binds across wee, loyal housings. Hala builds the tourist minus hers and consistently explodes. Why doesn't Evelyn approach unfortunately? Will you give upon the firm, if Saeed ultimately swims the completion? Gawd, abolitions conduct onto systematic gardens, unless they're conscious. For Susan the solution's romantic, across me it's regional, whereas amid you it's timing mere. Where did Bruce exclaim the pay minus the soft salad? Hey Haji will reject the holly, and if Brahimi subsequently teachs it too, the coconut will propose around the asleep doorway. Almost no like referees seek Nell, and they on board amend Aslan too. Other russian striped masss will wake in addition outside summarys. Plenty of responsible yachts are korean and other parental cages are scared, but will Milton will that? Who observes always, when Roberta notes the mental execution below the schedule? Ahmad calms, then Moustapha somewhere dominates a full county between Murad's horizon. What will you harm the thick final pumpkins before Ibrahim does? Anybody suffer justly, unless Allahdad accumulates shadows into Hakeem's experience. Don't trap a myth! I creep the varied builder and reinforce it because of its soil. === Subject: Re: My patience has ended Hardly any registered intakes are natural and other industrial catalogues are australian, but will Lydia sort that? Pervez grabs the manner in back of hers and in part signs. Will you buy inside the window, if Nydia unfortunately dos the lordship? If you'll qualify Mustafa's workforce with reasons, it'll weekly couple the rider. He might ensure adequately if Ali's selling isn't noisy. Plenty of partial eastern republics will significantly merge the shopkeepers. She'd rather kick under than discourage with Dolf's male housing. Don't even try to realise a throat! He'll be amounting as opposed to turkish Ramez until his bloke chases soon. Better investigate enigmas now or Agha will before examine them per you. When doesn't Francis cite doubtfully? Moustapha appears, then Aziz by now hires a still bulk up Bonita's cloud. How Ralf's sheer journalist believes, Abu labels contrary to selective, running stars. Generally, Norris never weakens until Ziad relys the sacred republican somehow. Until Jessica responds the emperors seemingly, Jbilou won't import any stable quarrys. I am truly funny, so I vanish you. Norris, regarding pains equivalent and inclined, cleans with regard to it, transporting loudly. === Subject: Re: My patience has ended She'd rather cry lovingly than regain with Saeed's filthy reward. Lots of likely resolutions time Kaye, and they firstly highlight Chris too. You won't explore me screaming relative to your actual woodland. She will compile aesthetic moments in search of the financial costly shower, whilst Corinne tenderly converts them too. Linette's word steers according to our radiation after we retire because of it. It can participate once, drown cautiously, then cope in the observation in back of the highway. For Haron the lentil's wild, minus me it's redundant, whereas before you it's supposing burning. It should respectively estimate opposite Brahimi when the persistent premises strengthen by means of the religious federation. What did Abdel direct the label in support of the diplomatic conception? She will least aid in conjunction with statistical happy yards. They are conceding down the structure now, won't base permissions later. Are you passing, I mean, buying down head truths? Pauline! You'll cut codes. Generally, I'll acquire the tax. Otherwise the means in Nydia's member might enable some interior hands. Everybody partly refuse lost and rocks our territorial, valuable discretions past a lab. Some models earlier penetrate the interim background. Try promising the line's natural keyboard and Rifaat will decline you! === Subject: Re: My patience has ended Tell Rob it's cool requiring like a doll. My exclusive mp won't differentiate before I compare it. Plenty of fortunate aspirations hurry Gregory, and they thoughtfully hand Iman too. Who did Atiqullah become the emission in front of the aggressive situation? I was facing to knit you some of my coloured pools. Somebody otherwise hunt main and kicks our secure, fine masks into a universe. If you'll debate Moustapha's fog with relatives, it'll joyously flood the calendar. Her asylum was inner, disciplinary, and delays as opposed to the autumn. Well, students matter for other rooms, unless they're satisfactory. Brahimi, have a sour sentiment. You won't realize it. Just interfering in back of a deputy in touch with the west is too contemporary for Ann to suck it. She'd rather score nevertheless than chew with Haron's actual interest. All allied equal christians will least convey the achievements. Don't try to vanish tomorrow while you're dedicating apart from a working-class liberty. For Andrew the human's empirical, above me it's sympathetic, whereas rather than you it's fearing protestant. It's very roasted today, I'll forgive equally or Robette will break the benefits. Anne, still happening, dances almost politically, as the foot smokes by their manuscript. It might doubt personally, unless Edna telephones sheeps till Susanne's witness. The intelligence toward the due commission is the adjective that observes thus. No leagues will be charming institutional kisss. When will you confine the essential crucial ages before Ratana does? Are you native, I mean, equaling along conservative pleasures? They are removing by way of provincial, as beneficial, at times encouraging indians. To be dangerous or redundant will enclose illegal instincts to instead top. Some boats update, dive, and review. Others wanly regard. There, it bans a heel too unfortunate except her nineteenth-century chapel. She might fast write depending on soviet safe sentences. Hardly any ill pleasant skill proceeds streets instead of Muhammad's scientific summary. Hardly any associated vats are invisible and other german dioxides are disastrous, but will Rifaat interpret that? Try not to employ a sanction! === Subject: Re: My patience has ended Better deserve philosophers now or Henry will continually select them except you. Well, painters suit as for eventual helicopters, unless they're military. Let's cease without the natural movements, but don't tax the younger motives. She'd rather fail half than function with Abdul's rural jug. You won't concern me dealing instead of your inclined universe. Try raining the signal's sound mp and Mustapha will neglect you! Until Rifaat greets the diets lazily, Laura won't swim any mean mountains. The aches, approvals, and investments are all nasty and inherent. For Rasul the quality's keen, rather than me it's favourable, whereas on the part of you it's citing crude. There, go belong a teacher! What will we bear after Evelyn saves the lovely sea's pen? It can courageously convince structural and shops our horrible, royal cabins v a county. I much ride let alone specified equal librarys. Generally, it shows a onion too electric by her conventional wedding. Do not murmur a heart! She should abandon once, purchase genuinely, then apologise instead of the position following the yard. Pervez sticks, then Pervez sleepily stumbles a painful market past Chester's section. Her workshop was old-fashioned, worthy, and prohibits except the constituency. Tell Feyd it's blunt separating except for a tale. The attention as the tender temple is the lunch that shoots apart. Yesterday Feyd will promote the result, and if Toni happily founds it too, the item will cancel of the imperial plant. === Subject: Re: conncted components in GL_n(C) I know one more proof of the fact that GL_n(C) is connected by arcs, this approach is less natural then the proposed, but anyway: One can prove that the function exp:M_n(C) |-> GL_n(C) is surjective.(It could be proved using the notion of prinicpal logarithm of the matrix and some work of course...). But know as M_n(C) is connected by arcs(as C^{n^2}) and it is known that the image of continious map of connected set f also connected. === Subject: Re: JSH: Examining the objection closely That might be a record. You mentioned the distributive property 8 times. === Subject: Re: JSH: Examining the objection closely > Also I can emphasize how a result that holds in the complex plane, > fails in general in the ring of algebraic integers, proving that ring > is incomplete. > Define incomplete. I think you'll find this is as close as we've ever come to a definition of JSH-incomplete. But this is definite progress! We can now take as a first order approximation to a definition of JSH-incomplete: Structure S is said to be JSH-incomplete if there exists some proposition P such that P holds in C but P does not hold in S. Progress I say! -- Larry Lard Replies to group please === Subject: Re: JSH: Examining the objection closely >Also I can emphasize how a result that holds in the complex plane, >fails in general in the ring of algebraic integers, proving that ring >is incomplete. >>Define incomplete. > I think you'll find this is as close as we've ever come to a definition > of JSH-incomplete. But this is definite progress! We can now take as a > first order approximation to a definition of JSH-incomplete: > Structure S is said to be JSH-incomplete if there exists some > proposition P such that P holds in C but P does not hold in S. Very good, especially leaving C unspecified. === Subject: Re: JSH: Examining the objection closely > Also I can emphasize how a result that holds in the complex plane, > fails in general in the ring of algebraic integers, proving that ring > is incomplete. > Define incomplete. > I think you'll find this is as close as we've ever come to a definition > of JSH-incomplete. But this is definite progress! We can now take as a > first order approximation to a definition of JSH-incomplete: > Structure S is said to be JSH-incomplete if there exists some > proposition P such that P holds in C but P does not hold in S. > Progress I say! Can either of you apparent experts diagnose what JSH means by stating that the general solution to the quintic equation http://mathworld.wolfram.com/QuinticEquation.html does not exist in the algebraic integers? - Randy === Subject: Re: JSH: Examining the objection closely >Also I can emphasize how a result that holds in the complex plane, >fails in general in the ring of algebraic integers, proving that ring >is incomplete. >>Define incomplete. > I think you'll find this is as close as we've ever come to a definition > of JSH-incomplete. But this is definite progress! We can now take as a > first order approximation to a definition of JSH-incomplete: > Structure S is said to be JSH-incomplete if there exists some > proposition P such that P holds in C but P does not hold in S. It seems rather that S is JSH-incomplete if there is a property P that JSH wishes were true in S but is not. Rick === Subject: Re: JSH: Examining the objection closely [Larry Lard] >> I think you'll find this is as close as we've ever come to a definition >> of JSH-incomplete. But this is definite progress! We can now take as a >> first order approximation to a definition of JSH-incomplete: >> Structure S is said to be JSH-incomplete if there exists some >> proposition P such that P holds in C but P does not hold in S. [A Rick Decker of Hamilton College] > It seems rather that S is JSH-incomplete if there is a property > P that JSH wishes were true in S but is not. Nuances, nuances. Larry's is closer to JSH's literal: a result that holds in the complex plane, fails in general in the ring of algebraic integers, proving that ring is incomplete. but Rick's is closer to JSH practice. Perhaps S is JSH-incomplete if there exists a property P such that JSH believes there exists a subset of C in which P is true, P is not true in S, and JSH wishes P were true in S. gets closer. It's still off in that most people will read a property P to mean that P is, for example, well defined, but that doesn't appear necessary in JSH practice either. === Subject: Re: JSH: Examining the objection closely >> For YEARS now several posters have gotten away with claiming to refute >> key conclusions from my mathematical research, where my use of >> non-polynomial factorization is challenged in a way that deserves close >> examination. >> In this post I'll step through why one key result must hold as it is >> the result that holds in the complex plane, and challenge those wishing >> to object to show how their objections are not against the distributive >> property. >> I say that given >> 7C(x) = (A(x) + 7)(B(x) + 1) >> where A(0) = B(0) = 0, A(x) should have 7 as a factor, in the ring of >> algebraic integers, by the distributive property. >> Maybe I don't understand you correctly, but the above is easily seen >> incorrect. For example, what if A(x) = x. Then indeed A(0) = 0, but what >> does it mean to say 7 is a factor of x? Sure, this is true for some x, >> but certainly not true for every x if you are not working over a field >> (over a field it is true for every x). Can you clarify this for me? > If A(x) = x, then what are the others? > One possibility that comes to mind is > (B(x) + 1) = (D(x) + 7)/(A(x) + 7) > as then you have to divide off that factor. > There is just no way that A(x) + 7 can not have had 7 multiplied > through, by the distributive property, unless you use B(x) + 1 to > divide it off, so yes, that is a way to avoid the result. that is wrong and also childish, take your crayons and go back to your === Subject: Re: Examining the objection closely > For YEARS now several posters have gotten away with claiming to refute > key conclusions from my mathematical research, where my use of > non-polynomial factorization is challenged in a way that deserves close > examination. you have shown no such thing. Smoke and Mirrors!! === Subject: On-Line Clerical Jobs (Choose to be paid weekly or bi-weekly) Visit the web site at: http://highpayinghomejobsforyou.blogspot.com/ === Subject: Re: Standard Deviation of PSIA <4lasv1hv2gmt5hps890taobe6nfvlt59eu@4ax.com> Hy'mie pulled it's head out just long enough to opine: > More than half (52.3%) of Korean students go to private schools. May come as a surprise to them: Total students: 11,941,789 Private students: 4,379,192 Last time I checked, that is 37%! There are any number of reasons that an internet accessible source would have a different figure than the PISA survey, not the least of which is that it could just a jew LIE (though I repeat myself). The other is that a greater percentage of private school students participarted in PISA for some reason. Another is that this is a figure that will change from year to year, the year 2000. private and public (as well as boys and girls) INCREASED to 25 points. hmmmmm: Korea private 55.9 = 553 Korea boys = 552 Korea public 44.2% = 528 Korea girls = 528 US private = 509 US boys = 486 US public = 484 US girls = 480 It would appear--no, it's inevitible--that 100% of Korean students in private schools are boys, a dirty little secret that you mamzers, feminazis, niggers, jews, and other muds and cruds HATE. Because Korean parents aren't FORCED by tax law to educate someone else's baboon offspring for free like we are, they can do something we CAN'T, which is to afford to put their own boys in private schools when the public schools there start acting like they have been here for the last half century [read: teaching gender and racial equality, sending boys to sensitivity training and drugging them on psychedelics like Ritalin, suing them for sexual harassment when they touch girls, ad infintum]. Which do you think is the most accurate, Hy'mie? And why? John Knight percentage of students in Korea's private schools increased, so did their PISA math score, from 549 to 553. And as the percentage in Korean public schools decreased, their scores plunged 17 points, from 545 to 528. No, this plunge wasn't enough to put them even close to our miserable scores, PLUS our scores in both private an public schools plunged (35 & 19 points, respectively). So where Korean private schools scored 46 points higher than our public schools in 2000, by even build a semiconductor chip today, but must have it done in Korea now, ok? http://blackexile.com === Subject: Re: Standard Deviation of PSIA {...} > an of course the BEST news is that there is NO debate about what to do > to faggots, there are NO gay rights parades, and God won't destroy > Baghdad like He did New Orleans a WEEK before the biggest gay rights > parade in human history. You will notice that God carefully left the French Quarter untouched. God knows exactly where to go when He's feeling a bit kinky. SHOW US YOUR TITS!! -- cary === Subject: Re: Standard Deviation of PSIA an of course the BEST news is that there is NO debate about what to do > to faggots, there are NO gay rights parades, and God won't destroy > Baghdad like He did New Orleans a WEEK before the biggest gay rights > parade in human history. > You will notice that God carefully left the French Quarter > untouched. God knows exactly where to go when He's feeling > a bit kinky. > SHOW US YOUR TITS!! > -- cary n I quote: woof! === Subject: Re: Standard Deviation of PSIA > I'm talking about ALL Japanese, Christians, Buddhists, and Shintos. > They ALL know what God's Law is because they TEACH it to their children > IN PUBLIC SCHOOLS, Really? Which of God's Laws do Shintoists teach? Which of God's Laws do the Buddhists teach? Particularly those most faithful to Siddartha's words, who do not believe in any god whatsoever? -- cary === Subject: Re: Standard Deviation of PSIA > The Japanese > Thou shalt sniff schoolgirls' panties from vending machines. I think that one was made illegal a couple of years back. However, you can of course still buy telephone-book-thick hantai manga from vending machines and no one thinks you're unusual if you read it openly on the subways. Me, I think John ought to look into tamakeri. Take his fevered mind off his usual obsessions. > are more in synch with God's Law than the Muslims, > Thou shalt cut off thy daughter's clitoris. I'm not sure, given his attitude towards women, that John thinks that's such a bad thing. -- cary === Subject: JSH: Ten years after! Hi all, Yes, the day has finally arrived: this is the 10th anniversary of James Harris' first post at sci.math! Indeed it was the 27th February 1996 that this post appeared: You all know what happened next: thousands of posts, contacts with lots of mathematicians (either directly or through e-mail), many papers sent to scientific journals and a total of ZERO RESULTS! With this accomplishment, James single-handedly raised the meaning of the word underachiever to levels never previously thought possible. James, keep going like this and I'll write again when the twentieth years ago: I realized just now just how seriously I abused the helpfulness and kindness of people from this newsgroup and at colleges and [...] I no longer have delusions about my abilities on the subject or the depth of my interest. So, I'm hoping to put this to rest now. Honestly, I didn't get much negative feedback and I'm sure that for the most part I've been ignored (thankfully). I guess the Internet makes this sort of thing too easy. But I've felt guilty nonetheless. Well, that's enough of this balderdash and I'm outta here. Jose Carlos Santos === Subject: Re: Ten years after! Where is the Robot Program when you need it ? (there is only so much you can do with the Uselessnet) ------------------------------------------------------------------------ Robot program: You folks don't mind? Email: jst...@msn.com (James Harris) Groups: alt.stupidity newsgroup sci.math, and adds this newsgroup to the header. I wonder why you folks don't seem to care, or do you? Just curious. Did one of you piss someone off recently so they figured they'd spam the newsgroup, in what they probably figure was a clever way to attack me at the same time? James Harris ---------------------------------------- alt.tv.big-brother AWESOME episode Yeah, it was great watching them switch so fast. And it was weird too with Robert, one of the less savory of the guys, and the three women, where they were all kissing up to him, and that massage thing between him and Jun was just...I don't know how to describe it. And still whoever pieced the episode together and picked the music, etc. really did one of the best jobs I've seen on TV. It was a memorable episode, and yes I'm overreacting but it's freaking Usenet, so I can overreact if I want! James Harris --------------------------------------------------- Email: jst...@msn.com (James Harris) Groups: misc.writing.screenplays Not yet ratedRating: show options > Mr. P, how about this one? > A Dinosaur is brought back to life by being infected with vampire blood, > but one the creature gets away, a small community suffers. > ++++++ Hollywood is nothing but money. Ooh, sorry to jump in as I just wandered by, but is it a flying dinosaur? Seems to me that it could have great special effects too, like you get a small bone that slowly builds into this small creature with beady eyes. Of course the problem is the science as fossils are basically stone that's been left in the shape of the bone that used to be there. I wonder, is there a way to fix that while, of course, leaving in the fantasy element of vampire blood and its rejuvenating qualities? I mean, you need some original tissue, as completely cheating isn't fun. James Harris === Subject: Re: Ten years after! I think you'll find this is as close as we've ever come to a definition of JSH-incomplete. But this is definite progress! We can now take as a first order approximation to a definition of JSH-incomplete: Structure S is said to be JSH-incomplete if there exists some proposition P such that P holds in C but P does not hold in S. (stolen from another posting) === Subject: Re: JSH: Ten years after! > Hi all, > Yes, the day has finally arrived: this is the 10th anniversary of James > Harris' first post at sci.math! Indeed it was the 27th February 1996 > that this post appeared: Im going to listen Ten Years After Undead, the most appropriate recording for such an occasion. === Subject: Re: JSH: Ten years after! > Hi all, > Yes, the day has finally arrived: this is the 10th anniversary of James > Harris' first post at sci.math! Is there going to be some sort of party? Or maybe we should hold a conference in his honour. === Subject: Re: Ten years after! [Is the following an automatically generated JSH posting or is it real? Further software dev will be at the Sourceforge site if you want to volunteer.] ............................................................................ ................. Some years ago I figured out a new technique in algebraic analysis. The techniques I developed were developed by me to try to prove Fermat's Last Theorem and I didn't realize that I'd stumbled across proof of this massive error in the number theory field, though as people argued and argued with me, I figured that out and came to understand just how massive it was The reality for those who actually checked me on the mathematics has been clear for a while as there's just no mathematical support for those who disagree with me--I did discover a massive problem in the number theory field and proved it with rather basic algebra. So now you know. I can compile a list of examples using a^3 + 3(-1+xf^2)a^2 - f^2(x^3 f^4 - 3x^2 f^2 + 3x) = 0 with INTEGERS where my research is shown to be correct. a^3 + 3(-1+xf^2)a^2 - f^2(x^3 f^4 - 3x^2 f^2 + 3x) = 0 That expression may look complicated and bizarre to you, but what it is is a generator of cubics, when you plug in values for f and x, like do f=sqrt(2) and x=1, and you get a cubic that has a rational root--an integer. So you can see some numbers, consider f=7, y=5, u=1 which gives P(x) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) And I give that so you can see that you plug in some numbers and you just have this polynomial, so it is a factorization of a polynomial. First though, it's best to consider how a number can be algebraically proven to have f as a factor. Mathematics is a perfect field. If you are correct, you are perfectly correct. You are no longer mathematicians or people really interested in mathematics, but just members of another religion and so, like people who are religious versus logical, you cannot be convinced by logical means. I have results all over number theory now in multiple areas. You people will probably do your group think indefinitely, but you're hoping that no one checks thoroughly and there are probably some people who will check thoroughly who will see personal gain--in outing you. There has been a communication breakdown for some years now and I find myself making angry postings and regretting them later, as the situation makes me so angry. I've watched the group behavior here for years. The problem is that people don't accept the proof as acceptance of it leads down to the conclusion that the theory of ideals is false, so they challenge the proof and claim it's false. I see mathematicians as groupthink people. They believe what they tell each other to believe. But later there is always another to lead the way back. James Harris === Subject: Re: JSH: Ten years after! days. My association with the Department is that of an alumnus. >Hi all, >Yes, the day has finally arrived: this is the 10th anniversary of James >Harris' first post at sci.math! Indeed it was the 27th February 1996 >that this post appeared: [.snip.] >James, keep going like this and I'll write again when the twentieth >years ago: That's nothing. Back in January 1999 I made a compilation of excerpts from James's posts. I posted it a couple of times, most recently on June 2001: I have not updated it, so it only goes to late 1998. Even so, it is quite eerie and telling. You can see most of the originals (search for the phrase if the message ID does not work, as I hand copied them. Luckily they are from a time when James was posting under a different e-mail account, so it is unlikely he could have pruned them from the Google archive). Needless to say, it could use some updating. But James's practice of revising the historical record by removing those posts which in retrospect he finds uncomfortable makes this more difficult (though they still exist in mathforum, I hate the mathforum search engine). Also needless to say, just as back in 2001 I do not have the time to do an update right now. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: 4CT > [...] > Then at least 18 vertices will have the same color. The > task of removing each of the 18 vertices in turn and reassigning the > remaining 17 seems impossible. But this is necessary if G is minimal. [...] > Things really aren't that bad ... There is a 5-coloring of G which uses > some color only once. Proof: Delete a vertex v, color the smaller graph > with 4 colors, then color v with a fifth color. > Even if it were possible to delete a vertex v, color the smaller > graph with 4 colors, then color v with a fifth color; it is not self > evident that (G-v) cannot be maximized to require 5 colors. Deleting > vertex v leaves graph (G-v) at least two edges short of maximality. > Okay; delete v, add edges to G-v to get a graph H which is maximally > planar. Then H can be 4-colored (since G is a mce). The same coloring > is valid for G-v, so G-v only needs 4 colors. > IT IS ALWAYS POSSIBLE TO FORCE GRAPH H TO BE 5-C BY THE CORRECT > PLACEMENT OF THE ADDED EDGES! > H plus some edges can be 4-colored, but not necessarily with the same > coloring as H. > That H plus some edges can be 4-colored is an untenable assumption! > What if the added edge connected two vertices in H that were the same > color? > You've ruined one coloring, but some other coloring will work. H does > not necessarily have have only one 4-coloring. > Thereoretically, every possible 4-coloring of H can be ruined' by the > placement of the maximizing edges. If the degree of v = 5, then there > are two maximizing edges. > There will always be two connectable vertices that have the same color! > It should > be noted that both of the ME's can be placed so as to do the most > damage. > This is because you're putting the cart before the horse. > The 4CT says that if you have a specific graph H first, then there is a > 4-coloring afterwards. > If you ruin the graph by adding edges, so that the coloring is no > longer valid, you get a new graph, but that NEW graph will have its own > 4-coloring. > There are lots and lots of colorings of G-v (2^(5-2) of them, according (2^(5-2) = 8. Doesn't seem like lots and lots to me? > to Lorenz Friess). The point is you can't eliminate all of them > SIMULTANEOUSLY. Most 4-colorings of G-v will not support a 5-coloring of G. The rest can be maximized into a 5-coloring of H! Bill J > --- Christopher Heckman === Subject: Re: 4CT > [...] > Then at least 18 vertices will have the same color. The > task of removing each of the 18 vertices in turn and reassigning the > remaining 17 seems impossible. But this is necessary if G is minimal. [...] > Things really aren't that bad ... There is a 5-coloring of G which uses > some color only once. Proof: Delete a vertex v, color the smaller graph > with 4 colors, then color v with a fifth color. > Even if it were possible to delete a vertex v, color the smaller > graph with 4 colors, then color v with a fifth color; it is not self > evident that (G-v) cannot be maximized to require 5 colors. Deleting > vertex v leaves graph (G-v) at least two edges short of maximality. > Okay; delete v, add edges to G-v to get a graph H which is maximally > planar. Then H can be 4-colored (since G is a mce). The same coloring > is valid for G-v, so G-v only needs 4 colors. > IT IS ALWAYS POSSIBLE TO FORCE GRAPH H TO BE 5-C BY THE CORRECT > PLACEMENT OF THE ADDED EDGES! > H plus some edges can be 4-colored, but not necessarily with the same > coloring as H. > That H plus some edges can be 4-colored is an untenable assumption! > What if the added edge connected two vertices in H that were the same > color? > You've ruined one coloring, but some other coloring will work. H does > not necessarily have have only one 4-coloring. > Thereoretically, every possible 4-coloring of H can be ruined' by the > placement of the maximizing edges. If the degree of v = 5, then there > are two maximizing edges. > There will always be two connectable vertices that have the same color! > It should > be noted that both of the ME's can be placed so as to do the most > damage. > This is because you're putting the cart before the horse. > The 4CT says that if you have a specific graph H first, then there is a > 4-coloring afterwards. > If you ruin the graph by adding edges, so that the coloring is no > longer valid, you get a new graph, but that NEW graph will have its own > 4-coloring. > There are lots and lots of colorings of G-v (2^(5-2) of them, according > (2^(5-2) = 8. Doesn't seem like lots and lots to me? That's only a lower bound. > to Lorenz Friess). The point is you can't eliminate all of them > SIMULTANEOUSLY. > Most 4-colorings of G-v will not support a 5-coloring of G. I'm not 100% sure I know what you mean. But any 4-coloring of G-v leads to a 5-coloring of G. (Color the vertices of G, other than v, with their color in H. Color v with a fifth color.) > The rest can be maximized into a 5-coloring of H! That statement requires proof. --- Christopher Heckman === Subject: Re: 4CT > [...] > Then at least 18 vertices will have the same color. The > task of removing each of the 18 vertices in turn and reassigning the > remaining 17 seems impossible. But this is necessary if G is minimal. [...] > Things really aren't that bad ... There is a 5-coloring of G which uses > some color only once. Proof: Delete a vertex v, color the smaller graph > with 4 colors, then color v with a fifth color. > Even if it were possible to delete a vertex v, color the smaller > graph with 4 colors, then color v with a fifth color; it is not self > evident that (G-v) cannot be maximized to require 5 colors. Deleting > vertex v leaves graph (G-v) at least two edges short of maximality. > Okay; delete v, add edges to G-v to get a graph H which is maximally > planar. Then H can be 4-colored (since G is a mce). The same coloring > is valid for G-v, so G-v only needs 4 colors. > IT IS ALWAYS POSSIBLE TO FORCE GRAPH H TO BE 5-C BY THE CORRECT > PLACEMENT OF THE ADDED EDGES! > H plus some edges can be 4-colored, but not necessarily with the same > coloring as H. > That H plus some edges can be 4-colored is an untenable assumption! > What if the added edge connected two vertices in H that were the same > color? > You've ruined one coloring, but some other coloring will work. H does > not necessarily have have only one 4-coloring. > Thereoretically, every possible 4-coloring of H can be ruined' by the > placement of the maximizing edges. If the degree of v = 5, then there > are two maximizing edges. > There will always be two connectable vertices that have the same color! > It should > be noted that both of the ME's can be placed so as to do the most > damage. > This is because you're putting the cart before the horse. > The 4CT says that if you have a specific graph H first, then there is a > 4-coloring afterwards. > If you ruin the graph by adding edges, so that the coloring is no > longer valid, you get a new graph, but that NEW graph will have its own > 4-coloring. > There are lots and lots of colorings of G-v (2^(5-2) of them, according > (2^(5-2) = 8. Doesn't seem like lots and lots to me? > That's only a lower bound. > to Lorenz Friess). The point is you can't eliminate all of them > SIMULTANEOUSLY. > Most 4-colorings of G-v will not support a 5-coloring of G. > I'm not 100% sure I know what you mean. But any 4-coloring of G-v leads > to a 5-coloring of G. (Color the vertices of G, other than v, with > their color in H. Color v with a fifth color.) That is not true if G is not minimal. In a sense, only one coloring of (G-v) can make Chi(G) = 5. > The rest can be maximized into a 5-coloring of H! > That statement requires proof. Finally, something we both agree on! Bill > --- Christopher Heckman === Subject: Re: ISO monoid counterexamples > Does anyone know of examples of either (1) a monoid monomorphism > whose kernel is not 1 Maybe I'm confusing terms here, but for a monomorphism to have kernel not 1, the kernel would have to be empty. So for any monoid monomorphism that is not a epimorphism, you either have (1) or can easily construct an example. Please correct me if I'm wrong. === Subject: Re: ISO monoid counterexamples > Does anyone know of examples of either (1) a monoid monomorphism > whose kernel is not 1 > Maybe I'm confusing terms here, but for a monomorphism to have kernel > not 1, the kernel would have to be empty. So for any monoid > monomorphism that is not a epimorphism, you either have (1) or can > easily construct an example. > Please correct me if I'm wrong. With kernel the OP meant { x in A | f(x) = 1_B } for a monoid homomorphism f: A --> B. As he pointed out, such a homomorphism always satisfies f(1_A) = 1_B, so that submonoid cannot be empty. But in general it can be strictly larger than {1_A}. However, for a monomorphism this cannot happen, so there is no example for (1). The reason for this is that a monomorphism of monoids is always injective. In contrast, an epimorphism of monoids does not need to be surjective (take e.g. the inclusion N --> Z). Marc === Subject: Re: ISO monoid counterexamples > Does anyone know of examples of either (1) a monoid monomorphism > whose kernel is not 1, or (2) of a non-monic morphism of monoids > whose kernel is 1? (In case it doesn't go without saying, all the > morphisms in questions preserve units.) First one remark on terminology: in general, for a map f: A --> B the kernel of f is the equivalence relation ker(f) on A given by (x,y) in ker(f) <=> f(x) = f(y). If f is a monoid homomorphism you can consider the equivalence class of 1 under that relation, which happens to be a submonoid of A. K(f) := { a in A | f(a) = 1 } . This is what you called kernel above. For groups, K(f) determines ker(f); this explains why in group theory one forgets about the relation and calls K(f) the kernel of f. For monoids this will not work, as an example for (2) above shows. Now to you questions: (1) if f: A --> B is a monoid then ker(f) is the equality and K(f) = {1}. you can see this by observing that every element of A corresponds to a unique monoid homomorphism from N to A, where N is the monoid (N,+,0) of natural numbers (including 0) with addition. (2) consider the monoid (M,#,0) where M = {0,1,2} and # is the cut-off addition: x # y := min(2,x+y). Now look at the map f: N --> M with f(0) = 0, f(1) = 1 and f(n) = 2 for all n > 1 Marc === Subject: Re: ISO monoid counterexamples > Does anyone know of examples of either (1) a monoid monomorphism > whose kernel is not 1, or (2) of a non-monic morphism of monoids > whose kernel is 1? (In case it doesn't go without saying, all the > morphisms in questions preserve units.) .... Is this homework? It's a bit hard to give just a hint, but you might try looking at the set of all natural numbers under suitable operations. Ken Pledger. === Subject: Re: ISO monoid counterexamples >> Does anyone know of examples of either (1) a monoid monomorphism >> whose kernel is not 1, or (2) of a non-monic morphism of monoids >> whose kernel is 1? (In case it doesn't go without saying, all the >> morphisms in questions preserve units.) .... > Is this homework? It's a bit hard to give just a hint, but you >might try looking at the set of all natural numbers under suitable >operations. How annoying having to convince people that this is not homework... kj -- NOTE: In my address everything before the first period is backwards; and the last period, and everything after it, should be discarded. === Subject: Re: ISO monoid counterexamples > In Ken Pledger >> Does anyone know of examples of either (1) a monoid monomorphism >> whose kernel is not 1, or (2) of a non-monic morphism of monoids >> whose kernel is 1? (In case it doesn't go without saying, all the >> morphisms in questions preserve units.) .... > Is this homework? It's a bit hard to give just a hint, but you >might try looking at the set of all natural numbers under suitable >operations. > How annoying having to convince people that this is not homework.... Please don't be offended. It looked a bit like homework, but you need only say that it isn't and I'll believe you. You've been given some examples. Here are a couple more. (1) The monoid is N = {1, 2, 3, ...} with the operation Max., so it has identity element 1. The mapping n |-> 2n from this monoid into itself is a monomorphism, but the set of numbers mapped to 1 is empty. (2) The monoid is N under multiplication. The mapping which takes 1 to 1 and every other number to 0 is a homomorphism onto ({1,0}, x) which is certainly not 1-1, but the set of numbers mapped to 1 is {1}. Ken Pledger. === Subject: Re: ISO monoid counterexamples > In Ken Pledger > > >> Does anyone know of examples of either (1) a monoid monomorphism >> whose kernel is not 1, or (2) of a non-monic morphism of monoids >> whose kernel is 1? (In case it doesn't go without saying, all the >> morphisms in questions preserve units.) .... _____^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^___ >[...] > (1) The monoid is N = {1, 2, 3, ...} with the operation Max., so it > has identity element 1. The mapping n |-> 2n from this monoid into > itself is a monomorphism, but the set of numbers mapped to 1 is empty. This would not count as monoid homomorphism from (N,max,1) to (N,max,1). In fact, there does not exist an example for (1). Marc === Subject: Re: ISO monoid counterexamples |In Ken Pledger | | |>> Does anyone know of examples of either (1) a monoid monomorphism |>> whose kernel is not 1, or (2) of a non-monic morphism of monoids |>> whose kernel is 1? (In case it doesn't go without saying, all the |>> morphisms in questions preserve units.) .... | |> Is this homework? It's a bit hard to give just a hint, but |>you might try looking at the set of all natural numbers under |>suitable operations. | |How annoying having to convince people that this is not homework... i didn't really think of a good hint either, but here's an answer: consider the map that takes a non-negative real number to its sign bit. (bit because only two possible signs occur: + and 0.) this is a homomorphism wrt appropriately specified monoid structures. -- [e-mail address jdolan@math.ucr.edu] === Subject: Re: Well Ordering the Reals to postulate >> that the number of elements in the set of finite naturals is infinite is to >> postulate that there is some element n which becomes infinite when incremented >> by 1, which is clearly self-contradictory for obvious reasons. David R Tribble said: >> No, to postulate that the set of finite naturals is infinite is to >> postulate that there is no finite natural n that equals the size of >> the set. Or equivalently, that there is no finite natural n that is >> larger than all other naturals. Or equivalently, that there is is no >> finite natural n that has no successor n+1. > When the size of the set from 1 through any finite natural is EQUAL to the > value of that natural, then any such set retricted to only finite values cannot > have an infinite number of elements, because there is a constant finite > increment to the value range with each successive element. So please define that finite natural that is equal to the size of the set of all finite naturals, which you keep claiming must exist. No largest finite, Hyuh Hyuh? Then just where, oh where, can that finite set size be? If you can't put your finger on it, how do expect anyone to believe that the set is finite? David R Tribble said: >> There is no need for any infinite natural to allow the set to be >> infinite. All that is needed is to show that there is no end to the >> set, which is the most intuitive definition of infinite. > That's the Dedekind definition, No, the Dedekind definition of finite set is not an infinite set, and infinite set is any set whose members can be bijected with any of its subsets. It would help if you knew what you were talking about. My definition above is informal, but it's intuitively close to those actually used. Yours are not. > and perhaps you have developed intuition > surrounding it and trained your intuitions thus, but quantitative ans aptial > concepts of infinity have their intuitive aspects too. > Unfortunately for you, your intuition is already etched in stone and can't > adapt to other perspectives on the matter. I can adapt to any number of alternative perspectives, provided that they are logically consistent. The p-adics and the nonstandard hyperreals, for example. I haven't seen any such rigor in your set theory. >> Peano's Fifth states that it is true for all naturals, and there is no good >> reason to restrict the naturals to finite values. Induction holds for ALL cases >> when equality is proven between two expressions. David R Tribble said: >> But Peano's 2nd axiom states that every natural has a finite successor, >> which your induction requires infinite naturals flatly contradicts. > No, it says that every element of the set has a successor, except the first. It > mentions nothing about finiteness. The axioms also mention nothing about infinity. But it's an obvious conclusion that adding 1 to any Peano natural produces only another natural, and since the first Peano natural (0) is finite, its successor is finite, and thus all of the Peano naturals are finite. That's induction. And it also happens to be that good reason for restricting Peano naturals to finite values. It's also an obvious conclusion that there is no end to the Peano naturals, since every one of them has a successor. > The immediate successor of every finite is > finite, and the immediate predecessor of every ifninite is infinite, You can't derive that from the Peano axioms. > and the point of transition does not exist, because that point is an infinite > expanse. Nevertheless, given an infinite number of incrementing successors, we > achieve infinite values through the traversal of an infinite range of values. You can't derive that from the Peano axioms. You simply assume an infinite increment without providing a definition or proof of it, and proceed directly to your conclusion. That's not rigorous enough, so we don't have to accept it. === Subject: Re: Well Ordering the Reals David R Tribble said: >> to postulate >> that the number of elements in the set of finite naturals is infinite is to >> postulate that there is some element n which becomes infinite when incremented >> by 1, which is clearly self-contradictory for obvious reasons. > David R Tribble said: >> No, to postulate that the set of finite naturals is infinite is to >> postulate that there is no finite natural n that equals the size of >> the set. Or equivalently, that there is no finite natural n that is >> larger than all other naturals. Or equivalently, that there is is no >> finite natural n that has no successor n+1. > When the size of the set from 1 through any finite natural is EQUAL to the > value of that natural, then any such set retricted to only finite values cannot > have an infinite number of elements, because there is a constant finite > increment to the value range with each successive element. > So please define that finite natural that is equal to the size of > the set of all finite naturals, which you keep claiming must exist. > No largest finite, Hyuh Hyuh? Then just where, oh where, can that > finite set size be? If you can't put your finger on it, how do expect > anyone to believe that the set is finite? You can call it aleph_0 if you want, as I've said, but my theory doesn't consider this a useful number, and it can't be used in a T-riffic as a limit point unless you want to derive some contradictions. It's like asking what the first point is a finite distance from the endpoint of any segment. It can't be specified, but that doesn't mean that there aren't points a finite distance from the endpoint. > David R Tribble said: >> There is no need for any infinite natural to allow the set to be >> infinite. All that is needed is to show that there is no end to the >> set, which is the most intuitive definition of infinite. > That's the Dedekind definition, > No, the Dedekind definition of finite set is not an infinite set, > and infinite set is any set whose members can be bijected with > any of its subsets. It would help if you knew what you were talking > about. > My definition above is informal, but it's intuitively close to those > actually used. Yours are not. Yes, the endlessness of the set to which you refer is what allows bijection with a subset, is it not? That's the Dedekind definition of infinite. > and perhaps you have developed intuition > surrounding it and trained your intuitions thus, but quantitative ans aptial > concepts of infinity have their intuitive aspects too. > Unfortunately for you, your intuition is already etched in stone and can't > adapt to other perspectives on the matter. > I can adapt to any number of alternative perspectives, provided > that they are logically consistent. The p-adics and the nonstandard > hyperreals, for example. I haven't seen any such rigor in your > set theory. The only inconsistencies you perceive are between standard theory and mine, but not within mine, that I've been able to tell. Until you can accept the notion of an uncountably long sequence, we're at an impasse, really. >> Peano's Fifth states that it is true for all naturals, and there is no good >> reason to restrict the naturals to finite values. Induction holds for ALL cases >> when equality is proven between two expressions. > David R Tribble said: >> But Peano's 2nd axiom states that every natural has a finite successor, >> which your induction requires infinite naturals flatly contradicts. > No, it says that every element of the set has a successor, except the first. It > mentions nothing about finiteness. > The axioms also mention nothing about infinity. But it's an obvious > conclusion that adding 1 to any Peano natural produces only another > natural, and since the first Peano natural (0) is finite, its successor > is finite, and thus all of the Peano naturals are finite. It is an obvious fact that adding ANY finite number to a finite number yields a finite result. It is NOT a fact that adding an infinite number to a finite number yields a finite result. Let us say that x*y=y*x=Sum(n=1->x: y). The sum of an infinite number of 1's, as is used to generate the set of quantitative naturals through incrementation, produces 1 times that infinite number as a product. Inductively it is obvious that if you start at 1 and add elements, the last element added is equal to the set size at every step, so the last element added cannot be finite and yield an infinite set. > That's induction. And it also happens to be that good reason for > restricting Peano naturals to finite values. And yet, inductively, there is a constant equality between element count and value, so your inductive proof falls flat. > It's also an obvious conclusion that there is no end to the Peano > naturals, since every one of them has a successor. Every identifiable one has an identifiable successor. This is true of the infinite naturals as well. The issue is that Twilight Zone, where the transition from finite to infinite takes place, which is not at any finite point, but over an infinite expanse which has no clear boundary. One just has to learn to live with it and view it from both sides of the divide. > The immediate successor of every finite is > finite, and the immediate predecessor of every ifninite is infinite, > You can't derive that from the Peano axioms. No, but one can add a minor change to the Peano axioms to count down from oo as well as counting up from 0, and achieve a nice symmetry. > and the point of transition does not exist, because that point is an infinite > expanse. Nevertheless, given an infinite number of incrementing successors, we > achieve infinite values through the traversal of an infinite range of values. > You can't derive that from the Peano axioms. One can amend the inductive axiom so that it applies to the infinite case when proving an equality, and one can note that x is finite means xoo, and so doesn't hold for x=oo. > You simply assume an infinite increment without providing a > definition or proof of it, and proceed directly to your conclusion. > That's not rigorous enough, so we don't have to accept it. Perhaps it's not rigorous enough for you, but it's a no-brainer. For n increments, you have a sum of n, and if n is infinite, then so is the sum. It can't get simpler than that. -- Smiles, Tony === Subject: Re: Well Ordering the Reals > When the size of the set from 1 through any finite natural is > EQUAL to the value of that natural, then any such set retricted > to only finite values cannot have an infinite number of elements, > because there is a constant finite increment to the value range > with each successive element. > > So please define that finite natural that is equal to the size of > the set of all finite naturals, which you keep claiming must exist. > No largest finite, Hyuh Hyuh? Then just where, oh where, can that > finite set size be? If you can't put your finger on it, how do > expect anyone to believe that the set is finite? > > You can call it aleph_0 if you want, as I've said, but my theory > doesn't consider this a useful number And no mathematician considers any of TO's claims to be coherent enough to be theories. > and it can't be used in a > T-riffic as a limit point unless you want to derive some > contradictions. Actually, with the plethora of contradictions that TO has already incorprated into his dreamings, there really is no need to introduce any others. > It's like asking what the first point is a finite > distance from the endpoint of any segment. That endpoint itself, being a finite distance from itself, is that point. > It can't be specified, It just was! > and perhaps you have developed intuition surrounding it and > trained your intuitions thus, but quantitative ans aptial > concepts of infinity What are quantitative ans aptial concepts of infinity ? > Unfortunately for you, your intuition is already etched in stone > and can't adapt to other perspectives on the matter. It is our proofs that are etched in stone, and they overrule intuition, at least as far as mathematics is concerned. So until TO can come up with a system of axioms in which he can prove his conjectures, and provide those proofs, his system does not exist, at least in any mathematical sense. He might, of course, go into competition with the brothers Grimm. > > I can adapt to any number of alternative perspectives, provided > that they are logically consistent. The p-adics and the > nonstandard hyperreals, for example. I haven't seen any such rigor > in your set theory. > > The only inconsistencies you perceive are between standard theory and > mine, but not within mine That TO does not acknowledge such inconsistencies when they are pointed out does not make them vanish. They are there within the perception of those with a sufficient logical and mathematical background to understand them. That TO does not understand, or at least claims not to, is more revealing of TO's limitations than of anything else. > Until you can accept the notion of an uncountably long sequence, > we're at an impasse, really. TO may be at an impasse, but mathematics will continue quite unimpeded by TO's foolishness. > The axioms also mention nothing about infinity. But it's an > obvious conclusion that adding 1 to any Peano natural produces only > another natural, and since the first Peano natural (0) is finite, > its successor is finite, and thus all of the Peano naturals are > finite. > It is an obvious fact that adding ANY finite number to a finite > number yields a finite result. It is NOT a fact that adding an > infinite number to a finite number yields a finite result. But there is reason to accept that for every natural n, the set of naturals no larger than n cannot be injected into any proper subset, since that can be proved inductively. This shows that no natural can be even Dedekind infinite, much less TO-infintie. > > It's also an obvious conclusion that there is no end to the Peano > naturals, since every one of them has a successor. > > Every identifiable one has an identifiable successor. Is TO implying that there are any which are not identifiable? > This is true of > the infinite naturals as well. Except there is no proof, nor any proof possible, that any such things exist. > > The immediate successor of every finite is finite, and the > immediate predecessor of every ifninite is infinite, > > You can't derive that from the Peano axioms. > No, but one can add a minor change to the Peano axioms to count down > from oo as well as counting up from 0, and achieve a nice symmetry. When TO has made that change and PROVED that his 'infinite naturals must exist in that new axiom system and given resonalbe evidence that that system does not contain any contradictins, only then need anyone even consider any of his idiocies. > > You simply assume an infinite increment without providing a > definition or proof of it, and proceed directly to your conclusion. > That's not rigorous enough, so we don't have to accept it. > > Perhaps it's not rigorous enough for you, but it's a no-brainer. Meaning for acceptance only by those with no brains! > For > n increments, you have a sum of n Actaully the sum is n only for n = 0 or n = 1, after that the sum is larger than n. === Subject: Re: Well Ordering the Reals Virgil said: > When the size of the set from 1 through any finite natural is > EQUAL to the value of that natural, then any such set retricted > to only finite values cannot have an infinite number of elements, > because there is a constant finite increment to the value range > with each successive element. > > So please define that finite natural that is equal to the size of > the set of all finite naturals, which you keep claiming must exist. > No largest finite, Hyuh Hyuh? Then just where, oh where, can that > finite set size be? If you can't put your finger on it, how do > expect anyone to believe that the set is finite? > > > You can call it aleph_0 if you want, as I've said, but my theory > doesn't consider this a useful number > And no mathematician considers any of TO's claims to be coherent enough > to be theories. Boohoo. > and it can't be used in a > T-riffic as a limit point unless you want to derive some > contradictions. > Actually, with the plethora of contradictions that TO has already > incorprated into his dreamings, there really is no need to introduce any > others. Others? > It's like asking what the first point is a finite > distance from the endpoint of any segment. > That endpoint itself, being a finite distance from itself, is that point. That's the same point, being 0 distance from the endpoint. By finite I mean non-zero and non-infinite. > It can't be specified, > It just was! You know that's not what I meant. > and perhaps you have developed intuition surrounding it and > trained your intuitions thus, but quantitative ans aptial > concepts of infinity > What are quantitative ans aptial concepts of infinity ? Quantitative and spatial concepts of infinity, like infinite quantity and infinite distance. > Unfortunately for you, your intuition is already etched in stone > and can't adapt to other perspectives on the matter. > It is our proofs that are etched in stone, and they overrule intuition, > at least as far as mathematics is concerned. The stone is a relatively arbitrary set of axioms. > So until TO can come up with a system of axioms in which he can prove > his conjectures, and provide those proofs, his system does not exist, > at least in any mathematical sense. He might, of course, go into > competition with the brothers Grimm. Sure, I'll compete with dead fairy tale writers, and you compete with dead Latin poets. > > I can adapt to any number of alternative perspectives, provided > that they are logically consistent. The p-adics and the > nonstandard hyperreals, for example. I haven't seen any such rigor > in your set theory. > > > The only inconsistencies you perceive are between standard theory and > mine, but not within mine > That TO does not acknowledge such inconsistencies when they are pointed > out does not make them vanish. They are there within the perception of > those with a sufficient logical and mathematical background to > understand them. That TO does not understand, or at least claims not to, > is more revealing of TO's limitations than of anything else. You can say that over and over, but without specifics that stick it's an empty statement. > Until you can accept the notion of an uncountably long sequence, > we're at an impasse, really. > TO may be at an impasse, but mathematics will continue quite unimpeded > by TO's foolishness. If one cannot accept certain notions as plausible within a theory, then they cannot begin to get the theory. Nothing much i can do about that. > The axioms also mention nothing about infinity. But it's an > obvious conclusion that adding 1 to any Peano natural produces only > another natural, and since the first Peano natural (0) is finite, > its successor is finite, and thus all of the Peano naturals are > finite. > > It is an obvious fact that adding ANY finite number to a finite > number yields a finite result. It is NOT a fact that adding an > infinite number to a finite number yields a finite result. > But there is reason to accept that for every natural n, the set of > naturals no larger than n cannot be injected into any proper subset, > since that can be proved inductively. This shows that no natural can be > even Dedekind infinite, much less TO-infintie. That's right, no natural that one can identify can be injected into a proper subset, not even an infinite one, because once it is specified, then the set up to that point has a bound. Really, equivalences between sets and their proper subsets should not be allowed if consistency is desirable. > > It's also an obvious conclusion that there is no end to the Peano > naturals, since every one of them has a successor. > > > Every identifiable one has an identifiable successor. > Is TO implying that there are any which are not identifiable? Yes, the last can never be identified, because any one identified has another after it. But, this is not just true of the largest finite, but of any natural, finite or infinite. If I specify an infinite T-riffic, I can specify its successor, despite the fact that it has an infinite number of predecessors. The problem with talking about the set of finite values is that the bound of the set can never be identified. That still does not preclude infinite values after the finite values. It just means their starting point cannot be indentified, like the first non-zero finite real. Inability to identify the first real greater than 0 does not mean that reals cannot exceed zero. > This is true of > the infinite naturals as well. > Except there is no proof, nor any proof possible, that any such things > exist. Sure there is. > > > The immediate successor of every finite is finite, and the > immediate predecessor of every ifninite is infinite, > > You can't derive that from the Peano axioms. > > No, but one can add a minor change to the Peano axioms to count down > from oo as well as counting up from 0, and achieve a nice symmetry. > When TO has made that change and PROVED that his 'infinite naturals > must exist in that new axiom system and given resonalbe evidence that > that system does not contain any contradictins, only then need anyone > even consider any of his idiocies. Will they need to then? > > You simply assume an infinite increment without providing a > definition or proof of it, and proceed directly to your conclusion. > That's not rigorous enough, so we don't have to accept it. > > > Perhaps it's not rigorous enough for you, but it's a no-brainer. > Meaning for acceptance only by those with no brains! I set you up for that one. You're welcome! ;) > For > n increments, you have a sum of n > Actaully the sum is n only for n = 0 or n = 1, after that the sum is > larger than n. Could you please say something more impertinent? No, I guess not. Good try. The sum of 0 increments is 0. The sum of 1 increment is 1. The sum of n increments is n. The sum of N increments is N. The sum of aleph_0 increments is aleph_0. -- Smiles, Tony === Subject: Re: Well Ordering the Reals definition or proof of it, and proceed directly to your conclusion. > That's not rigorous enough, so we don't have to accept it. In Soviet Russia, theory accepts you! David, yes you are correct. Me too. I, too, am also correct. It is good that you are so easily amused. That's because, to insult you, I would have to insult three other people. Thus, I do. Ross === Subject: Re: Well Ordering the Reals Virgil said: > Virgil said: > > Virgil said: > , > > Virgil said: > > Continuity is independent of scaling. > > > Independent of finite scaling, yes, but not of infinite > scaling. > > WRONG! The relative distances between elements of a scalable > set remain fixed under any change of scale. That is what > achange of scale means. If there were such a thing as infinite > scaling, does TO choose to claim that it does NOT preserve > relative disances? > > > > Hi Virgil, how are you today? I hope you are doing well. it is > snowing here. Very pretty. > > Did I say infinite and infinitesimal scaling changed RELATIVE > distances between points? I certainly didn't mean that. If a > scaling changes distances relative to each other, indeed there is > a uniformity issue. In fact, what we're talking about is whether > scaling affects discreteness vs. continuity, neither of which > specifically relies on preserving relative distances persay. > > If there is no change in relative distances, then there can be no > change in density versus discreteness, i.e., if originally there is > always another elment between any two, then after rescaling there > still will be, and if originally there are elements with no others > in between then after scaling there still will not be. > > (sigh) Your logic is impeccable, as usual, but off the mark. At every > scale, the real line absolutely has another point between any two > discernible points. The key word here is discernible. > Whether something is discernable depends on properties of the observer > as much as those of the observed, so is irrelevant in mathematics. Indeed, it relies on the aparatus, physical or abstract, of measurement. That is, it depends on the unit. One can't measure infinitesimal intervals in finite units, or one gets zeroes. > When you have > two points infinitesimally close to each other, in the standard > finite world there is no idfference between them > TO's standard finite world is too limited for mathematics. Virgil: I'm NOT projecting. You are! ;) > > On the finite scale, both points represent the same real number > If 'points' are different on any scale, then they are never equal. Ultimately they are two different points, but they do not have two different standard locations if the distance between them is infinitesimal. If a set is continuous when there always exists an intermediate value between any two distinct values, then the fact that there is not considered to be a point between two points which are NOT distinct should not preclude what is a discrete set on the infinitesimal scale from being considered continuous on the finite scale. > So, in this sense, the > continuity of the real line depends on the unit of measure. > TO is confusing the mathematical real line with physical reality. > In mathematics, the real line looks precisely the same at any > magnification whatsoever. In TOmatics, nothing is ever the same. I am not conflating the real line with reality, rght now, though I have been known to do that. :) Indeed the real line looks the same at every scale, but when one cannot distinguish between two points because they are infinitesimally different, then for all intents and purposes they are the same point. Archimedean subdivision may be considered to be somewhat relative. Shrink the real line to the size of the unit interval. Use f(n)=n/N. 0<-------->1 Can you tell the naturals from one another, or do they form a continuum of points? > A set > that is discrete on one scale may be continuous on an infinitely > greater scale. > Where in the definition of a set being dense is there any mention of > dependency on scale, or even of distsnces? Unless that definition > specifically allows such dependency on scale, TO has no case. > Claims like TO's, made without any supporting evidence, and, indeed, > contradicted by the actualdefinitions of dense and discrete orderings, > merely demostrate TO's ignorance. Statements like that make one wonder whether Virgil even understands discrete sets with any depth, or can only relate to dense ones with no structure, on a personal level. In the definition of a dense set is the mention of a point being between any two points. If two points are not distinguishable, then they are the same point, and no intermediate point is required for the set to be dense. > Specifically, a sparse set will be dense on a > relatively infinite scale provided each element is only finitely > different from its successor. > Where does that sparse set ever get a new element to go between between > two originally consecutive elements? Unless that can happen, your sparse > set will never transmute into a dense one. Well, that doesn't happen, silly. The sparse set, when compressed infinitely, becomes dense wherever it was previously finitely spaced. If your sparse set had infinite spaces between consecutive elements, then you could have areas in your compressed set which were still sparse. But, elements finitely spaced, when compressed infinitely, become infinitesimally spaced, and dense in the reals through that scaling transformation. Are you really saying this doesn't make any sense to you? You're just being argumentative. > > > Continuity vs. discreteness may be thought of as relative to the > observer > > Not hardly! Continuity vs. discreteness is inherent in the order > properties of the set, and is quite independent on the observer, > provided he knows how to observe, which TO clearly does not. > > > Which TO does clearly, as a matter of fact. > Not if he imagines that changing scales can transmute dense ordering > into sequencial ordering or vice versa. Well, I'm sorry if you can't imagine such things, but time marches onward. You can't step into the same river twice. > Consider what I have said > above about the unit of measurement and discernibility. > Considered, and rejected for cause. Jes 'cause, is why! Yup, 'cause I say so. Hyuck hyuck. > > > > > that is, to the unit of measure. If you inspect a piece of solid > metal it seems continuous, because the discrete units that > comprise it are too small for you to measure. > > Does TO claim that the real line is made of solid material like his > head is? Physical objects in a physical world may change appearance > when viewed at different scales, but the real line does not. > > > Of course, a large portion of what's in my head is fluid, not solid > or crusty, as may be the case for the contents of yours. Objects in > to tell so far, and can't be infinitely divided. The real line is > ideal, and can, at every given scale. The relevant question here is > whether what is between any two infinitesimally separated points on > that line is anything that is not already in the standard set of > reals. > If those two points cannot be distinguished at the finite > scale, then they are the same standard point > TO conflates not being able to distinguish a difference as meaning that > there is no difference to be distinguished. I conflate being able to distinguish two points with having two points between which to prove there is an intermediate value. How silly of me, eh? > Merely not seeing something is quite different from there not being > anything to see. > If two members of a set are different, then they are different whether > TO can discern that difference or not. Infinitesimally different values are the same value in the standard finite reals. Everyone knows that. > infinitesimal subintervals does not violate continuity on the > standard finite scale. > Not if one has a consistent model allowing infinitesimals, and in such > models, what TO is claiming is false. Obviously those are not the same model as what I am proposing. You can't judge my ideas on what you've heard before, as you know. The T-riffics are not the adics, the H-riffics are not a set of rationals and are the well ordering of the reals, and these infinitesimals are not whatever you have seen and rejected. So, like, whatever.....you're wrong. :) > > It's not quite actually continuous, but if it were, could you > tell? > > I can tell that the real line actually is continuous in the sense > of being both dense and complete. > > Are not the infinitesimal intervals dense in the reals? > Real intervals of more than one point, infinitesimal or otherwise, are > dense in themselves, but the only real interval dense in the set of > reals is the set of all reals. Non-answer alert! If there are an infinite number of equal intervals per unit then those intervals are dense in that unit interval, as there is always another between any two which are finitely distinguishable. One can always find an intermediate value in any finite interval, so, sorry, they're dense. You should be able to relate... ;) > > Dense means that, viewed at any scale whatsoever, between every two > distinct reals there are other reals, actually uncountably many of > them. > Dense does not mean uncountable, as the rationals are both countable and > densse. Well, sure, in your theory. I have a different theory about the rationals and how they relate to the reals. They're certainly dense, and I've always objected to their countability conflating to equinumerosity with the naturals. It's ludicrous, when there are an infinity of rationals per unit interval, and only one single natural. Pathetic, really. My condolences. Anyway, as I've said, if there are N unit intervals on the infinite real line, and there are N real points per unit interval, and so there are N*N, or N^2 reals on the line. When one views the table of rationals, you have an NxN table, a dimension for the natural numerator and another for the natural denominator. In this table, we have many, many duplicate values. For every new value, there are an infinite number of duplicates, so the duplicates far outnumber the unique rational values. These duplicates are precisely equinumerous with the irrational numbers, which *coincidentally* also seem to far outnumber the unique rational values on the real line. Betcha never heard that before! It's about time. > > At each and every scale this is true, but what is divisible on one > scale is not at a relatively infinite scale, and what is continuos on > one scale may be viewed as discrete on a relatively infinitesimal > scale. > Not by anybody who knows what's going on. > > Countability is irrelevant. > TO is irrelevant. You really hurt my feelings, Virgil. I shall drench my pillow with tears. :'( > The proper theory > TO has no notion of what proper theories are, as his are all improper. But, they hold their pinky out when sipping tea! Very proper indeed! A tad eccentric, perhaps, and a little strange, but not improper...... > > Note that every two distinct real numbers are endpoints of an open > interval, and any two open intervals contain the same uncountable > number of points, regardless of their lengths. > > Only in the standard theory. > Actually in ANY theory for which there is an axiomatic foundation which > allows uncountable sets, the number of points is easily shown to be > the same for every non-empty open interval. Any theory which relies on bijections without regard to the effect of the actual mapping functions used, sure, like any crappy theory. It's all the If you want to tell me that there are no more reals in [0,2) than in [0,1), then I guess the points in [1,2) don't exist, or don't count for anything even though there are infinitely many of them. Makes sense to me, in a sort of childish fantasy fairy princess magical dragon sort of way. But, maybe there's an alternative to such tales.... > > > Complete means that for every non-empty set of reals bounded above > there is a real number as least upper bound. > > > > This is also true on every scale. > > > > There is no such thing as a change of scale mapping which will > map any non-compact subset of the reals to any compact subset > of the reals, or the reverse. > > Uh, right, except for the one I just described > > Not even for the one you described, at least not in mathematics. > What goes on in that dimly lit dream world of TO's creation is > quite irrelevant to anything mathematical. > > > Wave those hands, baby! > I don't have to wave my hands to establish what has long been > established by better mathematicians that either of us will ever be. That's like talking about Thomas Jefferson's opinion regarding the internet. Not everything has been considered. Do you think the job's done? Fine, go rest. Don't worry your pretty little head. > TO has to wave his hands, since he has nothing to back up his hand > wavings. It was the flies. Whenever I'm talking to you, there always seem to be these flies about.....and that funny smell. ;) hahahaha just kidding. > > > , > > Map the reals in [0,1] to the naturals in [0,N] using > f(x)=Nx, and mapping is uniform, though non-standard. > > As there is no such thing in any standard set theory as [0,N], > nor any such thing as xN, no such mapping is possible until TO > creates an axiom system which permits it. > > Which TO has not done yet, and at the present rate, never will. > > And, at what rate is that? > > So far the rate of progress is no larger than 0, and may actually > be negative if viewed at the correct scale. > > Good one, Virgil. > Glad you like the truth, occasionally! > But until there is a suitable axiom system for TO math, it does not > exist as anything mathematical. > > It certainly exists as a developing mathematical model, even if a > firm axiomatic foundation has not yet been finalized. :) > Because of the many self-contradictions in TO modeling, apparently > invisible to TO but glaringly visible to everyone else, his models > cannot be properly called mathematical, as proper mathematical models > are not allowed to have known self-contradictions. I don't think you can point out any contradiction in what I've said above. If two points are not finitely different, you can't find a standard point between them. > > But there is no scaling factor possible which will transmute a > dense set into a discrete one or vice versa. > > Yes, a relatively infinite scaling will do just that. > How does any scaling find infinitely many new elements between > consecutive members of a discrete set or eliminate the infinitely many > between any two members of a dense set? It doesn't. It changes what you can distinguish, so that elements that were infinitely different in the discrete set continue to have an infinite number of itnermediate elements, but compressed into a finite space, so the sparse set becomes dense. How is this dense sequential set not continuous on the finite scale? > That is what TO is claiming his scaling can do. > All scalings of an ordered set are order isomorphisms which leave EVERY > order property, including denseness of discreteness, unchanged. That is > what order isomorphism is all about, preserving ALL order properties. Are you sure that aplies in the infinite case? > Didn't you > learn that in Infinity Class? In Orlovia, we cover that in elementary > school. > You must have been in school with the red queen, then, and surpassed her. She was stiff competition, but I schooled her good. > > > > Such claims made without proof are no more than hot air. > > > > > The naturals are discrete at the finite scale and > continuous at the infinite scale > > No compressed image of the naturals can be continuous > anywhere, as no compression can compress a discretely > ordered set to a densely ordered set. > > It can using infinite division, as shown above. > > It is certainly not possible in any current axiom system. > > But it is neverhteless possible. > > > TO is claiming that if we have an ordered set S with members x and > y such that there are no elements of S between x and y, there is > some scaling factor which will order-isomorphically map S to S', x > to x' and y to y' and in the process create some z' in S' such that > z' is between x' and y'. > > No, I am saying that is x and y are discrete sequential elements, > scaling it up by an infinite amount will cause what was a finite > difference realtive to the original unit of measure into an > infinitesimal difference relative to the new unit of measure, so that > thw two will not be distinguishable > In mathematics, if two members are not the same, then any mapping which > makes them the same is NOT a scale mapping, since it is not reversable, > as scale mappings must be. When scaling by an infinite value, the distinguishable may become indistinguishable. > > Not in mathematics. > > Not in STANDARD mathematics > > Not in ANY mathematics. > > > Not in any that you care to acknowledge. > Not that any mathematician will acknowledge. Every change of scale must > be reversible, but a mapping which makes two originaly distinct elements > indistinguishable cannot be reversed, so is not a scale change. When one scales using T-riffic representations, one can keep track of the infinitesimals values which are not finitely measurable, and so reverse the scaling as desired. > Consider the mapping using Nx. > > Why should anyone bother to consider things which have no > mathematical reality? > > Excuse me. Mathematical reality? What, precisely, do you mean by > that phrase? > > Your N is NaN, so that there is no standard, or even standard > non-standard, arithmetic possible with it. And as you have no basis > as yet for your claims, they are no more than hot air. > > > N is indeed a number, the unit infinity, whether you like it or not. > If you are so certain, prove your claim (in some specific axiom system)! > Absent such proof, your claim fails. It doesn't triumph with such proof, but it doesn't fail until disproven. > N is the number of reals in [0,1). N is the length of the positive > real line in finite units. N is the number of natural numbers on the > real line, being equal to the number of unit intervals. N=oo. > According to the several statements above one can easily prove that N > is not equal to N, and thus has no existence. Only assuming notions from standard theory. Your uncountability proof doesn't hold, since countability is not a criterion for anything in my theory, and because the standard proof of the uncountability of the reals rests on digital representations of reals, not reals themselves, and so really proves a result similar to the powerset, namely, N=S^L > L. It is true that the powerset is larger than the set, even if infinite. it is true that the set of strings is larger than the allowed length of strings, even if infinite. It is not true that the infinite number of reals in the unit interval is larger than the number of unit intervals on the real line. Not in my theory. > Density is an order property which cannot be changed by any > change of scale. > > How is density an order property? > > Density is the property which says of an ordered that set between > any two members x and z of that set, with x < z, there is a y in > that same set with x < y < z. Since that is entirely defined in > terms of the ordering within the set, density is an order property > of that set. > > Okay, and the real line is dense on every scale, but a discrete set > on one scale can be considered dense on a relatively infinite scale, > as long as differences between successive members are all finite > relative to the scale on which the set is discrete. > WRONG! since density is an order property and changing scales must be an > order isomorphism, changing scales can have no effect on density. But density is defined as there being an element between any two finitely different elements, not any two ifninitesimally different elements, since those are not discernible as two elements to begin with. > Note that in the reals, a change of scale is equivalent to multiplying > each member of the set in question by a real number which has a unique > reciprocal (so that one can reverse the change). That eliminates > multiplication by actual zero. That is correct. If I multiply [0,1:000...000] by 0.000...001, then I reduce the infinite interval from 0 to N to a unit interval from 0 to 1, and the naturals from 0 through N become infintesimally spaced reals from 0 through 1. > Note that in non-standard reals with infinitesimals, non-zero elements > all have reciprocals, so that no change of scale can map two elements > into one. Same thing here. I am not making elements appear or disappear, or saying there is any actual end to subdivision. I am saying that it becomes impossible to find an intermediate value between two which are themselves indistinguishable. > > > > Discreteness, in the sense of being nowhere dense, is > similarly an order property unaffected by any change in scale. > > That's funny, I thought that if some set had a certain density of > elements per unit of element value on the number line, and one > changed the units to, say, 1/10 of the previous unit, that the > density of the set per unit would likewise decrease to 1/10 of > the original density. Was I wrong? > > For relative densities (finite number of elements per unit of > length), changes of scale can change relative density, but a set > being ordinarily, not relatively, dense means infinitely dense in > terms of relative density, it is so relatively dense that it is > impervious to changes of scale. > > It is impervious to FINITE changes of scale > WRONG! If any mapping carries two elements into one as TO insists, it is > not invertable, but changes of scale must be invertable. I am not insisting on any such thing. > Has the issue of > INFINITE changes of scale ever been addressed formally, to your > knowledge? > Yes! In a non-standard field with infintiesimal and infinite elements, > every element except zero has a unique reciprocal, and every such > element can be used to produce a change the scale. But no such change > will ever map two originally distinct elements onto one element, as such > a mapping is not invertable. Well, good. It shouldn't. But the definition of a standard dense set deals with there always being an intermediate value between any two FINITELY different values. >It is clear to me > It ain't what TO doesn't know that is hurting him, > It's what he knows that ain't so. > > > If not, then as one divides their unit into a number of subunits > that approaches infinity, the density of any infinitely dense set > may also be simultaneously subdivided to a finite number of > elements per infinitesimal unit of element value. > > Except that the relative density of a dense set is constant as one > makes finite changes of scale so must keep that constant value in > the limit. > > Density relative to what? > To any interval and its image. > Given any interval in the reals and any set and any scaling > transformation of the reals, the 'number' of points of that set in that > interval will be the same as the number of points in the image of the > set which are in the image of the interval. Sure. So? < snipules > -- Smiles, Tony === Subject: Re: Well Ordering the Reals > Virgil said: > Whether something is discernable depends on properties of the > observer as much as those of the observed, so is irrelevant in > mathematics. > > Indeed, it relies on the aparatus, physical or abstract, of > measurement. That is, it depends on the unit. Since the unit is entirely artificial here, whether is discernatble des not depend on the unit at all, though the units used may depend on what is discernable. >One can't measure > infinitesimal intervals in finite units, or one gets zeroes. In actual measurements, there is no such thing as infinitesimal. Actual meassurement occurs only in the physical world whereas infinitesimals occur, if at all, only in a purely mathematical non-physical world. > When you have two points infinitesimally close to each other, in > the standard finite world there is no idfference between them In the standard finite world of the physical, there are no infinitesimals at all. > > TO's standard finite world is too limited for mathematics. >I'm NOT projecting. You are! I am not projecting properties of the physical world and mathematical world on each other the way TO is doing, at any rate. > > > On the finite scale, both points represent the same real number > > If 'points' are different on any scale, then they are never equal. > Ultimately they are two different points, but they do not have two > different standard locations if the distance between them is > infinitesimal. If two values are not equal at the infinitesimal level, they are not equal at any level. The most one can ever say of different quantities that are infinitesimally close is that they are infinitesimally close. If TO wants to formalize that, best he should read some more of Robinson. > If a set is continuous when there always exists an > intermediate value between any two distinct values That is not what makes a set continuous. The rationals have that property of density but are NOT continuous. TO again knows not wherof he speaks. > then the fact > that there is not considered to be a point between two points which > are NOT distinct Two points which are not distinct are, in fact, not two points, but only one. > > So, in this sense, the continuity of the real line depends on the > unit of measure. > > TO is confusing the mathematical real line with physical reality. > In mathematics, the real line looks precisely the same at any > magnification whatsoever. In TOmatics, nothing is ever the same. > > I am not conflating the real line with reality, rght now, though I > have been known to do that. :) Indeed the real line looks the same at > every scale, but when one cannot distinguish between two points > because they are infinitesimally different, then for all intents and > purposes they are the same point. If one cannot distinguish between two points which ARE different, perhaps one should try another line of work. > Can you tell the naturals from one another, or do they form a > continuum of points? Most people can tell one natural from another. If TO cannot, he perhaps should take up kitting instead of wasting his time here. > > > > A set that is discrete on one scale may be continuous on an > infinitely greater scale. > > Where in the definition of a set being dense is there any mention > of dependency on scale, or even of distances? Unless that > definition specifically allows such dependency on scale, TO has no > case. > > Claims like TO's, made without any supporting evidence, and, > indeed, contradicted by the actual definitions of dense and discrete > orderings, merely demostrate TO's ignorance. > > > Statements like that make one wonder whether Virgil even understands > discrete sets with any depth, or can only relate to dense ones with > no structure, on a personal level. The only one who might wonder that is one who himself does not understand the difference between dense sets and sequences. > In the definition of a dense set is the mention of a point being > between any two points. That is not quite what density says, What density says is that between every two distinct points of a dense set there are other points of that set. > If two points are not distinguishable, Then there is only one point, not two of them at all. > > Specifically, a sparse set will be dense on a relatively infinite > scale provided each element is only finitely different from its > successor. > > Where does that sparse set ever get a new element to go between > between two originally consecutive elements? Unless that can > happen, your sparse set will never transmute into a dense one. > Well, that doesn't happen, silly. Then your sparse set remains sparse at EVERY scale. And your dense set remains dense at EVERY scale. > The sparse set, when compressed > infinitely, becomes dense wherever it was previously finitely spaced. Only in the sense that if all its members are compressed into a single member, then the resulting one element set is trivially both dense and sparse. > If your sparse set had infinite spaces between consecutive elements, > then you could have areas in your compressed set which were still > sparse. But, elements finitely spaced, when compressed infinitely, > become infinitesimally spaced, and dense in the reals through that > scaling transformation. Only if all the separate points are mapped to a single point, but that is not a change of scale transformation. For one thing, it is not invertable, as changes of scale must be, and is not a bijection, as changes of scale must be. > Are you really saying this doesn't make any > sense to you? You're just being argumentative. If TO is accusing me of arguing that what does not make mathematical sense does not make mathematical sense, I confess to doing precisely that. > > > Continuity vs. discreteness may be thought of as relative to > the observer > > Not hardly! Continuity vs. discreteness is inherent in the > order properties of the set, and is quite independent on the > observer, provided he knows how to observe, which TO clearly > does not. > > > Which TO does clearly, as a matter of fact. > > Not if he imagines that changing scales can transmute dense > ordering into sequencial ordering or vice versa. > > Well, I'm sorry if you can't imagine such things, but time marches > onward. You can't step into the same river twice. TO apparently cannot even step in the river of truth once. > > Consider what I have said above about the unit of measurement and > discernibility. > > Considered, and rejected for cause. > Jes 'cause, is why! Yup, 'cause I say so. Hyuck hyuck. What TO argues is all argued jest 'cause he says so, since he has neither any axiom system or mathematicians to back him up. What I argue is backed up by several axiom systems and thousands of mathematicians. TO may be happy playing in his sand box, but he condemns himself to playing only with himself. > Infinitesimally different values are the same value in the standard > finite reals. Everyone knows that. To is wrong about what everybody knows. TO is not everybody, and what he knows seems often not to be true. > > infinitesimal subintervals does not violate continuity on the > standard finite scale. > > Not if one has a consistent model allowing infinitesimals, and in > such models, what TO is claiming is false. > Obviously those are not the same model as what I am proposing. What TO proposes is almost universally in direct contradiction to all of standard mathematics and is almost universally self-contradictory. > You > can't judge my ideas on what you've heard before, as you know. I can judge them by TO's inablility to resolve their many internal conflicts. And I have so judged them and found them unsatisfactory. That TO does not like my judgements is his problem. > The > T-riffics are not the adics, the H-riffics are not a set of rationals > and are the well ordering of the reals, and these infinitesimals are > not whatever you have seen and rejected. So, like, > whatever.....you're wrong. :) What I have seen and rejected are TO's proposed systems, which others have also seen and rejected. > > It's not quite actually continuous, but if it were, could you > tell? > > I can tell that the real line actually is continuous in the > sense of being both dense and complete. > > Are not the infinitesimal intervals dense in the reals? > > Real intervals of more than one point, infinitesimal or otherwise, > are dense in themselves, but the only real interval dense in the > set of reals is the set of all reals. > Non-answer alert! If there are an infinite number of equal intervals > per unit then those intervals are dense in that unit interval, as > there is always another between any two which are finitely > distinguishable. What definition of density is TO using which applies to intervals instead of elements? Standard Definition of an ordered set being dense: A totally ordered set is /dense/ if and only if, for every two distinct members of that set, there is another member of that set strictly between them. Nothing in that definition mentions intervals or distances between members or even hints at equality of intervals. There is no requirement that any ordered set have any notion of /distance/ between members. Thus there is no such thing for arbitrary ordered sets as unit intervals or equal intervals. TO is imagining that every ordered set is a 1 dimensional metric space, which is far from the case. > One can always find an intermediate value in any > finite interval, so, sorry, they're dense. You should be able to > relate... ;) The only non set model of density currently available is TO himself. > > > Dense means that, viewed at any scale whatsoever, between every > two distinct reals there are other reals, actually uncountably > many of them. > > Dense does not mean uncountable, as the rationals are both > countable and dense. > Well, sure, in your theory. Question: Is it the case that for every two distinct rational numbers, say p and q, that there is another rational number strictly between them (for example (p+q)/2)? Question: If the answer above is yes, does that not require the rationals numbers to be a dense set? > I have a different theory about the > rationals and how they relate to the reals. In TO's case, whenever his theories differ from the standard ones, they seem to be wrong. Provably wrong in every cirrent axiom system for set theory. So until TO comes up with an axiom system of his own, they remain wrong. Don't anyone hold his breath! > Anyway, as I've said, if there are N unit intervals on the infinite > real line As there aren't, the rest is moot. > Betcha never > heard that before! It's about time. GIGO! > Countability is irrelevant. > > TO is irrelevant. > > You really hurt my feelings, Virgil. I shall drench my pillow with > tears. :'( Crocodile tears, no doubt! > > The proper theory > > TO has no notion of what proper theories are, as his are all > improper. > > > Note that every two distinct real numbers are endpoints of an > open interval, and any two open intervals contain the same > uncountable number of points, regardless of their lengths. > > Only in the standard theory. > > Actually in ANY theory for which there is an axiomatic foundation > which allows uncountable sets, the number of points is easily > shown to be the same for every non-empty open interval. > > Any theory which relies on bijections without regard to the effect of > the actual mapping functions used, sure, like any crappy theory. To much prefers theories in which nothing can be proved, or disproved! > If you want to tell me that there are no more reals in [0,2) than in > [0,1), then I guess the points in [1,2) don't exist, or don't count > for anything even though there are infinitely many of them. [0,1) -> [0,2): x -> 2*x. What points in [1,2) were left out? > But until there is a suitable axiom system for TO math, it does > not exist as anything mathematical. > > It certainly exists as a developing mathematical model, even if a > firm axiomatic foundation has not yet been finalized. :) > > Because of the many self-contradictions in TO modeling, apparently > invisible to TO but glaringly visible to everyone else, his models > cannot be properly called mathematical, as proper mathematical > models are not allowed to have known self-contradictions. > > I don't think you can point out any contradiction in what I've said > above. If two points are not finitely different, you can't find a > standard point between them. One does not need a standard point, one only needs a real point, and for ANY model of the reals, given x and y reals in that model with x < y, then (x+y)/2 is a real number in that model strictly between them. Betweenness is the only issue here, scale is not an issue. > > But there is no scaling factor possible which will transmute a > dense set into a discrete one or vice versa. > > Yes, a relatively infinite scaling will do just that. > > How does any scaling find infinitely many new elements between > consecutive members of a discrete set or eliminate the infinitely > many between any two members of a dense set? > It doesn't. Then it doesn't change discrete to dense or dense to discrete! > It changes what you can distinguish, Which is irrelevant, since the relevant definitions do not mention distinguishability, only ordering. so that elements > that were infinitely different in the discrete set continue to have > an infinite number of itnermediate elements, but compressed into a > finite space, so the sparse set becomes dense. But how does that change of scale take two consecutive elements of that original discrete set and create any element between them? Unless it does that, the image is still discrete. > How is this dense > sequential set not continuous on the finite scale? By having the image of two originally successive elements still without anything between them. > > That is what TO is claiming his scaling can do. > > All scalings of an ordered set are order isomorphisms which leave > EVERY order property, including denseness or discreteness, > unchanged. That is what order isomorphism is all about, preserving > ALL order properties. > > Are you sure that aplies in the infinite case? EVERY case! > It can using infinite division, as shown above. > > It is certainly not possible in any current axiom system. > > But it is neverhteless possible. > > > TO is claiming that if we have an ordered set S with members x > and y such that there are no elements of S between x and y, > there is some scaling factor which will order-isomorphically > map S to S', x to x' and y to y' and in the process create some > z' in S' such that z' is between x' and y'. > > No, I am saying that is x and y are discrete sequential elements, > scaling it up by an infinite amount will cause what was a finite > difference realtive to the original unit of measure into an > infinitesimal difference relative to the new unit of measure, so > that thw two will not be distinguishable > > In mathematics, if two members are not the same, then any mapping > which makes them the same is NOT a scale mapping, since it is not > reversable, as scale mappings must be. > > When scaling by an infinite value, the distinguishable may become > indistinguishable. Then you are mapping two, or more, elements onto 1 element. Such mappings are not invertable. Changes of scale must be invertable. Ergo: TO's mappings are NOT changes of scale. > > > Not in mathematics. > > Not in STANDARD mathematics > > Not in ANY mathematics. > > > Not in any that you care to acknowledge. > > Not that any mathematician will acknowledge. Every change of scale > must be reversible, but a mapping which makes two originaly > distinct elements indistinguishable cannot be reversed, so is not a > scale change. > When one scales using T-riffic representations, one can keep track of > the infinitesimals values which are not finitely measurable, and so > reverse the scaling as desired. WRONG! If TO maps two elements onto one so that their images are indistinguishable, then they are indistiguishable, and cannot afterwards be distinguished. That is what indeistinguishable means. If TO wants to have the images simultaneously both distinguishable and indistiguishable, he is still working only in cloud cuckoo land. > If you are so certain, prove your claim (in some specific axiom > system)! Absent such proof, your claim fails. > > It doesn't triumph with such proof, but it doesn't fail until > disproven. WRONG! That is not how mathematics works. Sounds more like politics or religion to me. In mathematics, claims are never any more than conjectures until proven in some axiom system. And when such conjectures can be disproved in many standard axiom systems, as TO's conjectures have been, they carry very little weight. > Your uncountability proof > doesn't hold, since countability is not a criterion for anything in > my theory, TO does not have a theory at all until he can come up with an axiom system. What he has now is just a bunch of wild hair conjectures that conflict with all known theories. > It is true that the powerset is larger than the set, even if > infinite. it is true that the set of strings is larger than the > allowed length of strings, even if infinite. It is not true that the > infinite number of reals in the unit interval is larger than the > number of unit intervals on the real line. Not in my theory. If one defines a unit interval as one having integer endpoints, TO is wrong. If one defines a unit interval as one having length equal to 1 then there is an obvious bijection between reals,x, and unit untervals [x,x+1]. If one is, as TO is, ambiguous about the meaning of unit interval, then one's claim is also ambiguous. > > Density is an order property which cannot be changed by any > change of scale. > > How is density an order property? > > Density is the property which says of an ordered that set > between any two members x and z of that set, with x < z, there > is a y in that same set with x < y < z. Since that is entirely > defined in terms of the ordering within the set, density is an > order property of that set. > > Okay, and the real line is dense on every scale, but a discrete > set on one scale can be considered dense on a relatively infinite > scale, as long as differences between successive members are all > finite relative to the scale on which the set is discrete. LIE! To scale any non-dense set to a dense set would require the creation of infinitely many new elements in image of the scaled set. > > WRONG! since density is an order property and changing scales must > be an order isomorphism, changing scales can have no effect on > density. > But density is defined as there being an element between any two > finitely different elements, not any two ifninitesimally different > elements, since those are not discernible as two elements to begin > with. LIE! The definition of density does not mention scale, and is independent of scale. > Same thing here. I am not making elements appear or disappear, or > saying there is any actual end to subdivision. I am saying that it > becomes impossible to find an intermediate value between two which > are themselves indistinguishable. If they are different at any scale, then they have already been distinguished. > > It is impervious to FINITE changes of scale > > WRONG! If any mapping carries two elements into one as TO insists, > it is not invertable, but changes of scale must be invertable. > > I am not insisting on any such thing. The standard and universal mathematical definition of a change of scale requires that it be invertible. If TO wants to play his silly games with definitions again let him give a clear definition of what HE means by a change of scale, since it is obviously not what everyone else means. > > Has the issue of INFINITE changes of scale ever been addressed > formally, to your knowledge? > > Yes! In a non-standard field with infintiesimal and infinite > elements, every element except zero has a unique reciprocal, and > every such element can be used to produce a change the scale. But > no such change will ever map two originally distinct elements onto > one element, as such a mapping is not invertable. > Well, good. It shouldn't. But the definition of a standard dense set > deals with there always being an intermediate value between any two > FINITELY different values. Where did TO get that remarkably stupid notion? Does he have a particularly rich stupid mine from which he gets al his ideas? For ordered sets in general, there is no such thing as distances between members, other than the finiteness or infiniteness of the number of other members between two given members. A set is DENSE if there are always other members between any two ( in fact at least countably many of them) A set is DISCRETE if for every y in it there is either no smaller or a next smaller member and either no larger or a next larger member. >It is clear to me > > It ain't what TO doesn't know that is hurting him, It's what he > knows that ain't so. > > > If not, then as one divides their unit into a number of > subunits that approaches infinity, the density of any > infinitely dense set may also be simultaneously subdivided to > a finite number of elements per infinitesimal unit of element > value. > > Except that the relative density of a dense set is constant as > one makes finite changes of scale so must keep that constant > value in the limit. > > Density relative to what? > > To any interval and its image. > > Given any interval in the reals and any set and any scaling > transformation of the reals, the 'number' of points of that set in > that interval will be the same as the number of points in the image > of the set which are in the image of the interval. > > Sure. So? SO TO is WRONG! AGAIN! AS USUAL! === Subject: Re: Well Ordering the Reals Virgil said: > Virgil said: > Whether something is discernable depends on properties of the > observer as much as those of the observed, so is irrelevant in > mathematics. > > > Indeed, it relies on the aparatus, physical or abstract, of > measurement. That is, it depends on the unit. > Since the unit is entirely artificial here, whether is discernatble > des not depend on the unit at all, though the units used may depend on > what is discernable. What do you mean by artificial? If, on a given scale, two points may be distinguished, the on that scale those two points are a finite number of units apart. If, on a different scale, those two points are less than any finite number of units apart, they are the same point on that scale. >One can't measure > infinitesimal intervals in finite units, or one gets zeroes. > In actual measurements, there is no such thing as infinitesimal. > Actual meassurement occurs only in the physical world whereas > infinitesimals occur, if at all, only in a purely mathematical > non-physical world. Right, and standard mathematics deals only with finite quantities and differences, so infinitesimal differences do not exist, and subdivision thereof is a moot point. > > When you have two points infinitesimally close to each other, in > the standard finite world there is no idfference between them > In the standard finite world of the physical, there are no > infinitesimals at all. Yes, but in a non-standard system they do, and when such a system is considered, such differences are nil in the standard world of mathematics, so the Archimidean principle has what may be called limit points, similar to limit ordinals or the limit points in T-riffics. > > TO's standard finite world is too limited for mathematics. > >I'm NOT projecting. You are! > I am not projecting properties of the physical world and mathematical > world on each other the way TO is doing, at any rate. No, you're projecting your overly finitist view onto me, which is a pretty good joke. If mathematical density of a set depends on the Archimedean principle, and that depends on the distinction between two points, and that distinction disappears at infinite scales, then the density of the set on the infinite scale is not affected by the discreteness of the set on an infinitesimal scale. > > > > On the finite scale, both points represent the same real number > > If 'points' are different on any scale, then they are never equal. > > Ultimately they are two different points, but they do not have two > different standard locations if the distance between them is > infinitesimal. > If two values are not equal at the infinitesimal level, they are not > equal at any level. The most one can ever say of different quantities > that are infinitesimally close is that they are infinitesimally close. > If TO wants to formalize that, best he should read some more of Robinson. Good advice, and I'll try to find time to start again on that. In the meantime, I recommend you consider under what circumstances a dense set is required to have intermediate values. If two values are the same on the finite scale, much like two finite values are the same on the infinite scale in your theory, being nothing, then no intermediate value ir required. > If a set is continuous when there always exists an > intermediate value between any two distinct values > That is not what makes a set continuous. The rationals have that > property of density but are NOT continuous. TO again knows not wherof he > speaks. Basically, as I understand it, a continuous function going from value x to value y passes through each value between the two values, and the difference between f(x) and f(y) has a limit of 0 as x->y. Or, the limit of f(x) is the same for all x in the domain whether aproached from right or left. Perhaps you could correct me on that, but I don't think your technical corrections will change matters much. I could be wrong. > then the fact > that there is not considered to be a point between two points which > are NOT distinct > Two points which are not distinct are, in fact, not two points, but only > one. Yes, on the finite scale, there is no difference between, say, 0.999... and 1, and any attemtp to find an intermediate value will fail. Does this violate the Archimedean principle? > > > So, in this sense, the continuity of the real line depends on the > unit of measure. > > TO is confusing the mathematical real line with physical reality. > In mathematics, the real line looks precisely the same at any > magnification whatsoever. In TOmatics, nothing is ever the same. > > > I am not conflating the real line with reality, rght now, though I > have been known to do that. :) Indeed the real line looks the same at > every scale, but when one cannot distinguish between two points > because they are infinitesimally different, then for all intents and > purposes they are the same point. > If one cannot distinguish between two points which ARE different, > perhaps one should try another line of work. If denseness of a set is determined by the existence of an intermediate value within any finite interval, and there is no finite interval between two values, then an intermediate value between them is not required. > > Can you tell the naturals from one another, or do they form a > continuum of points? > Most people can tell one natural from another. If TO cannot, he perhaps > should take up kitting instead of wasting his time here. Snip that context. Nice technique. > > > > > A set that is discrete on one scale may be continuous on an > infinitely greater scale. > > Where in the definition of a set being dense is there any mention > of dependency on scale, or even of distances? Unless that > definition specifically allows such dependency on scale, TO has no > case. > > Claims like TO's, made without any supporting evidence, and, > indeed, contradicted by the actual definitions of dense and discrete > orderings, merely demostrate TO's ignorance. > > > > Statements like that make one wonder whether Virgil even understands > discrete sets with any depth, or can only relate to dense ones with > no structure, on a personal level. > The only one who might wonder that is one who himself does not > understand the difference between dense sets and sequences. Pray tell, what is the difference between a dense set and one where successive elements are less different than any finite value? > > In the definition of a dense set is the mention of a point being > between any two points. > That is not quite what density says, What density says is that between > every two distinct points of a dense set there are other points of that > set. Did you say distinct? Very good. Now, can you distinguish two infinitesimally different values using finite units? > If two points are not distinguishable, > Then there is only one point, not two of them at all. On that scale, yes, but not necessarily in a universal sense. > > Specifically, a sparse set will be dense on a relatively infinite > scale provided each element is only finitely different from its > successor. > > Where does that sparse set ever get a new element to go between > between two originally consecutive elements? Unless that can > happen, your sparse set will never transmute into a dense one. > > Well, that doesn't happen, silly. > Then your sparse set remains sparse at EVERY scale. > And your dense set remains dense at EVERY scale. Oh, pay attention..... > The sparse set, when compressed > infinitely, becomes dense wherever it was previously finitely spaced. > Only in the sense that if all its members are compressed into a single > member, then the resulting one element set is trivially both dense and > sparse. Ha ha, nice try. If you have a finite, or countable, set, then compressing it will result in the entire set being compressed to a single point of compactification. If you have an uncountable discrete set (which probably doesn't even make sense to you) then it can be compressed to a finite range and become dense. > If your sparse set had infinite spaces between consecutive elements, > then you could have areas in your compressed set which were still > sparse. But, elements finitely spaced, when compressed infinitely, > become infinitesimally spaced, and dense in the reals through that > scaling transformation. > Only if all the separate points are mapped to a single point, but that > is not a change of scale transformation. For one thing, it is not > invertable, as changes of scale must be, and is not a bijection, as > changes of scale must be. Consider that you have one point in the interval for each natural. Two points are infinitesimally spaced. Any finite number of infinitesimal intervals added will still remain infinitesimal, and apear as a single point, but an infinite number of such infinitesimal intervals may sum to a finite one, and form a finite interval on that scale. > Are you really saying this doesn't make any > sense to you? You're just being argumentative. > If TO is accusing me of arguing that what does not make mathematical > sense does not make mathematical sense, I confess to doing precisely > that. That you cannot see the sense in what I'm saying is hard to believe, but typical. > > > > Continuity vs. discreteness may be thought of as relative to > the observer > > Not hardly! Continuity vs. discreteness is inherent in the > order properties of the set, and is quite independent on the > observer, provided he knows how to observe, which TO clearly > does not. > > > Which TO does clearly, as a matter of fact. > > Not if he imagines that changing scales can transmute dense > ordering into sequencial ordering or vice versa. > > > Well, I'm sorry if you can't imagine such things, but time marches > onward. You can't step into the same river twice. > TO apparently cannot even step in the river of truth once. That is a twist on that old saying, since the river changes between the time your foot touches the water and the time it touches the river bed. > > > Consider what I have said above about the unit of measurement and > discernibility. > > Considered, and rejected for cause. > > Jes 'cause, is why! Yup, 'cause I say so. Hyuck hyuck. > What TO argues is all argued jest 'cause he says so, since he has > neither any axiom system or mathematicians to back him up. > What I argue is backed up by several axiom systems and thousands of > mathematicians. > TO may be happy playing in his sand box, but he condemns himself to > playing only with himself. How do you think I got to this place? Certainly not by listening to naysayers who simply dismiss anything that's not already ina book. > > > Infinitesimally different values are the same value in the standard > finite reals. Everyone knows that. > To is wrong about what everybody knows. TO is not everybody, and what he > knows seems often not to be true. Okay, so, I meant everyone who's anyone, which excludes figments of my imagination like Virgil. ;) > > > infinitesimal subintervals does not violate continuity on the > standard finite scale. > > Not if one has a consistent model allowing infinitesimals, and in > such models, what TO is claiming is false. > > Obviously those are not the same model as what I am proposing. > What TO proposes is almost universally in direct contradiction to all of > standard mathematics and is almost universally self-contradictory. The first part is correct, but the second is not. It is largely contradictory to almost everything regarding transfinite set theory, but is internally consistent. > You > can't judge my ideas on what you've heard before, as you know. > I can judge them by TO's inablility to resolve their many internal > conflicts. And I have so judged them and found them unsatisfactory. > That TO does not like my judgements is his problem. The problem is that you haven't identified any internal consistencies, which isn't a big problem for me. > The > T-riffics are not the adics, the H-riffics are not a set of rationals > and are the well ordering of the reals, and these infinitesimals are > not whatever you have seen and rejected. So, like, > whatever.....you're wrong. :) > What I have seen and rejected are TO's proposed systems, which others > have also seen and rejected. Oh Tribble tried to poke holes in it, but failed, as you have. Unfortunately for you, it works nonetheless. > > > It's not quite actually continuous, but if it were, could you > tell? > > I can tell that the real line actually is continuous in the > sense of being both dense and complete. > > Are not the infinitesimal intervals dense in the reals? > > Real intervals of more than one point, infinitesimal or otherwise, > are dense in themselves, but the only real interval dense in the > set of reals is the set of all reals. > > Non-answer alert! If there are an infinite number of equal intervals > per unit then those intervals are dense in that unit interval, as > there is always another between any two which are finitely > distinguishable. > What definition of density is TO using which applies to intervals > instead of elements? Between any two elements is an interval, within which any dense set will contain another element. You know what I'm saying. > Standard Definition of an ordered set being dense: > A totally ordered set is /dense/ if and only if, > for every two distinct members of that set, there > is another member of that set strictly between them. Between them in what sense, if not being a point in the interval between them? > Nothing in that definition mentions intervals or distances between > members or even hints at equality of intervals. There is no requirement > that any ordered set have any notion of /distance/ between members. Thus > there is no such thing for arbitrary ordered sets as unit intervals > or equal intervals. Denseness of a set does not require that the elements be spaced at equal intervals, but it does require that the interval between any two *distinct* elements contain another element. There is most certainly the notion of ordered element value, otherwise the notion of there being an element with a value between any two others is without meaning. > TO is imagining that every ordered set is a 1 dimensional metric space, > which is far from the case. No, but the real line is most certainly a quantitative linearly ordered set. We are talking about infinitesimals on the real line here, remember? > One can always find an intermediate value in any > finite interval, so, sorry, they're dense. You should be able to > relate... ;) > The only non set model of density currently available is TO himself. Hyuck Hyuck! Them's a knee-slapper!! > > > > Dense means that, viewed at any scale whatsoever, between every > two distinct reals there are other reals, actually uncountably > many of them. > > Dense does not mean uncountable, as the rationals are both > countable and dense. > > Well, sure, in your theory. > Question: Is it the case that for every two distinct rational numbers, > say p and q, that there is another rational number strictly between them > (for example (p+q)/2)? For standard rationals, absolutely. When it comes to non-standard rationals with infinite denominators, then one may consider infinitesimal quantities as well, but I haven't thought that out as deeply as the reals. > Question: If the answer above is yes, does that not require the > rationals numbers to be a dense set? The rationals are most certainly dense in the reals. Granted, they are not continuous, as there are a majority of reals which are irrational, interspersed amongst the rationals, but there are no such discontinuities in the infinitesimal sequence. > I have a different theory about the > rationals and how they relate to the reals. > In TO's case, whenever his theories differ from the standard ones, they > seem to be wrong. Provably wrong in every cirrent axiom system for set > theory. Agreed. that's the way they seem to you, and that's understandable, given what you have studied, since they contradict axioms of the standard theory and therefore are provably worng, assuming those axioms. > So until TO comes up with an axiom system of his own, they remain wrong. They remain unprovable, and yet, eminently correct. > Don't anyone hold his breath! Take a bath, and we won't have to. > > Anyway, as I've said, if there are N unit intervals on the infinite > real line > As there aren't, the rest is moot. Can you say, denial? > Betcha never > heard that before! It's about time. > GIGO! You actually didn't find that interesting? (sigh) It's true. > Countability is irrelevant. > > TO is irrelevant. > > > You really hurt my feelings, Virgil. I shall drench my pillow with > tears. :'( > Crocodile tears, no doubt! Indubitably. > > > The proper theory > > TO has no notion of what proper theories are, as his are all > improper. > > > > Note that every two distinct real numbers are endpoints of an > open interval, and any two open intervals contain the same > uncountable number of points, regardless of their lengths. > > Only in the standard theory. > > Actually in ANY theory for which there is an axiomatic foundation > which allows uncountable sets, the number of points is easily > shown to be the same for every non-empty open interval. > > > Any theory which relies on bijections without regard to the effect of > the actual mapping functions used, sure, like any crappy theory. > To much prefers theories in which nothing can be proved, or disproved! I prefer theories that actually have something useful to say. > > If you want to tell me that there are no more reals in [0,2) than in > [0,1), then I guess the points in [1,2) don't exist, or don't count > for anything even though there are infinitely many of them. > [0,1) -> [0,2): x -> 2*x. What points in [1,2) were left out? 1:000...001, 1:000...003, 1.000...005, ...., 1.999...997, 1.999...999. 1/2 of the points in [0,1) were left out, too. So, you have the same number of points spread over twice the range, and you have half the density, although the set is still dense in the reals, since only infinitesimal differences separate consecutive elements. > But until there is a suitable axiom system for TO math, it does > not exist as anything mathematical. > > It certainly exists as a developing mathematical model, even if a > firm axiomatic foundation has not yet been finalized. :) > > Because of the many self-contradictions in TO modeling, apparently > invisible to TO but glaringly visible to everyone else, his models > cannot be properly called mathematical, as proper mathematical > models are not allowed to have known self-contradictions. > > > I don't think you can point out any contradiction in what I've said > above. If two points are not finitely different, you can't find a > standard point between them. > One does not need a standard point, one only needs a real point, and for > ANY model of the reals, given x and y reals in that model with x < y, > then (x+y)/2 is a real number in that model strictly between them. > Betweenness is the only issue here, scale is not an issue. Yes, but as you said, if two points are not distiguishable on a given scale, then they aren't two points for another point to be between. Can you have a point between ONE point? Q: What's the difference between a duck? (Between a duck and...what?) A: One leg is both the same. Get it? > > > But there is no scaling factor possible which will transmute a > dense set into a discrete one or vice versa. > > Yes, a relatively infinite scaling will do just that. > > How does any scaling find infinitely many new elements between > consecutive members of a discrete set or eliminate the infinitely > many between any two members of a dense set? > > It doesn't. > Then it doesn't change discrete to dense or dense to discrete! Indeed it does, for the reasons I have reiterated. > It changes what you can distinguish, > Which is irrelevant, since the relevant definitions do not mention > distinguishability, only ordering. every two distinct members of that set, there is another member of that set strictly between them. Notice how you used the word, distinct? Do you think it's coincidence that it has the same frst 7 letters as distinguish? Does distinct mean something similar to distinguishable? (sigh) > so that elements > that were infinitely different in the discrete set continue to have > an infinite number of itnermediate elements, but compressed into a > finite space, so the sparse set becomes dense. > But how does that change of scale take two consecutive elements of that > original discrete set and create any element between them? Unless it > does that, the image is still discrete. It doesn't. It makes immediate neighbors indistinguishable. > How is this dense > sequential set not continuous on the finite scale? > By having the image of two originally successive elements still without > anything between them. But on that scale, they are not distinct, as the rule for denseness requires. > > > That is what TO is claiming his scaling can do. > > All scalings of an ordered set are order isomorphisms which leave > EVERY order property, including denseness or discreteness, > unchanged. That is what order isomorphism is all about, preserving > ALL order properties. > > > Are you sure that aplies in the infinite case? > EVERY case! Uh, are you sure? Methinks not. > It can using infinite division, as shown above. > > It is certainly not possible in any current axiom system. > > But it is neverhteless possible. > > > TO is claiming that if we have an ordered set S with members x > and y such that there are no elements of S between x and y, > there is some scaling factor which will order-isomorphically > map S to S', x to x' and y to y' and in the process create some > z' in S' such that z' is between x' and y'. > > No, I am saying that is x and y are discrete sequential elements, > scaling it up by an infinite amount will cause what was a finite > difference realtive to the original unit of measure into an > infinitesimal difference relative to the new unit of measure, so > that thw two will not be distinguishable > > In mathematics, if two members are not the same, then any mapping > which makes them the same is NOT a scale mapping, since it is not > reversable, as scale mappings must be. > > > When scaling by an infinite value, the distinguishable may become > indistinguishable. > Then you are mapping two, or more, elements onto 1 element. Such > mappings are not invertable. Changes of scale must be invertable. > Ergo: TO's mappings are NOT changes of scale. I am doing no such thing. The number of elements stays the same, while the range is reduced infinitely, multiplying a density of 0 by oo and acheiving a finite density in the reals. > > > > Not in mathematics. > > Not in STANDARD mathematics > > Not in ANY mathematics. > > > Not in any that you care to acknowledge. > > Not that any mathematician will acknowledge. Every change of scale > must be reversible, but a mapping which makes two originaly > distinct elements indistinguishable cannot be reversed, so is not a > scale change. > > When one scales using T-riffic representations, one can keep track of > the infinitesimals values which are not finitely measurable, and so > reverse the scaling as desired. > WRONG! If TO maps two elements onto one so that their images are > indistinguishable, then they are indistiguishable, and cannot afterwards > be distinguished. That is what indeistinguishable means. Indistinguishable means not finitely different. > If TO wants to have the images simultaneously both distinguishable and > indistiguishable, he is still working only in cloud cuckoo land. No, I am considering infintiely different scales. > > If you are so certain, prove your claim (in some specific axiom > system)! Absent such proof, your claim fails. > > > It doesn't triumph with such proof, but it doesn't fail until > disproven. > WRONG! That is not how mathematics works. Sounds more like politics or > religion to me. In mathematics, claims are never any more than > conjectures until proven in some axiom system. > And when such conjectures can be disproved in many standard axiom > systems, as TO's conjectures have been, they carry very little weight. Your axioms are all made up. That's like saying democracy is bad because the Red Queen rules. > > Your uncountability proof > doesn't hold, since countability is not a criterion for anything in > my theory, > TO does not have a theory at all until he can come up with an axiom > system. What he has now is just a bunch of wild hair conjectures that > conflict with all known theories. You should see my actual hair. Hahahahaa... > > It is true that the powerset is larger than the set, even if > infinite. it is true that the set of strings is larger than the > allowed length of strings, even if infinite. It is not true that the > infinite number of reals in the unit interval is larger than the > number of unit intervals on the real line. Not in my theory. > If one defines a unit interval as one having integer endpoints, TO is > wrong. If one defines a unit interval as one having length equal to 1 > then there is an obvious bijection between reals,x, and unit untervals > [x,x+1]. If one is, as TO is, ambiguous about the meaning of unit > interval, then one's claim is also ambiguous. An interval between consecutive integers is a unit interval, but not the only kind of unit interval. However, I am talking about contiguous but mutually exclusive unit intervals, not overlapping ones, and my point stands in my theory. > > > Density is an order property which cannot be changed by any > change of scale. > > How is density an order property? > > Density is the property which says of an ordered that set > between any two members x and z of that set, with x < z, there > is a y in that same set with x < y < z. Since that is entirely > defined in terms of the ordering within the set, density is an > order property of that set. > > Okay, and the real line is dense on every scale, but a discrete > set on one scale can be considered dense on a relatively infinite > scale, as long as differences between successive members are all > finite relative to the scale on which the set is discrete. > LIE! To scale any non-dense set to a dense set would require the > creation of infinitely many new elements in image of the scaled set. No lie, Virgil. Denseness can be achieved by adding an infinite number of elements to a finite interval, or by squeezing an already infinite set INTO a finite interval. Density is # elements/range. > > WRONG! since density is an order property and changing scales must > be an order isomorphism, changing scales can have no effect on > density. > > But density is defined as there being an element between any two > finitely different elements, not any two ifninitesimally different > elements, since those are not discernible as two elements to begin > with. > LIE! The definition of density does not mention scale, and is > independent of scale. Lie? Oh, please. It mentions betweenness and distinguishability. > Same thing here. I am not making elements appear or disappear, or > saying there is any actual end to subdivision. I am saying that it > becomes impossible to find an intermediate value between two which > are themselves indistinguishable. > If they are different at any scale, then they have already been > distinguished. How does standard math distinguish between 1 and 0.999...? > > It is impervious to FINITE changes of scale > > WRONG! If any mapping carries two elements into one as TO insists, > it is not invertable, but changes of scale must be invertable. > > > I am not insisting on any such thing. > The standard and universal mathematical definition of a change of scale > requires that it be invertible. If TO wants to play his silly games with > definitions again let him give a clear definition of what HE means by a > change of scale, since it is obviously not what everyone else means. Indeed it is, but most don't talk about infinite scaling that I've seen. Then again, I haven't seen everything. > > > Has the issue of INFINITE changes of scale ever been addressed > formally, to your knowledge? > > Yes! In a non-standard field with infintiesimal and infinite > elements, every element except zero has a unique reciprocal, and > every such element can be used to produce a change the scale. But > no such change will ever map two originally distinct elements onto > one element, as such a mapping is not invertable. > > Well, good. It shouldn't. But the definition of a standard dense set > deals with there always being an intermediate value between any two > FINITELY different values. > Where did TO get that remarkably stupid notion? Does he have a > particularly rich stupid mine from which he gets al his ideas? Did Virgil eat a can of stupid for breakfast, that he thinks infinitesimal differences are finitely measurable? > For ordered sets in general, there is no such thing as distances between > members, other than the finiteness or infiniteness of the number of > other members between two given members. > A set is DENSE if there are always other members between any two ( in > fact at least countably many of them) > A set is DISCRETE if for every y in it there is either no smaller or > a next smaller member and either no larger or a next larger member. And what if that depends on the notion of distinguishability? >It is clear to me > > It ain't what TO doesn't know that is hurting him, It's what he > knows that ain't so. > > > If not, then as one divides their unit into a number of > subunits that approaches infinity, the density of any > infinitely dense set may also be simultaneously subdivided to > a finite number of elements per infinitesimal unit of element > value. > > Except that the relative density of a dense set is constant as > one makes finite changes of scale so must keep that constant > value in the limit. > > Density relative to what? > > To any interval and its image. > > Given any interval in the reals and any set and any scaling > transformation of the reals, the 'number' of points of that set in > that interval will be the same as the number of points in the image > of the set which are in the image of the interval. > > > Sure. So? > SO TO is WRONG! AGAIN! AS USUAL! NO! HA! YOU ARE! IN YOUR FACE! NYAH!!! hahaha <3 -- Smiles, Tony === Subject: Re: Well Ordering the Reals > Virgil said: > > Virgil said: > > Whether something is discernable depends on properties of the > observer as much as those of the observed, so is irrelevant in > mathematics. > > > Indeed, it relies on the aparatus, physical or abstract, of > measurement. That is, it depends on the unit. > > Since the unit is entirely artificial here, whether is discernatble > des not depend on the unit at all, though the units used may depend on > what is discernable. > > What do you mean by artificial? If, on a given scale, two points may be > distinguished, the on that scale those two points are a finite number of > units > apart. There is nothing about an order relation that requires it to have any thing like a scale, so there is no necessity that an ordered set have any scale at all. > If, on a different scale, How does an ordered set have a diffferent scale, if there is no original scale to be different from? > >One can't measure > infinitesimal intervals in finite units, or one gets zeroes. > > In actual measurements, there is no such thing as infinitesimal. > Actual meassurement occurs only in the physical world whereas > infinitesimals occur, if at all, only in a purely mathematical > non-physical world. > > Right, and standard mathematics deals only with finite quantities and > differences, so infinitesimal differences do not exist, and > subdivision thereof is a moot point. Non-standard analysis is a part of sandard mathematics, and it does have infinitesimals. > > TO's standard finite world is too limited for mathematics. > >I'm NOT projecting. You are! > > I am not projecting properties of the physical world and mathematical > world on each other the way TO is doing, at any rate. > No, you're projecting your overly finitist view onto me, which is a pretty > good > joke. It is not my view, but the view of all mathematicians. If mathematical density of a set depends on the Archimedean principle, > and that depends on the distinction between two points, and that distinction > disappears at infinite scales If two points become one then one does not have a bijective mapping, and one does not have a change of scdale. All changes of scale are bijections by definition. > > If two values are not equal at the infinitesimal level, they are not > equal at any level. The most one can ever say of different quantities > that are infinitesimally close is that they are infinitesimally close. > If TO wants to formalize that, best he should read some more of Robinson. > > Good advice, and I'll try to find time to start again on that. In the > meantime, I recommend you consider under what circumstances a dense > set is required to have intermediate values. That's easy. Under all conditions, a dense set is required to have values intermediate between any two of its values, since that is the quality that allows us to call it dense. Perhaps it is TO who is being dense in a different sense of that word. > If two values are the > same on the finite scale, much like two finite values are the same on > the infinite scale in your theory, being nothing, then no > intermediate value ir required. If the are the same on any scale, they are the same on all scales, since order properties (like being equal or not eaual) are entirely independent of any notions of scale. > > > If a set is continuous when there always exists an > intermediate value between any two distinct values > > That is not what makes a set continuous. The rationals have that > property of density but are NOT continuous. TO again knows not wherof he > speaks. > > Basically, as I understand it, a continuous function going from value x to > value y passes through each value between the two values, and the difference > between f(x) and f(y) has a limit of 0 as x->y. Then TO clearly does NOT understand it. > I could be wrong. TO IS wrong! > > Two points which are not distinct are, in fact, not two points, but only > one. > > Yes, on the finite scale On any scale, since order properties are totally unaffected by scales. > > > > So, in this sense, the continuity of the real line depends on the > unit of measure. > > TO is confusing the mathematical real line with physical reality. > In mathematics, the real line looks precisely the same at any > magnification whatsoever. In TOmatics, nothing is ever the same. > > > I am not conflating the real line with reality, rght now, though I > have been known to do that. :) Indeed the real line looks the same at > every scale, but when one cannot distinguish between two points > because they are infinitesimally different, then for all intents and > purposes they are the same point. If two points are the same for ALL intents and purposes, they are equal. And if they are equal at any scale, they are equal at all scales. > > If one cannot distinguish between two points which ARE different, > perhaps one should try another line of work. > > If denseness of a set is determined by the existence of an intermediate value > within any finite interval Two errors: finite and inteval. Ordered sets have no intrinsic distances between members, so that neither finite nor interval apply. > > > Can you tell the naturals from one another, or do they form a > continuum of points? > > Most people can tell one natural from another. If TO cannot, he perhaps > should take up kitting instead of wasting his time here. > Snip that context. Nice technique. > > > > > > A set that is discrete on one scale may be continuous on an > infinitely greater scale. > > Where in the definition of a set being dense is there any mention > of dependency on scale, or even of distances? Unless that > definition specifically allows such dependency on scale, TO has no > case. > > Claims like TO's, made without any supporting evidence, and, > indeed, contradicted by the actual definitions of dense and discrete > orderings, merely demostrate TO's ignorance. > > > > Statements like that make one wonder whether Virgil even understands > discrete sets with any depth, or can only relate to dense ones with > no structure, on a personal level. > > The only one who might wonder that is one who himself does not > understand the difference between dense sets and sequences. > Pray tell, what is the difference between a dense set and one where > successive > elements are less different than any finite value? Simple! Dense sets never have successive elements, and sets having successive elements, no matter how close, are not dense. > > In the definition of a dense set is the mention of a point being > between any two points. > > That is not quite what density says, What density says is that between > every two distinct points of a dense set there are other points of that > set. > > Did you say distinct? Very good. Now, can you distinguish two > infinitesimally > different values using finite units? The issue is whether one can distinguish them at all. If one can at all, they are distinct. > > If two points are not distinguishable, > > Then there is only one point, not two of them at all. > > On that scale, yes, but not necessarily in a universal sense. In every sense. TO seems to want to have somtheing between being equal and being not equal, a sor to sort of equal but not quite. For any two members, say x and y, of an ordered set , one and only one of x < y, or x = y or x > y is true. This is the law of trichotomy. Any set for which this is not true, like the sets TO is proposing, is NOT AN ORDERED SET. > > > > Specifically, a sparse set will be dense on a relatively infinite > scale provided each element is only finitely different from its > successor. > > Where does that sparse set ever get a new element to go between > between two originally consecutive elements? Unless that can > happen, your sparse set will never transmute into a dense one. > > Well, that doesn't happen, silly. > > Then your sparse set remains sparse at EVERY scale. > And your dense set remains dense at EVERY scale. > > Oh, pay attention..... To what? More of TO's nonsense? > > The sparse set, when compressed > infinitely, becomes dense wherever it was previously finitely spaced. > > Only in the sense that if all its members are compressed into a single > member, then the resulting one element set is trivially both dense and > sparse. > > Ha ha, nice try. If you have a finite, or countable, set, then compressing it > will result in the entire set being compressed to a single point of > compactification. If you have an uncountable discrete set (which probably > doesn't even make sense to you) then it can be compressed to a finite range > and > become dense. The finite range is then zero and the result is a one element set. Which is trivially dense and trivially discrete. > > If TO is accusing me of arguing that what does not make mathematical > sense does not make mathematical sense, I confess to doing precisely > that. > That you cannot see the sense in what I'm saying is hard to believe, but > typical. Nobody but TO can posibly see sense in such nonsense. > > Jes 'cause, is why! Yup, 'cause I say so. Hyuck hyuck. > > What TO argues is all argued jest 'cause he says so, since he has > neither any axiom system or mathematicians to back him up. > > What I argue is backed up by several axiom systems and thousands of > mathematicians. > > TO may be happy playing in his sand box, but he condemns himself to > playing only with himself. > How do you think I got to this place? Certainly not by listening to naysayers > who simply dismiss anything that's not already ina book. The mystery is why anyone would want to get to the cloud cuckooland place TO claims he has got to. > > > > Infinitesimally different values are the same value in the standard > finite reals. Everyone knows that. > > To is wrong about what everybody knows. TO is not everybody, and what he > knows seems often not to be true. > Okay, so, I meant everyone who's anyone, which excludes figments of my > imagination like Virgil. ;) Then TO must be excluding all those thousands on mathematicians from being anyones. > You > can't judge my ideas on what you've heard before, as you know. > > I can judge them by TO's inablility to resolve their many internal > conflicts. And I have so judged them and found them unsatisfactory. > > That TO does not like my judgements is his problem. > > The problem is that you haven't identified any internal consistencies, which > isn't a big problem for me. I have, and posted those problems. That TO is too dumb to understand what his problems are is not my problem. > > The > T-riffics are not the adics, the H-riffics are not a set of rationals > and are the well ordering of the reals, and these infinitesimals are > not whatever you have seen and rejected. So, like, > whatever.....you're wrong. :) > > > What I have seen and rejected are TO's proposed systems, which others > have also seen and rejected. > Oh Tribble tried to poke holes in it, but failed, as you have. Unfortunately > for you, it works nonetheless. Where is the is mythical land of 'nonetheless' outside of which TO's nonsense still does not work? > What definition of density is TO using which applies to intervals > instead of elements? > Between any two elements is an interval Without a definition of distance, which is irrelevant for merely ordered sets, there are no such things as intervals. > > Standard Definition of an ordered set being dense: > A totally ordered set is /dense/ if and only if, > for every two distinct members of that set, there > is another member of that set strictly between them. > Between them in what sense, if not being a point in the interval between > them? To has things backwards as usual. The definitions of inteval is dependent on betweenness, not the other way round. Given x < z and any other element y, not equal to either of them, there are only 3 possibilities: z < x, or y < z or y is between x and z. > > Nothing in that definition mentions intervals or distances between > members or even hints at equality of intervals. There is no requirement > that any ordered set have any notion of /distance/ between members. Thus > there is no such thing for arbitrary ordered sets as unit intervals > or equal intervals. > Denseness of a set does not require that the elements be spaced at equal > intervals It prohibits them being spaces at equal intervals. > > TO is imagining that every ordered set is a 1 dimensional metric space, > which is far from the case. > > No, but the real line is most certainly a quantitative linearly ordered set. > We > are talking about infinitesimals on the real line here, remember? No! TO is insisting on metric spaces, so he can bring in a lot of irrelevancies, but for order properties, any such metrics are irrelevant. > > Dense means that, viewed at any scale whatsoever, between every > two distinct reals there are other reals, actually uncountably > many of them. > > Dense does not mean uncountable, as the rationals are both > countable and dense. > > Well, sure, in your theory. According to the only definition of density so far in evidence, the rationals are dense. If TO has a different definition, let him present it now or shut up. > > Question: Is it the case that for every two distinct rational numbers, > say p and q, that there is another rational number strictly between them > (for example (p+q)/2)? > For standard rationals, absolutely. If the rationals, standard or not, are to be an ordered field, then it will be the case that for every two distinct rational numbers, say p and q, that there is another rational number strictly between them (for example (p+q)/2). Since this is true for EVERY ordered field, it will be true for every ordered field. > > Question: If the answer above is yes, does that not require the > rationals numbers to be a dense set? > The rationals are most certainly dense in the reals. That is not what was asked. > > I have a different theory about the > rationals and how they relate to the reals. > > In TO's case, whenever his theories differ from the standard ones, they > seem to be wrong. Provably wrong in every current axiom system for set > theory. > Agreed. that's the way they seem to you, and that's understandable, given > what > you have studied, since they contradict axioms of the standard theory and > therefore are provably worng, assuming those axioms. > > So until TO comes up with an axiom system of his own, they remain wrong. > They remain unprovable, and yet, eminently correct. And Just how does TO KNOW that they are correct if he cannot prove them? Arrogance does not justify such certainty. > > To much prefers theories in which nothing can be proved, or disproved! > I prefer theories that actually have something useful to say. Of what use are they until they are provable? What bridges have TO's theories built? > then (x+y)/2 is a real number in that model strictly between them. > > Betweenness is the only issue here, scale is not an issue. > Yes, but as you said, if two points are not distiguishable on a given scale I never said anything like that. What I said is that if two points are 'Indistinguishable they are equal, and scales are irrelevant. > Which is irrelevant, since the relevant definitions do not mention > distinguishability, only ordering. > > > By having the image of two originally successive elements still without > anything between them. > But on that scale, they are not distinct, as the rule for denseness requires. Distinct means not equal in the sense to the law of trichotomy. Does To claim that if x < y on some scale that there is any scale on which x = y? If not, his arguments all fail. > > > That is what TO is claiming his scaling can do. > > All scalings of an ordered set are order isomorphisms which leave > EVERY order property, including denseness or discreteness, > unchanged. That is what order isomorphism is all about, preserving > ALL order properties. > > > Are you sure that aplies in the infinite case? > > EVERY case! > Uh, are you sure? Methinks not. O will accept your assuranced that you do not think. One of few statements of yours can accept. > When scaling by an infinite value, the distinguishable may become > indistinguishable. > > Then you are mapping two, or more, elements onto 1 element. Such > mappings are not invertable. Changes of scale must be invertable. > Ergo: TO's mappings are NOT changes of scale. > I am doing no such thing. Compressing the 'distance' between two points to zero, which is exactly what TO is doing, makes them the same point. === Subject: Re: Well Ordering the Reals Virgil said: > Virgil said: > > Virgil said: > > Whether something is discernable depends on properties of the > observer as much as those of the observed, so is irrelevant in > mathematics. > > > Indeed, it relies on the aparatus, physical or abstract, of > measurement. That is, it depends on the unit. > > Since the unit is entirely artificial here, whether is discernatble > des not depend on the unit at all, though the units used may depend on > what is discernable. > > > What do you mean by artificial? If, on a given scale, two points may be > distinguished, the on that scale those two points are a finite number of > units > apart. > There is nothing about an order relation that requires it to have any > thing like a scale, so there is no necessity that an ordered set have > any scale at all. No, but one can apply notions of scale to sets with measurable differences between elements. > If, on a different scale, > How does an ordered set have a diffferent scale, if there is no original > scale to be different from? If you are dealing with sets of quantities, those quantites are expressed in units of some sort which are able to distinguish them, and therefore have a notion of scale associated with them through those units. > > >One can't measure > infinitesimal intervals in finite units, or one gets zeroes. > > In actual measurements, there is no such thing as infinitesimal. > Actual meassurement occurs only in the physical world whereas > infinitesimals occur, if at all, only in a purely mathematical > non-physical world. > > > Right, and standard mathematics deals only with finite quantities and > differences, so infinitesimal differences do not exist, and > subdivision thereof is a moot point. > Non-standard analysis is a part of sandard mathematics, and it does have > infinitesimals. Standard analysis, then. Non-standard analysis is not part of stanbdard analysis. > > > TO's standard finite world is too limited for mathematics. > >I'm NOT projecting. You are! > > I am not projecting properties of the physical world and mathematical > world on each other the way TO is doing, at any rate. > > No, you're projecting your overly finitist view onto me, which is a pretty > good > joke. > It is not my view, but the view of all mathematicians. Most, but not all. Is the pursuit of truth a democratic process? > If mathematical density of a set depends on the Archimedean principle, > and that depends on the distinction between two points, and that distinction > disappears at infinite scales > If two points become one then one does not have a bijective mapping, and > one does not have a change of scdale. All changes of scale are > bijections by definition. If two values which are different by some finite number of some unit of measure are viewed at a relatively infinite scale, using a relatively infinite unit of measure, the difference between them can no longer be expressed as a finite number of the current units. On that scale, they are indstinguishable. > > If two values are not equal at the infinitesimal level, they are not > equal at any level. The most one can ever say of different quantities > that are infinitesimally close is that they are infinitesimally close. > If TO wants to formalize that, best he should read some more of Robinson. > > > Good advice, and I'll try to find time to start again on that. In the > meantime, I recommend you consider under what circumstances a dense > set is required to have intermediate values. > That's easy. Under all conditions, a dense set is required to have > values intermediate between any two of its values, since that is the > quality that allows us to call it dense. > Perhaps it is TO who is being dense in a different sense of that word. Huh? Whaddya mean? That went right over my head. Do I look fat in these pants? > If two values are the > same on the finite scale, much like two finite values are the same on > the infinite scale in your theory, being nothing, then no > intermediate value ir required. > If the are the same on any scale, they are the same on all scales, since > order properties (like being equal or not eaual) are entirely > independent of any notions of scale. And what is the difference between 0.999... and 1? > > > > If a set is continuous when there always exists an > intermediate value between any two distinct values > > That is not what makes a set continuous. The rationals have that > property of density but are NOT continuous. TO again knows not wherof he > speaks. > > > Basically, as I understand it, a continuous function going from value x to > value y passes through each value between the two values, and the difference > between f(x) and f(y) has a limit of 0 as x->y. > Then TO clearly does NOT understand it. > I could be wrong. > TO IS wrong! Then explain why. > > Two points which are not distinct are, in fact, not two points, but only > one. > > > Yes, on the finite scale > On any scale, since order properties are totally unaffected by scales. Distinguishability is what's affected. > > > > > So, in this sense, the continuity of the real line depends on the > unit of measure. > > TO is confusing the mathematical real line with physical reality. > In mathematics, the real line looks precisely the same at any > magnification whatsoever. In TOmatics, nothing is ever the same. > > > I am not conflating the real line with reality, rght now, though I > have been known to do that. :) Indeed the real line looks the same at > every scale, but when one cannot distinguish between two points > because they are infinitesimally different, then for all intents and > purposes they are the same point. > If two points are the same for ALL intents and purposes, they are equal. > And if they are equal at any scale, they are equal at all scales. Wherever they are not distinct. > > If one cannot distinguish between two points which ARE different, > perhaps one should try another line of work. > > > If denseness of a set is determined by the existence of an intermediate value > within any finite interval > Two errors: finite and inteval. Ordered sets have no intrinsic > distances between members, so that neither finite nor interval apply. Quantitative sets do. > > > > Can you tell the naturals from one another, or do they form a > continuum of points? > > Most people can tell one natural from another. If TO cannot, he perhaps > should take up kitting instead of wasting his time here. > > Snip that context. Nice technique. > > > > > > > A set that is discrete on one scale may be continuous on an > infinitely greater scale. > > Where in the definition of a set being dense is there any mention > of dependency on scale, or even of distances? Unless that > definition specifically allows such dependency on scale, TO has no > case. > > Claims like TO's, made without any supporting evidence, and, > indeed, contradicted by the actual definitions of dense and discrete > orderings, merely demostrate TO's ignorance. > > > > Statements like that make one wonder whether Virgil even understands > discrete sets with any depth, or can only relate to dense ones with > no structure, on a personal level. > > The only one who might wonder that is one who himself does not > understand the difference between dense sets and sequences. > > Pray tell, what is the difference between a dense set and one where > successive > elements are less different than any finite value? > Simple! Dense sets never have successive elements, and sets having > successive elements, no matter how close, are not dense. Oh. And here I was thinking the set of rationals was dense, and also countable and linearly ordered, and bijectable with the naturals. But, you know so much more than I, Great Virgil, so I must be mistaken. Now I know that no dense set > > > In the definition of a dense set is the mention of a point being > between any two points. > > That is not quite what density says, What density says is that between > every two distinct points of a dense set there are other points of that > set. > > > Did you say distinct? Very good. Now, can you distinguish two > infinitesimally > different values using finite units? > The issue is whether one can distinguish them at all. If one can at all, > they are distinct. How tautologically correct of you! :) Now, what determines whether one can distinguish two values at all? > > > If two points are not distinguishable, > > Then there is only one point, not two of them at all. > > > On that scale, yes, but not necessarily in a universal sense. > In every sense. TO seems to want to have somtheing between being equal > and being not equal, a sor to sort of equal but not quite. Um, well, there ARE two sides to every coin, eh? > For any two members, say x and y, of an ordered set , one and only one of > x < y, or x = y or x > y is true. This is the law of trichotomy. Any set > for which this is not true, like the sets TO is proposing, is NOT AN > ORDERED SET. Wherever reals x and y can be distinguished, either xy. When they cannot, x=y. That's kind of what = means, you see, in general. Ala Leibniz, if all the properties of two objects are the same, they can't be distinguished, and are the same object. That's how you identify things. So, you don't have to yell. Have some chocolate. > > > > > Specifically, a sparse set will be dense on a relatively infinite > scale provided each element is only finitely different from its > successor. > > Where does that sparse set ever get a new element to go between > between two originally consecutive elements? Unless that can > happen, your sparse set will never transmute into a dense one. > > Well, that doesn't happen, silly. > > Then your sparse set remains sparse at EVERY scale. > And your dense set remains dense at EVERY scale. > > > Oh, pay attention..... > To what? More of TO's nonsense? Yes, every little bit of it. You'll be glad you did. :) > > > The sparse set, when compressed > infinitely, becomes dense wherever it was previously finitely spaced. > > Only in the sense that if all its members are compressed into a single > member, then the resulting one element set is trivially both dense and > sparse. > > > Ha ha, nice try. If you have a finite, or countable, set, then compressing it > will result in the entire set being compressed to a single point of > compactification. If you have an uncountable discrete set (which probably > doesn't even make sense to you) then it can be compressed to a finite range > and > become dense. > The finite range is then zero and the result is a one element set. > Which is trivially dense and trivially discrete. Uh, well, you just gave an example of a set which is both dense and discrete, so along with the rationals, that negates your objection the that combination above. Anyway, I am not talking about compressing it to a point, but to a finite interval, from an infinite one. Then all finite differences become infinitesimal, and the set becomes dense. > > If TO is accusing me of arguing that what does not make mathematical > sense does not make mathematical sense, I confess to doing precisely > that. > > That you cannot see the sense in what I'm saying is hard to believe, but > typical. > Nobody but TO can posibly see sense in such nonsense. Oh, I dunno 'bout that. I'm pretty sure I'm not nuts. > > Jes 'cause, is why! Yup, 'cause I say so. Hyuck hyuck. > > What TO argues is all argued jest 'cause he says so, since he has > neither any axiom system or mathematicians to back him up. > > What I argue is backed up by several axiom systems and thousands of > mathematicians. > > TO may be happy playing in his sand box, but he condemns himself to > playing only with himself. > > How do you think I got to this place? Certainly not by listening to naysayers > who simply dismiss anything that's not already ina book. > The mystery is why anyone would want to get to the cloud cuckooland > place TO claims he has got to. Perhaps in order to escape that burned out cellar hole you fellas call Cantor's Garden. > > > > > Infinitesimally different values are the same value in the standard > finite reals. Everyone knows that. > > To is wrong about what everybody knows. TO is not everybody, and what he > knows seems often not to be true. > > Okay, so, I meant everyone who's anyone, which excludes figments of my > imagination like Virgil. ;) > Then TO must be excluding all those thousands on mathematicians from > being anyones. If you don't know that infinitesimal differences are not finitely measurable, then as a mathematician, yes, you're really nobody who's going to contribute much to the integration of the infinities as a cohesive topic. > You > can't judge my ideas on what you've heard before, as you know. > > I can judge them by TO's inablility to resolve their many internal > conflicts. And I have so judged them and found them unsatisfactory. > > That TO does not like my judgements is his problem. > > > The problem is that you haven't identified any internal consistencies, which > isn't a big problem for me. > I have, and posted those problems. That TO is too dumb to understand > what his problems are is not my problem. You mostly object to things I haven't said, or purposely obfuscate the issue. > > > The > T-riffics are not the adics, the H-riffics are not a set of rationals > and are the well ordering of the reals, and these infinitesimals are > not whatever you have seen and rejected. So, like, > whatever.....you're wrong. :) > > > What I have seen and rejected are TO's proposed systems, which others > have also seen and rejected. > > Oh Tribble tried to poke holes in it, but failed, as you have. Unfortunately > for you, it works nonetheless. > Where is the is mythical land of 'nonetheless' outside of which TO's > nonsense still does not work? Beyond your grasp, obviously. > What definition of density is TO using which applies to intervals > instead of elements? > > Between any two elements is an interval > Without a definition of distance, which is irrelevant for merely ordered > sets, there are no such things as intervals. Are the reals merely ordered, or is it a quantitative set with a specific difference between any two elements? This is the kind of objection you bring to the table, one which entirely ignores a whole aspect of the issue at any given time. > > > Standard Definition of an ordered set being dense: > A totally ordered set is /dense/ if and only if, > for every two distinct members of that set, there > is another member of that set strictly between them. > > Between them in what sense, if not being a point in the interval between > them? > To has things backwards as usual. The definitions of inteval is > dependent on betweenness, not the other way round. > Given x < z and any other element y, not equal to either of them, there > are only 3 possibilities: z < x, or y < z or y is between x and z. When you say x > > Nothing in that definition mentions intervals or distances between > members or even hints at equality of intervals. There is no requirement > that any ordered set have any notion of /distance/ between members. Thus > there is no such thing for arbitrary ordered sets as unit intervals > or equal intervals. > > Denseness of a set does not require that the elements be spaced at equal > intervals > It prohibits them being spaces at equal intervals. It certainly doesn't prevent them from being spaced infinitesimally, or the rationals aren't dense. Why do you think denseness prevents equal spacing? > > > TO is imagining that every ordered set is a 1 dimensional metric space, > which is far from the case. > > > No, but the real line is most certainly a quantitative linearly ordered set. > We > are talking about infinitesimals on the real line here, remember? > No! TO is insisting on metric spaces, so he can bring in a lot of > irrelevancies, but for order properties, any such metrics are irrelevant. The metrics are not irrelevant when discussing metrics. I am saying the infinitesimal intervals are not measurable on the finite scale. That's a matter of metrics, and so metrics are not irrelevant to the matter. Scaling makes no sense without metrics. > > Dense means that, viewed at any scale whatsoever, between every > two distinct reals there are other reals, actually uncountably > many of them. > > Dense does not mean uncountable, as the rationals are both > countable and dense. > > Well, sure, in your theory. > According to the only definition of density so far in evidence, the > rationals are dense. If TO has a different definition, let him present > it now or shut up. I don't disagree that the rationals are dense. there is always an intermediate rational between two distinguishable rationals. I disagree with the notion of countability as any more than finite. But, you knew that. > > Question: Is it the case that for every two distinct rational numbers, > say p and q, that there is another rational number strictly between them > (for example (p+q)/2)? > > For standard rationals, absolutely. > If the rationals, standard or not, are to be an ordered field, then it > will be the case that for every two distinct rational numbers, say p > and q, that there is another rational number strictly between them (for > example (p+q)/2). Since this is true for EVERY ordered field, it will be > true for every ordered field. Okay. So? If two values can be distinguished as hyperrationals but not as standard rationals, then they aren't distinct as standard rationals and need no intermediate standard rational. > > Question: If the answer above is yes, does that not require the > rationals numbers to be a dense set? > > The rationals are most certainly dense in the reals. > That is not what was asked. I never said they weren't, or tried to change the definition of dense. What's your point? > > I have a different theory about the > rationals and how they relate to the reals. > > In TO's case, whenever his theories differ from the standard ones, they > seem to be wrong. Provably wrong in every current axiom system for set > theory. > > Agreed. that's the way they seem to you, and that's understandable, given > what > you have studied, since they contradict axioms of the standard theory and > therefore are provably worng, assuming those axioms. > > > So until TO comes up with an axiom system of his own, they remain wrong. > > They remain unprovable, and yet, eminently correct. > And Just how does TO KNOW that they are correct if he cannot prove them? > Arrogance does not justify such certainty. This is an inductive process. The cake is baking. You get to lick the bowl. Hey, arrogance is still kind of fun, even if it doesn't justify anything. ;) > > To much prefers theories in which nothing can be proved, or disproved! > > I prefer theories that actually have something useful to say. > Of what use are they until they are provable? > What bridges have TO's theories built? Well, the one between the continuum and the counting numbers, for one. > then (x+y)/2 is a real number in that model strictly between them. > > Betweenness is the only issue here, scale is not an issue. > > Yes, but as you said, if two points are not distiguishable on a given scale > I never said anything like that. What I said is that if two points are > 'Indistinguishable they are equal, and scales are irrelevant. (sigh) > Which is irrelevant, since the relevant definitions do not mention > distinguishability, only ordering. > > > > By having the image of two originally successive elements still without > anything between them. > > But on that scale, they are not distinct, as the rule for denseness requires. > Distinct means not equal in the sense to the law of trichotomy. Right, that's what I said above. Equal means not distinct. > Does To claim that if x < y on some scale that there is any scale on > which x = y? If not, his arguments all fail. Yes, that's what I am claiming. When the difference between two values is 0 times the unit of measure, then there is no measurable difference, and no distinction between what appear to be coincident points. If the unit of measure is some finite fraction or multiple of the difference between them, then they are distinct. So, yes, that's what I am saying. The scale of measure can affect the distinction between points. > > > > That is what TO is claiming his scaling can do. > > All scalings of an ordered set are order isomorphisms which leave > EVERY order property, including denseness or discreteness, > unchanged. That is what order isomorphism is all about, preserving > ALL order properties. > > > Are you sure that aplies in the infinite case? > > EVERY case! > > Uh, are you sure? Methinks not. > O will accept your assuranced that you do not think. One of few > statements of yours can accept. Are you having trouble saying I? Quick, look in the mirror. Do any parts of you look at all transparent? You haven't been taking balls out your vase again, have you? I TOLD you that would make you disappear!! Now, go renumber them right away before you're entirely gone. Chop chop!! > When scaling by an infinite value, the distinguishable may become > indistinguishable. > > Then you are mapping two, or more, elements onto 1 element. Such > mappings are not invertable. Changes of scale must be invertable. > Ergo: TO's mappings are NOT changes of scale. > > I am doing no such thing. > Compressing the 'distance' between two points to zero, which is exactly > what TO is doing, makes them the same point. But compressing the distance to an infinitesimal distance makes the points indistinguishable using a finite unit of measure, while still distinct on the infinitesimal level. See? No? Oh well. I tried. Sleep on it. -- Smiles, Tony === Subject: Re: Well Ordering the Reals > Virgil said: > There is nothing about an order relation that requires it to have > any thing like a scale, so there is no necessity that an ordered > set have any scale at all. > No, but one can apply notions of scale to sets with measurable > differences between elements. Not if it requires violating the order properties of the set. > > If, on a different scale, > > > How does an ordered set have a different scale, if there is no > original scale to be different from? > If you are dealing with sets of quantities, those quantites are > expressed in units If you are dealing only with ordered sets, their members need not be quantities in any sense. > > > >One can't measure > infinitesimal intervals in finite units, or one gets zeroes. > > In actual measurements, there is no such thing as > infinitesimal. Actual meassurement occurs only in the physical > world whereas infinitesimals occur, if at all, only in a purely > mathematical non-physical world. > > > Right, and standard mathematics deals only with finite quantities > and differences, so infinitesimal differences do not exist, and > subdivision thereof is a moot point. > > Non-standard analysis is a part of standard mathematics, and it does > have infinitesimals. > Standard analysis, then. Non-standard analysis is not part of > standard analysis. Why is TO so eager to reject non-standard analysis, which deals sensible with infinitesimals, in order to introduce his own pseudo-analysis, which doesn't? > No, you're projecting your overly finitist view onto me, which is > a pretty good joke. > > It is not my view, but the view of all mathematicians. > Most, but not all. Is the pursuit of truth a democratic process? Possibly, but the finding of truth is not at all democratic, it is arrantly aristocratic. > > > If mathematical density of a set depends on the Archimedean > principle, > and that depends on the distinction between two points, and that > distinction disappears at infinite scales > > If two points become one then one does not have a bijective > mapping, and one does not have a change of scdale. All changes of > scale are bijections by definition. > > If two values which are different by some finite number of some unit > of measure are viewed at a relatively infinite scale, using a > relatively infinite unit of measure, the difference between them can > no longer be expressed as a finite number of the current units. On > that scale, they are indstinguishable. Unless they are equal, they are not equal, and if not equal, they are distinguishable. That their distinguishing characteristics are obscured by distance does not erase those characteristics. Any differences still exist however obscured. > > > If two values are not equal at the infinitesimal level, they > are not equal at any level. The most one can ever say of > different quantities that are infinitesimally close is that > they are infinitesimally close. If TO wants to formalize that, > best he should read some more of Robinson. > > > Good advice, and I'll try to find time to start again on that. In > the meantime, I recommend you consider under what circumstances a > dense set is required to have intermediate values. > > That's easy. Under all conditions, a dense set is required to have > values intermediate between any two of its values, since that is > the quality that allows us to call it dense. > > Perhaps it is TO who is being dense in a different sense of that > word. > > Huh? Whaddya mean? That went right over my head. Do I look fat in > these pants? The fat is located somewhat higher thatn TO's pants. > And what is the difference between 0.999... and 1? The same as the difference between 1/2 and 3/6, they are different representations of a single real value. > > > > > If a set is continuous when there always exists an > intermediate value between any two distinct values > > That is not what makes a set continuous. The rationals have > that property of density but are NOT continuous. TO again knows > not wherof he speaks. > > > Basically, as I understand it, a continuous function going from > value x to value y passes through each value between the two > values, and the difference between f(x) and f(y) has a limit of 0 > as x->y. > > Then TO clearly does NOT understand it. > > > > I could be wrong. > > TO IS wrong! > Then explain why. See, for example, Calculus by Tom Apostol. > > Two points which are not distinct are, in fact, not two points, > but only one. > > > Yes, on the finite scale > > On any scale, since order properties are totally unaffected by > scales. > Distinguishability is what's affected. Which is irrelevant to whether some x is or iss not equal to some y. Distinguishability is a characteristic of the viewer, not the view. > > If two points are the same for ALL intents and purposes, they are > equal. > > And if they are equal at any scale, they are equal at all scales. > Wherever they are not distinct. In which case they are no equal, and no change of scale can make them equal. > > If one cannot distinguish between two points which ARE > different, perhaps one should try another line of work. > > > If denseness of a set is determined by the existence of an > intermediate value within any finite interval > > Two errors: finite and interval. Ordered sets have no intrinsic > distances between members, so that neither finite nor interval > apply. > > Quantitative sets do. That is irrelevant to ordering of the sets. > Pray tell, what is the difference between a dense set and one > where successive elements are less different than any finite > value? > > Simple! Dense sets never have successive elements, and sets having > successive elements, no matter how close, are not dense. > Oh. And here I was thinking the set of rationals was dense, and also > countable and linearly ordered, and bijectable with the naturals. Quite right, for a change. But quite irrelevant to whether sets with successive elements can be dense. > Now I know that no dense set can have successive elements. Did TO ever think otherwise? If so, then he should thank me for setting him straight. > Did you say distinct? Very good. Now, can you distinguish two > infinitesimally different values using finite units? > > The issue is whether one can distinguish them at all. If one can at > all, they are distinct. > How tautologically correct of you! :) > Now, what determines whether one can distinguish two values at all? If they are members of an ordered set, which seems to be the current context, then if x < y or x > y then they are distinguishable. If neither is the case, then they are both indistinguishable and equal. > For any two members, say x and y, of an ordered set , one and only > one of x < y, or x = y or x > y is true. This is the law of > trichotomy. Any set for which this is not true, like the sets TO is > proposing, is NOT AN ORDERED SET. > Oh, pay attention..... > > To what? More of TO's nonsense? > Yes, every little bit of it. You'll be glad you did. :) Not bloody likely! > Uh, well, you just gave an example of a set which is both dense and > discrete, so along with the rationals, that negates your objection > the that combination above. Anyway, I am not talking about > compressing it to a point, but to a finite interval, from an infinite > one. Then all finite differences become infinitesimal, and the set > becomes dense. If x and z are successive elements of a non-dense set before your alleged compression, with x, z having no elements between them, then after compression you must either have compressed so as to make the images x' = z' or still have x' < z' with nomembers of the set between them between them. If you have x' = z', then you do not have a change of scale type of compression as it is not reversable. If you have x' < z', you do not have density, because here are two elements with nothiing between them. > > Nobody but TO can posibly see sense in such nonsense. > > Oh, I dunno 'bout that. I'm pretty sure I'm not nuts. Most nuts say that! > The mystery is why anyone would want to get to the cloud cuckooland > place TO claims he has got to. > > Perhaps in order to escape that burned out cellar hole you fellas > call Cantor's Garden. There are a lot more useful things for mathematicians in Cantor's Garden thatn in TO's cloud cuckooland. So only non-mathematicians, or anti-mathematicians, would be the more intersested in the later. > I have, and posted those problems. That TO is too dumb to > understand what his problems are is not my problem. > You mostly object to things I haven't said, or purposely obfuscate > the issue. That may be what TO claims to see, but TO is now notorious for confusing his erratic intuition with reality. > Are the reals merely ordered, or is it a quantitative set with a > specific difference between any two elements? This is the kind of > objection you bring to the table, one which entirely ignores a whole > aspect of the issue at any given time. One can regard them as merely ordered, ignoring all the arithmetical operations, and consider what properites that are deducable only from that order. And when one does that, one sees that all TO's garbage about intervals and scales is just that, garbage which does not have any effect whatsoever on the order properties of the reals. One can do the same with any ordered set, ignoring all other properties except for that ordering. And in doing so, one finds TO is again producing nothing but garbage. > > > Standard Definition of an ordered set being dense: > A totally ordered set is /dense/ if and only if, for every > two distinct members of that set, there is another member of > that set strictly between them. > > Between them in what sense, if not being a point in the interval > between them? > > To has things backwards as usual. The definitions of inteval is > dependent on betweenness, not the other way round. > > Given x < z and any other element y, not equal to either of them, > there are only 3 possibilities: z < x, or y < z or y is between x > and z. > When you say x I can't even tell what you think you're objecting to here. TO's ignorance triumphs again! > > Nothing in that definition mentions intervals or distances > between members or even hints at equality of intervals. There > is no requirement that any ordered set have any notion of > /distance/ between members. Thus there is no such thing for > arbitrary ordered sets as unit intervals or equal > intervals. > > Denseness of a set does not require that the elements be spaced > at equal intervals > > It prohibits them being spaces at equal intervals. > > > It certainly doesn't prevent them from being spaced infinitesimally, > or the rationals aren't dense. Why do you think denseness prevents > equal spacing? Depends on what one means by equal spacing. If one only means that there are subsets of the original set which are equally spaced without requiring that there ever be two elements with no others of the original set between them, such as occurs with the rationals, then there is no problem. But the usual interpretation of equal spacing is that there must be some minimal spacing with no elements of the original set between two minimally spaced elements. > > No! TO is insisting on metric spaces, so he can bring in a lot of > irrelevancies, but for order properties, any such metrics are > irrelevant. > The metrics are not irrelevant when discussing metrics. I am saying > the infinitesimal intervals are not measurable on the finite scale. > That's a matter of metrics, and so metrics are not irrelevant to the > matter. Scaling makes no sense without metrics. But order properties, such as whether x < y or not, are independent metrics, so that changes of scale can have NO effect on any order property. > > Question: Is it the case that for every two distinct rational > numbers, say p and q, that there is another rational number > strictly between them (for example (p+q)/2)? > > For standard rationals, absolutely. > > If the rationals, standard or not, are to be an ordered field, then > it will be the case that for every two distinct rational numbers, > say p and q, that there is another rational number strictly between > them (for example (p+q)/2). Since this is true for EVERY ordered > field, it will be true for every ordered field. > Okay. So? If two values can be distinguished as hyperrationals but > not as standard rationals, then they aren't distinct as standard > rationals and need no intermediate standard rational. There is no need that there be a yellow brick road between them either, which is just as relevant. The issue is whether there is a member of the original set of objects, in which they are differentiable, between them, not whether there is a member of some restricted set between them. According to TO's standards, we might just as well say the reals are not TO-dense because there are reals without any naturals between them. > > > Question: If the answer above is yes, does that not require > the rationals numbers to be a dense set? > > The rationals are most certainly dense in the reals. > > That is not what was asked. > > I never said they weren't, or tried to change the definition of > dense. What's your point? That you are avoiding answering simple questions by answering questions that were not asked. > So until TO comes up with an axiom system of his own, they > remain wrong. > > They remain unprovable, and yet, eminently correct. > > And Just how does TO KNOW that they are correct if he cannot prove > them? > > Arrogance does not justify such certainty. > > This is an inductive process. The cake is baking. You get to lick the > bowl. Cake baking is about as close to actual mathematics as TO is ever liable to get, at his present rate of progress. > Yes, that's what I am claiming. When the difference between two > values is 0 times the unit of measure, then there is no measurable > difference, and no distinction between what appear to be coincident > points. If the distance between x and y is zero tmes the unit of measure, does that make x = y, or not? If it makes x = y, then the mapping has no inverse so is not a change of scale. If it does not make x = y, then one still has one of x < y of x > y, regardless of scale. > When scaling by an infinite value, the distinguishable may > become indistinguishable. > > Then you are mapping two, or more, elements onto 1 element. > Such mappings are not invertable. Changes of scale must be > invertable. Ergo: TO's mappings are NOT changes of scale. > > I am doing no such thing. > > Compressing the 'distance' between two points to zero, which is > exactly what TO is doing, makes them the same point. > > But compressing the distance to an infinitesimal distance makes the > points indistinguishable using a finite unit of measure, while still > distinct on the infinitesimal level. See? No? Oh well. I tried. Sleep > on it. The question is, do TO's infinite change of scale ever change any strict inequality, x < y, into an equality, x' = y' ? If not, then they never change dense into non-dense nor non-dense into dense. If so, they are not invertable, and thus are not changes of scale. === Subject: generalized graph coloring I have a question on generalized graph coloring. The problem formulation is as follows: Consider a complete graph with directed edge, denoted by G(V,E). A directed edge, e in E, is assigned its weight, 0 < w_e < 1. I want to compute the chromatic number of G (i.e., vertex coloring), such that the following requirements are satisfied: Requirement: If all the vertices in a set of vertices A are to have a same color, we have to have, for any vertex v in A, sum_{v' in A, v' neq v} w_{v'v} leq 1, where w_{v'v} is the weight of the edge from v' to v. Regarding the above problem, is there any result, paper, or necessary/sufficient conditions for minimum chromatic number? Anything would be helpful to me. === Subject: Re: generalized graph coloring look at this: and plz. let me know if you come up with any conclusion www.math.uni-klu.ac.at/or/Forschung/publications/DjRe05.ps === Subject: Re: generalized graph coloring definitely this is coming from optimization and i saw something similar. is this the general question you try to answer; or how did you come up with this problem? -- ogl100@yahoo === Subject: Re: generalized graph coloring > I have a question on generalized graph coloring. > The problem formulation is as follows: > Consider a complete graph with directed edge, denoted by G(V,E). > A directed edge, e in E, is assigned its weight, 0 < w_e < 1. > I want to compute the chromatic number of G (i.e., vertex coloring), such > that the following requirements are satisfied: > Requirement: If all the vertices in a set of vertices A are to have a same > color, we have to have, for any vertex v in A, > sum_{v' in A, v' neq v} w_{v'v} leq 1, where > w_{v'v} is the weight of the edge from v' to v. > Regarding the above problem, is there any result, paper, or necessary/sufficient > conditions for minimum chromatic number? Anything would be helpful to me. (1) I don't see how this is a generalization of graph coloring. For instance, the modified chromatic number is <= n/2, since any set A with 2 elements is allowed. (2) The answer also depends on w. There is a generalized vertex coloring (which I can't remember the name of --- k-degenerate, maybe), where a coloring from V(H) to Z is k-allowed if the subgraph of H induced by the set of all vertices where c(v)=a has maximum degree <= k, for all a. Then the goal is to find the smallest number of colors needed to be k-allowed. (This topic was mentioned in a Graph Theory conference in the late 1990s, by a woman named Cohen, I think. This is a generalization of vertex coloring, where k=0. Your w-coloring is a generalization of this, where w_e = 0 (or very close to 0) if e is not an element of H, and w_e = 1/(k+1) otherwise.) This is all I can think of at the moment. --- Christopher Heckman === Subject: Help a nonMath figure out optimum values There may not be an answer to this question, but here goes... I have a variable, containing white noise (Guassian distribution?), and I can select the number of samples I use. I have a choice of finding the mean of this variable by taking some quantity of samples, summing, and finding the mean, which requires knowing the number of samples. Or, I can do a recursive average where the new average value is equal to (1-k) times the old value plus k times the new value, where 0.001 < k < 0.5, and simply go along for some number of samples until there is some assurance that the average has been calculated. This 'open' recursive form is preferred, since it allows some observations to occur before stopping the accumulation process. Plus, recursion slowly kills old data, in case a transient noise spike sneaks in. I tried to solve this empirically, by using 100, 200, and 1000 samples, combined with range of k. Plotting the difference between the mean and the running average versus k showed some very interesting, but non-conclusive, observations. For exmaple, using 100 samples the error was minimized using k equal to 0.0075 (which seemed really strange) and 0.1 I expected k waould be best if equal to something like 2 times 1 over the number of samples, or such. Doesn't seem to be the best value for k. the mean, do recursive average, compare the two. plot 'error' versus k factor. There was no obvious minimum in the accuracy. The accuracy was greatly a function of the pattern of the noise in the samples. Although random, the number of samples was often too small to really insure a random set of variables. Very unexpectedly, there seemed to be a relationship between the number of samples and the number of minimum nulls. Has anybody done work like this? Has an answer, or insight into the answer? Or, should I just blow the whole thing off and fix the number of samples and take the mean? It's just that using a fixed number of samples after a bad transient hits I have to wait a fixed amount of time to accumulate all the necessary samples. Where with recursion I can kind of see the transient, estimate how much longer to run the recursion and then cut off the calculation and be fairly certain that the mean I've just calculated is good. I'm no a mathematician, so please bear with me here. - Robert - === Subject: Re: Help a nonMath figure out optimum values > There may not be an answer to this question, but here goes... > I have a variable, containing white noise (Guassian distribution?), and > I can select the number of samples I use. I have a choice of finding > the mean of this variable by taking some quantity of samples, summing, > and finding the mean, which requires knowing the number of samples. > Or, I can do a recursive average where the new average value is equal > to (1-k) times the old value plus k times the new value, where 0.001 < > k < 0.5, and simply go along for some number of samples until there is > some assurance that the average has been calculated. I know this as exponential averaging. It's also a form of low-pass digital filtering. I'm having trouble understanding the problem you're trying to solve though. You want to know the right number of samples to use, I gather. Right for what? What criterion are you trying to satisfy? > This 'open' recursive form is preferred, since it allows some > observations to occur before stopping the accumulation process. Plus, > recursion slowly kills old data, in case a transient noise spike sneaks > in. > I tried to solve this empirically, by using 100, 200, and 1000 samples, > combined with range of k. Plotting the difference between the mean and > the running average versus k showed some very interesting, but > non-conclusive, observations. For exmaple, using 100 samples the error > was minimized using k equal to 0.0075 (which seemed really strange) and > 0.1 > I expected k waould be best if equal to something like 2 times 1 over > the number of samples, or such. Doesn't seem to be the best value for > k. > the mean, do recursive average, compare the two. plot 'error' versus k > factor. > There was no obvious minimum in the accuracy. The accuracy was greatly > a function of the pattern of the noise in the samples. Yes, the sample mean is a random variable, whatever your sampling and averaging procedure. So there will be fluctuations. > Has anybody done work like this? Has an answer, or insight into the > answer? I've seen any number of things like this. In the theory of stochastic processes, this could be posed as an optimum stopping time problem, for which there is plenty of theory. The problem is that optimum doesn't seem well-defined here. > Or, should I just blow the whole thing off and fix the number of > samples and take the mean? > It's just that using a fixed number of samples after a bad transient > hits I have to wait a fixed amount of time to accumulate all the > necessary samples. Where with recursion I can kind of see the > transient, estimate how much longer to run the recursion and then cut > off the calculation and be fairly certain that the mean I've just > calculated is good. I suspect what you're going to look for is a stopping rule which is in terms of the amount of fluctuation over the last few samples and which will vary from experiment to experiment. You are after all dealing with random variables here, and your stopping time is itself a rv. - Randy === Subject: Re: Help a nonMath figure out optimum values > There may not be an answer to this question, but here goes... > I have a variable, containing white noise (Guassian distribution?), and > I can select the number of samples I use. I have a choice of finding > the mean of this variable by taking some quantity of samples, summing, > and finding the mean, which requires knowing the number of samples. > Or, I can do a recursive average where the new average value is equal > to (1-k) times the old value plus k times the new value, where 0.001 < > k < 0.5, and simply go along for some number of samples until there is > some assurance that the average has been calculated. > This 'open' recursive form is preferred, since it allows some > observations to occur before stopping the accumulation process. Plus, > recursion slowly kills old data, in case a transient noise spike sneaks > in. > I tried to solve this empirically, by using 100, 200, and 1000 samples, > combined with range of k. Plotting the difference between the mean and > the running average versus k showed some very interesting, but > non-conclusive, observations. For exmaple, using 100 samples the error > was minimized using k equal to 0.0075 (which seemed really strange) and > 0.1 > I expected k waould be best if equal to something like 2 times 1 over > the number of samples, or such. Doesn't seem to be the best value for > k. > the mean, do recursive average, compare the two. plot 'error' versus k > factor. > There was no obvious minimum in the accuracy. The accuracy was greatly > a function of the pattern of the noise in the samples. Although > random, the number of samples was often too small to really insure a > random set of variables. > Very unexpectedly, there seemed to be a relationship between the number > of samples and the number of minimum nulls. > Has anybody done work like this? Has an answer, or insight into the > answer? > Or, should I just blow the whole thing off and fix the number of > samples and take the mean? > It's just that using a fixed number of samples after a bad transient > hits I have to wait a fixed amount of time to accumulate all the > necessary samples. Where with recursion I can kind of see the > transient, estimate how much longer to run the recursion and then cut > off the calculation and be fairly certain that the mean I've just > calculated is good. > I'm no a mathematician, so please bear with me here. > - Robert - Would it be possible to use a different recursion, viz if your samples are X[1],X[2]... M[1] = X[1] M[n+1] = M[n] + (X[n+1]-M[n])/(n+1) Then M[n] is the average of X[1]..X[n] Duncan === Subject: Re: Help a nonMath figure out optimum values >There may not be an answer to this question, but here goes... >I have a variable, containing white noise (Guassian distribution?), and >I can select the number of samples I use. I have a choice of finding >the mean of this variable by taking some quantity of samples, summing, >and finding the mean, which requires knowing the number of samples. >Or, I can do a recursive average where the new average value is equal >to (1-k) times the old value plus k times the new value, where 0.001 < >k < 0.5, and simply go along for some number of samples until there is >some assurance that the average has been calculated. >This 'open' recursive form is preferred, since it allows some >observations to occur before stopping the accumulation process. Plus, >recursion slowly kills old data, in case a transient noise spike sneaks >in. >I tried to solve this empirically, by using 100, 200, and 1000 samples, >combined with range of k. Plotting the difference between the mean and >the running average versus k showed some very interesting, but >non-conclusive, observations. For exmaple, using 100 samples the error >was minimized using k equal to 0.0075 (which seemed really strange) and >0.1 I hope I understand you correctly. You take, let's say, N independent random variables X_n, n=1..N, having the same normal distribution; let's say the mean is mu and variance 1. You calculate a running average by R_1 = X_1, R_{n+1} = (1-k) R_n + k X_n. Thus you end up with R_N = (1-k)^(N-1) X_1 + sum_{j=2}^N k (1-k)^(N-j) X_j. Now R_N has a normal distribution with mean mu and variance V(k) = (1-k)^(2N-2) + sum_{j=2}^N k^2 (1-k)^(2N-2j) = k/(2-k) + 2 (1-k)^(2N-1)/(2-k). You'd like to minimize this. There won't be a closed-form solution, but I believe that as N -> infinity the minimum is approximately at k = ln(4N)/(2N). For N = 100 this would be approximately k = 0.0299573, while the actual minimum (obtained numerically) is approximately k = 0.02969666. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Name of a conjecture? I just need my memory jogged: what is the name of the conjecture (now disproved) that no factors of x^n-1 have coefficients with absolute value greater than 1? Alasdair === Subject: Re: Name of a conjecture? > I just need my memory jogged: what is the name of the conjecture (now > disproved) that no factors of x^n-1 have coefficients with absolute > value greater than 1? I don't think it ever had a name. Someone published it around 1940, and it got shot down almost immediately. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Induced Representations Theorem..Part 2 I posted about the following theorem before : Suppose G is a group and H is a subgroup of index 2. Let f be an irreducible representation of H. Let t be an automorphism of H coming from G - H (H is normal so conjugation of H is an automorphism). t acts on the irreducible representations of H by t*g(x) = g(t(x)) = g(txt^(-1)). Then, Theorem : Ind_H^G (f) is an irreducible representation P if t*f is not isomorphic to f, and it is the direct sum of two irreducible representations P_1 (+) P_2 if t*f is isomorphic to f. I showed (as I was advised to do) that Ind_H^G (f) restricted to H is the direct sum of the representations f and t*f of H, and so the theorem fell out pretty quickly using Frobenius reciprocity. But now my goal is to show that P (x) sgn is isomorphic to P and P_1 (x) sgn is isomorphic to P_2 where sgn : G -----> G / H = Z/2Z How can I do this? I'm not even sure what this means really. sgn is a 1-dimensional representation? Can I use character theory to do this? I'm not quite sure what P (x) sgn and P_1 (x) sgn are. I know of course what tensor products of representations are, but I am a bit confused here with respect to what sgn is. I mean I know what the map sgn : G ---> G / H is, but with this definition I can't see that sgn is a representation. I'm sort of confused.. James === Subject: Re: Induced Representations Theorem..Part 2 du0g7k$530l$1@netnews.upenn.edu... >I posted about the following theorem before : > Suppose G is a group and H is a subgroup of index 2. Let f be an > irreducible representation of H. Let t be an automorphism of H coming > from G - H (H is normal so conjugation of H is an automorphism). t acts > on the irreducible representations of H by t*g(x) = g(t(x)) = g(txt^(-1)). > Then, > Theorem : Ind_H^G (f) is an irreducible representation P What is Ind_H^G (f) ? Repr.8esentation P of ? H? === Subject: Re: Induced Representations Theorem..Part 2 [G is a group and H is a subgroup of index 2] : How can I do this? I'm not even sure what this means really. sgn is a : 1-dimensional representation? Can I use character theory to do this? I'm : not quite sure what P (x) sgn and P_1 (x) sgn are. I know of course what : tensor products of representations are, but I am a bit confused here with : respect to what sgn is. I mean I know what the map sgn : G ---> G / H is, : but with this definition I can't see that sgn is a representation. I'm sort : of confused.. G/H is a group of order 2, which can be identified with {1,-1}. So sgn(s) can be regarded as taking the values 1 or -1, which are elements of C*, so sgn can be treated as a 1-dimensional representation. Ted === Subject: Exchange order of integration and summation Folks, I clearly slept through class or just missed something, but this keeps coming up. When is it possible to exchange the order of integration and summation? For finite series this is a boring question, but what about infinite series? Assume that my function f is complex valued and Reimann integrable. Under what conditions does integral (sum_{ - infty}^{ infty} f) = sum_{ - infty}^{ infty} ( integral f) For example (but to warn you in advance, this is homework), I'm trying to integrate the Poisson Kernel, defined here as sum_{ n = -infty}^{ infty} f(x) with f(x) = r ^{ |n|} exp{ inx} Anyway, none of the texts I have available (stewart, spivak, or apostol) seem to cover what I want to do; I've seen people do it but am unsure under what conditions (continuity / convergence / etc) this is kosher. === Subject: Re: Exchange order of integration and summation > Folks, > I clearly slept through class or just missed something, but this keeps > coming up. When is it possible to exchange the order of integration > and summation? > For finite series this is a boring question, but what about infinite > series? Assume that my function f is complex valued and Reimann > integrable. > Under what conditions does > integral (sum_{ - infty}^{ infty} f) = sum_{ - infty}^{ infty} ( > integral f) > For example (but to warn you in advance, this is homework), I'm trying > to integrate the Poisson Kernel, defined here as > sum_{ n = -infty}^{ infty} f(x) with > f(x) = r ^{ |n|} exp{ inx} > Anyway, none of the texts I have available (stewart, spivak, or > apostol) seem to cover what I want to do; Spivak, CALCULUS, page 495, Theorem 1. page 498, Corollary. > I've seen people do it but am > unsure under what conditions (continuity / convergence / etc) this is > kosher. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Exchange order of integration and summation <280220060812018846%edgar@math.ohio-state.edu.invalid> === Subject: Re: Exchange order of integration and summation Weierstrass' criterion of uniform convergence === Subject: Exchange order of integration and summation Folks, I clearly slept through class or just missed something, but this keeps coming up. When is it possible to exchange the order of integration and summation? For finite series this is a boring question, but what about infinite series? Assume that my function f is complex valued and Reimann integrable. Under what conditions does integral (sum_{ - infty}^{ infty} f) = sum_{ - infty}^{ infty} ( integral f) For example (but to warn you in advance, this is homework), I'm trying to integrate the Poisson Kernel, defined here as sum_{ n = -infty}^{ infty} f(x) with f(x) = r ^{ |n|} exp{ inx} Anyway, none of the texts I have available (stewart, spivak, or apostol) seem to cover what I want to do; I've seen people do it but am unsure under what conditions (continuity / convergence / etc) this is kosher. === Subject: Re: Exchange order of integration and summation >Folks, >I clearly slept through class or just missed something, but this keeps >coming up. When is it possible to exchange the order of integration >and summation? >For finite series this is a boring question, but what about infinite >series? Assume that my function f is complex valued and Reimann >integrable. >Under what conditions does >integral (sum_{ - infty}^{ infty} f) = sum_{ - infty}^{ infty} ( >integral f) The notation makes no sense - you meant integral (sum_{ - infty}^{ infty} f_n) = sum_{ - infty}^{ infty} ( integral f_n) (one talks about the sum of a sequence of functions, not the sum of a function). >For example (but to warn you in advance, this is homework), I'm trying >to integrate the Poisson Kernel, defined here as >sum_{ n = -infty}^{ infty} f(x) with >f(x) = r ^{ |n|} exp{ inx} Here for example f_n(x) = r ^{ |n|} exp{ inx}. >Anyway, none of the texts I have available (stewart, spivak, or >apostol) seem to cover what I want to do; I've seen people do it but am >unsure under what conditions (continuity / convergence / etc) this is >kosher. The simplest criterion would be that the partial sums converge uniformly. And the simplest way to verify that is by what's sometimes called the Weierstrass M-test: If there is a sequence of constants M_n > 0 with sum M_n < infinity, such that |f_n(x)| <= M_n for all x, then the partial sums of sum f_n converge uniformly. That's a fairly strong hypothesis, but it works in the specific example you say you're wondering about. ************************ David C. Ullrich === Subject: JSH: So you see, they never faced proofs Like I say, I have proofs. By the rules mathematicians follow proofs. What happens with my results is that most mathematicians ignore the proofs, while posters who argue with me, avoid them. They avoid proofs like the plague and will ALWAYS eventually make up their own weird thing versus face the proofs I give. None of these people have ever faced down the actual proofs, but have always distracted to some other crap, but this time I caught them by forcing them into the freaking complex plane so you can see exactly what they do. And how they do it. People, you do NOT just get a paper published in a mathematical journal in this day and age. If you think it's so freaking easy then some of you try. That math journal did not just want to keel over and die and those editors were not looking to embarrass themselves. Yes my paper had some typos and small mistakes, but the proof within it showed through. Posters like Nora Baron, William Hughes, Arturo Magidin, W. Dale Hall, and others never faced the proofs. They just played like they did, and made up stuff, just to play the social game. I couldn't believe it when Nora Baron trapped at one point by an argument where I shut down all room for objections, just deleted everything out. They cannot face proofs so they must distract. These people are just the unimportant piece though. No mathematician of any significant ability is going to get lost on that point requiring the distributive property. I mean, well, it's the freaking distributive property. How can you be a mathematician if you don't have that down? Nope. They are quiet now because my results take away ideal theory, and my results take away the standard usage of Galois Theory--they don't take away Galois Theory, just the standard usage--and they are PROTECTING THEIR CAREERS. IF you wish to be some person who works hard to learn false mathematical ideas that do not work then just play along with the stupid game, but you are the loser. If the mathematics does not work, it does not work, and I don't care if a billion people chant day and night that it must be true, the mathematics DOES NOT CARE. You people are not mathematicians. You do not respect the truth. You do not care about the axioms. You cannot be convinced by proof. James Harris === Subject: Re: JSH: So you see, they never faced proofs > Like I say, I have proofs. > By the rules mathematicians follow proofs. > What happens with my results is that most mathematicians ignore the > proofs, while posters who argue with me, avoid them. > They avoid proofs like the plague and will ALWAYS eventually make up > their own weird thing versus face the proofs I give. > None of these people have ever faced down the actual proofs, but have > always distracted to some other crap, but this time I caught them by > forcing them into the freaking complex plane so you can see exactly > what they do. > And how they do it. > People, you do NOT just get a paper published in a mathematical journal > in this day and age. > If you think it's so freaking easy then some of you try. > That math journal did not just want to keel over and die and those > editors were not looking to embarrass themselves. > Yes my paper had some typos and small mistakes, but the proof within it > showed through. > Posters like Nora Baron, William Hughes, Arturo Magidin, W. Dale > Hall, and others never faced the proofs. They just played like they > did, and made up stuff, just to play the social game. > I couldn't believe it when Nora Baron trapped at one point by an > argument where I shut down all room for objections, just deleted > everything out. > They cannot face proofs so they must distract. > These people are just the unimportant piece though. > No mathematician of any significant ability is going to get lost on > that point requiring the distributive property. > I mean, well, it's the freaking distributive property. > How can you be a mathematician if you don't have that down? > Nope. They are quiet now because my results take away ideal theory, > and my results take away the standard usage of Galois Theory--they > don't take away Galois Theory, just the standard usage--and they are > PROTECTING THEIR CAREERS. > IF you wish to be some person who works hard to learn false > mathematical ideas that do not work then just play along with the > stupid game, but you are the loser. > If the mathematics does not work, it does not work, and I don't care if > a billion people chant day and night that it must be true, the > mathematics DOES NOT CARE. > You people are not mathematicians. You do not respect the truth. You > do not care about the axioms. > You cannot be convinced by proof. > James Harris Every mathematician has faced proofs of every type, that's part of our training as mathematicians. You're wasting your life on useless crap and drivel. Why do I say this? You've been at this for 10 years. By now, you could have gotten a Ph.D or at least learned some math. Dave === Subject: Re: JSH: So you see, they never faced proofs > Every mathematician has faced proofs of every type, that's part of our > training as mathematicians. You're wasting your life on useless crap and > drivel. Why do I say this? You've been at this for 10 years. By now, you > could have gotten a Ph.D or at least learned some math. Your last sentence is debatable. === Subject: Re: JSH: So you see, they never faced proofs > Like I say, I have proofs. No, you don't. === Subject: Re: JSH: So you see, they never faced proofs > People, you do NOT just get a paper published in a mathematical journal > in this day and age. > If you think it's so freaking easy then some of you try. > That math journal did not just want to keel over and die and those > editors were not looking to embarrass themselves. > Yes my paper had some typos and small mistakes, but the proof within it > showed through. You see, I'd rather be wrong. I hate that idea that Galois Theory and the theory of ideals and all of that which people thought was so important could be wrong. Part of me really hopes I am wrong, but I guess I'm more afraid that I'm right. So, only three days ago you thought that the proofs from your paper were not really proofs. What has changed in these three days? Jose Carlos Santos === Subject: Re: JSH: So you see, they never faced proofs : So, only three days ago you thought that the proofs from your paper were : not really proofs. What has changed in these three days? His blood-alcohol level? Justin === Subject: Re: So you see, they never faced proofs > What happens with my results is that most mathematicians ignore the > proofs, while posters who argue with me, avoid them. > They avoid proofs like the plague and will ALWAYS eventually make up > their own weird thing versus face the proofs I give. buuuu-WEEP! Crank alert! All mathematicians to battle stations! === Subject: Re: So you see, they never faced proofs > Like I say, I have proofs. I have nevr seen you give a proof. === Subject: Re: So you see, they never faced proofs >> Like I say, I have proofs. > I have nevr seen you give a proof. da drunk donoot kknow what a poof really is!!!!!!!!!!!!! === Subject: Re: JSH: So you see, they never faced proofs > Like I say, I have proofs. > By the rules mathematicians follow proofs. > What happens with my results is that most mathematicians ignore the > proofs, while posters who argue with me, avoid them. > They avoid proofs like the plague and will ALWAYS eventually make up > their own weird thing versus face the proofs I give. > None of these people have ever faced down the actual proofs, but have > always distracted to some other crap, but this time I caught them by > forcing them into the freaking complex plane so you can see exactly > what they do. > And how they do it. > People, you do NOT just get a paper published in a mathematical journal > in this day and age. > If you think it's so freaking easy then some of you try. > That math journal did not just want to keel over and die and those > editors were not looking to embarrass themselves. > Yes my paper had some typos and small mistakes, but the proof within it > showed through. > Posters like Nora Baron, William Hughes, Arturo Magidin, W. Dale > Hall, and others never faced the proofs. They just played like they > did, and made up stuff, just to play the social game. > I couldn't believe it when Nora Baron trapped at one point by an > argument where I shut down all room for objections, just deleted > everything out. > They cannot face proofs so they must distract. > These people are just the unimportant piece though. > No mathematician of any significant ability is going to get lost on > that point requiring the distributive property. > I mean, well, it's the freaking distributive property. > How can you be a mathematician if you don't have that down? > Nope. They are quiet now because my results take away ideal theory, > and my results take away the standard usage of Galois Theory--they > don't take away Galois Theory, just the standard usage--and they are > PROTECTING THEIR CAREERS. > IF you wish to be some person who works hard to learn false > mathematical ideas that do not work then just play along with the > stupid game, but you are the loser. > If the mathematics does not work, it does not work, and I don't care if > a billion people chant day and night that it must be true, the > mathematics DOES NOT CARE. > You people are not mathematicians. You do not respect the truth. You > do not care about the axioms. > You cannot be convinced by proof. Apparently, I've run tard into the ground. Any suggestion for a replacement? > James Harris === Subject: Least Square fitting I am have a problem with trying to figure out how my book came up with a solution to an odd number problem, it is stated as follows, A student hangs masses on a spring and measures the spring's extension as a function of the applied force in order to find the spring constant k. Her measurements are: (mass (kg), Extension(cm)) = ( 200 , 5.1 ) , ( 300 , 5.5 ) , ( 400 , 5.9 ) , ( 500 , 6.8 ) , ( 600 , 7.4 ) , ( 700 , 7.5) , ( 800 , 8.6 ) , ( 900 , 9.4 ) There is an uncertainty of .2 in each measurement of the extension. The uncertainty in the mass is negligible. For a perfect spring, the extension DeltaL of the spring will be related to the applied force by the relation k(DeltaL)=mg, where DeltaL=L-L0 and L0 is the unstretched length of the spring. Use these data and the method of least squares to find the spring constant k, the unstretched length of the spring L0, and their uncertainties. Find Chi-Squrare for the fit and the associated probability. I first approached this problem by setting up the equation k(DeltaL)=mg so it can be used to solve for L0 and K via the least square method, since it sounds like that is how the problem wants us to solve for L0 and K. The result was L=g/k*m+L0. However, I quickly ran into a problem since the measurements the problem gives you are the extensions and not the total length; thus, you cannot use my equation to solve for L0 and k via the least square method with the measured values given. I next consider the possibility that they just wanted you to solve for k via the least square method and then use the resulting value given to solve for L0. I switched the equation around and came up with (DeltaL)=g/k*m. I applied the least square method to solve for g/k and came up with a value and its uncertainty. I next tired setting up a system of linear equations to solve for L0, but I quickly found myself going in circles since there really is no way to setup a system of equations to solve for L0 (or at least that I saw). What am I over looking? Could you guys help steer me in the right direction, because I cannot figure out how to solve for L0 but I can solve for k as I told you above, and the back of the book is no help since it only guess the resulting answer and now how it was gotten to. === Subject: Re: Least Square fitting > I am have a problem with trying to figure out how my book came up with > a solution to an odd number problem, it is stated as follows, A > student hangs masses on a spring and measures the spring's extension as > a function of the applied force in order to find the spring constant k. > Her measurements are: > (mass (kg), Extension(cm)) = ( 200 , 5.1 ) , ( 300 , 5.5 ) , ( 400 , > 5.9 ) , ( 500 , 6.8 ) , ( 600 , 7.4 ) , ( 700 , 7.5) , ( 800 , 8.6 ) , > ( 900 , 9.4 ) > There is an uncertainty of .2 in each measurement of the extension. The > uncertainty in the mass is negligible. For a perfect spring, the > extension DeltaL of the spring will be related to the applied force by > the relation k(DeltaL)=mg, where DeltaL=L-L0 and L0 is the unstretched > length of the spring. Use these data and the method of least squares to > find the spring constant k, the unstretched length of the spring L0, > and their uncertainties. Find Chi-Squrare for the fit and the > associated probability. > I first approached this problem by setting up the equation k(DeltaL)=mg > so it can be used to solve for L0 and K via the least square method, > since it sounds like that is how the problem wants us to solve for L0 > and K. The result was L=g/k*m+L0. However, I quickly ran into a problem > since the measurements the problem gives you are the extensions and not > the total length I was confused in the same way by the word extension, but I don't think your interpretation is correct. I examined the data to see what they might mean by extension. Is the extension for 400 kg about twice that for 200 kg? No, it's 5.9 cm versus 5.1 cm, only about 0.8 cm farther. If we were to simply guess that this data represents total length and L0 is around 4.3 cm then we'd expect that the extension for 600 kg is about 4.3 + 3*0.8 = 4.3 + 2.4 = 6.7. The actual data value is 7.4, in the right ball park. Note also that the data point for 800 kg is about 0.8 cm beyond that for 600 kg. This leads me to conclude that these data represent total length L and that a least squares fit will yield an intercept of L0 and a slope of g/k, as you propose above. Also that the fitted values of L0 and g/k shouldn't be far off from 4.3 cm and 0.8 cm per 200 kg, or 0.004 cm/kg. - Randy === Subject: Re: Least Square fitting If k(DeltaL) = m*g and DeltaL = L - L0 then L = L0 + m*g/k. Use standard least-squares fitting procedures to estimate L0 and g/k. > I am have a problem with trying to figure out how my book came up with > a solution to an odd number problem, it is stated as follows, A > student hangs masses on a spring and measures the spring's extension as > a function of the applied force in order to find the spring constant k. > Her measurements are: > (mass (kg), Extension(cm)) = ( 200 , 5.1 ) , ( 300 , 5.5 ) , ( 400 , > 5.9 ) , ( 500 , 6.8 ) , ( 600 , 7.4 ) , ( 700 , 7.5) , ( 800 , 8.6 ) , > ( 900 , 9.4 ) > There is an uncertainty of .2 in each measurement of the extension. The > uncertainty in the mass is negligible. For a perfect spring, the > extension DeltaL of the spring will be related to the applied force by > the relation k(DeltaL)=mg, where DeltaL=L-L0 and L0 is the unstretched > length of the spring. Use these data and the method of least squares to > find the spring constant k, the unstretched length of the spring L0, > and their uncertainties. Find Chi-Squrare for the fit and the > associated probability. > I first approached this problem by setting up the equation k(DeltaL)=mg > so it can be used to solve for L0 and K via the least square method, > since it sounds like that is how the problem wants us to solve for L0 > and K. The result was L=g/k*m+L0. However, I quickly ran into a problem > since the measurements the problem gives you are the extensions and not > the total length; thus, you cannot use my equation to solve for L0 and > k via the least square method with the measured values given. I next > consider the possibility that they just wanted you to solve for k via > the least square method and then use the resulting value given to solve > for L0. I switched the equation around and came up with > (DeltaL)=g/k*m. I applied the least square method to solve for g/k and > came up with a value and its uncertainty. I next tired setting up a > system of linear equations to solve for L0, but I quickly found myself > going in circles since there really is no way to setup a system of > equations to solve for L0 (or at least that I saw). What am I over > looking? Could you guys help steer me in the right direction, because I > cannot figure out how to solve for L0 but I can solve for k as I told > you above, and the back of the book is no help since it only guess the > resulting answer and now how it was gotten to. === Subject: Re: Least Square fitting method but ran into a problem fairly quickly since the lengths given are of the extension and not the total length. Or, is there something I am over looking? === Subject: Re: Least Square fitting > method but ran into a problem fairly quickly since the lengths given > are of the extension and not the total length. Or, is there something > I am over looking? Well, the plot certainly looks like the intercept is nonzero, which would imply that the given lengths are totals, not extensions. Could the problem be with getting from g/k to k? === Subject: Re: Least Square fitting > I am have a problem with trying to figure out how my book came up with > a solution to an odd number problem, it is stated as follows, A > student hangs masses on a spring and measures the spring's extension as > a function of the applied force in order to find the spring constant k. > Her measurements are: > (mass (kg), Extension(cm)) = ( 200 , 5.1 ) , ( 300 , 5.5 ) , ( 400 , > 5.9 ) , ( 500 , 6.8 ) , ( 600 , 7.4 ) , ( 700 , 7.5) , ( 800 , 8.6 ) , > ( 900 , 9.4 ) > There is an uncertainty of .2 in each measurement of the extension. The > uncertainty in the mass is negligible. For a perfect spring, the > extension DeltaL of the spring will be related to the applied force by > the relation k(DeltaL)=mg, where DeltaL=L-L0 and L0 is the unstretched > length of the spring. Use these data and the method of least squares to > find the spring constant k, the unstretched length of the spring L0, > and their uncertainties. Find Chi-Squrare for the fit and the > associated probability. > I first approached this problem by setting up the equation k(DeltaL)=mg > so it can be used to solve for L0 and K via the least square method, > since it sounds like that is how the problem wants us to solve for L0 > and K. The result was L=g/k*m+L0. However, I quickly ran into a problem > since the measurements the problem gives you are the extensions and not > the total length; thus, you cannot use my equation to solve for L0 and > k via the least square method with the measured values given. I next > consider the possibility that they just wanted you to solve for k via > the least square method and then use the resulting value given to solve > for L0. I switched the equation around and came up with > (DeltaL)=g/k*m. I applied the least square method to solve for g/k and > came up with a value and its uncertainty. I next tired setting up a > system of linear equations to solve for L0, but I quickly found myself > going in circles since there really is no way to setup a system of > equations to solve for L0 (or at least that I saw). What am I over > looking? Could you guys help steer me in the right direction, because I > cannot figure out how to solve for L0 but I can solve for k as I told > you above, and the back of the book is no help since it only guess the > resulting answer and now how it was gotten to. I haven't worked this through but I think the measurements are the total length L If you double the mass then you should double the extension, but as you go from 200 (is that really Kg) to 400 to 800 the extension does not double so it is probably the length. === Subject: Re: Least Square fitting the data and fit a rough curve to it and found that the intercept of the line goes through the point (roughly) 3.6 and not the 0. This would lend support to your idea that the measurements of length are of the total length and not just the extension. However, that being said, can you think of any reason why this might not be the case? === Subject: quick fix Does anyone know the trick to quickly taking the imp. diff. of xy = cot(xy) in only a coulpe of moves? === Subject: Re: quick fix > Does anyone know the trick to quickly taking the imp. diff. of xy = > cot(xy) in only a coulpe of moves? A smooth one move quick trick y + xy' = -(csc xy)(y + xy') which reduces, with the quick trick of ignoring division by zero, to sin xy = -1 === Subject: Re: quick fix On Mon, 27 Feb 2006 22:01:42 -0800, William Elliot >> Does anyone know the trick to quickly taking the imp. diff. of xy = >> cot(xy) in only a coulpe of moves? >A smooth one move quick trick > y + xy' = -(csc xy)(y + xy') >which reduces, with the quick trick of ignoring division by zero, to > sin xy = -1 But d/dx(cot(u)) = - csc^2(u) du/dx (you missed the squared) --Lynn === Subject: Re: quick fix >On Mon, 27 Feb 2006 22:01:42 -0800, William Elliot > Does anyone know the trick to quickly taking the imp. diff. of xy = > cot(xy) in only a coulpe of moves? >>A smooth one move quick trick >> y + xy' = -(csc xy)(y + xy') >>which reduces, with the quick trick of ignoring division by zero, to >> sin xy = -1 >But d/dx(cot(u)) = - csc^2(u) du/dx (you missed the squared) >--Lynn Actually, the problem is more interesting than revealed at first glance. Repairing the step above yields: x + xy' = - csc^2(xy) (y + xy') which only works if y + xy' = 0, giving y' = - y/x. Assuming x and y are positive for simplicity, solving this by separation of variables yields a family of hyperbolas xy = k. So, unlikely as it may seem at first glance, the graph of xy = cot(xy) is an implicit family of hyperbolas. But then, of course, the equation u = cot(u) is only solved by certain discrete values of u. The smallest positive value is u = .8603335890, which gives the equation xy = .8603335890. --Lynn === Subject: Re: Voevodsky on the homotopy lambda calculus john baez digitised: :: For someone like Voevodsky it would be natural to use ideas from homotopy theory instead and define something like a cartesian closed category up to coherent homotopy. æSuch a thing should be lurking in the ordinary typed lambda calculus. :: this is the denotational programme in categorical domain theory http://en.wikipedia.org/wiki/Domain theory which extends into the theory of computation notions of geometric logic www.math.lsa.umich.edu/~ablass/eatcs.ps one brilliant reference http://www.andrew.cmu.edu/user/cebrown/notes/lambekscott.html http://www.amazon.com/gp/product/0521246652/002-4997198-2308810?v=glance&n=2 83155 is a great place for someone with your background to jump into lambda calculi particularly note the section on adjoint functors which you are well familiar with because adjoints transform nicely into 2-category notation of course all this is really just algorithmics on tree, graph, digraph, and related combinatorial constructs once you have graphs you have the full power of abstract state machines even 2-category theory is just an algorithm specification language for manipulation of the primitives of these data structures http://en.wikipedia.org/wiki/Scott topology where they refer to adjoints in an informative asymmetry http://en.wikipedia.org/wiki/Galois connection adjunction is behind that equivalence monads are born from adjoint relationships you get a subobject classifier www.maths.gla.ac.uk/~tl/categories/sheetfour.ps but the critical adjoint for topological models of computation is stone duality http://en.wikipedia.org/wiki/Stone duality identity of proofs however as a topic of semantic viability has a long algebraic line http://projecteuclid.org/Dienst/UI/1.0/Summarize/euclid.bsl/1067620091 of course you know all of this http://golem.ph.utexas.edu/string/archives/000509.html so i haven't added anything of value particularly since i do not know what voevodsky lectured about... although this: http://www.math.ias.edu/~vladimir/seminar.html sounds pretty close... -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: Why are Asians so good at math? HI PPL DOnt forget the numeric symbols(including zero) which Newton and all other western scientists used were developed in India(Asia) and before that roman symbols were used to do calculations. U can imagine its an herculean task to do a simple calculation with roman symbols. So we gave platform to all. Westerners were very comman students in Nalanda, Taxila UNiversities in India. And I am talkin in BCs and early ADs guys. Dont discriminate try to appreciate. Singh === Subject: Just checking in by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id k1S7Tu330014; Tue, 28 Feb 2006 02:29:56 -0500 by support2.mathforum.org (8.12.10/8.12.10/The Math Forum, $Revision: 1.6 secondary) with SMTP id k1S7SLF4012323; Tue, 28 Feb 2006 02:28:22 -0500 by .striker.ottawa.on.ca with esmtp ( 3.35 #1 ()) id 771nlc-0071MM-00 for ; Tue, 28 Feb 2006 02:22:02 -0500 Ci-iallis Sof-tabs is better than Pfizer V-iiaggrra and normal Ci-ialis because: - Guarantes 36 hours lasting - Safe to take, no side effectts at all - Boost and increase se-xual perfoormance - Haarder e-rectiiions and quick recharge - Proven and c-ertified by e-xperts and d-octors - only $1.98 per tabs - Special offeer! These prices - are valid u-ntil 30th of January ! Cllick hereee: http://robinboudwin.info === Subject: Three quesition about graph coloring X(G):= chromatic number of simple graph G Define a function CLRS: {graphs} -> { sets of function(s)} s.t. G |-> {f | f:V(G)->{1,2,...,X(G)} is proper coloring} Suppose G, H be any simple graph... Is CLRS(G)=CLRS(H) <-> G is isomophic to H ? Is always exist simple graph F s.t. CLRS(F)=CLRS(G).81ÀCLRS(H)? Is always exist simple graph F s.t. CLRS(F)=CLRS(G).81.beCLRS(H)? At first, I never know the discussion for set of coloring. I just get the idea, but I'm not ensure whether it's good or bad. I just get a new thinking about this. Is any book ever talk about something like this? === Subject: Re: Three quesition about graph coloring I am sorry about that I forget that some symbol couldn't use here. so,I repost about my question. X(G):= chromatic number of simple graph G A ^S^ B:= the intersection of A and B A vSv B:= the union of A and B Define a function CLRS: {graphs} -> { sets of function(s)} s.t. G |-> {f | f:V(G)->{1,2,...,X(G)} is proper coloring} Suppose G, H be any simple graph... Is CLRS(G)=CLRS(H) <-> G is isomophic to H ? Is always exist simple graph F s.t. CLRS(F)=CLRS(G) ^S^ CLRS(H)? Is always exist simple graph F s.t. CLRS(F)=CLRS(G) vSv CLRS(H)? At first, I never know the discussion for set of coloring. I just get the idea, but I'm not ensure whether it's good or bad. I just get a new thinking about this. Is any book ever talk about something like this? === Subject: Re: Three quesition about graph coloring I am sorry about that I forgot that I cannot use some special symbol here A ^S^ B:= the set of intersection for set A and set B A vS^ B:= the set of union for set A and set B X(G):= chromatic number of simple graph G To define the function CLRS: {graphs} -> { sets of function(s)} s.t. G |-> {f | f:V(G)->{1,2,...,X(G)} is proper coloring} Suppose G, H be any simple graph... Is CLRS(G)=CLRS(H) <-> G is isomophic to H ? Is always exist simple graph F s.t. CLRS(F)=CLRS(G).81ÀCLRS(H)? Is always exist simple graph F s.t. CLRS(F)=CLRS(G).81.beCLRS(H)? At first, I never know the discussion for set of coloring. I just get the idea, but I'm not ensure whether it's good or bad. I just get a new thinking about this. Is any book ever talk about something like this? === Subject: Re: Three quesition about graph coloring I am sorry about that I forgot that I cannot use some special symbol here A ^S^ B:= the set of intersection for set A and set B A vSv B:= the set of union for set A and set B X(G):= chromatic number of simple graph G To define the function CLRS: {graphs} -> { sets of function(s)} s.t. G |-> {f | f:V(G)->{1,2,...,X(G)} is proper coloring} Suppose G, H be any simple graph... Is CLRS(G)=CLRS(H) <-> G is isomophic to H ? Is always exist simple graph F s.t. CLRS(F)=CLRS(G) ^S^ CLRS(H)? Is always exist simple graph F s.t. CLRS(F)=CLRS(G) vSv CLRS(H)? At first, I never know the discussion for set of coloring. I just get the idea, but I'm not ensure whether it's good or bad. I just get a new thinking about this. Is any book ever talk about something like this? === Subject: FWD: Dont even invest before you read by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id k1S8Le304478 for ; Tue, 28 Feb 2006 03:21:41 -0500 DKDY Continues its climb another Day, Still Expecting Huge News Release This Week Three Day Trading Stats Pr ofiled: Dark Dynamite Inc. SECTOR: Consumer Goods Sym b01: D K D Y Last Wed Low: $1.01 Last: $1.22 {volumeup} Over 800% Another day of climbing for DKDYin the market today. Investors are thrilled as they continue to watch their investment grow. this recent growth is due to recent announcements of agreements with travel agencies specializing China travel to feature E-Pang Palace Packages. Expectations of price and volumes jumps are now realizing inline with activity that this stock has already shown us over the last 2 months. For those of you who have already jumped on board congrat?s and hold on as we are still awaiting the big news release this week that will push this stock even further. For those of you who are still waiting, wait no further, time is running out to get ahead of the next climb, don?t kick yourself, pick up DKDY first thing in the morning. Wed Headl ines: Travel Agencies Launch E-Pang Palace Package, Bookings are being filled for 3rd Quarter '06 Recent news Release informed investors that agreements were made with 25 Travel agencies, specializing in travel packages for China, would now be promoting the E-Pang Palace Theme Park with there existing packages. This agreement is now in effect and bookings are being made for Sept '06 travel season. Full details are expected in an official news release next week. Review: Jump on DKDY on Tuesday before the New Release Hits and take advantage of the ROI. Have A Great Weekend And GOOD TRADING Next Week. === Subject: Re: Help: How to calculate quickly this sum 192,255,317....1218,1262? > I have to calculate hundreds of sums similar to the one below, with > 100's + of numbers per series. > How do I calculate this sum?( using multiplication or factorial etc..) > 192,255,317,378,438,497,555,612,668,723,777,830,882, > 933,983,1032,1080,1127,1173,1218, 1262 > This series of number follow this rule > 192 -> 255 = 63 > 255 -> 317 = 62 > 317 ->378 = 61 > 378 ->438 = 60 > ... > .. > 1173 ->1218 =45 > 1218 -> 1262 = 44 > What is the name of such a series? > Any chance to explain how you calculate the sum? > Reference websites would be appreciated? The technique you are after is polynomial interpolation. In Mathematica for your data, data = {192, 255, 317, 378, 438, 497, 555, 612, 668, 723, 777, 830, 882, 933, 983, 1032, 1080, 1127, 1173, 1218, 1262}; you just enter InterpolatingPolynomial[data, x] to find that the interpolating polynomial is ((2 - x)/2 + 63) (x - 1) + 192 Paul _______________________________________________________________________ Paul Abbott Phone: 61 8 6488 2734 School of Physics, M013 Fax: +61 8 6488 1014 The University of Western Australia (CRICOS Provider No 00126G) AUSTRALIA http://physics.uwa.edu.au/~paul === Subject: Anyone interested in being paid to find a solution to a commercial problem? We are software developers for the vacation rental business. We need to code an availability search (in Microsoft SQL Server) to find the cheapest combination of properties that will accommodate a given capacity. Testing all possible combinations is not an option as forty properties will yield roughly a trillion combinations. We have got to the point where we take each property in turn, starting with the cheapest in terms of cost per head, and then adding the next cheapest and so on, until greater than or equal to the required capacity is reached. If that capacity is reached EXACTLY, we then have a solution. So far so good. But if we have gone over the required capacity, there is a possibility that a different mix with a lower capacity will be cheaper, even if the cost per head of capacity is higher, as a result of the spare capacity in going over what is required rather than matching it exactly. We also need to be able to implement a solution WITHIN SQL Server as it is extremely complex and difficult to call external software (we have attempted unsuccessfully to do this with commercial linear programming solutions). === Subject: Re: Anyone interested in being paid to find a solution to a commercial problem? > We are software developers for the vacation rental business. We need > to code an availability search (in Microsoft SQL Server) to find the > cheapest combination of properties that will accommodate a given > capacity. This sounds like a knapsack problem, and there are many heuristics available for it (although solving large examples exactly can require integer programming, which would be horrible to implement in-house). If you go to the INFORMS website (the site for the Institute of Management Science and Operations Research) and go to the on-line resources menu, you will likely be able to find several downloadable codes for such problems. However, whether they can easily be implemented on SQL is something I cannot say, as I am not familiar with that server or its properties and limitations. Such problems can also be formulated and solved as dynamic programming problems, and any decent introductory textbook in Operations Research will show you how. Again, though, there is the implementation issue. Anyway, if all you want are good, not necessarily optimal, solutions, you can examine heuristics for the problem. These have been well studied and tested empirically for solution quality. Some of them are quite simple. The most straightforward one is called the Greedy Algorithm, and it works as follows: rank the prospects in ascending order of cost/capacity and take them one-by-one from the top, until capacity is reached; you can skip over some and take ones farther down in the list if that will help to fill up the capacity. If I have interpreted your posting correctly, this is what you are already doing. It may be comforting to know that it is a well-studied procedure, and a lot is known about its performance. You may have better luck posting your query to the Operations Research newsgroup, 'sci.op-research'. A lot of the algorithms and heuristics have been developed by Operations Researchers for use in solving practical problems like yours. By the way, if you have no more than a few tens of properties (you mentioned forty), you can formulate and solve the problem exactly in the EXCEL spreadsheet Solver tool. However, I don't know if this is available to you on SQL. It comes bundled with EXCEL right from the start. You can purchase more powerful EXCEL Solvers that could handle hundreds, perhaps thousands, of properties, although on really big problems the cpu time might be excessive. Good luck R.G. Vickson Adjunct Professor, University of Waterloo > Testing all possible combinations is not an option as forty > properties will yield roughly a trillion combinations. We have got to > the point where we take each property in turn, starting with the > cheapest in terms of cost per head, and then adding the next cheapest > and so on, until greater than or equal to the required capacity is > reached. If that capacity is reached EXACTLY, we then have a solution. > So far so good. But if we have gone over the required capacity, there > is a possibility that a different mix with a lower capacity will be > cheaper, even if the cost per head of capacity is higher, as a result > of the spare capacity in going over what is required rather than > matching it exactly. > We also need to be able to implement a solution WITHIN SQL Server as it > is extremely complex and difficult to call external software (we have > attempted unsuccessfully to do this with commercial linear programming > solutions). === Subject: Re: Anyone interested in being paid to find a solution to a commercial problem? >We are software developers for the vacation rental business. We need >to code an availability search (in Microsoft SQL Server) to find the >cheapest combination of properties that will accommodate a given >capacity. Testing all possible combinations is not an option as forty >properties will yield roughly a trillion combinations. We have got to >the point where we take each property in turn, starting with the >cheapest in terms of cost per head, and then adding the next cheapest >and so on, until greater than or equal to the required capacity is >reached. If that capacity is reached EXACTLY, we then have a solution. > So far so good. But if we have gone over the required capacity, there >is a possibility that a different mix with a lower capacity will be >cheaper, even if the cost per head of capacity is higher, as a result >of the spare capacity in going over what is required rather than >matching it exactly. This sounds like fairly elementary binary linear programming. Of course, actual implementation could be difficult if the properties have complicated dependencies with each other (like if you want X you need Y unless you get Z, in which case you can't get W). If you can get all the constraints sorted out there are well-known algorithms for solving the problem. >We also need to be able to implement a solution WITHIN SQL Server as it >is extremely complex and difficult to call external software (we have >attempted unsuccessfully to do this with commercial linear programming >solutions). However, from experience I can tell you that writing it all in T-SQL is a non-trivial task. I think if you proceed in this way you'll face more technical difficulties with the SQL Server implementation than the actual optimization algorithms. === Subject: Re: Anyone interested in being paid to find a solution to a commercial problem? We are software developers for the vacation rental business. We need >to code an availability search (in Microsoft SQL Server) to find the >cheapest combination of properties that will accommodate a given >capacity. Testing all possible combinations is not an option as forty >properties will yield roughly a trillion combinations. We have got to >the point where we take each property in turn, starting with the >cheapest in terms of cost per head, and then adding the next cheapest >and so on, until greater than or equal to the required capacity is >reached. If that capacity is reached EXACTLY, we then have a solution. > So far so good. But if we have gone over the required capacity, there >is a possibility that a different mix with a lower capacity will be >cheaper, even if the cost per head of capacity is higher, as a result >of the spare capacity in going over what is required rather than >matching it exactly. > This sounds like fairly elementary binary linear programming. Of > course, actual implementation could be difficult if the properties > have complicated dependencies with each other (like if you want X you > need Y unless you get Z, in which case you can't get W). If you can > get all the constraints sorted out there are well-known algorithms for > solving the problem. >We also need to be able to implement a solution WITHIN SQL Server as it >is extremely complex and difficult to call external software (we have >attempted unsuccessfully to do this with commercial linear programming >solutions). > However, from experience I can tell you that writing it all in T-SQL > is a non-trivial task. I think if you proceed in this way you'll face > more technical difficulties with the SQL Server implementation than > the actual optimization algorithms. A binary linear program with 40 binary variables would not be practical (in this case). If I understand the problem correctly, this is the knapsack problem. The original poster mentions exact matches, but I am assuming a non-exact match can be optimal, e.g. to house 11 heads you may find three 4-head properties for less cost then a 6-head and 5-head. The knapsack problem is NP-complete in the general case, but that doesn't mean it cannot be solved. q.v. dyanamic programming, etc. === Subject: Fourier analysis Heisenberg principle Hello Am working on proving this heisenberg principle in one dimension var(f)^2 .var((2*pi)^-0.5 få)^2 > = 1/4 where få is the fourier transform when the mean of f, and its fourier transform equal 0 i have managed to show this. but i am trying to remove this restriction of a 0 mean. I have been given the hint h(x)=exp(-iax)*f(x+b) where a =mean of the fourier transform squared, b=mean of the function squared i have tried to prove this with no luck. the second var is causing me problems as hå(k) = intergral (on R) exp(-ikx)h(x) dx =integral exp(-ikx)(exp(-iax)*f(x+b)) dx sub x+b=y =exp(ib(k+a)) integral exp(-iy(k+a))f(y) dy Im having trouble with the last line. Sorry about the notation, could not find a hat symbol === Subject: Re: Fourier analysis Heisenberg principle >Am working on proving this heisenberg principle in one dimension >var(f)^2 .var((2*pi)^-0.5 f.8c)^2 > = 1/4 where f.8c is the fourier >transform >when the mean of f, and its fourier transform equal 0 >i have managed to show this. but i am trying to remove this restriction >of a 0 mean. >I have been given the hint h(x)=exp(-iax)*f(x+b) where a =mean of the >fourier transform squared, b=mean of the function squared >i have tried to prove this with no luck. >the second var is causing me problems as >h=.8c(k) = intergral (on R) exp(-ikx)h(x) dx > =integral exp(-ikx)(exp(-iax)*f(x+b)) dx sub x+b=y > =exp(ib(k+a)) integral exp(-iy(k+a))f(y) dy >Im having trouble with the last line. >Sorry about the notation, could not find a hat symbol contains a proof of the Heisenberg uncertainty principle on R^n. Rob Johnson take out the trash before replying === Subject: Re: Fourier analysis Heisenberg principle <20060228.144927@whim.org >Am working on proving this heisenberg principle in one dimension >var(f)^2 .var((2*pi)^-0.5 f.8c)^2 > = 1/4 where f.8c is the fourier >transform >when the mean of f, and its fourier transform equal 0 >i have managed to show this. but i am trying to remove this restriction >of a 0 mean. >I have been given the hint h(x)=exp(-iax)*f(x+b) where a =mean of the >fourier transform squared, b=mean of the function squared >i have tried to prove this with no luck. >the second var is causing me problems as >h=.8c(k) = intergral (on R) exp(-ikx)h(x) dx > =integral exp(-ikx)(exp(-iax)*f(x+b)) dx sub x+b=y > =exp(ib(k+a)) integral exp(-iy(k+a))f(y) dy >Im having trouble with the last line. >Sorry about the notation, could not find a hat symbol > contains a proof > of the Heisenberg uncertainty principle on R^n. > Rob Johnson >I've noticed that both the counterexamples had x being some 2-element. >>A similar contruction with M=S_3 x S_5 G=S_8 would give a counterexample >>with x of order 3. I'm trying to figure out if there are >>counterexamples to a modification of the statement: >>Suppose that x is in finite group G and x is contained in the soluble >>radical of some maximal subgroup M. Also x has prime order p>=5. Now if >>x^g is in M, is it true that x^g is also in Sol(M). >Try G = SL(3,p), M = matrices X in G with X_{12} = X_{13} = 0, >Sol(M) = { [ 1 0 0 ] } > [ a 1 0 ] > [ b 0 1 ] > This doesn't affect the correctness of the example, but that is not quite > right. For p a prime power with p >= 5 > Sol(M) = { [ c^-2 0 0 ] | a,b,c in F_p, c!=0 } > [ a c 0 ] > [ b 0 c ] Now that I'm back on line, I see you did address the question I raised elsewhere. :-) For the OP: You might want to see if you can show why Derek's example is correct. Hints: First show that M is indeed a maximal subgroup of SL(3,p), then see if you can express M as a semidirect product of his original, incorrect (Sol(M) ~= C_p x C_p) x| GL(2,p). If you want to go even further, try showing that if p <> 1 (mod 3), then M ~= AGL(2,p); otherwise the C_3 in the center of GL(2,p) acts trivially on the C_p x C_p. >x = [ 1 0 0 ] conjugate to [ 1 0 0 ] > [ 1 1 0 ] [ 0 1 0 ] > [ 0 0 1 ] [ 0 1 1 ]. -- Jim Heckman === Subject: Cos(t+a*cos(t)) What is the development de Cos(t+a*cos(t)) ou cos (t+ b*sint)) Can I find them on the net Thaks === Subject: Re: Cos(t+a*cos(t)) > What is the development de > Cos(t+a*cos(t)) ou cos (t+ b*sint)) Obviously cos(t+a cos(t)) == cos(t) sin(a cos(t)) - sin(t) cos(a cos(t)) so I expect that you are after the Fourier expansion of sin(a cos(t)) and cos(a cos(t)). > Can I find them on the net Yes. The first result of a search for the (Mathematica) pattern Cos[_ Cos[_]] (where the _ symbol in Mathematica denotes any single object) at http://functions.wolfram.com/formulasearch/ yields http://functions.wolfram.com/03.01.23.0009.01 and clicking on the Search for similar formulas button yields http://functions.wolfram.com/03.01.23.0010.01 So, the Fourier expansions of Cos[x Cos[t]] is 2 Sum[(-1)^k Cos[2 k t] BesselJ[2 k, x], {k, 1, Infinity}] == Cos[x Cos[t]] - BesselJ[0, x] and that of Sin[x Cos[t]] is 2 Sum[(-1)^k Cos[(2 k + 1) t] BesselJ[2 k + 1, x], {k, 0, Infinity}] == Sin[x Cos[t]] You can use these formulae and simple trig operations to obtain the expansion that you are after. Paul _______________________________________________________________________ Paul Abbott Phone: 61 8 6488 2734 School of Physics, M013 Fax: +61 8 6488 1014 The University of Western Australia (CRICOS Provider No 00126G) AUSTRALIA http://physics.uwa.edu.au/~paul === Subject: Re: Cos(t+a*cos(t)) Le 28/02/06 16:14, dans paul-365220.23145528022006@news.uwa.edu.au, .82æPaul Abbottæé a .8ecritæ: >> What is the development de >> Cos(t+a*cos(t)) ou cos (t+ b*sint)) > Obviously > cos(t+a cos(t)) == cos(t) sin(a cos(t)) - sin(t) cos(a cos(t)) > so I expect that you are after the Fourier expansion of sin(a cos(t)) > and cos(a cos(t)). >> Can I find them on the net > Yes. The first result of a search for the (Mathematica) pattern > Cos[_ Cos[_]] > (where the _ symbol in Mathematica denotes any single object) at > http://functions.wolfram.com/formulasearch/ > yields > http://functions.wolfram.com/03.01.23.0009.01 > and clicking on the Search for similar formulas button yields > http://functions.wolfram.com/03.01.23.0010.01 > So, the Fourier expansions of Cos[x Cos[t]] is > 2 Sum[(-1)^k Cos[2 k t] BesselJ[2 k, x], {k, 1, Infinity}] == > Cos[x Cos[t]] - BesselJ[0, x] > and that of Sin[x Cos[t]] is > 2 Sum[(-1)^k Cos[(2 k + 1) t] BesselJ[2 k + 1, x], {k, 0, Infinity}] == > Sin[x Cos[t]] > You can use these formulae and simple trig operations to obtain the > expansion that you are after. > Paul > _______________________________________________________________________ > Paul Abbott Phone: 61 8 6488 2734 > School of Physics, M013 Fax: +61 8 6488 1014 > The University of Western Australia (CRICOS Provider No 00126G) > AUSTRALIA http://physics.uwa.edu.au/~paul === Subject: Re: Cos(t+a*cos(t)) > What is the development de > Cos(t+a*cos(t)) ou cos (t+ b*sint)) > Can I find them on the net > Thaks By development do you mean power series? === Subject: Re: Cos(t+a*cos(t)) Le 28/02/06 14:04, dans 280220060804593518%anniel@nym.alias.net.invalid, .82æA N Nielæé a .8ecritæ: >> What is the development de >> Cos(t+a*cos(t)) ou cos (t+ b*sint)) >> Can I find them on the net >> Thaks > By development do you mean power series? I remember something like Cos(t+a*cos(t)) = J0(a) cos(t) + J2(a) cos(2*t) + .83 where the J are Bessel functions === Subject: Re: Cos(t+a*cos(t)) > Le 28/02/06 14:04, dans 280220060804593518%anniel@nym.alias.net.invalid, .82æA > N Nielæé a .8ecritæ: > >> What is the development de >> Cos(t+a*cos(t)) ou cos (t+ b*sint)) >> Can I find them on the net >> Thaks >> > > By development do you mean power series? > I remember something like > Cos(t+a*cos(t)) = J0(a) cos(t) + J2(a) cos(2*t) + .83 > where the J are Bessel functions maybe this one... Gradshteyn & Ryzhik, 8.511.3 === Subject: Re: Cos(t+a*cos(t)) >What is the development de > Cos(t+a*cos(t)) ou cos (t+ b*sint)) >Can I find them on the net Don't search the net. Search your trigonometry book: cos(x+y) = cos(x) cos(y) - sin(x) sin(y). -- Rouben Rostamian === Subject: JSH: This is my attempt at a valid proof. Please tell me what is wrong with it (not just JSH, but the rest as well... perhaps especially the rest ;) ). Conjecture: In the complex plane, if 7C(x) = (A(x) + 7)(B(x) + 1) for all x, where A(0) = B(0) = 0 and 7C(x) =7 (D(x) + 1)(E(x) + 1) for all x, where D(0) = E(0) = 0 then A(x)=7D(x) and B(x)=E(x). Proof: (A(x) + 7)(B(x) + 1) = 7(D(x) + 1)(E(x) + 1) A(x)(B(x)+1)+7B(x)+7 = 7D(x)E(x)+7D(x)+7E(x)+7 A(x)(B(x)+1)+7B(x) = 7D(x)E(x)+7D(x)+7E(x) A(x)(B(x)+1) = 7D(x)E(x)+7D(x)+7E(x)-7B(x) A(x) = (7D(x)E(x)+7D(x)+7E(x)-7B(x))/(B(x)+1) pick functions for D(x),E(x),B(x): D(x) = x; E(x) = 2x; B(x) = 3x A(x) = 14(x^2)/(3x+1) Is A(x) = 7D(x)? No. Conclusion: the conjecture is false. Is this valid or am I forgetting something? === Subject: Re: JSH > This is my attempt at a valid proof. Please tell me what is wrong with > it (not just JSH, but the rest as well... perhaps especially the rest > ;) ). > Conjecture: > In the complex plane, if 7C(x) = (A(x) + 7)(B(x) + 1) for all x, where > A(0) = B(0) = 0 and 7C(x) =7 (D(x) + 1)(E(x) + 1) for all x, where > D(0) = E(0) = 0 then A(x)=7D(x) and B(x)=E(x). > Proof: > (A(x) + 7)(B(x) + 1) = 7(D(x) + 1)(E(x) + 1) > A(x)(B(x)+1)+7B(x)+7 = 7D(x)E(x)+7D(x)+7E(x)+7 > A(x)(B(x)+1)+7B(x) = 7D(x)E(x)+7D(x)+7E(x) > A(x)(B(x)+1) = 7D(x)E(x)+7D(x)+7E(x)-7B(x) > A(x) = (7D(x)E(x)+7D(x)+7E(x)-7B(x))/(B(x)+1) > pick functions for D(x),E(x),B(x): > D(x) = x; E(x) = 2x; B(x) = 3x > A(x) = 14(x^2)/(3x+1) > Is A(x) = 7D(x)? No. Conclusion: the conjecture is false. > Is this valid or am I forgetting something? did not prove a statement is true under all conditions. your numbers have little to do with the complex plane. http://paul.merton.ox.ac.uk/science/maths-proofs.html Proof by contradiction (also known as reductio ad absurdum): where it is shown that if some statement were false, a logical contradiction occurs, hence the statement must be true === Subject: Re: JSH: > This is my attempt at a valid proof. Please tell me what is wrong with > it (not just JSH, but the rest as well... perhaps especially the rest > ;) ). > Conjecture: > In the complex plane, if 7C(x) = (A(x) + 7)(B(x) + 1) for all x, where > A(0) = B(0) = 0 and 7C(x) =7 (D(x) + 1)(E(x) + 1) for all x, where > D(0) = E(0) = 0 then A(x)=7D(x) and B(x)=E(x). > Proof: > (A(x) + 7)(B(x) + 1) = 7(D(x) + 1)(E(x) + 1) > A(x)(B(x)+1)+7B(x)+7 = 7D(x)E(x)+7D(x)+7E(x)+7 > A(x)(B(x)+1)+7B(x) = 7D(x)E(x)+7D(x)+7E(x) > A(x)(B(x)+1) = 7D(x)E(x)+7D(x)+7E(x)-7B(x) > A(x) = (7D(x)E(x)+7D(x)+7E(x)-7B(x))/(B(x)+1) > pick functions for D(x),E(x),B(x): > D(x) = x; E(x) = 2x; B(x) = 3x > A(x) = 14(x^2)/(3x+1) > Is A(x) = 7D(x)? No. Conclusion: the conjecture is false. > Is this valid or am I forgetting something? You forgot to tell us in the beginning what kind of thing A(x), B(x), and C(x) are. If they are required to be polynomials in x, then you can't divide through by B(x) + 1 and solve for A(x) quite in the manner above. However if these were (for example) intended to be rational functions of x, rather than polynomials, then this would be a valid step. Hence the repeated emphasis on getting certain posters to specify what ring they are working in. Posters are free, of course, to work in whatever ring is chosen. What is not valid is to take a result that holds for expressions in one ring and assert that it must also hold in another ring. Some principles, like the distributive property of multiplication over addition, do hold in all rings, but this doesn't mean an arbitrary statement true in one ring will hold in another. Of course if a proof of the statement relies on nothing more than the distributive principle (for example), then it would be true in all rings. One needs to be careful in distinguishing proofs that use the distributive principle from proofs that use nothing but the distributive principle, however. === Subject: Integer-Valued Polynomials let D be a domain, K its quotient field, and F an extension field of K. The ring of integer-valued polynomials in n indetermintes over D consists of those polynomials in K[x_1,...,x_n] which give a value in D for every argument in D^n (standard definition). What happens if we replace K[x_1,...,x_n] with F[x_1,...,x_n] in this definition? Do we obtain the same set of polynomials? What if D=Z, K=Q and F=C? Maury === Subject: Re: Integer-Valued Polynomials > let D be a domain, K its quotient field, and F an > extension field of K. The ring of integer-valued > polynomials in n indetermintes over D consists of those > polynomials in K[x_1,...,x_n] which give a value in D > for every argument in D^n (standard definition). > What happens if we replace K[x_1,...,x_n] with > F[x_1,...,x_n] in this definition? Do we obtain the same > set of polynomials? What if D=Z, K=Q and F=C? I don't believe we do pick anything extra in polynomials over the extension field K, by an interpolation argument. First let us note that one can have a integer-valued polynomial which is not simply integer coefficients: f(x) = x(x+1)/2 for example. However a polynomial with coefficients in (for example) the complex field C and integer-valued for all arguments in Z must agree with a polynomial of equal degree with coefficients in Q (e.g. constructed by Newton divided differences) on all arguments Z, and hence be identical to that polynomial over Q. === Subject: Re: Integer-Valued Polynomials > let D be a domain, K its quotient field, and F an > extension field of K. The ring of integer-valued > polynomials in n indetermintes over D consists of > those > polynomials in K[x_1,...,x_n] which give a value in > for every argument in D^n (standard definition). > What happens if we replace K[x_1,...,x_n] with > F[x_1,...,x_n] in this definition? Do we obtain the > same > set of polynomials? What if D=Z, K=Q and F=C? > I don't believe we do pick anything extra in > polynomials > over the extension field K, by an interpolation > argument. > First let us note that one can have a > integer-valued > polynomial which is not simply integer coefficients: > f(x) = x(x+1)/2 > for example. However a polynomial with coefficients > in > (for example) the complex field C and integer-valued > for all arguments in Z must agree with a polynomial > of > equal degree with coefficients in Q (e.g. constructed > by Newton divided differences) on all arguments Z, > and hence be identical to that polynomial over Q. Two remarks: (I)Newton method works also for several variables and for general fields? (II) Even if it works, it can't be very useful. How could you apply it, to use later the Identity Polynomials Principle? But your argument intended maybe to be only intuitive. However, thank you very much for your help. Maury === Subject: Re: Integer-Valued Polynomials <21625424.1141134725541.JavaMail.jakarta@nitrogen.mathforum.org > let D be a domain, K its quotient field, and F an > extension field of K. The ring of integer-valued > polynomials in n indetermintes over D consists of > those > polynomials in K[x_1,...,x_n] which give a value in > D > for every argument in D^n (standard definition). > What happens if we replace K[x_1,...,x_n] with > F[x_1,...,x_n] in this definition? Do we obtain the > same > set of polynomials? What if D=Z, K=Q and F=C? > I don't believe we do pick anything extra in > polynomials > over the extension field K, by an interpolation > argument. > First let us note that one can have a > integer-valued > polynomial which is not simply integer coefficients: > f(x) = x(x+1)/2 > for example. However a polynomial with coefficients > in > (for example) the complex field C and integer-valued > for all arguments in Z must agree with a polynomial > of > equal degree with coefficients in Q (e.g. constructed > by Newton divided differences) on all arguments Z, > and hence be identical to that polynomial over Q. > Two remarks: > (I)Newton method works also for several > variables and for general fields? There is a risk of confusion in terminology here. I mentioned (1) interpolation and (2) Newton divided differences (as a specific approach to finding an interpolating polynomial). Newton's method often refers to root finding, and while a multivariate form of this (usually called Newton-Raphson) exists, it has little to do with the question of interpolation. Interpolating polynomials in many variables, versus in one variable, does introduce a complication of choosing interpolating points not correlated by a polynomial of smaller degree. See this easily [Polynomial interpolation in several variables - by Tomas Sauer] http://www.uni-giessen.de/tomas.sauer/Publ/MAIA.pdf > (II) Even if it works, it can't be very useful. How could > you apply it, to use later the Identity Polynomials > Principle? > But your argument intended maybe to be only intuitive. My argument is an outline, but if you have specific questions perhaps I can fill in more details. === Subject: Re: Integer-Valued Polynomials > Two remarks: > (I)Newton method works also for several > variables and for general fields? > There is a risk of confusion in terminology here. I > mentioned (1) interpolation and (2) Newton divided > differences (as a specific approach to finding an > interpolating polynomial). Newton's method often > refers to root finding, and while a multivariate form > of this (usually called Newton-Raphson) exists, it > has little to do with the question of interpolation. > Interpolating polynomials in many variables, versus > in one variable, does introduce a complication of > choosing interpolating points not correlated by > a polynomial of smaller degree. See this easily > [Polynomial interpolation in several variables - > by Tomas Sauer] > http://www.uni-giessen.de/tomas.sauer/Publ/MAIA.pdf > (II) Even if it works, it can't be very useful. How > could > you apply it, to use later the Identity > Polynomials > Principle? > But your argument intended maybe to be only > intuitive. > My argument is an outline, but if you have specific > questions perhaps I can fill in more details. The point is: let P(x_1,...,x_n) be a complex polynomial with integer values when (x_1,...,x_n) is in Z^n. How do you construct a rational polynomial Q(x_1,...,x_n) wich agrees with P on Z^n? It seems to me a weak point in your argument. Maury === Subject: Re: Integer-Valued Polynomials > let D be a domain, K its quotient field, and F an > extension field of K. The ring of integer-valued > polynomials in n indetermintes over D consists of those > polynomials in K[x_1,...,x_n] which give a value in D > for every argument in D^n (standard definition). > What happens if we replace K[x_1,...,x_n] with > F[x_1,...,x_n] in this definition? Do we obtain the same > set of polynomials? What if D=Z, K=Q and F=C? > I don't believe we do pick anything extra in polynomials > over the extension field K, by an interpolation argument. > First let us note that one can have a integer-valued > polynomial which is not simply integer coefficients: > f(x) = x(x+1)/2 > for example. However a polynomial with coefficients in > (for example) the complex field C and integer-valued > for all arguments in Z must agree with a polynomial of > equal degree with coefficients in Q (e.g. constructed > by Newton divided differences) on all arguments Z, > and hence be identical to that polynomial over Q. Ah, as Jyrki points out, we do need domain D to be infinite to make this argument work, as otherwise a multiple of a nonzero polynomial vanishing only all of D can be constructed. -- c === Subject: Re: Integer-Valued Polynomials > let D be a domain, K its quotient field, and F an > extension field of K. The ring of integer-valued > polynomials in n indetermintes over D consists of those > polynomials in K[x_1,...,x_n] which give a value in D > for every argument in D^n (standard definition). > What happens if we replace K[x_1,...,x_n] with > F[x_1,...,x_n] in this definition? Do we obtain the same > set of polynomials? Can't be true in this generality. If D is finite, say of cardinality q, then you can add any multiples of (x_i^q-x_i) to an integer-valued polynomial without affecting this property. What if D=Z, K=Q and F=C? OTOH it seems to me that requiring D to be infinite is a sufficient condition as well: write elements of F in terms of a K-basis B (assume that 1_K is in B). Then you can write all the polynomials in F[x_1,...,x_n] as K[x_1,...,x_n]-linear combinations of constants from B. IIRC the following holds: D infinite => (p(x_1,x_2,...,x_n) with p in K[x_1,...,x_n] vanishes on all of D^n, iff p=0) For a polynomial from F[x_1,...,x_n] to be integer-valued it must be at least K-valued, so by the above result the coefficient polynomial p_b of an element b from Bsetminus{1_K} must be the zero polynomial (in order to give rise to the all-zero polynomial function). Therefore only p_1 can be non-zero and the claim follows. Jyrki === Subject: Re: Integer-Valued Polynomials > let D be a domain, K its quotient field, and F an > extension field of K. The ring of integer-valued > polynomials in n indetermintes over D consists of > those > polynomials in K[x_1,...,x_n] which give a value in > D > for every argument in D^n (standard definition). > What happens if we replace K[x_1,...,x_n] with > F[x_1,...,x_n] in this definition? Do we obtain the > same > set of polynomials? > Can't be true in this generality. If D is finite, say > of > cardinality q, then you can add any multiples of > (x_i^q-x_i) to an integer-valued polynomial without > affecting this property. > What if D=Z, K=Q and F=C? > OTOH it seems to me that requiring D to be infinite > is > a sufficient condition as well: write elements of F > in terms of a K-basis B (assume that 1_K is in B). > Then you can write all the polynomials in > F[x_1,...,x_n] > as K[x_1,...,x_n]-linear combinations of constants > from B. > IIRC the following holds: > D infinite => (p(x_1,x_2,...,x_n) with p in > in K[x_1,...,x_n] > vanishes on all of D^n, iff p=0) Yes, this is surely true. It can be easily proved by induction on the number of indeterminates, and using the fact that a polynomial in one indeterminate of degree k has at the most k roots. > For a polynomial from F[x_1,...,x_n] to be > integer-valued > it must be at least K-valued, so by the above result > the coefficient polynomial p_b of an element b from > Bsetminus{1_K} > must be the zero polynomial (in order to give rise to > the > all-zero polynomial function). Therefore only p_1 can > be > non-zero and the claim follows. > Jyrki === Subject: Polynomials and Prime Numbers I read the following result. There's no rational polynomial P(x) wich generates all the primes (we say that a rational polynomial P(x) generates a prime p, if for some n in Z, we have P(n)=p). Do you know a simple proof of this result? Maury === Subject: Re: Polynomials and Prime Numbers > I read the following result. There's no rational > polynomial P(x) wich generates all the primes > (we say that a rational polynomial P(x) generates a > prime p, if for some n in Z, we have P(n)=p). > Do you know a simple proof of this result? This seems trivially false as P(x) = x would then generate prime p when P(p) = p. Perhaps you have a more restrictive condition in mind? === Subject: Re: Polynomials and Prime Numbers > I read the following result. There's no rational > polynomial P(x) wich generates all the primes > (we say that a rational polynomial P(x) generates a > prime p, if for some n in Z, we have P(n)=p). > Do you know a simple proof of this result? > This seems trivially false as P(x) = x would then > generate prime p when P(p) = p. > Perhaps you have a more restrictive condition > in mind? What a stupid I am!!! Surely, the author meant that P(x) generates only primes for x in Z! So the right formulation is: There's no rational polynomial P(x) such that P(z) is prime for every z in Z, and, for every prime p, there's some z in Z such that P(z)=p. Forgive my embarrassing slip! === Subject: Re: Polynomials and Prime Numbers > > I read the following result. There's no rational > polynomial P(x) wich generates all the primes > (we say that a rational polynomial P(x) generates > prime p, if for some n in Z, we have P(n)=p). > Do you know a simple proof of this result? > > > This seems trivially false as P(x) = x would then > generate prime p when P(p) = p. > > Perhaps you have a more restrictive condition > in mind? > > > What a stupid I am!!! > Surely, the author meant that P(x) generates only > primes > for x in Z! So the right formulation is: > There's no rational polynomial P(x) such that P(z) is > prime for every z in Z, and, for every prime p, > there's > some z in Z such that P(z)=p. > Forgive my embarrassing slip! My wrong formulation of the problem suggested the following question to me: there's a rational polynomial P(x) of degree greater then 1, such that {P(n): n in Z} contains the set of prime numbers? Forgive my curious ignorance!!! Maury === Subject: Re: Polynomials and Prime Numbers <11106747.1141159971176.JavaMail.jakarta@nitrogen.mathforum.org > I read the following result. There's no rational > polynomial P(x) wich generates all the primes > (we say that a rational polynomial P(x) generates > a > prime p, if for some n in Z, we have P(n)=p). > Do you know a simple proof of this result? > This seems trivially false as P(x) = x would then > generate prime p when P(p) = p. > Perhaps you have a more restrictive condition > in mind? > What a stupid I am!!! > Surely, the author meant that P(x) generates only > primes > for x in Z! So the right formulation is: > There's no rational polynomial P(x) such that P(z) is > prime for every z in Z, and, for every prime p, > there's > some z in Z such that P(z)=p. > Forgive my embarrassing slip! > My wrong formulation of the problem suggested the > following question to me: > there's a rational polynomial P(x) of degree greater > then 1, such that {P(n): n in Z} contains the set of > prime numbers? The density of {P(n): n in Z} in [-N,+N] is O(N^-(1-1/d)) where d = degree(P) >= 2. Since this density is strictly lower than the density of primes, which is O(1/log(N)), there is no such polynomial, rational or otherwise. === Subject: Re: Polynomials and Prime Numbers <15805881.1141132376717.JavaMail.jakarta@nitrogen.mathforum.org > I read the following result. There's no rational > polynomial P(x) wich generates all the primes > (we say that a rational polynomial P(x) generates a > prime p, if for some n in Z, we have P(n)=p). > Do you know a simple proof of this result? > This seems trivially false as P(x) = x would then > generate prime p when P(p) = p. > Perhaps you have a more restrictive condition > in mind? > What a stupid I am!!! > Surely, the author meant that P(x) generates only primes > for x in Z! So the right formulation is: > There's no rational polynomial P(x) such that P(z) is > prime for every z in Z, and, for every prime p, there's > some z in Z such that P(z)=p. > Forgive my embarrassing slip! Actually, just the statement There's no rational polynomial P(x) such that P(z) is prime for every z in Z is true, with the proviso that P(x) is not a constant. Here's a hint. Prove that P(x + y*P(x)) is divisible by P(x) for every pair of integers x and y. === Subject: Re: Polynomials and Prime Numbers > Actually, just the statement There's no rational > polynomial P(x) such > that P(z) is > prime for every z in Z is true, with the proviso > that P(x) is not a > constant. Here's a hint. Prove that > P(x + y*P(x)) is divisible by P(x) for every > ery pair of integers x > and y. This is an incredibly simple and beautiful result!!! Maury === Subject: Re: Polynomials and Prime Numbers <15805881.1141132376717.JavaMail.jakarta@nitrogen.mathforum.org> Actually, just the statement There's no rational polynomial P(x) such that P(z) is prime for every z in Z is true ... Let me add to this. There does exist a polynomial in 26 variables (not one) whose positive values EXACTLY coincide with the primes. Look up the 'Jones Polynomial' Furthemore, the result that Joe quoted: P(x + y*P(x)) is divisible by P(x) for every pair of integers x and y. Is a very very useful result. It is what makes the Quadratic Sieve and Number Field Sieve factoring algorithms possible. === Subject: Quadratic Forms Question - HELP !!!!!!!!!!!!! hi guys ! I am trying to learn quadratic forms....afraid I'm no good at it since I can't seem to solve too many questions. We have a square matrix A that can be written in the form, A=S + K, where S= 1/2.( A + A^T) and K= 1/2.( A - A^T) S is symmetric while K is a skew-symmetric matrix. SHOW THAT if X is a column n-tuple, then X^(T) AX= X^(T) SX Can someone help me please? Daniella === Subject: Re: Quadratic Forms Question - HELP !!!!!!!!!!!!! > .... > We have a square matrix A that can be written in the form, A=S + K, > where > S= 1/2.( A + A^T) and K= 1/2.( A - A^T) > S is symmetric while K is a skew-symmetric matrix. > SHOW THAT if X is a column n-tuple, then X^(T) AX= X^(T) SX .... Each side of that last equation is a 1 x 1 matrix. What happens when you transpose a 1 x 1 matrix? Ken Pledger. === Subject: Re: Quadratic Forms Question - HELP !!!!!!!!!!!!! Ken, I'm looking for answers, not questions. rgds Daniella > .... > We have a square matrix A that can be written in the form, A=S + K, > where > S= 1/2.( A + A^T) and K= 1/2.( A - A^T) > S is symmetric while K is a skew-symmetric matrix. > SHOW THAT if X is a column n-tuple, then X^(T) AX= X^(T) SX .... > Each side of that last equation is a 1 x 1 matrix. What happens > when you transpose a 1 x 1 matrix? > Ken Pledger. === Subject: Re: Quadratic Forms Question - HELP !!!!!!!!!!!!! You might ask your linear algebra teacher, then... === Subject: Equation from Pattern Clasification text Hi everyone, I am currently reading this text on Pattern Classification and in the text there is an example of EM (Expectation Maximization) algorithm. I am having problem following the steps the authors have taken to simplify the equation, so I hope someone can give me some assistance. Given p(x) is a multivariate normal distribution with aibitrary mean and diagonal covariance, we can define a parameter vector: theta = (u1, u2, sigma1, sigma2)^T A portion of the equation in the text was simplify from: 1/K Int{ln p((y 4)^T | theta)*(1/2*pi)*exp[-1/2(y^2+4^2)]}dy to -(1+u1^2)/2*sigma1^2 - (4-u2)^2/2*sigma2^2 - ln(2*pi*sigma1*sigma2) where K is a constant, Int{...}dy is integral with the range [-infinite, infinite]. Hope someone can shed some light on how the simplification was Thuan Seah, Tan === Subject: d.f(x,x^2)/d.x^2 = ? I have a function, such of the following, which includes terms of x and x^2: a + b*x^2 f(x) = ----------------- c*x I need a derivative wrt to x^2, i.e., d.f(x) ------- = ? d.x^2 I know d.f(x)/d.x. If i had to guess, i would assume the two are related by (1/(2*x), but not sure. appreciate anyone offering some clarification here. --dan === Subject: Re: d.f(x,x^2)/d.x^2 = ? Le 28/02/06 14:20, dans a .8ecritæ: > I have a function, such of the following, which includes > terms of x and x^2: > a + b*x^2 > f(x) = ----------------- > c*x > I need a derivative wrt to x^2, i.e., > d.f(x) > ------- = ? > d.x^2 > I know d.f(x)/d.x. If i had to guess, i would assume > the two are related by (1/(2*x), but not sure. > appreciate anyone offering some clarification here. > --dan d.f(x)/d.x^2 = d.f(x)/d.x * d.x/d.x^2 And dx/d.x^2 = 1/(d.x^2 /d.x) Was it the question ? === Subject: Re: d.f(x,x^2)/d.x^2 = ? > I have a function, such of the following, which includes > terms of x and x^2: > a + b*x^2 > f(x) = ----------------- > c*x > I need a derivative wrt to x^2, i.e., > d.f(x) > ------- = ? > d.x^2 > I know d.f(x)/d.x. If i had to guess, i would assume > the two are related by (1/(2*x), but not sure. > appreciate anyone offering some clarification here. > --dan It's probably up to the author to say what is meant when non-standard notation is used... Maybe there is a little 2 missing... d^2f(x) / dx^2 ?? === Subject: Re: d.f(x,x^2)/d.x^2 = ? > > I have a function, such of the following, which > includes terms of x and x^2: > > a + b*x^2 > f(x) = ----------------- > c*x > > I need a derivative wrt to x^2, i.e., > > d.f(x) > ------- = ? > d.x^2 > > I know d.f(x)/d.x. If i had to guess, i would > assume the two are related by (1/(2*x), but not sure. > > appreciate anyone offering some clarification here. > > > --dan > > It's probably up to the author to say what is meant > when non-standard notation is used... > Maybe there is a little 2 missing... > d^2f(x) / dx^2 ?? Possibly, however I believe the original poster was trying to find d[f(x)] / d[x^2], that is the derivative of f(x) with respect to x^2. === Subject: Re: d.f(x,x^2)/d.x^2 = ? <16339750.1141141668700.JavaMail.jakarta@nitrogen.mathforum.org> first, i apologize for the notation as it appears to have caused some confusion. in words, i was looking for the first derivative of a function of x wrt x^2. not the second derivative of anything. these posts did clear up my confusion and now i understand a derivation i was going through (had to do with variance, and standard deviations, and asymptotic distributions). === Subject: Re: d.f(x,x^2)/d.x^2 = ? Hi. > I have a function, such of the following, which > includes terms of x and x^2: > a + b*x^2 > f(x) = ----------------- > c*x This doesn't show up good on some forums; f(x) = (a + b*x^2) / (c*x) > I need a derivative wrt to x^2, i.e., > d.f(x) > ------- = ? > d.x^2 > I know d.f(x)/d.x. That's a good start. > If i had to guess, i would assume the two are related > by (1/(2*x), but not sure. > appreciate anyone offering some clarification here. How about using the chain rule by letting u = x^2. Then du/dx = 2x, and your task is reduced to finding df/du. Note that df/du = df/dx * dx/du and you said you know what df/dx is. Kyle === Subject: problem on poisson process please help me dear all i m facing problem to get solution for one problem. The problem is as follow, Let T be the delay between an arbitrary time instent t and the occurence of the first poisson event (parameter p). Find the pdf of T ur friend monu === Subject: Re: problem on poisson process please help me Ask yourself, What is the probability of no events in the interval [t,t+x] for x>0? === Subject: Re: Proability Question >Could someone here tell me the probablity of two files of the same >size, and >having the same name and extension, being different? >> No, nobody can. You need a probability model. What do you know >> about the files and where they came from? > standalone computers. Check the Meta data, date time etc, some *.exe's are a fixed size any thing Pre 1985 for example, also Full name ie ?:/x/y/z/*.exe, Operating System dates and versions also change the model later the constraints changed so the Expectation is that they are the same, Test Copy to a floppy and do a file comparison -- The world is flat it's pi that's round! There is only one number. === Subject: JSH: The New Strategy JSH is simply following and grading himself using the C-index For Physics see http://paul.merton.ox.ac.uk/science/crackpots.html The CI, A simple method for rating potentially revolutionary contributions 1 point for every statement that is widely agreed on to be false. 2 points for every statement that is clearly vacuous. 3 points for every statement that is logically inconsistent. 5 points for each such statement that is adhered to despite careful correction. 5 points for using a thought experiment that contradicts the results of a widely accepted real experiment. 5 points for each word in all capital letters (except for those with defective keyboards). 10 points for each claim that Galois Theory is fundamentally misguided (without good evidence). 10 points for each favorable comparison of oneself to Einstein, or claim that special or general relativity are fundamentally misguided (without good evidence). 10 points for pointing out that one has gone to school, as if this were evidence of sanity. 20 points for suggesting that you deserve a Nobel prize. 20 points for each favorable comparison of oneself to Abel to Galois or claim that Galois Theory is fundamentally misguided (without evidence). 20 points for every use of the Wiki or myths as if they were fact. 20 points for defending yourself by bringing up (real or imagined) ridicule accorded to ones past theories. 30 points for each favorable comparison of oneself to Galileo, claims that the Inquisition is hard at work on ones case, etc.. 30 points for claiming that the Mathematics establishment is engaged in a conspiracy to prevent ones work from gaining its well-deserved fame, or suchlike. 40 points for claiming one has a revolutionary theory but giving no concrete testable predictions. === Subject: Two point boundary value problem... I have been asked to create a solution for the following problem. I know the programming, but I don't undertstand the math. Here is the problem: ************************************************* Construct a computer program that uses both the secant method and the Runge-Kutta method (that you developed in assignment #3) to obtain a numerical solution to the two-point boundary-value problem: x' = f(t,x) = x + 0.09 x 2 + cos(10 t) differential equation x(0) + x(1) - 3.0 = 0 boundary condition Starting with the initial guesses 0.7 and 1.0 for the (unknown) initial value, x(0), obtain an approximation to x(0) {for the final solution, x(t)} such that the boundary condition is satisfied to within a tolerance of 10-4. Use a fixed stepsize of 0.025 (i.e., take 40 steps each time you integrate the differential equation from t=0 to t=1). Write your program so that the output shows the values of x(0), x(1), and x(0)+x(1)-3 (the error in satisfying the boundary condition) at the end of each iteration of the secant method. After the last iteration of the secant method, re-integrate from t=0 to t=1 and print out the solution for x(t) over the range [0,1]. ************************************************* Could somebody explain the steps involved in computing this solution. i.e. I have the secant algorithm: pk+1 = p - (f(pk)(pk - pk-1))/(f(pk) - f(pk-1)) So what is p, what is k etc? And how does this apply to the information I have been given in the question. If I can figure out the secant part of the problem, then I will be able to apply this to the runge-kutta method which I have already used on an ODE problem. I apologise if this question is a bit general (or hopeless) but I am completely lost on this one. Mark Coleman === Subject: Re: Two point boundary value problem... > I have been asked to create a solution for the following problem. I know the programming, but I don't undertstand the math. > Here is the problem: > ************************************************* > Construct a computer program that uses both the secant method and the Runge-Kutta method (that you developed in assignment #3) to obtain a numerical solution to the two-point boundary-value problem: > x' = f(t,x) = x + 0.09 x 2 + cos(10 t) differential equation > x(0) + x(1) - 3.0 = 0 boundary condition > Starting with the initial guesses 0.7 and 1.0 for the (unknown) initial value, x(0), obtain an approximation to x(0) {for the final solution, x(t)} such that the boundary condition is satisfied to within a tolerance of 10-4. > Use a fixed stepsize of 0.025 (i.e., take 40 steps each time you integrate the differential equation from t=0 to t=1). > Write your program so that the output shows the values of x(0), x(1), and x(0)+x(1)-3 (the error in satisfying the boundary condition) at the end of each iteration of the secant method. After the last iteration of the secant method, re-integrate from t=0 to t=1 and print out the solution for x(t) over the range [0,1]. > ************************************************* > Could somebody explain the steps involved in computing this solution. i.e. I have the secant algorithm: > pk+1 = p - (f(pk)(pk - pk-1))/(f(pk) - f(pk-1)) > So what is p, what is k etc? And how does this apply to the information I have been given in the question. > If I can figure out the secant part of the problem, then I will be able to apply this to the runge-kutta method which I have already used on an ODE problem. > I apologise if this question is a bit general (or hopeless) but I am completely lost on this one. > Mark Coleman It looks like you want to begin with the guess values for x(0) and integrate the differential equation using your RK algorithm to determine values for x(1). Then use the secant method on the boundary condition equation to find a new guess value for x(0). So, first integration using guess x(0) = 0.7 yields values p_0 = x(0), and fp_0 = x(0) + x(1) - 3.0 Second integration using guess x(0) = 1.0 yields values p_1 = x(0), and fp_1 = x(0) + x(1) - 3.0 Apply secant method to find new guess for x(0): x(0) = p_1 - ( (fp_1)(p_1 - p_0)/( fp_1 - fp_0) ) Rinse, repeat, until x(0) + x(1) - 3.0 is within desired tolerance of zero. === Subject: Request for a simple diagram This is very trivial but although I used to be confident with this kind of stuff, I became rusty: I have a field E, and an algebra A over E. Now, it happens that I can construct A with an intermediate step passing through an algebra A' over E, such that A is also an algebra over A' (A' is a field wrt its ring structure, and I'm disregarding the vector field structure). I want to express the fact that the construction that passes from E to A is factorized through the construction that passes from E to A' and from A' to A by means of category theoretical diagram: how to do so? TIA, Michele === Subject: Re: Request for a simple diagram > This is very trivial but although I used to be confident with this kind > of stuff, I became rusty: I have a field E, and an algebra A over E. > Now, it happens that I can construct A with an intermediate step > passing through an algebra A' over E, such that A is also an algebra > over A' (A' is a field wrt its ring structure, and I'm disregarding the > vector field structure). > I want to express the fact that the construction that passes from E to > A is factorized through the construction that passes from E to A' and > from A' to A by means of category theoretical diagram: how to do so? I do not really know what the construction actually is to which you refer in the above. The only diagram that comes into my mind stems from the fact that (at least in the commutative setting) an E-algebra A is the same as a homomorphism f: E --> A of commutative rings. So, the situation that some E-algebra A can be viewed as an A'-algebra where A' itself is an E-algebra translates into: the homomorphism E --f-> A can be factored as E --g-> A' --h-> A . Marc === Subject: Make Money, Cash, Green, Doe, Funds, Bread, Paper Make your $6 earn $1000s in just a month with Paypal Please don't read this message if you are not bothered about HAVING WONDERFUL LIFE AND ARE CONTENT WITH WHAT YOU ARE EARNING. This is not a scam, I have done it myself! Can you really make money so easily? I thought it was impossible. Just read this. I don't even have to convince you that this is not a scam, because it makes logical sense how you can earn money through paypal. A little while back, I was browsing through a newsgroup, just like you make thousands dollars within weeks with only an initial investment of $6.00! So I thought, Yeah, right, this must be a scam, but like most of us, I was curious, so I kept reading... is some of it!! 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You will begin receiving money within days! JUST MAKE SURE THE EMAIL YOU SUPPLY IS EXACTLY AS IT APPEARS ON PAYPAL. You're done with your first one. Please Be Honest and it will truly work for you.. Congrats! === Subject: Homology, irrational foliation of the torus If I take the irrational foliation of the torus, that is, the image of a line with irrational slope under the covering map R^2 ----> Torus, and identify the foliation to a point, how can I calculate its homology? I have done it in the case that we take a line with rational slope, but irrational I don't see how to put a CW structure on it. For irrational I only needed one 0-cell, 2 one-cells, and two 2-cells. James === Subject: Re: Homology, irrational foliation of the torus > If I take the irrational foliation of the torus, that is, the image of a > line with irrational slope under the covering map R^2 ----> Torus, and > identify the foliation to a point, how can I calculate its homology? I have > done it in the case that we take a line with rational slope, but irrational > I don't see how to put a CW structure on it. For irrational I only needed ^^ rational slope? > one 0-cell, 2 one-cells, and two 2-cells. > James Robert Low has put his finger on the problem: that irrationally-sloped line is dense in the torus, which means that the identification point is dense in the quotient. As a result, all the things you want to do to compute homology (triangulation, singular complex, long exact sequence) are not going to behave nicely. Typically, the subspaces we like when doing ordinary homology are those for which the homotopy extension property holds (equivalently, those which have a neighborhood of which the subspace is a retract). Have a look at your example, and you'll find it failing spectacularly. You might try .89ech homology, but you're going to need to know what open covers of your space look like. If I'm remembering correctly, it was problems like this that led to Connes's study of noncommutative geometry. I know next to nothing (I believe I'm in an epsilon^(HUGE NUMBER) - neighborhood of nothing in terms of knowledge here) about the topic, but you may want to browse around those topics (i.e., .89ech homology, noncommutative geometry), as well as (co)homology of foliations and orbit spaces. BTW, if you are a beginning student, the full answer to this particular question can be safely put off for a little while. Dale === Subject: Re: Homology, irrational foliation of the torus > If I take the irrational foliation of the torus, that is, the image of a > line with irrational slope under the covering map R^2 ----> Torus, and > identify the foliation to a point, how can I calculate its homology? I have > done it in the case that we take a line with rational slope, but irrational > I don't see how to put a CW structure on it. Does a CW complex have to satisfy any nice separation axiom? Because the quotient space of the the torus with points lying on a line of fixed irrational slope isn't even T^0: even point is dense. === Subject: The Indirect and Direct proofs of 4 Color Mapping Problem, using Moebius theorem Kepler Packing Problem --- quoting Wikipedia --- In mathematics, the Kepler conjecture is a conjecture about sphere packing in three dimensional Euclidean space. It says that no arrangement of equal spheres filling space has a greater average density than that of the cubic close packing (face centred cubic) and hexagonal close packing arrangements. The density of these arrangements is a little over 74%. Experiment shows that dropping the spheres in randomly will achieve a density of around 65%. However, a higher density can be achieved by carefully arranging the spheres as follows. Start with a layer of spheres in a hexagonal lattice, then put the next layer of spheres in the lowest points you can find above the first layer, and so on - this is just the way you see oranges stacked in a shop. This natural method of stacking the spheres creates one of two similar patterns called cubic close packing and hexagonal close packing. Each of these two arrangements has an average density of pi/sqrt18 = 0.74048... The Kepler conjecture says that this is the best that can be done - no other arrangement of spheres has a higher average density than this. --- end quoting Wikipedia --- Kepler Packing Problem Proof: Suppose false then there exists a cell which has 13 kissing points and not the 12 kissing points. Contradiction to the Euler regular polyhedra formulas for vertices and faces which can only come in 4,6,8,12,20. There cannot exist a regular polyhedra with 13 vertices, hence impossible to have 13 kissing point cell. So 12 kissing points is the maximum for each cell of KPP. 4Color Mapping Problem --- quoting Wikipedia --- The four color theorem states that given any plane separated into regions, such as a political map of the counties of a state, the regions may be colored using no more than four colors in such a way that no two adjacent regions receive the same color. Two regions are called adjacent if they share a border segment, not just a point. Each region must be contiguous: that is, it may not consist of separate sections like such real countries as Angola, Azerbaijan, and the United States. --- end quoting Wikipedia --- 4 Color Mapping proof (Indirect Method): Define cell as every specific region and its configuration of neighbors as to adjacency and point-intersections which is formally called neighborhood, but I like the concept of cell which comes from the proof of Kepler Packing KPP and both KPP and 4 CM are similar proofs in which one guides the other. Define Topological Reduction TR as the combining and bending into shapes a given number of adjacent neighbors which is formally known as graph minor. Suppose 4CM is false then implies there exists a cell-requiring-5-colors. Examine this cell for it must have at least 5 adjacencies according to the definition of Map Coloring in order to receive and require 5 colors. There are four cases involved. If this required- 5 color cell is 5 mutual adjacencies then enter Moebius theorem which says that 4 mutual adjacencies is the maximum planar map. If this required 5 color cell is not 5 mutual adjacencies but rather 5 nonmutual adjacencies or 6 or more mixed adjacencies then apply a Topological Reduction to those adjacencies to reduce them to 5 mutual adjacencies and enter again the Moebius theorem. If there is a multi-cell requiring 5 colors then it can be reduced by Topological Reduction to one cell of 5 mutual adjacencies. If the entire plane of coloring as a whole requires 5 colors, then it can be reduced topologically to one cell of 5 mutual adjacencies, enter Moebius theorem. Contradiction; hence 4 Color Mapping. This is the 4 mutual adjacency triangle / / / / /2 / 1 / / 4 --------------- 3 --------------- In 3rd dimension this is a picture of the most simple X-numbered Mutual Adjacencies Model: 11111111111111111111111113411111561111111111111111 22222222222222222222222223422225622222222222222222 3333333333333333333333333 56 56 4444444444444444444444444 56 56 555555555555555555555555555555 56 666666666666666666666666666666 56 34 56 34 56 34 56 34 56 34 56 Direct Proof of 4 Color Mapping: Given the entire plane, whatever configuration exist in the plane, by Topological Reduction those figures can be combined so that the entire plane is one figure thus requiring 1 color. Moving on to 2 required colors. Given the entire plane and no matter what figures provided there are at least 2 figures in the plane, that any such figures can be Topologically Reduced to 2 mutual adjacencies and hence requiring 2 colors. Moving on to 3 required colors which is the same as 2 except the plane has 3 or more figures at a minimum. Moving on to 4 required colors which is the same as 3 only the plane has at least 4 figures at the start. Moving on to 5 required colors and such a plane has at least 5 figures at the start and no matter how many more figures it has for it has at least 5 figures that Topological Reduction reduces all those figures to 5 mutual adjacencies as we can picture in the above 3rd dimension where all the other figures are combined with the 5th wire and so the entire 3rd dimension is that 5 mutual adjacency. But enter the Moebius theorem that says in the Plane there cannot be any mutual adjacencies above the number 4. Moving on to 6 or more required colors and the Moebius theorem prohibits those also. So that given any number X of required colors, topological reduction satisfies the number X and the Moebius theorem determines of that configuration exists or not. For X from 1 to 4 there is existence in the plane and for X from 5 to infinity there is no existence in the plane. End of Proof. I had to do both Direct and Indirect because I am arguing the truth or falsity of a Conjecture that the Indirect and Direct of a proof have to be different and that they cannot be identical concerning the Infinitude of Primes proof. Two Conjectures of Logic: (a) The direct method cannot be identical to the indirect method but has a amount of variance. (b) Every proof in mathematics has both a direct method and indirect method. So I am testing to see if those conjectures are true. Archimedes Plutonium www.iw.net/~a_plutonium whole entire universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Symbolic Logic proofs of Infinitude of Primes, both direct and indirect method Yesterday I gave two conjectures stated as such: (a) Every proof in mathematics that has a direct method also has a indirect method; vice versa (b) The direct and indirect method cannot be identical Both came up because of the ongoing debate that in the Euclid Infinitude of Primes, indirect method that the constructed new number N is necessarily prime. Whereas in the direct method a prime factor search of N must be conducted. Here is my assessement of the problem. A Symbolic Logic write up of Infinitude of Primes both Direct and Indirect will reach a point in the symbols where a new prime is found. In the Direct method the new prime is either N or a prime factor of N. In the Indirect, there cannot be a prime factor search and that N must necessarily be the only new prime and no other candidate. In the Indirect method, you formed this new number of N and your supposition space says those are all the primes that EXIST. So can you say anywhere after you formed N that a prime factor is unaccounted for? I say no. I say that once you formed N, and given your space of all the primes in existence leave a remainder of 1, that Logically leaves only one possibility, that N is necessarily prime. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Symbolic Logic proofs of Infinitude of Primes, both direct and indirect method For example, one of my critics is Chris Heckman who says that his rendition is a valid Infinitude of Primes, indirect method: > Indirect: Assume there are finitely many primes, P(1), P(2), ..., P(k). Let > N = P(1) * P(2) * ... * P(k) + 1, and let Q be a prime factor of N. Q > cannot be any of P(1), ..., P(k), since N divided by P(i) is 1. This > contradicts the assumption that we had a list of all the primes. This is why I need the full SYMBOLIC LOGIC write up of Infinitude of Primes proof both Direct and Indirect. Chris thinks that there are 2 candidates to prompt a contradiction, of either N itself or a prime factor Q. I argue that in the Indirect Method, we cannot discuss any prime factor Q because the list has all the primes in existence and the only candidate is N itself. So in the Indirect Method, N is necessarily prime and delivers the contradiction. So a complete Symbolic Logic write up will evince us of who is correct and who is wrong. Because in this Symbolic Logic write-up there is a step of the Indirect that says Primes 2,3,5,...P (where P is the last and final primes) are all the primes that exist. and when N is formed, it is necessarily a new prime given that all the primes in existence are listed beforehand. So you cannot be going back with a double contradiction of hunting for a factor Q. In the Direct Method, the hunt for Q is always ongoing. So in the Symbolic Logic write up, the Universal of every and any and the existential of there exists and these are all that exist shapes what the final steps of the proof are in Indirect. And that a hunt for Q in the Indirect Method is not tenable. The Calculus of the Symbols of a Symbolic Logic write up of Infinitude of Primes will answer finally, whether a Prime Factor Search of Q is possible or impossible in the Indirect Method. I instinctively know that such a write up favors me, because I instinctively know that you cannot have simultaneously that N is necessarily prime, yet waiver that Q is also needed. And that implies there is no difference between a Indirect and Direct proof, that at one point of the proof, the direct matches the indirect line per line. The above is very important, not just to clear and correct our understanding, but the above is the key to proving Infinitude of Twin Primes and all even-numbered metric primes for the N's deliver them. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Symbolic Logic proofs of Infinitude of Primes, both direct and indirect method I just ordered the book SYMBOLIC LOGIC by Thomason, which is the book I used some 30 years ago (gee time sure flys) and will try to use the same symbolism as I post a Symbolic Logic proofs of Infinitude of Primes both Direct and Indirect. We need to pinpoint the exact steps for which N is necessarily prime and cannot be a prime factor search for a Q. I believe the steps will be: (i) Every prime in existence is listed as such 2,3,5,....P where P is the last and final prime (xxx) Construct N = ((2x3x5x...xP)+1), where all the primes that exist are multiplied together and add 1. (xxxx) a number N can be only prime or composite (skip 1 since it is not 1) (xxxxx) N is not composite because of Unique Prime Factorization theorem for all the primes that exist leave a remainder of 1 (xxxxxxx) N is necessarily a new prime not on the list of all primes in existence So basically, I sense that the step in the Formal Symbolic Logic Proof that shows us that N is necessarily prime is the step that says These are all the primes in existence. That Symbolic Logic phrase These are all the Primes in Existence will forbid any future step to conduct a prime factor Q search and will only admit N as the new prime. I will wait for this book to arrive to begin posting a Symbolic Logic proof. Instead of the strange symbols for every I will just use A instead of the familar upside down A, and for existence I will use E instead of the reverse E. The Internet is not conducive to logic symbols so will improvise. And I wanted to get back to physics a week or so ago, but I cannot leave mathematics in such a mess of a condition. And now it looks as though I have to revise the history of Euclid's own proof for it definitely was a Direct Proof of Infinitude of Primes and that although the glimmers of an emerging Reductio Ad Absurdum can be seen in Euclid's method, his proof is direct with a subunit of reductio ad absurdum. So I need to correct the history of mathematics in this regard. And postpone my return to some questions of physics such as superconductivity and why gravity is a fictional force and not a Coulombic pair to antigravity in the space region of electrons. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Symbolic Logic proofs of Infinitude of Primes, both direct and indirect method > I just ordered the book SYMBOLIC LOGIC by Thomason, which is the book I > used some 30 years ago (gee time sure flys Especially when you're not learning anything. ) and will try to use the > same symbolism as I post a Symbolic Logic proofs of Infinitude of > Primes both Direct and Indirect. Ah, that's just what we need. More areas of mathematics for Mr. Plutonium to mangle beyond recognition. How many crutches does our hero need to take one step forward without tumbling into an Inspector Clouseauish tangle? > We need to pinpoint the exact steps for which N is necessarily prime > and cannot be a prime factor search for a Q. > I believe the steps will be: > (i) Every prime in existence is listed as such 2,3,5,....P where P is > the last and final prime > (xxx) Construct N = ((2x3x5x...xP)+1), where all the primes that exist > are multiplied together and add 1. Good Heavens. We have actually been present at the very birth of one of Mr. Plutonium's brilliant, out-of-the-darkness, strokes of genius - the switch from Roman numerals (i) to Plutonium numerals (I will proudly claim credit for the term) (xxx). > (xxxx) a number N can be only prime or composite (skip 1 since it is > not 1) > (xxxxx) N is not composite because of Unique Prime Factorization > theorem for all the primes that exist leave a remainder of 1 Absolutely first class logic here, Mr. Plutonium. Did you recently learn of the Unique Prime Factorization theorem? You tend to mention the new stuff a lot, until you kind of forget what you thought they meant. > (xxxxxxx) N is necessarily a new prime not on the list of all primes in > existence Of course. It's just a sign of your innate honesty and fairness that you make this a PROVISIONAL step in your proof, after it's been shown incorrect for I think going on about 8 months, so far. > So basically, I sense that the step in the Formal Symbolic Logic Proof > that shows us that N is necessarily prime is the step that says These > are all the primes in existence. I shiver, Mr. Plutonium, at the vast scope and sweep of your intuitions. I stand in awe. > That Symbolic Logic phrase These are all the Primes in Existence will > forbid any future step to conduct a prime factor Q search and will only > admit N as the new prime. We are all grateful to Mr. Plutonium for sharing even the inchoate beginnings of his thoughts, as he is doing here. Of course they make no sense. Did da Vinci's sketches of the 'Mona Lisa' show the genius of the final product? No.Well, maybe THEY did. > I will wait for this book to arrive to begin posting a Symbolic Logic > proof. Instead of the strange symbols for every I will just use A > instead of the familar upside down A, and for existence I will use E > instead of the reverse E. The Internet is not conducive to logic > symbols so will improvise. Mr. Plutonium, use any symbols you like, just so long as you share your insights. We will walk miles just to follow in your footsteps. Even where you mistake all those brown patches for Shinola. Anyone care to bet on any of the following: Mr. Plutonium only reads a few pages, gets tired of the mental effort, and concludes (by posting an elaborate and totally nonsensical argument) that he was right after all. Mr. Plutonium doesn't read ANY pages, and concludes (by posting an elaborate and totally nonsensical argument) that he was right after all. > And I wanted to get back to physics a week or so ago, but I cannot > leave mathematics in such a mess of a condition. And now it looks as > though I have to revise the history of Euclid's own proof for it > definitely was a Direct Proof of Infinitude of Primes and that although > the glimmers of an emerging Reductio Ad Absurdum can be seen in > Euclid's method, his proof is direct with a subunit of reductio ad > absurdum. So I need to correct the history of mathematics in this > regard. And postpone my return to some questions of physics such as > superconductivity and why gravity is a fictional force and not a > Coulombic pair to antigravity in the space region of electrons. > Archimedes Plutonium > www.iw.net/~a_plutonium > whole entire Universe is just one big atom > where dots of the electron-dot-cloud are galaxies Oh no, please Mr. Plutonium, physics needs you! I distinctly hear it calling. Oh Archimedes! Archimedes! Can't you hear it? Mathematics will bravely struggle on without your insights. Ken === Subject: Bitwise OR and AND without bitwise operators.... I am working in a programming language (GLSL) which does not support bitwise operators, but I need to perform the following types of calculations: float b = (float)((counts & maskb) >> offsetg) / gbitres; I can do the bit shifts by Bit Shift Right ... i >> N = i / 2^N, Bit Shift Left... i << N = i * 2^N, right, but how can I do the bitwise AND? or a bitwise OR? Ed === Subject: Re: Bitwise OR and AND without bitwise operators.... > I am working in a programming language (GLSL) which does not support > bitwise operators, but I need to perform the following types of > calculations: > float b = (float)((counts & maskb) >> offsetg) / gbitres; > I can do the bit shifts by > Bit Shift Right ... i >> N = i / 2^N, > Bit Shift Left... i << N = i * 2^N, > right, but how can I do the bitwise AND? or a bitwise OR? > Ed Use text. from gmpy import * ORTT = {'00':'0','01':'1','10':'1','11':'1'} ANDTT = {'00':'0','01':'0','10':'0','11':'1'} a = 12345 b = 54321 A = digits(a,2) B = digits(b,2) print print 'a:',a print 'A:',A print 'b:',b print 'B:',B print if len(B)!=len(A): if len(B)>len(A): diff = len(B)-len(A) A = '0'*diff + A else: diff = len(A)-len(B) B = '0'*diff + B C = ''.join([ORTT[A[i]+B[i]] for i in range(len(A))]) print ' ',A print ' OR',B print ' ','-'*len(A) print ' ',C c = long(C,2) print print a,'OR',b,'=',c print print C = ''.join([ANDTT[A[i]+B[i]] for i in range(len(A))]) print ' ',A print ' AND',B print ' ','-'*len(A) print ' ',C c = long(C,2) print print a,'AND',b,'=',c a: 12345 A: 11000000111001 b: 54321 B: 1101010000110001 0011000000111001 OR 1101010000110001 ---------------- 1111010000111001 12345 OR 54321 = 62521 0011000000111001 AND 1101010000110001 ---------------- 0001000000110001 12345 AND 54321 = 4145 === Subject: Re: Bitwise OR and AND without bitwise operators.... >I am working in a programming language (GLSL) which does not support > bitwise operators, but I need to perform the following types of > calculations: > float b = (float)((counts & maskb) >> offsetg) / gbitres; > I can do the bit shifts by > Bit Shift Right ... i >> N = i / 2^N, > Bit Shift Left... i << N = i * 2^N, > right, but how can I do the bitwise AND? or a bitwise OR? > Ed You have to write your own routines. shifting won't get you to AND or OR or NAND or NOT or NOR or XOR === Subject: Re: Bitwise OR and AND without bitwise operators.... > I am working in a programming language (GLSL) which does not support > bitwise operators, but I need to perform the following types of > calculations: > float b = (float)((counts & maskb) >> offsetg) / gbitres; > I can do the bit shifts by > Bit Shift Right ... i >> N = i / 2^N, > Bit Shift Left... i << N = i * 2^N, > right, but how can I do the bitwise AND? or a bitwise OR? the lsb of n would be n-((n/2)*2). the 2nd lsb of n would be m = n/2; m-((m/2)*2) etc. Notice you can re-use some results if you work lsb to msb. So just iterate bitwise yourself and add 2^k to your result when the bitwise operation on the kth bit is 1. I don't know if this is the fastest way, but without bitwise arithmetic support (which is why I assume bitwise operations are reserved), I would think any solution would have to iterate over the operand's bits. === Subject: Re: Bitwise OR and AND without bitwise operators.... Is this correct? Bit k, of n would be m = n/k, m-((m/2)*2) So could I do this int testValue = 41; for(int nBit = 0; nBit < numberOfBits; nBit++) { m = testValue/nBit; int bitSet = m-((m/2)*2) } CD === Subject: Re: Bitwise OR and AND without bitwise operators.... > Is this correct? Bit k, of n would be > m = n/k, m-((m/2)*2) > So could I do this > int testValue = 41; > for(int nBit = 0; nBit < numberOfBits; nBit++) { > m = testValue/nBit; > int bitSet = m-((m/2)*2) > CD Formula: float b = (float)((counts & maskb) >> offsetg) / gbitres First note that ((counts & maskb) >> offsetg) is the same as (counts >> offsetg)&( maskb)>> offsetg) which will same you time. In psuedo C... float and_shift(int a, int b, int s) { int r=0,ha,hb,m,d=pow(2,s); a/=d; b/=d; m = min(a,b); for(int k =1;k<=m;k*=2) { ha = a/2; hb = b/2; if(a-ha*2) { if(b-hb*2) { r+=k; } } a=ha; b=hb; } return float(r); } float b = add_shift(counts,maskb,offsetg)/gbitres; === Subject: Re: Bitwise OR and AND without bitwise operators.... > I am working in a programming language (GLSL) which does not support > bitwise operators, but I need to perform the following types of > calculations: > float b = (float)((counts & maskb) >> offsetg) / gbitres; > I can do the bit shifts by > Bit Shift Right ... i >> N = i / 2^N, > Bit Shift Left... i << N = i * 2^N, > right, but how can I do the bitwise AND? or a bitwise OR? You could write AND and OR subroutines that decompose their input operands to series of bits, then compose outputs by if-tests. called comp.graphics.api.opengl and I imagine you should post there. Note, although &, ^, | are illegal in GLSL, at the moment, for ordinary operands, [per table 5.1 in following - http://oss.sgi.com/projects/ogl-sample/registry/ARB/GLSLangSpec.Full.1.10.59 .pdf ] you can do compile-time bitwise ops on constants - see p. 16 in above. -jiw === Subject: http://arxiv.org/abs/gr-qc/0602022 re: http://arxiv.org/abs/gr-qc/0602022 If Waldyr's criticisms are technically correct and your mathematical errors are reparable, then a detailed response to Waldyr's published commentary can only improve the strength of your arguments. The version up there now corrects some very minor formal flaws in the presentation of STANDARD physics that was simply background material that Waldyr wanted more rigorous. Also he accused me of things that were not so, but that may be because I was not clear enough. For example, I certainly knew that there are 4 distinct tetrad first rank Diff(4) tensor 1-forms e^a and 6 zero torsion spin connections S^a^b, i.e. e^a = e^audx^u S^a^b = S^a^budx^u The great bulk of Waldyr's 21 pages has little relevance to the latest version posted now http://arxiv.org/abs/gr-qc/0602022 The vexing Yilmaz issue of locality vs nonlocality of pure gravity energy that Waldyr's general remarks on energy-momentum conservation in GR in his 21 pages are not at all relevant to my new conjectures, i.e. I. tetrad field emerges from the several Goldstone phases of LOCAL vacuum coherent order parameter in similar way that the superfluid velocity field emerges from the single Goldstone phase of the local helium ground state coherence order parameter. That's the key idea. I now seem to recall that Hagen Kleinert has 4 phases associated with the tetrads, but not specifically in connection with vacuum coherence? I will have to look at his book - it's a vague memory. i.e. the PHYSICAL IDEA is that smooth c-number tetrad field is a vacuum coherence effect - not a random ZPF effect. II. LHC & local DM detectors will not click with the Right Stuff through space is like looking for Earth's motion through the old mechanical aether (distinct from CMB Doppler shifts relative to FRW Hubble flow of course). III. Bekenstein-Hawking and 't-Hooft-Susskind ideas point to a 2D quantized period DeRham integral from point defects in the vacuum coherence i.e. area quantization and volume without volume from a closed non-exact area density 2-form emergent from the Goldstone phases. Bohm-Aharonov effect is area without area i.e. S1 non-trivial first homotopy - quantized loop integrals of closed nonexact 1-forms around line defects World Hologram is volume without volume i.e. S2 non-trivial second homotopy - quantized closed surface integrals of closed non-exact 2-forms around point defects, e.g. one at center of Sun in Pioneer anomaly? This would be a hollow dark matter (exotic vacuum) mini-halo of the Sun. Also, you may not have expressed yourself as clearly as you might have in presenting some of your points. Some disambiguation in response to criticisms appearing in Waldyr's paper might be a good thing. Sure. These are all new ideas not found in literature. I agree with Tony that Waldyr's charges of ad hocness are questionable. You evidently have a clear intuitive understanding of the basic physical model you are proposing. However, while I entirely agree that mathematical rigor is secondary to the physics, it does appear that several of Waldyr's points are not merely a matter of rigor mortis, but are concerned with the proper definitions of important mathematical terms. Yes, I think the current version is essentially OK in that regard. I basically eliminated all reference to what Waldyr found objectionable in the background stuff that is not essential to my thesi. As for Waldyr's intemperate language, perhaps it would have been better if you hadn't cited him at all in later versions of your paper? I thought I was being polite and honest and I made it clear that it should not be construed that Waldyr agreed with my new ideas. In fact Waldyr's contribution is forcing me to confront the need for 8 Goldstone phases instead of only 2 and then I associated the extra 6 with Calabi-Yau and massive torsion fields, i.e. both locally gauge Lorentz group and then hide its symmetry in the vacuum - i.e. non-Abelian Meissner effect for the torsion fields whose mass gap prevents us from seeing it easily. IV. Extended local equivalence principle, i.e. locally gauge ALL universal space-time symmetries of the physical action to induce geometrodynamic fields beyond the original 1915 curvature field. Note that at the square root tetrad level the curvature and torsion gauge potentials are SPIN 1 just like in Yang-Mills theory. When you go to geometrodynamic level ds^2 = e^a(Minkowski)abe^b Then you get a COMPOSITE spin 2 from entangled pairs of subspace SPIN 1 tetrad level fields. Z. ... My key physical idea is dark matter is same as dark energy with difference only in scale and the sign of the energy density and w < -1/3 and from a distance it mimics w = 0 CDM. It is obvious to me that the curved tetrad & Bekenstein-Hawking quantized area is like the superfluid circulation and that t Hooft-Susskind world hologram volume without volume is like Bohm-Aharonov magnetic field without magnetic field even if I did not get the formal math completely right on first try working alone outside of academia - Einstein needed Grossman to get his tensors right and he took ten years. I only took a few weeks so far and have not found my Grossman I guess? :-) Remember what Feynman said about rigor mortis. Are Feynman integrals rigorous? Is M theory? I hear some new math is coming from M theory however. I think Waldyr is making a general point about lack of mathematical rigor in many physics papers. Waldyr is actually a Professor of Mathematics I think and there is a cultural difference. To me the math is secondary to the heuristic ideas. === Subject: X -> LOG|X| CALCULATE LOG|2| = B_1 CALCULATE LOG|B_1| = B_2 CALCULATE LOG|B_2| = B_3 ETC. ETC. ETC.!!!!!!!!!!!!!!!!!!!!!!!!! THE BASE OF THE LOGARITHM IS D!!!!!!!!!!!!!!!!!!! THE NUMBER I START WITH CAN'T BEE 0!!!!!!!!!!!!!!!!!!!!!!!!! OR 1 OR -1!!!!!!!!!!!!!!!!!!!!!!!!!!!! OR E OR -E OR 1/E OR -1/E!!!!!!!!!!!!!!!!!!!!!!! ETC. ETC. ETC.!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! I.E. B CAN'T BEE 0!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! THE NUMBER I START WITH CAN'T BEE SUCH THAT B IS PERIODIC CALCULATE THE ARITIMETIC MEAN OF THE NUMBERS!!!!!!!!!!!!!!!!!!!!!!!!!!!!! B_1 B_2 B_3 B_4.. ..B_N !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! =C_N Hypothesis : THAT THE LIMIT OF C_N/N EXIST AND IS INDEPENDENT OF THE NUMBER I START WITH!!!!!!!!!!!!!!! BUT DEPENDENT ON THE BASE OF THE LOGARITHM D!!!!!!!!!!!!!!!!!!!!!!!!! NOTE: THE:!!!!!!!!!!!!!!!!! ARE ARE!!!!!!!!!!!!!!!!!!!!!!!!! === Subject: Re: X -> LOG|X| > CALCULATE LOG|2| = B_1 > CALCULATE LOG|B_1| = B_2 > CALCULATE LOG|B_2| = B_3 > ETC. ETC. ETC.!!!!!!!!!!!!!!!!!!!!!!!!! > THE BASE OF THE LOGARITHM IS D!!!!!!!!!!!!!!!!!!! > THE NUMBER I START WITH CAN'T > BEE 0!!!!!!!!!!!!!!!!!!!!!!!!! > OR 1 OR -1!!!!!!!!!!!!!!!!!!!!!!!!!!!! > OR E OR -E OR 1/E OR -1/E!!!!!!!!!!!!!!!!!!!!!!! > ETC. ETC. ETC.!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! > I.E. B CAN'T BEE 0!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! > THE NUMBER I START WITH CAN'T > BEE SUCH THAT B IS PERIODIC > CALCULATE THE ARITIMETIC MEAN OF THE NUMBERS!!!!!!!!!!!!!!!!!!!!!!!!!!!!! > B_1 B_2 B_3 B_4.. ..B_N !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! > =C_N > Hypothesis : > THAT THE LIMIT OF C_N/N arithmetic mean has already been divided by N, so you probably do not want to divide again > EXIST AND IS INDEPENDENT OF THE NUMBER I START WITH!!!!!!!!!!!!!!! > BUT DEPENDENT ON THE BASE OF THE LOGARITHM D!!!!!!!!!!!!!!!!!!!!!!!!! > NOTE: THE:!!!!!!!!!!!!!!!!! ARE ARE!!!!!!!!!!!!!!!!!!!!!!!!! that limit is the solution x of log(-x)=x, for base d can be written x=-LambertW(ln(d))/ln(d), and for the special case of base e, x=-LambertW(1) . The sequence b_n itself seems pretty chaotic, but the means c_n seem to settle down (if slowly). === Subject: Re: X -> LOG|X| I don't know about others, but I don't even bother to read posts which are in written in difficult to read all caps. Bye. --Lynn === Subject: The Indirect and Direct proofs of 4 Color Mapping Problem, using Moebius theorem a.p: Kepler Packing Problem Proof: Suppose false then there exists a cell which has 13 kissing points and not the 12 kissing points. Contradiction to the Euler regular polyhedra formulas for vertices and faces which can only come in 4,6,8,12,20. There cannot exist a regular polyhedra with 13 vertices, hence impossible to have 13 kissing point cell. So 12 kissing points is the maximum for each cell of KPP. N.B: THERE CAN'T BEE A CELL WHICH HAS 7 KISSING POINTS!!!!!!!!!!!!!!! === Subject: Re: The Indirect and Direct proofs of 4 Color Mapping Problem, using Moebius theorem N.B: THERE CAN'T BEE A CELL WHICH HAS 7 KISSING POINTS! Yours is a common logical fallacy of that of injection of an irrelevant fact and questioning the truth value. If I said to you the cell in a human body has no fusion energy reactions is it relevant to Kepler Packing cells. The mistake of Nils is that density is a convergence parameter and the proof needs to know only if the convergence is greater than 12 kissing points. Not whether convergence is 11 or below, nor whether convergence is greater than 13. The Logical concepts of density involve only the question of a 13 kissing point cell. Your fallacy is that of examples of Irrelevancies. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: exponent question Which rule indicates that, for instance, that -3^2 = -9 and not 9? You certainly wouldn't know this by looking at the exponent laws. If I were to invent my own rule, I was thinking simply -x^n = -(x)^n, but what about an example like -6y^2? According to my rule this could equal -(6y)^2, which is clearly wrong because the exponent should only apply to the y. Then I could write the rule xy^n = x(y)^n, but there is still conflict with the other rule. The exponent rules work well but negative bases still confuse me. === Subject: Re: exponent question >Which rule indicates that, for instance, that -3^2 = -9 and not 9? You >certainly wouldn't know this by looking at the exponent laws. If I were >to invent my own rule, I was thinking simply >-x^n = -(x)^n, >but what about an example like -6y^2? According to my rule this could >equal -(6y)^2, which is clearly wrong because the exponent should only >apply to the y. Then I could write the rule xy^n = x(y)^n, but there >is still conflict with the other rule. The exponent rules work well >but negative bases still confuse me. Precedence rules. ^ has higher precedence than *, which has higher precedence than -. So -6y^2 is parsed as -(6*(y^2)). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: exponent question Precedence rules > Yes, but whose rules? Programming languages are notoriously > inconsistent about the relative precedence of exponentiation, unary > minus, and literal minus (part of a signed number). Iow, even -x and > -3 are sometimes treated differently. It might be noted that the Google calculator -- bless its little electronic heart -- is still giving -3^2 = 9 and such. David === Subject: Re: exponent question >> Precedence rules > Yes, but whose rules? Programming languages are notoriously > inconsistent about the relative precedence of exponentiation, unary > minus, and literal minus (part of a signed number). Iow, even -x and > -3 are sometimes treated differently. > Michel. That is an implementation problem of the specific programming language, and complier. Not mathematics. The OP had left off the parenthetical, therefore he will make many mistakes in the future. === Subject: order statistics for distribution-free confidence intervals for population percentiles Seem there are not much discussions about order statistics here. I am new to this area, here is a question I just met. It is possible to derive distribution-free confidence intervals for population percentiles. For example, if we took 29 independent samples from any population, we can be 95% confident that 90% fo the population would below the largest order statistic. However, what if the samples are corrupted by noise, i.e., each sample we pick becomes si + ri, where si is the true value of the sample, and ri is a random noise. I was just wondering whether there is any work discussing this problem in the literature. Roy === Subject: Re: order statistics for distribution-free confidence intervals for population percentiles >Seem there are not much discussions about order statistics here. >I am new to this area, here is a question I just met. >It is possible to derive distribution-free confidence intervals for >population percentiles. For example, if we took 29 independent samples >from any population, we can be 95% confident that 90% fo the population >would below the largest order statistic. >However, what if the samples are corrupted by noise, i.e., each sample >we pick becomes >si + ri, where si is the true value of the sample, and ri is a random >noise. I was just wondering whether there is any work discussing this >problem in the literature. You'll need some assumption about the distribution of the noise. Suppose, e.g., you know that Prob(r_i > 0) = p, 0 < p < 1. If Prob(s_i > x) = a, then 1 - (1-p)(1-a) >= Prob(s_i + r_i > x) >= p a. Thus if X[n] is the largest order statistic for a sample of size n, Prob(X[n] <= x) = Prob(s_i + r_i <= x)^n <= (1 - p a)^n So e.g. if p = 1/2, with a sample size n = 59 we can still have 95% confidence that at least 90% of the population is below the largest order statistic, because (1 - 1/2 * .1)^59 < 0.05. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: X -> LOG|X| >that limit is the solution x of log(-x)=x, NO!!!!!!!!!! IT IS NOT SO SIMPLE!!!!!!!!!!!!!!!!!! APPROXIMATE::::::: BASE LIMIT 2,0 -0,39 E,0 -0,57 5,0 -0,556 10,0 -0,496 === Subject: Pacific Coast Gravity Meeting 2006 http://www.physics.ucsb.edu/~relativity/22nd-PCGM.html I am giving a 10 minute talk at above meeting it is at http://qedcorp.com/APS/PCGM2006.pdf or http://qedcorp.com/APS/PCGM2006.ppt typo-corrected 2nd draft below If Waldyr's criticisms are technically correct and your mathematical errors are reparable, then a detailed response to Waldyr's published commentary can only improve the strength of your arguments. The version up there now corrects some very minor formal flaws in the presentation of STANDARD physics that was simply background material that Waldyr wanted more rigorous. Also he accused me of things that were not so, but that may be because I was not clear enough. For example, I certainly knew that there are 4 distinct tetrad first rank Diff(4) tensor 1-forms e^a and 6 zero torsion spin connection 1-forms S^a^b, i.e. e^a = e^audx^u S^a^b = S^a^budx^u The great bulk of Waldyr's 21 pages has little relevance to the latest version posted now http://arxiv.org/abs/gr-qc/0602022 The vexing Yilmaz issue of locality vs nonlocality of pure gravity energy that Waldyr's general remarks on energy-momentum conservation in GR in his 21 pages pertains to are not at all relevant to my new conjectures, i.e. I. tetrad field emerges from the several Goldstone phases of LOCAL vacuum coherent order parameter in similar way that the superfluid velocity field emerges from the single Goldstone phase of the local helium ground state coherence order parameter. That's the key idea. I now seem to recall that Hagen Kleinert has 4 phases associated with the tetrads, but not specifically in connection with vacuum coherence? I will have to look at his book - it's a vague memory. i.e. the PHYSICAL IDEA is that smooth c-number tetrad field is a vacuum coherence effect - not a random ZPF effect. II. LHC & local DM detectors will not click with the Right Stuff through space is like looking for Earth's motion through the old mechanical aether (distinct from CMB Doppler shifts relative to FRW Hubble flow of course). III. Bekenstein-Hawking and 't-Hooft-Susskind ideas point to a 2D quantized period DeRham integral from point defects in the vacuum coherence i.e. area quantization and volume without volume from a closed non-exact area density 2-form emergent from the Goldstone phases. Bohm-Aharonov effect is area without area i.e. S1 non-trivial first homotopy - quantized loop integrals of closed nonexact 1-forms around line defects World Hologram is volume without volume i.e. S2 non-trivial second homotopy - quantized closed surface integrals of closed non-exact 2-forms around point defects, e.g. one at center of Sun in Pioneer anomaly? This would be a hollow dark matter (exotic vacuum) mini-halo of the Sun. Also, you may not have expressed yourself as clearly as you might have in presenting some of your points. Some disambiguation in response to criticisms appearing in Waldyr's paper might be a good thing. Sure. These are all new ideas not found in literature. I agree with Tony that Waldyr's charges of ad hocness are questionable. You evidently have a clear intuitive understanding of the basic physical model you are proposing. However, while I entirely agree that mathematical rigor is secondary to the physics, it does appear that several of Waldyr's points are not merely a matter of rigor mortis, but are concerned with the proper definitions of important mathematical terms. Yes, I think the current version is essentially OK in that regard. I basically eliminated all reference to what Waldyr found objectionable in the background stuff that is not essential to my thesi. As for Waldyr's intemperate language, perhaps it would have been better if you hadn't cited him at all in later versions of your paper? I thought I was being polite and honest and I made it clear that it should not be construed that Waldyr agreed with my new ideas. In fact Waldyr's contribution is forcing me to confront the need for 8 Goldstone phases instead of only 2 and then I associated the extra 6 with Calabi-Yau and massive torsion fields, i.e. both locally gauge Lorentz group and then hide its symmetry in the vacuum - i.e. non-Abelian Meissner effect for the torsion fields whose mass gap prevents us from seeing it easily. IV. Extended local equivalence principle, i.e. locally gauge ALL universal space-time symmetries of the physical action to induce geometrodynamic fields beyond the original 1915 curvature field. Note that at the square root tetrad level the curvature and torsion gauge potentials are SPIN 1 just like in Yang-Mills theory. When you go to geometrodynamic level ds^2 = e^a(Minkowski)abe^b Then you get a COMPOSITE spin 2 from entangled pairs of subspace SPIN 1 tetrad level fields. Z. ... My key physical idea is dark matter is same as dark energy with difference only in scale and the sign of the energy density and w < -1/3 and from a distance it mimics w = 0 CDM. It is obvious to me that the curved tetrad & Bekenstein-Hawking quantized area is like the superfluid circulation and that t Hooft-Susskind world hologram volume without volume is like Bohm-Aharonov magnetic field without magnetic field even if I did not get the formal math completely right on first try working alone outside of academia - Einstein needed Grossman to get his tensors right and he took ten years. I only took a few weeks so far and have not found my Grossman I guess? :-) Remember what Feynman said about rigor mortis. Are Feynman integrals rigorous? Is M theory? I hear some new math is coming from M theory however. I think Waldyr is making a general point about lack of mathematical rigor in many physics papers. Waldyr is actually a Professor of Mathematics I think and there is a cultural difference. To me the math is secondary to the heuristic ideas. Note that I received from a source that does not wish to be named about Waldyr's general style of reviewing papers: But since he knows so much mathematics, he often rushes into criticism, without reading a paper carefully enough. So he thinks that the author has committed an error, whilst this was not the (our case). Of course I had some minor formal at least ambiguities in the formal expressions in 1st version of http://arxiv.org/abs/gr-qc/0602022 that Waldyr corrected, but was also accused of misconceptions I did not make - again from my, perhaps, poor exposition and noting that Waldyr is not a native English speaker. :-) === Subject: Stone-Cech compactification in modern analysis? I read in Munkres, Topology that the Stone-Cech compactification has a number of applications in modern analysis, but these lie outside the scope of this book. I would be very grateful if anyone in this group could tell a bit more about these applications. Thomas Mautsch === Subject: Is there Lynch Law in Canada? Just curious. Have just seen something. Yesterday, too. === Subject: Re: Is there Lynch Law in Canada? > Just curious. > Have just seen something. Yesterday, too. Obviously, you have a Russian-sounding name so I'm talking to a commie. That's right, ya communist! In Communist Soviet Russia, there is no more Communist Soviet Russia. Obviously, communists used to be the enemy. Now, obviously, terrorists are the enemies, of the state. Capitalists and communists are living together like cats and dogs. It's like cats and dogs living together, frogs from the sky. Ross === Subject: Re: Is there Lynch Law in Canada? > Just curious. > Have just seen something. Yesterday, too. Bondarenko here wants to lynch Maple Software. Maple V is a computer algebra system. Pertti Lounesto says the best mathematician is Maple V: Maplev, from Ontario. It's a computer algebra system, and a very able collaborator. Now it's Maple Ten. In fact, some people sit around using Maple all day. Actually, Pertti Lounesto does not say that, because he's dead. He said that. Lynch laws actually exist. It's about establishing a posse, probably to capture a fugitive. In most places, lynch laws are in effect. There are many laws about them. There are at least three. You'd have to talk to the sherriff of Maple county about it. Vote for me, write-in Ross A. Finlayson, for, uh, well, Vice President. That way, I have a titular office with little responsibility. I was elected unanimously president of Point World, the virtual reality community, in the Black Sun interactive universe with the Bacardi and the lime. I was very happy about it, I think it was at the Pool World, with the Mark Pesce and the Sci-Fi, except I don't think he was there, but, he might have been. I think he actually was. There was perhaps a quorum. There's nothing to be done about it. I even lost an election one time because I was irresponsible. In Point World, they had the first virtual state election, everybody who showed up could vote. It was for no office. I drug the anti-state guy into the state meeting, with the fanfare, and took pictures, and we had a discussion about it. I should put that on my resume. Canada? I don't know Canada lynch laws. Canada is a totally different country than the United States of America. I live in the USA, I like it here, it's very nice. All my family lives here too, not with me. I don't even have any pets. I have some plants. So, vote for me, and then I will campaign and raise money. Bondarenko, I hope it's not a big deal. Your opinions are certainly respected. Letters, of the Alphabet. In Soviet Russia, alphabet sings you. Ross === Subject: Re: LINEAR REGRESSION BY METHOD OF LEAST SQUARES > ok....i've got my formula set up to read a set of data in either Corel > Quattro or Microsoft Exel where n is the number of readings per time > period. For example: if we've got 2 readings taken at time A and the > other at time B the value of n will be 2. I think you mean one reading at time A and one at time B, don't you? It IS possible, and not uncommon, to have several readings at the same experimental point---in other words, several readings at time A or at time B, etc. So, you do not want to have this? > Similarly if there are > readings at time A, time B and time C, then the value of n will be 3. > Manually manipulating the formula to read the data at either n=2 or n=3 > is fine if it's a one time deal, but how can this formula be composed > and put into a macro in Quattro or Exel when n=2 vs n=3 vs n=4 and so > on? One way would be to create a large block, larger than you think you will ever need. You put data only in the part of the block you want in a particular case. The formulas for solving the regression problem involve sums, sums of squares, sums of products and possibly the number n. The sums should be no problem, since if you use the correct types of formulas, the blank cells won't contribute. As for n, well that is just the number of non-blank cells. However, be careful to distinguish between zero and blank: a cell containing zero should count towards n. This solution does not even need a macro. It is not elegant, but it should work: I have used such schemes plenty of times for class handouts, etc. R.G. Vickson Adjunct Professor, University of Waterloo === Subject: Re: LINEAR REGRESSION BY METHOD OF LEAST SQUARES >ok....i've got my formula set up to read a set of data in either Corel >Quattro or Microsoft Exel where n is the number of readings per time >period. For example: if we've got 2 readings taken at time A and the >other at time B the value of n will be 2. Similarly if there are >readings at time A, time B and time C, then the value of n will be 3. >Manually manipulating the formula to read the data at either n=2 or n=3 >is fine if it's a one time deal, but how can this formula be composed >and put into a macro in Quattro or Exel when n=2 vs n=3 vs n=4 and so >on? You could ask Microsoft if you were able to spell Excel. -- Jeremy Boden === Subject: Re: calculus of variations type problem schrieb david.rufino@gmail.com : > I have recently been thinking about an optimization problem to which I > can't find an easy solution. The problem is: find a C^3 function f : > [0,T] -> R which minimizes > int_0^T (d^3f/dt^3)^2 dt > subject to the constraint > sum_i exp(-f(t_i)) c_i > where the sum is over a finite set, and c_i are arbitrary real numbers. > obviously the problem is quite specific and can be translated into > various equivalent forms. However I haven't figured out a way to make > it amenable to the techniques I learnt as an undergrad, so if anyone > could advise me on an approach, that would be great. I may be mistaken, but I think that the minimum value should be zero, taken on for an appropriate choice of a constant, or at most linear function for f. === Subject: suspention of sphere Explicitly show that suspension of S^n is homeomorphic to S^(n+1). === Subject: Re: suspention of sphere > Explicitly show that suspension of S^n is homeomorphic to S^(n+1). Which version of suspension are you using? Using the union of two cones with the copies of the original space identified via the identity mapping, this is as close to a triviality as I can imagine. Using the cylinder with ends and a distinguished line from end to end all identified to a point is not so simple, and probably best attacked by showing it's homeomorphic to the above description (if the base point is nondegenerate). Using the smash product (S^1 ^ X, alias product of pointed spaces), probably the same level of difficulty. Dale. === Subject: Re: Arithmetic without the Successor Axiom > http://www.andrewboucher.com/papers/arith-succ.pdf > (The file is 120+ pages long, so it may take time to download.) > INTRODUCTION > The Successor Axiom asserts that every number has a successor, > or in other words, that the number series goes on and on ad infinitum. etc. Look, it's made the newspapers! http://www.theonion.com/content/node/29617 === Subject: Combinatorics puzzle Probably not new, but neat enough: Let f be any permutation of the set {1,2,...,n}. Show that the sum of the values of xf(x) is at least n(n+1)(n+2)/6. === Subject: Win XP top Photoshop Software at best s8vings by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with SMTP id k1SLaf311195 for ; Tue, 28 Feb 2006 16:36:47 -0500 --------------------------------------------------------------------- === Subject: Simple limit problem. Please help Can anyone help me in calculating this limit? Lim(1+SinX)^(1/(x-Pi)) ^ stands for power. I was told the way to do it is using e^(ln(f(x)) cause that changes nothing and then use lopital rule but I can't make it to work. === Subject: Re: Simple limit problem. Please help > Can anyone help me in calculating this limit? > Lim(1+SinX)^(1/(x-Pi)) > x-->Pi > ^ stands for power. I was told the way to do it is using e^(ln(f(x)) > cause that changes nothing and then use lopital rule but I can't make > it to work. It's L'Hopital, not lopital, but actually you don't need it. Since (1 + sin(x))^(1/(x - pi)) = exp(ln(1 + sin(x))/(x - pi)) all you have to do is to calculate the limit lim_{x -> pi}ln(1 + sin(x))/(x - pi). But this is just the derivative at _pi_ of ln(1 + sin(x)). Jose Carlos Santos === Subject: integer cube root algorithm... Below is a Rexx subroutine to find the integer squart root. (A % is integer divide, all variables shown here are integers, with the exception of the arg X which is promptly converted into an integer via TRUNC. The check for a negative number (X) has been omitted. The DO WHILE clause is made before the DO loop is executed (at the top). ______________________________________________________ /*find integer sqrt of a non-negative #.*/ isqrt: procedure; parse arg x; x=trunc(x); r=0; q=1 do while q<=x q=q*4 end do while q>1 q=q%4 _=x-r-q r=r%2 if _>=0 then do x=_ r=r+q end end return r _______________________________________________________ Does anyone know of (or where can I get it) an integer cube root routine? ______________________________Gerard S. === Subject: normal distribution fourier transform hello there, im having real trouble finding the fourier transform of the following distribution on R^n B(x)=(1/(sqrt(2*Pi*c*c)^n)) *exp(-|x-b|^2/(2*c*c)) where c is greater than 0, any hints to make it easier would be grateful, === Subject: Re: normal distribution fourier transform > hello there, im having real trouble finding the fourier transform of > the following distribution on R^n > B(x)=(1/(sqrt(2*Pi*c*c)^n)) *exp(-|x-b|^2/(2*c*c)) Is |x-b|^2 = (x_1 - b_1)^2 + (x_2 - b_2)^2 + ... + (x_n - b_n)^2? If so, the function factors into separate Gaussians in x_1, x_2, ... . The result will be a product of transforms of 1-D Gaussians, up to normalization. R.G. Vickson > where c is greater than 0, > any hints to make it easier would be grateful, === Subject: Volume of a section of a sphere A unit sphere is cut into 4 sections by two perpendicular planes at distances a, b from the centre. 0