mm-1449 === Subject: Re: Logic question Ambrish > Do you mean you attempted this kind of proof: > ((1 & 3) => 2) => 1 ? > If so, that is incorrect. Some manipulations yield: > ((1 & 3) => 2) => 1 > <=> '((1 & 3) => 2) V 1 > <=> '('(1 & 3) V 2) V 1 > <=> ((1 & 3) & '2) V 1 > <=> ((1 & 3) & false) V 1 > <=> false V 1 > <=> 1 > So, a rather circular proof. ((1 & 3) => 2) => 1 is true only if (1) is > true. But you were trying to show that (1) is true. > Part of your problem may perhaps be the fact that (1) is not true. Try A > = 1, B = 3, P = 0, Q = 1. Then A > P and B > Q, but > 1 - 3 < 0 - 1 <=> -2 < -1 > which is false. === Subject: Re: Logic question No problem. I noticed some mistakes in my own post as well ;-) >> So, a rather circular proof. ((1 & 3) => 2) => 1 is true only if (1) >> is true. that should be if and only if. >> Part of your problem may perhaps be the fact that (1) is not true. Try >> A = 1, B = 3, P = 0, Q = 1. Then A > P and B > Q, but >> 1 - 3 < 0 - 1 <=> -2 < -1 >> which is false. And here I got lost somewhere. Try B = 1, Q = 0. Which results in: 0 < 0 -- Daniel Sj.9ablom === Subject: moment of inverse Wishart distribution Does any one know how to compute the following expectation involving the Wishart Distribution E{I+D'*H'*H*D}^{-1} where ' denote conjugate transpose, H is a random matrix with independent, zero mean, unit variance, gaussian distributed complex entries, hence H'H has central wishart distribution. D is an arbitrary matrix. I know there exists the simply expression for the moment of inverse wishart distribution. Does any one have any idea how to deal with an additional identity matrix I, === Subject: Re: what is this equal to any easy approximation other than the normal one would be based on approximation of binom{N}{i} (C^N_i as you call it). approximations of the binomial coefficient are plentiful... === Subject: Re: what is this equal to > Let C_N^i denote choose i out of N > then what's this equal to > sum_{i>T} C_N^i > I know sum_{i=1}^N C_N^i = 2^N > but if I'd like to know the sum of when i>T. There's no nice closed form. But T-(N/2) is comparable to sqrt(N), then a normal approximation will be useful. See Feller or any probability book. And to nitpick, the known sum should start at 0: sum_{i=0}^N C_N^i = 2^N Don Coppersmith === Subject: Re: what is this equal to >> Let C_N^i denote choose i out of N >> then what's this equal to >> sum_{i>T} C_N^i >> I know sum_{i=1}^N C_N^i = 2^N >> but if I'd like to know the sum of when i>T. >There's no nice closed form. Unless you think hypergeometric functions are nice: in Maple's notation it's binomial(N,T+1) hypergeom([1,-N+T+1],[2+T],-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: what is this equal to > <4730918.1116524983105.JavaMail.jakarta@nitrogen.mathf > orum.org>, >> Let C_N^i denote choose i out of N >> >> then what's this equal to >> >> sum_{i>T} C_N^i >> >> I know sum_{i=1}^N C_N^i = 2^N >> >> but if I'd like to know the sum of when i>T. >There's no nice closed form. > Unless you think hypergeometric functions are nice: > in Maple's > notation it's > binomial(N,T+1) hypergeom([1,-N+T+1],[2+T],-1) > Robert Israel > israel@math.ubc.ca > Department of Mathematics > http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, > BC, Canada I stand corrected. === Subject: Re: number theory proof > As I read the OP's post, the question at hand was to > show that > the prime factors of a are contained in b it seems maybe you should re-read the OP's post. === Subject: Why Maplesoft does not give you access to Maple 10 trial version? Hello Maple customers all over the world, You can get a fully functional, 15-day, save-disabled trial version of Mathematica 5.1.1 at http://www.wolfram.com/products/mathematica/trial.cgi You can get a free fully-functional 30 days trial Derive 6.1 version at http://education.ti.com/us/product/software/derive/features/features.html No problem, you can get a free fully-functional 30 days trial MuPAD 3.1.1 version at http://www.mupad.com/index.php?menu=6 Maplesoft also had trial versions until Maple 9. Then during preparations to selling Maple 9, the Maplesoft's owners and leaders realized that they were going to sell low quality stuff, a beta version. A big stink! And they gave us NO mode of trying Maple 9. After a flagrant fiasco with Maple 9, it would be minimally honest of Maplesoft to give us a trial version of Maple 10. Not the case! Today, again, Maplesoft proposes you: buy a pig in a poke. The faster the better, as if you do not buy the pig today, soon its price will increase. Case # 1. Maplesoft is not able to protect Maple 10 properly as Wolfram Research, Texas Instruments, and SciFace can do with their applications. And/Or Case # 2. Maplesoft's owners and leaders again realize that Maple 10 is a raw stuff... Buy Maple 10, support Maplesoft staff K-12 math education, Vladimir Bondarenko VM and GEMM architect Co-founder, CEO, Mathematical Director Cyber Tester, LLC 13 Dekabristov Str, Simferopol Crimea 95000, Ukraine tel: +38-(0652)-447325 tel: +38-(0652)-230243 tel: +38-(0652)-523144 fax: +38-(0652)-510700 http://www.cybertester.com/ http://maple.bug-list.org/ http://www.CAS-testing.org/ === Subject: Re: Why Maplesoft does not give you access to Maple 10 trial version? Do you actually do anything other than posting messages to sci.math.symbolic and the other forums in the subject line? Just curious. === Subject: Re: Why Maplesoft does not give you access to Maple 10 trial version? Click Go at http://maple.bug-list.org/ === Subject: Re: Twin Primes Conjecture >I am sure 10^6 is a typo, BUT: >You miss the point. Me?? I agreed with his point, despite the typo. >Why do you restrict your set to odd numbers? Because the stated prime count was rather large, I went as far down the seive as was required to know for certain the count was wrong. I can't speak for the OP, but I agree with Dave's point and your expoundation (is that a word?) on it. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Crossover is the Optimum Strategy VonNeumann Gametheory in Playing the Stockmarket; 300 free shares of BCE > I wonder whether Moodys' is being fair about its downgrade of BCE citing > reasons of debt structure. So I looked up some Reuters data of various > telecoms to compare debt structure. > Now SBC has 41 B in revenue and 27 B in debt for a debt structure of 66%. > When companies are said to be 'debt-heavy', the meaure in question is > debt/equity, not debt/revenue. At the next Olympics if someone answered how fast is the Marathon running going with the measure that his left leg was running slightly slower than right leg is hardly any measure at all, is it. Dinosaurish outmoded measures. Much of economics and finance has these dinosaurish and illogical measures and concepts. Most of economics theory is muddle headed concepts. For the Marathon runner we expect a measure to something fixed to the ground to measure speed, not the right leg compared to the left leg. So also with debt in corporate finances. We expect debt to be compared to income of say revenue or profit or even dividendpayout. But not debt to debt because equity is really on the debt side of the financial house. A concept of (Total debt)/ dividend payout is very intriguing. How many years would it take for BCE and for SBC and for BLS if they halted dividend payout and used it instead to pay off debt would they be debt free. The concept of debt/equity is dinosaurish and outmoded and is bereft of logic because both debt and equity are forms of debt. A meaningful concept would divide debt by some form of income, an illogical concept would divide debt by another form of debt (which is equity). A meaningful and logic concept would divide money going out by money coming in which is something like debt/revenue or debt/dividendpayout or debt/profit. I do not like the measure of profit much because companies can mask and hide and play tricks with profit but not so much with revenue or dividend payout. seems to have been lost in transmission due to electronic failure. If the earlier one appears then excuse this second post. Archimedes Plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Crossover is the Optimum Strategy VonNeumann Gametheory in Playing the Stockmarket; 300 free shares of BCE >> 'debt-heavy', the meaure in question is >> debt/equity, not debt/revenue. > The concept of debt/equity is dinosaurish and outmoded and is bereft of > logic because both debt and equity are forms of debt. A meaningful > concept would divide debt by some form of income, an illogical concept > would divide debt by another form of debt (which is equity). Equity includes all net income. === Subject: Re: TOE Via Cantor's Transfinite Arithmetic(2) 05/17/2005 at 07:19 PM, mzafrullah@usa.net said: >I am afraid that an authoritative NO would not cut it. I have >witnessed folks making a hard to read paper out of simple material Sure; you can see that in any discipline, but it is not the norm and it is not the current approach. >Ordinary folks did get into trouble by over-formalizing, na.95ve >notions, Not in any case that I'm aware of. What typically happens is that the first people formalizing a discipline do it in a complicated fashion but later they or their successors learn how to do it in a simpler fashion. >Can you name a branch of Mathematics, which started with no >application in mind? Number theory. Complex Analysis. Hyperbolic Geometry. >We are talking about Mathematics here. Yes, and Mathematics is fully analogous to Music with regard to learning new things. >Perhaps it was the add-on intricacies that pushed >the new generation away from classical music. Some popular music is more intricate than some classical music. >I am glad that you brought this one up. Does the following ring a >bell? Do you think God plays dice? >I hear the beloved old man of Physics would bring point after point >against the use of probability in the study of physics and had them >refuted one by one. Not quite; he found legitimate objections to the way that QM was presented and in several cases caused significant reevaluations. As for opposing probability and QM, Bohr had earlier attacked Einstein as too radical, and Einstein did major work in Statistical Mechanics (ever hear of Bose-Einstein condensations?). His objections were profound, and quite different from those of people who simply didn't understand the subject. >I am usually a peaceful man and now somewhat weak as >well; yet you call Einstein a crackpot I would at least try to knock >your head in. Well, that tells me more about you than about Einstein. >Poncare seems to have said some nasty things about >Cantor's calculus of the infinities, and/or over-abstraction you >dare not call him a crackpot in my presence. I'd be more inclined to call you an intolerant bully. >In the mean time my e-mail and web site have been knocked down twice >and my website has been totally isolated (apparently by the very >friendly Google) from the area I publish in and can offer advice in. Assault as a means of rational discourse? >I am beginning to have doubts about freedom of speech. Freedom of the press belongs to the man that owns the press. My local tabloid refuses to print what I write unless I pay them for the space. Does that violate my First Amendment rights? You want to use their services, you abide by their rules. >Try to make some sense How? Offering to knock the block off of posters who disagree with you? You don't explain what you were kicked off for. >But again whom am I talking to? You don't know, do you? But that's your fault, not mine. >You would just, half quote, twist and turn what I am saying so that I >look bad. Why would I bother, when you do so well at it yourself? >But if there is a part quote and any unnecessarily crude remark, I >may not respond. Have you no decency? You threaten physical violence and then whine about unnecessarily crude? -- 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 === Subject: Analysis question Any clue to show that there exists a unique bijective and differentiable mapping f from the set of positive reals to itself, and such that f '(x) = f^(-1)(x) ? -- Julien Santini === Subject: Re: Analysis question > Any clue to show that there exists a unique bijective and > differentiable mapping f from the set of positive reals to itself, and > such that f '(x) = f^(-1)(x) ? I think if you let b be the golden ratio, and a = b^(-b/(b+1)), then f(x) = a x^b works. For uniqueness, you have an IVP in the quadrant x> epsilon, y>epsilon and an argument that the function is increasing. Bart === Subject: Re: Analysis question > Any clue to show that there exists a unique > bijective and > differentiable mapping f from the set of positive > reals to itself, and > such that f '(x) = f^(-1)(x) ? > I think if you let b be the golden ratio, and a = > b^(-b/(b+1)), then > f(x) = a x^b works. > For uniqueness, you have an IVP in the quadrant x> > epsilon, > y>epsilon and an argument that the function is > increasing. > Bart How do you handle the presence of f^{-1}(x) in the defining equation? An unrelated idea is to take the unique positive fixed point f(z)=z, and expand f in a power series around z: f(z)=z, f'(z)=z, f''(z)=1/z, etc. (z=phi but we don't know that.) Or take two solutions f,g, and correlate regions where f>g with regions where g>f. I couldn't get either of these to prove uniqueness. Don Coppersmith === Subject: Re: Analysis question > Any clue to show that there exists a unique bijective > and differentiable > mapping f from the set of positive reals to itself, > and such that f '(x) = > f^(-1)(x) ? > -- > Julien Santini We can prove existence: try f(x)=A*x^B for suitable A,B. (I get B=phi, A=1/phi^(1/phi), not guaranteed) Don Coppersmith === Subject: Re: Analysis question > Any clue to show that there exists a unique bijective and differentiable > mapping f from the set of positive reals to itself, and such that f '(x) = > f^(-1)(x) ? > -- > Julien Santini If it is bijective and continuous on R, then it is monotone, so if it is also differentiable, then the derivative has only one sign. Now assume f '(x) = f^(-1)(x) . So either f^(-1) misses all the negative numbers, or it misses all the positive numbers, which is a curious property for a bijection... === Subject: Re: Analysis question > > Any clue to show that there exists a unique bijective and differentiable > mapping f from the set of positive reals to itself, and such that f '(x) = > f^(-1)(x) ? > If it is bijective and continuous on R, then it is monotone, > so if it is also differentiable, then the derivative has only one sign. > Now assume f '(x) = f^(-1)(x) . > So either f^(-1) misses all the negative numbers, or it misses > all the positive numbers, which is a curious property > for a bijection... It's a mapping from the set of positive reals to itself. === Subject: Re: What is a good artifact for a unit quantity of mass? > Now, however, if you could come up with a definition for a unit of > force that is NOT derived from a unit of mass, you could have some > standing. Can you, Don? He can. He has. And yet, he's forgotten! Don appears to have lost it. But don't let that fool you, he never had it! Here's Don's standard force. Take a pint of pure water at 39.2ÁF. Place it on Don's kitchen table. The force it exerts on the table is, by definition, one pound. Of course, as has been explained to Don ad nauseum, this standard is less precise, by a couple orders of magnitude, than the current standard. For some reason Don fails to understand the basic principle of metrology: Better precision, not worse, is required when standards are adopted. I think Don feels that the world is too precise, and that we need to get back to simpler times when things weren't so darned confusing. http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- === Subject: Re: What is a good artifact for a unit quantity of mass? <428cdf9b$1_2@spool9-west.superfeed.net> > Now, however, if you could come up with a definition for a unit of > force that is NOT derived from a unit of mass, you could have some > standing. Can you, Don? > Here's Don's standard force. Take a pint of pure water at 39.2ÁF. Place it > on Don's kitchen table. The force it exerts on the table is, by definition, > one pound. But wouldn't that again include a mass, since volume * density (which is mass per volume)? > Of course, as has been explained to Don ad nauseum, this standard is less > precise, by a couple orders of magnitude, than the current standard. Not only that, but a very similar definition of the kilogram was abandoned 100+ years ago. 1 kilogramm used to be defined as a certaim volume of water at a certain temperature and (what Don completely omits) at a certain pressure. However, pressure equals force per area, and force equals mass times acceleration. So, this definition is stuck in a vicious circle, too. I can't help myself, but why does the image of a dog chasing his own tail keeps popping up in my mind? > For some reason Don fails to understand the basic principle of metrology: > Better precision, not worse, is required when standards are adopted. I > think Don feels that the world is too precise, and that we need to get back > to simpler times when things weren't so darned confusing. Or he wants 95% of people living on this planet to accept the slug as standard unit for mass ;-) Good luck, A. Friend News==---- > http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups > ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- === Subject: Re: What is a good artifact for a unit quantity of mass? Not only that, but a very similar definition of the kilogram was abandoned 100+ years ago. 1 kilogramm used to be defined as a certaim volume of water at a certain temperature and (what Don completely omits) at a certain pressure. However, pressure equals force per area, and force equals mass times acceleration. So, this definition is stuck in a vicious circle, too. The real reason for changing from a fixed volume of water to an artifact was, and still is, precision. Weigh out several different cubic decimeters of water under identical circumstances and they will differ by about a few parts in 10^6. Construct an artifact and make several copies of it, checking by balance beanm that they all weigh the same.. Their weights will differ by about a few parts in 10^8. And now, this is the part that's hard for Don. Varying by a few parts in 10^6 is LESS PRECISE than varying by a few parts in 10^8. Better precision is needed for science, technology, and commerce. That's the reason for having standards. http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- === Subject: Re: What is a good artifact for a unit quantity of mass? >... >>Or he wants 95% of people living on this planet to accept the slug as >>standard unit for mass ;-) >I found a slug in my garden today. Was it standard? Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: What is a good artifact for a unit quantity of mass? >In message <9E8je.99$25.20530@news.uchicago.edu>, >>>... >>>> >>>>Or he wants 95% of people living on this planet to accept the slug as >>>>standard unit for mass ;-) >>>> >>>I found a slug in my garden today. >>Was it standard? >I don't know. >I didn't have a standard slug to compare it with. Send it to Shead, let him do the comparisons. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: What is a good artifact for a unit quantity of mass? >> Send it to Shead, let him do the comparisons. >Hell, no, don't touch it. Slugs must be really, really dangerous! Hmm... >You know, I just found one in my garden, too. But then I realized: My >slug was about 5cm long and about 1 cm wide. Assuming the shape is >roughly cylindrical, I estimated the volume of this slug to be around >5cm * (0,5cm)^2 * Pi or approximately 3.9 cm^3 or 3.9e-6 m^3. >However, as I already concluded, the mass of 1 slug = 32.174 lbm or >14.593 kg. Now, if this small object has a mass of 14.593 kg, the >density would be 14.593kg / 3,9e-6 m^3 which equates to an enormous >3,741,795 kg/m^3. >So this slug is about 175,000 times denser than platinum (which only >has a mass of 21,450 kg/m^3). >So, now I'm stuck with a big problem and need your help! >1. Are my equations correct? Look fine. >2. What material is a slug made of? Degenerated matter? Metallic >hydrogen? Or even (gasp!) neutronium? I would bet on degenerated matter. Well, something degenerated, anyway. That would explain why Shead is so fond of them. >3. Should I wear a radiation suit when examining a slug more closely? Yes, certainly. Safety first:-) >And last, but not least: >4. Will salt really help? What about beer traps? Not sure about salt, but beer always helps, when properly applied. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: What is a good artifact for a unit quantity of mass? >>> Send it to Shead, let him do the comparisons. >> >>Hell, no, don't touch it. Slugs must be really, really dangerous! >> Hmm... >> >>You know, I just found one in my garden, too. But then I realized: >>My slug was about 5cm long and about 1 cm wide. Assuming the shape >>is roughly cylindrical, I estimated the volume of this slug to be >>around 5cm * (0,5cm)^2 * Pi or approximately 3.9 cm^3 or >>3.9e-6 m^3. >> >>However, as I already concluded, the mass of 1 slug = 32.174 lbm >>or 14.593 kg. Now, if this small object has a mass of 14.593 kg, >>the density would be 14.593kg / 3,9e-6 m^3 which equates to an >>enormous 3,741,795 kg/m^3. >> >>So this slug is about 175,000 times denser than platinum (which >>only has a mass of 21,450 kg/m^3). >> >>So, now I'm stuck with a big problem and need your help! >> >>1. Are my equations correct? >> Look fine. >>2. What material is a slug made of? Degenerated matter? Metallic >>hydrogen? Or even (gasp!) neutronium? >> I would bet on degenerated matter. Well, something degenerated, >> anyway. That would explain why Shead is so fond of them. >I agree with degenerated matter. Beer traps really work. >After the drowned slugs rot a bit, the results are quite gross >and degenerate... I can believe that but it would be a pity to waste good beer in such way. The proper application of beer to any problem is apply internally till the problem goes away. >>3. Should I wear a radiation suit when examining a slug more >>closely? >> >> Yes, certainly. Safety first:-) >>And last, but not least: >> >>4. Will salt really help? What about beer traps? >> >> Not sure about salt, but beer always helps, when properly applied. >Jerry Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: What is a good artifact for a unit quantity of mass? >>1. Are my equations correct? >> Look fine. >Phew! And I thought this would sound as weird as Don... Don practices his weirdness for a very long time, you can't just jump in and outweird him:-) >>2. What material is a slug made of? Degenerated matter? Metallic >>hydrogen? Or even (gasp!) neutronium? >> I would bet on degenerated matter. Well, something degenerated, >> anyway. That would explain why Shead is so fond of them. >Oh, I start to see the picture. Degenerated matter, degenerated units, >degenerated... Don, you mean? Well, now that you mention it, yes, it adds up:-) >>3. Should I wear a radiation suit when examining a slug more >closely? >> >> Yes, certainly. Safety first:-) >So, using slugs IS dangerous. Good to know... >>And last, but not least: >> >>4. Will salt really help? What about beer traps? >> >> Not sure about salt, but beer always helps, when properly applied. >You mean, like, internally? External use seems to lead to some bad body >odour afterwards... Yes, of course internally, and in ample quantities. May not remove slugs but after a while you won't care. >Ok, so I'm sitting here in my radiation suit, my 14.593 kg slug safely >stowed away in a lead box and just poked a hole in my gas mask so I can >sip my beer with a straw... Well, better you than me (though I may join and help with the beer). >Yay, physics sure is fun! All for the greater glory of science. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: What is a good artifact for a unit quantity of mass? <428cdf9b$1_2@spool9-west.superfeed.net> <9E8je.99$25.20530@news.uchicago.edu> >> Send it to Shead, let him do the comparisons. >Hell, no, don't touch it. Slugs must be really, really dangerous! > Hmm... >You know, I just found one in my garden, too. But then I realized: >My slug was about 5cm long and about 1 cm wide. Assuming the shape >is roughly cylindrical, I estimated the volume of this slug to be >around 5cm * (0,5cm)^2 * Pi or approximately 3.9 cm^3 or >3.9e-6 m^3. >However, as I already concluded, the mass of 1 slug = 32.174 lbm >or 14.593 kg. Now, if this small object has a mass of 14.593 kg, >the density would be 14.593kg / 3,9e-6 m^3 which equates to an >enormous 3,741,795 kg/m^3. >So this slug is about 175,000 times denser than platinum (which >only has a mass of 21,450 kg/m^3). >So, now I'm stuck with a big problem and need your help! >1. Are my equations correct? > Look fine. >2. What material is a slug made of? Degenerated matter? Metallic >hydrogen? Or even (gasp!) neutronium? > I would bet on degenerated matter. Well, something degenerated, > anyway. That would explain why Shead is so fond of them. I agree with degenerated matter. Beer traps really work. After the drowned slugs rot a bit, the results are quite gross and degenerate... >3. Should I wear a radiation suit when examining a slug more >closely? > Yes, certainly. Safety first:-) >And last, but not least: >4. Will salt really help? What about beer traps? > Not sure about salt, but beer always helps, when properly applied. Jerry === Subject: Re: What is a good artifact for a unit quantity of mass? <428cdf9b$1_2@spool9-west.superfeed.net> <9E8je.99$25.20530@news.uchicago.edu> >1. Are my equations correct? > Look fine. Phew! And I thought this would sound as weird as Don... >2. What material is a slug made of? Degenerated matter? Metallic >hydrogen? Or even (gasp!) neutronium? > I would bet on degenerated matter. Well, something degenerated, > anyway. That would explain why Shead is so fond of them. Oh, I start to see the picture. Degenerated matter, degenerated units, degenerated... Don, you mean? >3. Should I wear a radiation suit when examining a slug more closely? > Yes, certainly. Safety first:-) So, using slugs IS dangerous. Good to know... >And last, but not least: >4. Will salt really help? What about beer traps? > Not sure about salt, but beer always helps, when properly applied. You mean, like, internally? External use seems to lead to some bad body odour afterwards... Ok, so I'm sitting here in my radiation suit, my 14.593 kg slug safely stowed away in a lead box and just poked a hole in my gas mask so I can sip my beer with a straw... Yay, physics sure is fun! Good luck, A. Friend === Subject: Re: What is a good artifact for a unit quantity of mass? <428cdf9b$1_2@spool9-west.superfeed.net> <9E8je.99$25.20530@news.uchicago.edu> >>1. Are my equations correct? >> Look fine. >Phew! And I thought this would sound as weird as Don... >>2. What material is a slug made of? Degenerated matter? Metallic >>hydrogen? Or even (gasp!) neutronium? >> I would bet on degenerated matter. Well, something degenerated, >> anyway. That would explain why Shead is so fond of them. >Oh, I start to see the picture. Degenerated matter, degenerated units, >degenerated... Don, you mean? >>3. Should I wear a radiation suit when examining a slug more >closely? >> >> Yes, certainly. Safety first:-) >So, using slugs IS dangerous. Good to know... >>And last, but not least: >> >>4. Will salt really help? What about beer traps? >> >> Not sure about salt, but beer always helps, when properly applied. >You mean, like, internally? External use seems to lead to some bad body >odour afterwards... So does internal use if.... >Ok, so I'm sitting here in my radiation suit, there's no exit in the suit. > ... my 14.593 kg slug safely >stowed away in a lead box and just poked a hole in my gas mask so I can >sip my beer with a straw... >Yay, physics sure is fun! You may want to back off a bit when the two-ton starling drops by in search of supper. /BAH >Good luck, >A. Friend Subtract a hundred and four for e-mail. === Subject: Re: What is a good artifact for a unit quantity of mass? <428cdf9b$1_2@spool9-west.superfeed.net> <9E8je.99$25.20530@news.uchicago.edu> > Send it to Shead, let him do the comparisons. Hell, no, don't touch it. Slugs must be really, really dangerous! You know, I just found one in my garden, too. But then I realized: My slug was about 5cm long and about 1 cm wide. Assuming the shape is roughly cylindrical, I estimated the volume of this slug to be around 5cm * (0,5cm)^2 * Pi or approximately 3.9 cm^3 or 3.9e-6 m^3. However, as I already concluded, the mass of 1 slug = 32.174 lbm or 14.593 kg. Now, if this small object has a mass of 14.593 kg, the density would be 14.593kg / 3,9e-6 m^3 which equates to an enormous 3,741,795 kg/m^3. So this slug is about 175,000 times denser than platinum (which only has a mass of 21,450 kg/m^3). So, now I'm stuck with a big problem and need your help! 1. Are my equations correct? 2. What material is a slug made of? Degenerated matter? Metallic hydrogen? Or even (gasp!) neutronium? 3. Should I wear a radiation suit when examining a slug more closely? And last, but not least: 4. Will salt really help? What about beer traps? Good luck, A. Friend > Mati Meron | When you argue with a fool, > meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: What is a good artifact for a unit quantity of mass? <428cdf9b$1_2@spool9-west.superfeed.net> <9E8je.99$25.20530@news.uchicago.edu> In message <9E8je.99$25.20530@news.uchicago.edu>, >>... >>>Or he wants 95% of people living on this planet to accept the slug as >>>standard unit for mass ;-) >>I found a slug in my garden today. >Was it standard? I don't know. I didn't have a standard slug to compare it with. -- Jeremy Boden === Subject: Re: What is a good artifact for a unit quantity of mass? <428cdf9b$1_2@spool9-west.superfeed.net> ... >Or he wants 95% of people living on this planet to accept the slug as >standard unit for mass ;-) I found a slug in my garden today. -- Jeremy Boden === Subject: Re: WHAT THE MODERN MATHEMATHICIEN ARE SHY TO TALK ABOUT <22834516.1116472523559.JavaMail.jakarta@nitrogen.mathforum.org> > About my presntation I want to say that all statements are correct. Are we supposed to take your word for it? >If you take a pen and go from statement to statement you may be able to >find what connects each other. Yes. May be able. And may be unable. Let's face it: you are too smart for us.What's trivial to you is incomprehensible to us. Why don't you humor us, stupid mathematicians, and do what we, mathematicians, do: make presentations of our proofs comprehensible to others, so that they wouldn't have to go from statement to statement in the hope that they may be able to find what connects each other'. > It is a concise proof and the links between steps are not shown but can be deduce. ?? Well, maybe instead of posting tons of unhelpful posts, you could instead take a couple of hours and make your proof less concise. But you don't even bother with re-reading the typos that you make. Like: >Therfore k=1 must be disible at least by 4 === Subject: Re: WHAT THE MODERN MATHEMATHICIEN ARE SHY TO TALK ABOUT Well.IT is my mistake .I should had said that it is only a OUTLINE of the elementary proof of FLT,as Fermat did in his notes about the proof Of FLT for n=4 george ghiata > About my presntation I want to say that all > statements are correct. > Are we supposed to take your word for it? >If you take a pen and go from statement to statement > you may be able > to >find what connects each other. > Yes. May be able. And may be unable. Let's face it: > you are too smart > for us.What's trivial to you is incomprehensible to > us. > Why don't you humor us, stupid mathematicians, and do > what we, > mathematicians, do: make presentations of our proofs > comprehensible to > others, so that they wouldn't have to go from > statement to statement > in the hope that they may be able to find what > connects each other'. > It is a concise proof and the links between steps > are not shown but > can be deduce. ?? > Well, maybe instead of posting tons of unhelpful > posts, you could > instead take a couple of hours and make your proof > less concise. > But you don't even bother with re-reading the typos > that you make. > Like: >Therfore k=1 must be disible at least by 4 === Subject: Re: WHAT THE MODERN MATHEMATHICIEN ARE SHY TO TALK ABOUT <31299088.1116534193798.JavaMail.jakarta@nitrogen.mathforum.org> OUTLINE? Which also happens to contain typos in most of the outlined formulas? Then why are you posting your OUTLINE to sci.math? Look, I have spent all day today writing OUTLINE (Introduction) Sections for 2 papers. When I add the Sections that contain actual proofs, I will make my results public. And shall ask other mathematicians to comment on the veracity of my results. But it would never occur to me to publish just the Introductions (Outlines) to my papers without the actual proofs and to expect others to verify my results just by looking at OUTLINES. How can anybody try to find a mistake in one of my calcualtions if all they see is an OUTLINE? Morevoer, you started this whole discussion by making the following provocative statement: >In another words the modern mathematicien can not recognise that >the great mathematiciem can miss elementary proofs for long time >and when are face with this proofs they become shy of talking about > them They fall silent. What PROOFS do they face? All you have presented is an OUTLINE. I like the way your outline looks! It is a beautiful outline. It has a lot of interesting Latin letters connected to each other by charming arithmetic operators. And the typos are very genuine. But there are no rigorous proofs, which can be analyzed for their veracity. When you produce readable PROOFS and not an OUTLINE, post them. Mathematicians will be very happy to point out to you all the places where you made mistakes in these proofs. But until you post PROOFS, nobody can look for mistakes in your proof, can they. > Well.IT is my mistake .I should had said that it is only a OUTLINE > of the elementary proof of FLT,as Fermat did in his notes about > the proof Of FLT for n=4 > george ghiata > About my presntation I want to say that all > statements are correct. > Are we supposed to take your word for it? >If you take a pen and go from statement to statement > you may be able > to >find what connects each other. > Yes. May be able. And may be unable. Let's face it: > you are too smart > for us.What's trivial to you is incomprehensible to > us. > Why don't you humor us, stupid mathematicians, and do > what we, > mathematicians, do: make presentations of our proofs > comprehensible to > others, so that they wouldn't have to go from > statement to statement > in the hope that they may be able to find what > connects each other'. > It is a concise proof and the links between steps > are not shown but > can be deduce. ?? > Well, maybe instead of posting tons of unhelpful > posts, you could > instead take a couple of hours and make your proof > less concise. > But you don't even bother with re-reading the typos > that you make. > Like: > Therfore k=1 must be disible at least by 4 === Subject: Re: An exact 1-D summation challenge - 4 >> iscont(1/(1+z^346), z= 0..1); >My prediction: Maple 10 keeps running after 100,000 seconds. It returns true in 5.359 seconds. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: An exact 1-D summation challenge - 4 http://mygate.mailgate.org/mynews/sci/sci.math.symbolic/c22470a6a4230698236d 5 d8d8ef9d81b.122484%40mygate.mailgate.org Alec Mihailovs addresses the following to Generalissimus Bondarenko: > Vladimir, in spite of many posts trying to explain that to you, you don't > want to see the problem. There are 2 different things. One - finding bugs > and posting it on your web site - good. Another one - flooding 5 newsgroups > with daily bug reports - bad. There are at least three different things. The third one is the behaviour of an - apparently adult - person, whose ego reached the stage needing a competent psychiatrist. The point is that usually people who need a psy, won't admit it... [The standard answer will be, as usual: it is YOU who need a psy!] It not only flooding of newsgroups which is bad, but the insulting slogans, as taken directly from an Ancien R.8egime era... Perhaps somebody would raise some funds and *pay* Bondarenko, so that he would stick to his - possibly - useful work, and remain quiet? J. Karczmarczuk -- === Subject: Re: An exact 1-D summation challenge - 4 > It not only flooding of newsgroups which is bad, but the insulting > slogans, as taken directly from an Ancien R.8egime era... Jerry, He is in my filter list, so I don't see his posts and had no idea about that - just the number of posts. I just read one his of posts to check that and found out that you are absolutely right. One can see significant change of style comparing with his posts about a year ago - then they were just bug reports. Alec === Subject: Re: An exact 1-D summation challenge - 4 JK> person, whose ego reached the stage needing a competent JK> psychiatrist. Yep... I readily admit publicly that I got fully deranged... small wonder... I use Maple 9/9.5... for many many hours a day... JK> a competent psychiatrist. 'competent'... oh this not will do... maybe only a psy star could relieve a bit the eerie tortures my old good (formerly) rigorous logics experiences having Maple 9/9.5 (now Maple 10 soon) constantly before my eyes... (Ah Jerzy, Jerzy, please realize, you are a lucky beggar. As of today, you even have no idea of how many there are thosands distinct Maple bugs... enjoy life... never read my posts... or you would run risk ending like me, a poor thing...) Yet another idea. Maybe Maplesoft would manage at last produce say, Maple 12, with a reasonably decent math correctness level? Best wishes, Vladimir Bondarenko P.S. Guys & gals, 'pon my word, 2 years ago, I was a healthy person... then Maple 9 was released... tears are running down my haggard cheeks... Warning: Do NOT visit these sites... if you have minimal math knowlenge, data there could be dangerous for your mental health..... http://www.cybertester.com/ http://maple.bug-list.org/ http://www.CAS-testing.org/ === Subject: Re: An exact 1-D summation challenge - 4 > COMMENT 1 Albert D. Rich, www.derive.com www.mulisp.com > COMMENT 2 Prof Dr Oleg Marichev, WRI www.functions.wolfram.com > COMMENT 3 Dr Anwar Shiekh, WRI Test Development Group Supervisor > COMMENT 4 Prof Dr Walter Oevel (SciFace GmbH) > COMMENT 8 Dr Michael Wester, www.math.unm.edu/~wester > COMMENT 11 Stefan Wehmeier, SciFace GmbH > COMMENT 13 Kelly Roach, http://home.att.net/~themission/ What a pity... maybe you would write them? Well... I seem to have already explained this a couple of times... Maple bugs: Backward compatibility - 1 Your words mean that, step by little step, I am reaching one of my grand goals which is creating the surroundings supported by powerful techniques and services within which it will be only possible to sell the customers competitive Maple and not a mathematical parody we, the Maple customers, have under the name of Maple 9/Maple 9.5... having paid a pretty penny. Also please realize that you seem to be not a Maple customer, while I am, and quite a thousands, to put it mildly, of persons are. And the issue I am starting touching is actually not about Maple quality & prospect & Maplesoft survival only, but also about Mathematica, MuPAD, AXIOM, Macsyma, Maxima and *any* computer algebra system. In fact, via the Maple case study, I am launching a dialogue about the quality of any large & influential computer algebra system, be it an existent one or a future system to be designed and implemented in 20 years, about how to develop such a system delivering its powers intact from version to version while, at worst, keeping the total number of bug manifestations not growing, how to reduce the overhead costs during the development and maintenance. Don't you see that the current status of computer algebra systems is a far cry from what one could expect seeing a great inspirational beginning of the 60s-70s, Dr Hearn's Reduce, your Macsyma? Don't you see that we got stuck in a mire of screwed buggy environments, each ridiculous it own way? Had the queen would be naked, it would be an attractive picture for my man's eye, - but the problem is that we have comical emperor's new clothes instead of noble eternal feminine nude. I am commencing a dialogue about expanding of computer algebra impact on industry & education, about more markets for computer algebra systems, and more customers for computer algebra manufacturers. Couldn't I indulge myself in a yet another self-quotation? Could I remind you that you are absolutely free to ignore all my messages... or read them... Well, or read part of them. Or read on Sunday, and ignore the rest of days etc. In a word, there seems to be quite a number of choices. Aha, I came near forgetting, or attempt solving my harder challenges coming. Please do as you wish. I understand that a person of significant drive like you not understanding, currently, the true reasons behind these messages is yawning your life away. By sending your message you maybe demand more decisive action from me. All in good time. If however you like playing chess you realize that there is stuff like positional game and combinative game, there is stuff like opening, endgame, gambit etc Anyway, I think that you are too clever to have real hope of shaking the serenity of mind of such a strong player like me via such a child's remark. Keep your chin up, life is good Vladimir I agree with you about the existence of the dichotomy... but it looks like I have a viewpoint that differs a bit from yours. So, how I see those 2 different things? One - Maple experts and customers who helps me, or at least do not make attempts to tell me what to do, in my carefully designed large-scale effort to force Maplesoft to go beyond these bells and whistles and produce, at least, something not so much loathsome as the bug-ridden Maple 9/9.5. Another one - please take the negation from 'One'. Good. Keep doing. Yet, this is NOT enough to resolve the task. Daily? Oh thank you for a fresh idea. To get serious, an integral part of my plan is maximal Maple bugs exposure. Maximal publicity. Let even each penguin in cool snow-white Antarctica learns about Maple bugs. (Now you were complaining about 5 newsgroups... you really underestimate me :) But all in good time. Best wishes, Vladimir http://www.cybertester.com/ http://maple.bug-list.org/ http://www.CAS-testing.org/ === Subject: Re: Math commentary from LA times > Stephen J. Herschkorn quotes: >Yet Philip Davis, >emeritus professor of mathematics at Brown University > I guess with lower-case letters that's technically correct; > but what he is, is Emeritus Professor of Applied Mathematics > at Brown University. Brown has (very!) separate departments > (for many years Wendell Fleming was the only overlap; there > may be, or have been, a very few others). At Johns Hopkins, the applied math department (called the Mathematical Sciences Department) is so separate, it's under the Engineering School. The pure math department is in the Arts & Sciences School. - Randy === Subject: Re: Math commentary from LA times In , on 05/17/2005 at 08:28 PM, lrudolph@panix.com (Lee Rudolph) said: >(The only pure--though he may fancy himself applied--mathematician >present who could compare satorially with the applied ones was A >tiyah. English tailoring.) Sewn on a Singer sewing machine? -- 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 === Subject: Relative distances question Originator: markc@chiark.greenend.org.uk ([193.201.200.170]) I've found a problem where my intuition suggests that there may be a simple solution, but I just can't see it. In three-dimensional Euclidean space we have two disjoint sets of points, P and Q. The points p1, p2 are members of P, and point q is a member of Q. One can define a distance d1 between p1 and q and a distance d2 between p2 and q. For each possible p1,p2,q triple I am told d1 - d2. I do not know where any of these points are from P or Q, but I have a suspicion that, with few enough P and many enough Q, if they aren't pathologically arranged (all coplanar or something), there is a way to use f where f(p1,p2,q) = d1-d2 to figure out the distance between pairs of points from P. Is this so? Do I really have enough information? If so, do you have any idea how I can figure out those distances, what the pathological conditions are, and how many Q I need for how many P? If not, what extra nugget of information would help? Hints and pointers would certainly be appreciated. Right now I'm drawing pictures of the two-dimensional version with three points in P to see if I can understand how those points are constrained by the d1-d2 from a handful of Q. -- Mark === Subject: Re: Relative distances question >In three-dimensional Euclidean space we have two disjoint sets of >points, P and Q. The points p1, p2 are members of P, and point q is a >member of Q. One can define a distance d1 between p1 and q and a >distance d2 between p2 and q. For each possible p1,p2,q triple I am >told d1 - d2. >I do not know where any of these points are from P or Q, but I have a >suspicion that, with few enough P and many enough Q, if they aren't >pathologically arranged (all coplanar or something), there is a way to >use f where f(p1,p2,q) = d1-d2 to figure out the distance between >pairs of points from P. Probably, yes. If you have |P| points in set P and |Q| points in Q, then for each q in Q you have |P|-1 independent equations of the form d_i - d_(i+1) = something given . (There's no extra information if you also tell me the value of e.g. d1-d3, except perhaps a check on the consistency of your data -- you should have d1-d3 = (d1-d2) + (d2-d3) of course.) So the data you supply amounts to |Q| (|P|-1) equations which you want to use to determine 3( |P| + |Q| ) unknowns (the coordinates of the points). Actually there are about six fewer unknowns than this because the distance data won't be changed if the coordinate axes are redrawn, so you can choose to draw the axes with q1 at the origin, q2 on the x-axis, and q3 on the x-y plane. So as long as |Q| (|P|-1) >= 3 (|P|+|Q|-2) then it is reasonable to think that you have enough data to locate all the points. (Note that this reasoning suggests that only some distance-difference measurements are really needed to find the points, so that the other distance differences should be calculable from them; as noted above, this means you run the risk of having inconsistent data if your numbers do not reall come from actual points.) Your wording suggests you want |P| small and |Q| large; well, if |P|=5 and |Q|=9, for example, then these estimates suggest there will be just enough data to determine the locations of all the points. Exactly how you will solve 36 equations in 36 unknowns is not very clear. You should also be aware that the solution set is likely to be _finite_, but need not contain just a _single_ 36-tuple of numbers. You would need at least one additional constraint to rule out all but one of the solutions to these 36 constraints. >Right now I'm drawing pictures of the two-dimensional version with >three points in P to see if I can understand how those points are >constrained by the d1-d2 from a handful of Q. The corresponding constraint is that you need |Q| (|P|-1) >= 2(|P|+|Q|)-3, and if |P|=3 this won't work for any |Q|. But with |P|=4 you should be able to solve this when |Q|>=5, if you can solve 15 non-polynomial equations in 15 unknowns! (Well, you CAN rewrite them as polynomial equations since d1-d2 = c implies d1^2 = (d2+c)^2, so d1^2-d2^2-c^2 = 2 c d2 and thus (d1^2 - d2^2 - c^2)^2 - 4 c^2 d2^2 = 0 and the left side involves only _squares_ of distances, so that it is a polynomial in the unknown coordinates of the points involved. So now you have 15 polynomial (actually quartic) equations in 15 unknowns....) I should stress that I am only thinking generically here. It is quite possible that N equations in N unknowns have a solution set that contains whole curves (or even bigger sets) rather than being a finite number of points, and it is conversely possible to write down fewer equations than unknowns and still have only a finite number of (real) solutions, e.g. x^2+y^2=0 forces x = y = 0. I am loathe to abandon the geometric origin of the equations, since that surely would provide hints for finding the solutions effectively, but if you choose to do so you can just throw some generic equation-solving tools at your problem. You can try for a numerical solution (especially if you have a ballpark estimate of what the solution coordinates are), or you can use elimination methods (e.g. Groebner Bases) to rewrite the system of equations in a better form, e.g. as a single equation in one unknown, together with another equation involving this and a second unknown, together with ... . You might start with 4+5 random points in the plane (one being (0,0) and another being of the form (x,0) ) and compute the 15 differences of distances, and then see how hard it is to recover the 15 remaining coordinates from those equations. dave === Subject: Re: Accelerator design question >> >> 2. Correct modeling of the centripetal motion. For >> relativistic electrons, is F = qv x B = dp/dt still correct? >> Yes. >> I guess F = mv^2/R would not be, since you probably >> derive that from F = dp/dt = d/dt (mv). >> Hartle says: >> R = m*v*gamma / q*B = sqrt[ E^2 - (m*c^2)^2 ] / q*B >OK, everybody seems to agree on that. Let me just check >if I can still do physics... >For velocity v, radius R, angular velocity = v/R. >dp/dt = p d(theta)/dt = pv/R = m*v^2*gamma/R > = F = qvB >So that gives R = m*v*gamma/q*B, voila! >OK, a 10 GeV electron has a gamma of 10 GeV/0.5 MeV = 20000. >q = 1.6 * 10^(-19) Coulomb, v = 3*10^8 m/sec, >let's take B = 10 Tesla, and m = 9*10^-31 kg >So that gives me R = 3.38 m from relativity. That >seems a little small. Are all my units right? Yes, they're. You're doing fine. >Anyway, let's do the same calculation non-relativistically. >My classical KE is 0.5*mv^2, so v = sqrt(2E/m) >Classically, F = mv^2/R = qvB so R = mv/qB. >R = mv/qB = m*sqrt(2E/m)/qB >E = 10 GeV = 10^10 * 1.6*10^(-19) Joules = 1.6*10^(-9) J >That gives me a velocity of 6*10^10 m/sec and a radius >of R = 0.0335 m, or 3.4 cm. Huh. The classical radius >is smaller. What do you know. >But these numbers seem small for interesting accelerators. >Let me try 500 GeV electrons, gamma = 10^6, KE = 8*10^8 J, >classical velocity = 4.2*10^11 m/sec. >Relativistic R = mc*gamma/qB = 169 m. >Newtonian R = m*4.2*10^11/qB = 0.24 m. >Am I missing something? I could have sworn that the required >relativistic one. No, you're not missing anything. The required field for same radius >At any rate they're clearly different so it's obvious >that relativistic physics gives you different design >specifications than Newtonian physics. Way, way different. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: double-step Pollard p-1 ? Here [1], [2] and there [3] are hints of a so-called double-step Pollard p-1 factorization algorithm, which is supposed to uncover a prime factor p even though p-1 has one large prime factor; on the face of it, this seems a tremendous improvement to the regular Pollard p-1 method. I can't find a description; neither online nor in [4]. Any pointer or hint ? Fran.8dois Grieu [1] Eric W. Weisstein. Pollard p-1 Factorization Method. http://mathworld.wolfram.com/Pollardp-1FactorizationMethod.html [2] John Brillhart, D.H. Lehmer, J.L. Selfridge, Bryant Tuckerman, and S.S. Wagstaff, Jr.: Factorizations of b^n+-1, b=2,3,5,6,7,10,11,12 up to high powers page lviii http://www.ams.org/online_bks/conm22 [3] Sjoerd.J.Schaper's PollarP1.bas in http://www.home.versatel.nl/vspickelen/Largefiles/LargeInt.zip [4] Bressoud, D.æM.: Factorization and Primality Testing. Springer-Verlag, 1989. === Subject: Re: Contradictory nature of Goldbach Conjecture >> This is NOT so. In fact, we can prove that for any system >> of axioms which are adequate to achieve the commonly used >> axioms of arithmetic, if there is no contradiction, this >> fact cannot be proved. Thus it is hardly trivial. >We can prove no such thing. What we can prove is that the commonly >studied theories of arithmetic can't prove their own consistency. This >has nothing to do with the fact that they can be proved consistent. >For example, take the finite ordinals and interprete '+' as ordinal >addition and '*' as ordinal multiplication and you'll find it's trivial >to show that the axioms of commonly studied theories of arithmetic are >true in the resulting structure and hence the theories are consistent. But the existence of the finite ordinals relies on axioms of set theory, like the axiom of infinity. I don't believe that there is any way that we can be absolutely certain that the axioms of arithemtic are consistent. Derek Holt. === Subject: Re: Contradictory nature of Goldbach Conjecture > But the existence of the finite ordinals relies on axioms of set theory, > like the axiom of infinity. I don't believe that there is any way that > we can be absolutely certain that the axioms of arithemtic are consistent. It's not a matter of absolute certainty in some vague philosophical sense, but of mathematics. It is a trivial mathematical fact that the commonly mentioned axioms of arithmetic are consistent. Whether you regard this and the other trivial mathematical facts as absolutely certain does nothing to alter the situation. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Contradictory nature of Goldbach Conjecture >> But the existence of the finite ordinals relies on axioms of set theory, >> like the axiom of infinity. I don't believe that there is any way that >> we can be absolutely certain that the axioms of arithemtic are consistent. >It's not a matter of absolute certainty in some vague philosophical >sense, but of mathematics. Really? >It is a trivial mathematical fact that the >commonly mentioned axioms of arithmetic are consistent. What you call trivial mathematical facts depend on the consistency of the axioms of set theory, which is a stronger assumption than the consistency of the axioms of arithmetic. Derek Holt. >Whether you >regard this and the other trivial mathematical facts as absolutely >certain does nothing to alter the situation. >-- >Aatu Koskensilta (aatu.koskensilta@xortec.fi) >Wovon man nicht sprechen kann, daruber muss man schweigen > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Contradictory nature of Goldbach Conjecture > What you call trivial mathematical facts depend on the consistency of the > axioms of set theory, which is a stronger assumption than the consistency > of the axioms of arithmetic. The consistency of arithmetic is easily provable using a bit of perfectly ordinary mathematics. It is of course open to you to doubt this mathematical theorem, or say that you don't have absolutely certain knowledge of its truth. But why pick out the trivial theorem PA is consistent in particular? You could with equal justification doubt any other mathematical theorem proved using similar mathematical assumptions - there are lots of more interesting candidates. === Subject: Re: Contradictory nature of Goldbach Conjecture >> What you call trivial mathematical facts depend on the consistency of the >> axioms of set theory, which is a stronger assumption than the consistency >> of the axioms of arithmetic. > The consistency of arithmetic is easily provable using a bit of >perfectly ordinary mathematics. It is of course open to you to doubt >this mathematical theorem, or say that you don't have absolutely >certain knowledge of its truth. But why pick out the trivial theorem >PA is consistent in particular? You could with equal justification >doubt any other mathematical theorem proved using similar mathematical >assumptions - there are lots of more interesting candidates. I have heard this attributed to Andre Weil: God exists because mathematics is consistent, and the Devil exists because it cannot be proved. -- 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: Contradictory nature of Goldbach Conjecture > I have heard this attributed to Andre Weil: > God exists because mathematics is consistent, > and the Devil exists because it cannot be proved. Well, mathematics is consistent is not a mathematical statement, unlike e.g. PA is consistent, which is also a simple mathematical theorem. === Subject: how powerful is the normalization? uniqueness of distributions... Hi all, I recenlty vaguely learned the notion of the uniqueness of distributions... But I am not very sure... can you help me clearify and give me some pointers? For example, the type of distributions f(x)=c*exp(-lamda1*x^2+lamda2*x+lamda3) is unique to Gaussian distribution, only the Gaussian distribution will have second order x^2 in the exponential, ... after some normalization, it must be turn out to Gaussian distribution finally. Another example: p(x)=c*exp(lamda*x)*p^(x-1)*(1-p), in order to make this p(x) to be a distribution, without even calculating C the scaling factor, we can immediately say that , p(x)=(p*exp(lamda))^(x-1)*(1-p*exp(lamda)) which is another Geometric distribution... thus all the tedious calculation of the scaling factor C can be skipped fastly... I hope I have vaguely captured the notion of uniqueness of distributions... I want to ask to what extent is this notion true and how powerful is the classification? Is there a name for this kind of argument? Any pointers? === Subject: Re: Orlow cardinality question > Hi John, > Please don't become angry with me. Please learn to quote the posts you respond to. > I defended here a physicist standpoint. Do you see that as bad from me? > You are not the only one in this world. You defended something that needed no defending against an attack that did not exist. You wasted words on nothing. If I thought I were the only one in this world, I probably would not be wasting time talking to you. Come to think of it... why am I? > Do you know what scimath means. Do you know what non sequitur means? John Briggs === Subject: Re: Orlow cardinality question Randy Poe said: > Randy Poe said: > > No, but in our natural view of counting (finite sets) we also > expect > other things - for example, that if one set includes another, > with > some > left over, then the one with the extra bits is larger. Choosing > to > throw away this intuition and keep the ordering one is somewhat > arbitrary. (Of course I know why: because Tony's idea ain't > actually > going anywhere, but that would be an abysmal reason for deterring > him > from trying.) > > OK, I'll stick with bigulosity. > > Now, I think it might actually be interesting to develop > a self-consistent bigulosity, but I don't think what > he's got is self-consistent, and I think it violates > other intuitions. > What other intuitions? > Some others you might hold. Perhaps ones you might hold. I am curious as to whether any of my derivations or conclusions violate your intuitions, in absence of Cantor. > Do you want to claim my system must satisfy your > intuitions, > No, but you are bothered by a system that doesn't satisfy > YOUR intuitions. There's no reason why it has to satisfy > anybody's. There's no reason why a naive intuition should > govern mathematics. But there is good reason why different mathematical approaches to a problem should not contradict each other logically, and derive different results for the same problems, don't you think? > However the idea of lining things up next to other things, > the counting model of set size, is the motivating intuition > behind set counting by bijection. Yes I understand that you try to use counting, which never bridges the gaps between the finite and infinite. I choose different approaches. > What I suspect is that YOU hold intuitions which are > contradicted by your bigulosity notions, that you can't > construct a system which simultaneously satisfies > all of your own intuitions. I rather think not. I have been working on this a long time, and in my conversations here I have been motivated to expand on them and state them clearly, which has been surprisingly easy. What intutitions do you think I might violate, in you or me? > when you don't even claim your systems satisfies yours? What ARE > your intuitions about infinite sets? > That they're bigger than finite sets. Well, no kidding. That's kind of the definition, isn't it? Is that all? > He says that the bigulosity of the rationals is |N|^2. OK, > fine. But is that bigger than |N|? Well, it certainly looks > bigger because it's the square of |N|, right? But he > hasn't worked out his infinite arithmetic yet. > It is clearly infinitely bigger, as it has one more infinite > dimension. > That is not obvious. You say it is clear that > oo x oo > oo, but you are assuming an arithmetic of > oo which is not yet constructed. It is for me. When you multiply two numbers m and n, it is represented as a rectangle of size m by n units, the product being the area in square units. If you multiply two infinities, it is not different. For instance, the grid or real numbers which Cantor tries (unsuccessfully) to traverse diagonally is N bits across, and R numbers long, so the number of digits in that grid is N*R. I don't see any limit to this construction. Do you? > That's > why it's displayed in a 2D grid of N by N terms, right? What does > that > contradict, in your view? I mean, you would conclude that one line of > N numbers > contains as many numbers as N such lines of N numbers? > Sure, because I can count them 1,2,3,... Are you counting the lines now? The columns? Oh, you're weaving back and forth diagonally. No matter how far you go, you can only have covered half of any grid whose corners on top and left have been reached. You never get to the bottom right corner. Does that ever bother you? Better to take each linear direction in the grid as a separate dimension, especially as each one corresponds to one natural number of the pair that constitutes each rational number. > As > a more trivial example, he wants to say that oo > is bigger than oo-1 and smaller than oo+1. But > for many informal proofs, we rely on an intuition > that oo+1 is equal to oo. For instance: > > x = 0.3333.... > 10x = 3.3333... = 3 + x > > Doesn't that rely on the notion that the decimal > part of 10x, which has an Orlow bigulosity of > oo-1, is equal to the decimal part of x? > The decimal part of x? That's a string representing a single value, > not a set. > Yes, very good. I spoke sloppily in an attempt to construct > an intuitive argument rather than a mathematical one. > Underlying 0.3333... is an infinite sequence, which > is a set of numbers. > 0.3333... = lim{0.3, 0.33, 0.333, 0.3333, ...} > Do you mean the set of all reals in [0,1]? > No. I mean this one infinitely length string. Can we assign > a number to its length, by numbering all of its digits? > Let us number them as a_1, a_2, a_3, ... > How many digits does 0.333... have past the decimal point? > I say it's |N|, because there's exactly one digit for > every n in N. Do you assign it a number? Yes, i also assign it N. > Now multiply it by 10. We've shifted it to the left by > one digit. So beyond the decimal place, instead of > the digits {a_1, a_2, a_3, ....}, we now have > the digits that were number {a_2, a_3, ...}, with > a_1 being on the other side of the decimal place. > How many digits are left on the right if we take one > of them and shift it to the right? N. You have just multiplied by the number base which increases the significant digit count by one. There are actually infinite digits to the left as well, but in the first they are all zeroes. By mutiplying by the number base, you turned a zero digit to the left to a non-zero digit. It didn't increase the overall infinity of digits. > How many digits are there in a string of length oo > if we delete the first one? oo-1, by that description, but that has nothing to do with multiplication. > How big is a set {a_1, a_2, a_3, ...} after we delete > the first element? one less. > How long is the sequence {0.3, 0.03, 0.003,...} if we > delete the first element to get {0.03, 0.003, ....} > (yes I know, that's not the same sequence I talked > about before, it's another one associated with the > same number). They all have infinite digits to either side. You are simply changing the values of the digits arithmetically. > - Randy -- Smiles, Tony === Subject: Re: Orlow cardinality question > But there is good reason why different mathematical approaches to a problem > should not contradict each other logically, and derive different results for > the same problems, don't you think? If I understand you correctly, then I think not. (It depends, delicately, on quite what you mean by contradict.) For example, in the reals, the polynomial x^3-x^2+4x-4 has one root. In the complex numbers it has two. In the rationals, 3 divided by 7 has a solution; in the integers it doesn't. In primary school, 7 into 3 won't go. These claims are all true in their own terms, yet mutually incompatible. One of the important features of real maths is that people choose terminology* carefully, so that lots of claims in different systems can be made without everyone losing track of what's being said. That's also why it is (I submit) a vastly better idea to call your ideas on set size something distinct, like 'bigulosity'. * (Often fanciful, which upsets a few zealots obsessed with True Meaning of words) Brian Chandler http://imaginatorium.org > However the idea of lining things up next to other things, > the counting model of set size, is the motivating intuition > behind set counting by bijection. > Yes I understand that you try to use counting, which never bridges the gaps > between the finite and infinite. I choose different approaches. > What I suspect is that YOU hold intuitions which are > contradicted by your bigulosity notions, that you can't > construct a system which simultaneously satisfies > all of your own intuitions. > I rather think not. I have been working on this a long time, and in my > conversations here I have been motivated to expand on them and state them > clearly, which has been surprisingly easy. What intutitions do you think I > might violate, in you or me? > when you don't even claim your systems satisfies yours? What ARE > your intuitions about infinite sets? > That they're bigger than finite sets. > Well, no kidding. That's kind of the definition, isn't it? Is that all? > He says that the bigulosity of the rationals is |N|^2. OK, > fine. But is that bigger than |N|? Well, it certainly looks > bigger because it's the square of |N|, right? But he > hasn't worked out his infinite arithmetic yet. > It is clearly infinitely bigger, as it has one more infinite > dimension. > That is not obvious. You say it is clear that > oo x oo > oo, but you are assuming an arithmetic of > oo which is not yet constructed. > It is for me. When you multiply two numbers m and n, it is represented as a > rectangle of size m by n units, the product being the area in square units. If > you multiply two infinities, it is not different. For instance, the grid or > real numbers which Cantor tries (unsuccessfully) to traverse diagonally is N > bits across, and R numbers long, so the number of digits in that grid is N*R. I > don't see any limit to this construction. Do you? > That's > why it's displayed in a 2D grid of N by N terms, right? What does > that > contradict, in your view? I mean, you would conclude that one line of > N numbers > contains as many numbers as N such lines of N numbers? > Sure, because I can count them 1,2,3,... > Are you counting the lines now? The columns? Oh, you're weaving back and forth > diagonally. No matter how far you go, you can only have covered half of any > grid whose corners on top and left have been reached. You never get to the > bottom right corner. Does that ever bother you? > Better to take each linear direction in the grid as a separate dimension, > especially as each one corresponds to one natural number of the pair that > constitutes each rational number. > As > a more trivial example, he wants to say that oo > is bigger than oo-1 and smaller than oo+1. But > for many informal proofs, we rely on an intuition > that oo+1 is equal to oo. For instance: > > x = 0.3333.... > 10x = 3.3333... = 3 + x > > Doesn't that rely on the notion that the decimal > part of 10x, which has an Orlow bigulosity of > oo-1, is equal to the decimal part of x? > The decimal part of x? That's a string representing a single value, > not a set. > Yes, very good. I spoke sloppily in an attempt to construct > an intuitive argument rather than a mathematical one. > Underlying 0.3333... is an infinite sequence, which > is a set of numbers. > 0.3333... = lim{0.3, 0.33, 0.333, 0.3333, ...} > Do you mean the set of all reals in [0,1]? > No. I mean this one infinitely length string. Can we assign > a number to its length, by numbering all of its digits? > Let us number them as a_1, a_2, a_3, ... > How many digits does 0.333... have past the decimal point? > I say it's |N|, because there's exactly one digit for > every n in N. Do you assign it a number? > Yes, i also assign it N. > Now multiply it by 10. We've shifted it to the left by > one digit. So beyond the decimal place, instead of > the digits {a_1, a_2, a_3, ....}, we now have > the digits that were number {a_2, a_3, ...}, with > a_1 being on the other side of the decimal place. > How many digits are left on the right if we take one > of them and shift it to the right? > N. You have just multiplied by the number base which increases the significant > digit count by one. There are actually infinite digits to the left as well, but > in the first they are all zeroes. By mutiplying by the number base, you turned > a zero digit to the left to a non-zero digit. It didn't increase the overall > infinity of digits. > How many digits are there in a string of length oo > if we delete the first one? > oo-1, by that description, but that has nothing to do with multiplication. > How big is a set {a_1, a_2, a_3, ...} after we delete > the first element? > one less. > How long is the sequence {0.3, 0.03, 0.003,...} if we > delete the first element to get {0.03, 0.003, ....} > (yes I know, that's not the same sequence I talked > about before, it's another one associated with the > same number). > They all have infinite digits to either side. You are simply changing the > values of the digits arithmetically. > > - Randy > > > -- > Smiles, > Tony === Subject: Re: Orlow cardinality question !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> But there is good reason why different mathematical approaches to a problem >> should not contradict each other logically, and derive different results for >> the same problems, don't you think? > If I understand you correctly, then I think not. (It depends, > delicately, on quite what you mean by contradict.) > For example, in the reals, the polynomial x^3-x^2+4x-4 has one > root. In the complex numbers it has two. Sputter. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Orlow cardinality question <85psvmezo8.fsf@lola.goethe.zz> > For example, in the reals, the polynomial x^3-x^2+4x-4 has one > root. In the complex numbers it has two. > Sputter. Ah, but you see it has two roots. I'm right. But it also has a third, nyah nyah! (Sorry, brain-finger disconnect) Brian Chandler http://imaginatorium.org === Subject: Re: Orlow cardinality question David Kastrup said: >> But there is good reason why different mathematical approaches to a problem >> should not contradict each other logically, and derive different results for >> the same problems, don't you think? > If I understand you correctly, then I think not. (It depends, > delicately, on quite what you mean by contradict.) > For example, in the reals, the polynomial x^3-x^2+4x-4 has one > root. In the complex numbers it has two. > Sputter. You need a new fuel filter? -- Smiles, Tony === Subject: Re: Orlow cardinality question > But there is good reason why different mathematical approaches to a > problem > should not contradict each other logically, and derive different > results for > the same problems, don't you think? > If I understand you correctly, then I think not. (It depends, > delicately, on quite what you mean by contradict.) > For example, in the reals, the polynomial x^3-x^2+4x-4 has one root. In > the complex numbers it has two. Yes, the real component is half of a complex number, along with the imaginary component. The reals are a subset of complex numbers. > In the rationals, 3 divided by 7 has a solution; in the integers it > doesn't. In primary school, 7 into 3 won't go. Yes, the integers are a subset of the rational numbers. The complex numbers are the most generalized of the number systems you mentioned. They include the integers, and every rational is equivalent to a real, and so is a subset as well. What might be true for certain subsets of the complex numbers may not hold for the most general case. If we say you can't take the square root of a negative numbers, we have restricted the complex plane to the real line. If you say you need to be able to count the value, then you have restricted it to whole numbers. If you say you can't divide by zero, then you have restricted your number set to finite values. 8D I was looking at Hilbert's axioms of geometry a little, and in the axioms of incidence, I notied that three or four could be boiled down to one axiom, if dimension were allowed to be a dependent variable. Also 1.7 is only true in 3 dimensions. Beyond 3D, two planes can intersect at a single point. I kind of think it would be nice to boil our axioms down and generalize them as much as possible, so we end up with as few and as powerful laws as we can manage, kind mathematics, and if that offends anyone, I'm sorry, but maybe it's about time. > These claims are all true in their own terms, yet mutually > incompatible. One of the important features of real maths is that > people choose terminology* carefully, so that lots of claims in > different systems can be made without everyone losing track of what's > being said. That's also why it is (I submit) a vastly better idea to > call your ideas on set size something distinct, like 'bigulosity'. Very well! I think I'll stick with Bigulosity, after all. Don't forget to capitalize it, or I'll claim not to know what you're talking about. heh heh ;) I do like terms such as equibigulous and relatively semibigulous, or even cobigulously complementary. Mathematics needs terms like that. Terms that go well with beer and nachos. > * (Often fanciful, which upsets a few zealots obsessed with True > Meaning of words) > Brian Chandler > http://imaginatorium.org -- Smiles, Tony === Subject: Re: Orlow cardinality question Ed van der Meulen said: > Hi Bob > Hou are you now? > You know that I never have heard of formalizing. > How could you know that? I have studied also proof theory. Also about our great friend Gí.a6del. > But you know me better than I know me? kind of annoying when people tell you about yourself when they hardly know you, isn't it? > You have a special brain for that > Please have a feast. I have a birthday feast tonight. Happy Birthday! O~ O~ O~ <- balloons > Are you so sour about my posting? Scientists are also people. Don't you know them? -- Smiles, Tony === Subject: Re: Orlow cardinality question Ed van der Meulen said: > Don't make it to difficult for Tony. He is a good guy. > You are right Randy. > Do you know what I would do Tony? > 1. I would write a paper about it. > 2. let that paper be commented by people who you trust > 3. rephrase your paper > 4. repeat this till you have a great result, > And you can everyone ask for help. > It's like the info analyzing phase from a project. > You have a kind of problem. You want to make it clear to everyone > I think that is a good way. > have a splendid day. The sun shines again, > ed I started out with a paper. Dave Rusin couldn't get past the first two pages, and insisted that i must accept Cantor. He cut me off right here in sci.math for not agreeing that a set of finite distinct naturals could be infinte. I submitted the paper to a mathematician friend of mine, but I think his computer couldn't handle the Word file. It is best for now to spar like this and prepare myself for all the complaints, and perhaps make a buzz. I will be getting back to the new axiom set soon, since people need to have those in order to use their bitwise minds, as well as work on my spatially-oriented visual proofs and demonstrations. The paper will come out eventually in better form. People will still hate it. I don't care. I still have my day job. And Istill feel accomplished, even if it's not recognized. That happens a lot. I'm not proud. -- Smiles, Tony === Subject: Re: Orlow cardinality question -- DO NOT REPLY TO THIS MESSAGE -- A message has been posted to the discussion sci.math. Author: TonyBones === Subject: Re: Orlow cardinality question Ed van der Meulen said: > Don't make it to difficult for Tony. He is a good guy. > You are right Randy. > Do you know what I would do Tony? > 1. I would write a paper about it. > 2. let that paper be commented by people who you trust > 3. rephrase your paper > 4. repeat this till you have a great result, > And you can everyone ask for help. > It's like the info analyzing phase from a project. > You have a kind of problem. You want to make it clear to everyone > I think that is a good way. > have a splendid day. The sun shines again, > ed I started out with a paper. Dave Rusin couldn't get past the first two pages, and insisted that i must accept Cantor. He cut me off right here in sci.math for not agreeing that a set of finite distinct naturals could be infinite. First lets establish this. You have the natural numbers and then say that is N and N is infinite. Do you mean that? We are going now very slow. Then we forget nothing. Step by step. Okay? The fist step. 1. But what does it mean you HAVE such an infinite line. In which form do we HAVE that. Do you understand me? What is having a line that we can take it and make a bundle of it. The line with numbers and stripes from your professor was an activity to do. You didn't HAVE that line. My peano staircase was also for go up the stairs. You didn't HAVE that Peano staircase. Could we now concentrate on this one notion of HAVING A LINE. How can we have that line in the way that we can work with it. So only the question how would you describe that notion of having a line. We are trying now to answer only this question. The rest has to wait. Let the sun shine Tony, ed === Subject: Re: Orlow cardinality question Ed van der Meulen said: > -- DO NOT REPLY TO THIS MESSAGE -- > A message has been posted to the discussion sci.math. > Author: TonyBones === > Subject: Re: Orlow cardinality question > Ed van der Meulen said: > Don't make it to difficult for Tony. He is a good guy. > > You are right Randy. > > Do you know what I would do Tony? > > 1. I would write a paper about it. > > 2. let that paper be commented by people who you trust > > 3. rephrase your paper > > 4. repeat this till you have a great result, > > And you can everyone ask for help. > > It's like the info analyzing phase from a project. > > You have a kind of problem. You want to make it clear to everyone > > I think that is a good way. > > have a splendid day. The sun shines again, > > ed > > I started out with a paper. Dave Rusin couldn't get past the first two pages, and insisted that i must accept Cantor. He cut me off right here in sci.math for not agreeing that a set of finite distinct naturals could be infinite. > First lets establish this. You have the natural numbers and then say that is N and N is infinite. Do you mean that? > We are going now very slow. Then we forget nothing. Step by step. Okay? > The fist step. > 1. But what does it mean you HAVE such an infinite line. In which form do we HAVE that. > Do you understand me? > What is having a line that we can take it and make a bundle of it. > The line with numbers and stripes from your professor was an activity to do. You didn't HAVE that line. My peano staircase was also for go up the stairs. You didn't HAVE that Peano staircase. > Could we now concentrate on this one notion of HAVING A LINE. > How can we have that line in the way that we can work with it. > So only the question how would you describe that notion of having a line. > We are trying now to answer only this question. The rest has to wait. > Let the sun shine Tony, > ed You are asking what an infinite line is? A set of points such that each point in the set is spatially contiguous to two other points in the line. Okay? -- Smiles, Tony === Subject: Re: Orlow cardinality question <18110088.1116530498186.JavaMail.jakarta@nitrogen.mathforum.org> > Ed van der Meulen said: > -- DO NOT REPLY TO THIS MESSAGE -- Yeah, right, whoever you are. Tony: > You are asking what an infinite line is? A set of points such that each point > in the set is spatially contiguous to two other points in the line. Okay? Really? If the point and one of its adjacent points are distinct, what is the distance between them? (Tony, remember you're going to overturn set theory: you have a lot of catching up to do first.) Brian Chandler http://imaginatorium.org === Subject: Re: Orlow cardinality question > Ed van der Meulen said: > -- DO NOT REPLY TO THIS MESSAGE -- > Yeah, right, whoever you are. > Tony: > You are asking what an infinite line is? A set of points such that > each point > in the set is spatially contiguous to two other points in the line. > Okay? > Really? If the point and one of its adjacent points are distinct, what > is the distance between them? > (Tony, remember you're going to overturn set theory: you have a lot of > catching up to do first.) > Brian Chandler > http://imaginatorium.org Why, the connection between points is an infinitely short line, or course! Just like on a circle, which is an infinity-sided regular polygon with finite diameter (measured to the center of each side of the polygon, of course, not to the vertices, for heaven's sake). -- Smiles, Tony === Subject: Re: Orlow cardinality question We are on our Way, Tony You say: You are asking what an infinite line is? A set of points such that each point in the set is spatially contiguous to two other points in the line. Okay? That is your first formulation. We store of course the history. Always nice to have it. This is a line. And now in what form can you have it. How is it given to you that you can use it as an input for a process so an output can be produced. In what way do you have that line. I know this isn't an easy question. Please think a moment about it. for instance... Can you put the line in your pocket. Can you throw it away. In what form do you have that line. And again this is already the most difficult question. Later comes that question about lazy programming. But first this. In what form do you have that line. You can always ask more questions. I have to attend the birthday party. Till tomorrow. ed === Subject: Re: Orlow cardinality question Ed van der Meulen said: > David, > I don't compare scientists with morons I have many friends who are scientist. And we talk a lot about each other interests. Ed, you missed his point. He was calling me a moron, and not a physicist, which I'm not. I am an information scientist with interest in many areas of science. I have never been called a moron by anyone that knows me, not even remotely. > For instance quantum scientists with their complex quantum theories, they like often another theory as well. > Peaple who work on quantum computing. Still in an early stage. Very interesting. I could use that for my artificial inteligence project. But I am too early for them. All open source as well. Checkable. > Theories are models for them. No more that that. > The word Moron doesn't suit in a normal discussion. > I know also cosmoligists. In need for many other theories. Everything looks completrelty new there. > So I don't undersatand you david. He's just being a jerk, Ed. > But surely I have a pure mathematical education as well. > So I understand Randy perfectly. He is right in all he has posted. > To Tony: > Tony > I mean it good for you. > You say: You (that is me) don't want the information. > I don't understand you in the form you present it. I don't talk about the content. You keep putting everything in terms of cardinality, which is not what I am talking about. This seems to be an almost universal problem here. > You want to bring a philosophic idea that has its own language and to put in an exact way in mathematics. It mathematical, Ed. You don't have to use my methods, but don't tell me they're wrong. > In CS you could use it neither in this way. Try to code it in a program. You would at least need a developing phase and often more analyzing as well. That has been going on for decades. > You have nothing of ideas that aren't realizable. Huh? > And the way to it is often rephrasing what your idea is not losing the content but adapting the form in which you present it. Using diagrams and other tools. And then arrives the developing phase. And in a real new project you can always come to surprises. > And you can trust people like Randy Poe and Dik de Winter. They know a lot of mathematics. More than you will learn. > So I don't know what your problem is with my postings. That you keep telling me I am wrong, and it doesn't seem like you have even considered what I've said, since you never refer to it specifically. > Have a nice drink, You too. > ed -- Smiles, Tony === Subject: Re: Orlow cardinality question Eckard Blumschein said: > Randy Poe said: > It is clearly infinitely bigger, as it has one more infinite dimension. That's > why it's displayed in a 2D grid of N by N terms, right? What does that > contradict, in your view? I mean, you would conclude that one line of N numbers > contains as many numbers as N such lines of N numbers? That is nonsensical bu > ANY intuition. > What you are calling intuition is noting but the stupid inability to > grasp the essence of the notion infinity. Cantor was also unable to do > so. He wonderd that the number of dimensions does not matter and the > whole universe does not contain more points than any tiny linear > interval: Je le vois, mais je ne le crois pas. It does. My notion of infinity must be different from yours. I say I can take each line of N numbers and treat them like a unit. If I have N such units, that is clearly infinitely larger than 1 such unit. Now, you can tell me I can't do this, but you had better have some justification, besides it's just not done. What is wrong with this idea, besides the fact that you didn't think of it yourself? > >> As >> a more trivial example, he wants to say that oo >> is bigger than oo-1 and smaller than oo+1. But >> for many informal proofs, we rely on an intuition >> that oo+1 is equal to oo. > This is not simply an intuition but it exactly describes the essence of > the notion infinity. it doesn't exactly describe anything. you say oo=oo+1=oo+oo=oo*oo. In other words, your position is not that arithmetic is different on infinite scales, but that it is simply irrelevant. That's simply not true. Infinite units may be defined and nested as much as desired, and finite differences can be considered, even ifinitely far from the origin. > This is some indication, perhaps, that > the rationals consitute some sort of enumeration on the reals, > Reals are uncountable, not just the number of reals but already every > single real number. In other words, reals are not exactly > enumerable. Otherwise they are not reals but rationals. Reals are enumerable across the entire range from -oo to oo. Each can be rrepresented as an ifninite string of N bits, exactly. Define what you mean by enumeration. On the subject of countability, it's irrelevant to my system, because my system doesn't rely on anything so ill-suited to studying infinities as counting. > Eckard Blumschein -- Smiles, Tony === Subject: Re: Orlow cardinality question I Define a set S Formally this is in a context. We could place braces around the texts. Talking in a meta language. This is usual in proof theory. N is a sleeping U, bottom left, from S. That means all natural numbers are also in S There is one element with name q such that all n element from N: n < q. In the James Bond Films was also a q. We have a set that includes a member infinity. Too easy. I know. But effective. Then we are looking at the consequences. Maybe we have to add more axioms. Beforehand we don't know it. So testing follows. And as soon as have q we see aleph_0 is the same. Nobody can have problems with this. Please don't discuss anymore the possibility that axioms can give infinity. Choice we have now. Would there be some scientist who isn't satisfied. Well, they can take the set without infinity. And if you want to use it. Great. When I would be a scientists. I would take... And I would reason computers are never precise. Floats they have with modulo. Losing information. A necessary consequence. And our pi has there only a limited precision, it has been reduced to a rational number. Boy oh boy those bad computers. But when you have to work with it. What then? I like also physics but I have a pure mathematical education. But I would do that in that way. Do you see that Bob with margins. Not at all precise. And they think, the throw away of information, isn't so bad. To-morrow we have a better model. How would you do that in a somewhat shabby way Bob. I have told you how I would do it. At least the first step. Shabby is also easier. I know that very well. But when something doesn't need to be precise, why not? For real analizing things you have to be precise. But in the adventurous landscape of many sciences. We walk through it, can you sing a song? And suddenly we have medicines, the plane, Missiles. Computers. And we can use them. They measure everything with a plus/mines, a margin. Who bakes our bread. What are all those people doing. Nobody can be missed. You neither, with your brain, Bob, be proud of it. Bob do you know how the models of macro-economists work. That is a real horror. Full of assumptions. Please go help them. For they advice the politicians. Maybe you are angry with the wrong people. === Subject: Re: Orlow cardinality question Ed van der Meulen said: > You are right Randy. > But Tony plunges the infinity notion as a number in it. The normal numbers are in fact ordinal numbers. And the size is a cardinal number. But Tony doesn't recognize the difference. > For me you are clear. But Tony sees it from a physicist side with physical notions. bring it back to reality and check your results periodically. There is nothing wrong with checking math against the world. > Like you are writing a computer program right away. Before thinking how the computer languages will work. I have been thinking about this for 25 years. > Tony could better post on a philosophy site. He likes to bring a general philosophy. Sure, so do you, with your lightning and waterfalls and painting. Not very mathematical subject matter. > So he will not understand this. > But your efforts were a good. > Do you know that in the cardinals Aleph-0^n = Aleph_0 for each natural n. No, and I don't agree. Aleph_0^n represents a set of n-tuples, each of which is a natural, like rational numbers are pairs of naturals, and therefore have aleph_0^2 elements in its grid. This is my perspective. > That Aleph_0 is the smallest infinity so the infinity of the even numbers is the same and also of the 1000-folds. You say that, but aleph_0 is the size of the counting numbers, and therefore a number equal to the upper bound of counting numbers, which is essentially an infinite sum of 1's. The harmonic series is also infinite, but every term is less than half the corresponding term in the sum of 1's, so that is a smaller infinity. The smallest infinity I can imagine is the sum of the inverses of primes. Gee, why didn't Cantor think of that? > The next infinity is Aleph-1. with nothing in between. That is the cardinality from R. So there's no 1-1 mapping from R to N. That is your opinion, or the continuum hypothesis, which is not derived logically from anything, but simply declared, and wrongly so. I have shown that there are many infinities between the size of the naturals and the size of the reals, and smaller than the size of the naturals, and bigger than the set of reals (but Cantor already knew that last part because he treated the continuum doifferently from the discrete infinity). > But Tony thinks the natural ordinals are the same as the infinite cardinals. Whole numbers are whole numbers. > I thought a cardinal number is normal thing. > You have cardinal numbers 1, 2 and 3 or three ones. > And ordinal numbers first, second, third and in math we write 1, 2, 3 but in fact we are just counting. Never coning to infinity. An ordinal number is just a cardinal used to refer to a position in a sequence, like an unsigned integer in a computer used as an array index. The math is the same. Ordinals define order, but cardinal derive their value from their place int he order of counting. I have never seen any important operational difference, but perhaps there is one. > You know of mathematics. But Tony speaks a physical language. That is not bad but the wrong language here. At least it's not Dutch. > Well oh well -- Smiles, Tony === Subject: Re: Orlow cardinality question > Like you are writing a computer program right away. > Before thinking how the computer languages will work. > I have been thinking about this for 25 years. I also haver given you a well meant advice. Okay. There was in the p20th century a grerat mathematisions Alan Church who developed the so called lambda calculus. Computer languages: Lisp and Haskel. (from Haskel B. Curry also a lambda calculas man) And the lamdas calculus is equivalent to the normal set theory. But now thge special thing. Chirch proved also that the lammda calculus equivalent was to the Turing machine. And I think you know what that means. This meansd when you could make a computer program af waht you want to do. That is equivalent to make it in mathematics. And that is a test as well. So could you make a software program that simulates what you want to reach. Then you have a consistent story as well. If not possible then you know that you have the way to rephrasing what you want. I could help you with that. But others can help you as well. So you have now two possibilities. I mean it good towards you. ed === Subject: Re: Orlow cardinality question Ed van der Meulen said: > Like you are writing a computer program right away. > Before thinking how the computer languages will work. > I have been thinking about this for 25 years. > I also haver given you a well meant advice. > Okay. There was in the p20th century a grerat mathematisions Alan Church who developed the so called lambda calculus. Computer languages: Lisp and Haskel. (from Haskel B. Curry also a lambda calculas man) > And the lamdas calculus is equivalent to the normal set theory. > But now thge special thing. Chirch proved also that the lammda calculus equivalent was to the Turing machine. And I think you know what that means. > This meansd when you could make a computer program af waht you want to do. That is equivalent to make it in mathematics. And that is a test as well. > So could you make a software program that simulates what you want to reach. Then you have a consistent story as well. If not possible then you know that you have the way to rephrasing what you want. I could help you with that. But others can help you as well. > So you have now two possibilities. > I mean it good towards you. > ed aTuring machine. It seems to have far more qualitative properties and relations ships than normal set theory. Can you say how they correspond, to you -- Smiles, Tony === Subject: Re: Orlow cardinality question For the real proof you have to read it. It's not a small proof from a few lines. But informal I can tell you this. Look at the Turing machine with shelves or tapes or whatever. Do you have that picture? Then think than all pieces of cake are trees that grow very fast. Each tick is a status of the whole machine and all the trees have grown and are to see. You can make a picture for each tick. Then the Turing machine goes from one state to the following. Can you follow me as well? But now this. Compare the states with grown trees. And apes jump from old grown trees to new grown trees. Are you following me? Well. Informal we are ready. Do things please always first informal. You will get a better touch with it. The Bourbaki school wanted to do it formal in all details and they didn't succeed. The apes are the lambda functions with an input state and an out put state tree. Drop the apes and the Turing machine works. Or drop the trees and you see the apes jump (together to a new state). And the same pattern we see then. It's like in photography. A negative and a positive picture. Or in paintings the figure and the rest figure. You can see it as a flip-flop. Turing-Lambda calculus. This is a link with people who work on the lambda calculus. A huge job as well and it cost many years. http://en.wikipedia.org/wiki/Lambda_calculus and http://ling.ucsd.edu/~barker/Lambda/ There's more on the net to find. When you want to make the Turing machine compared with the lambda calculus formal. Then you describe first exactly what happens with the Turing machine and the fast growing trees. and you are using just the same language. When you have done that you take a distance and you make a theory above this model and there you use a more mathematical and exact language. This is the real step we have to take when we make in formal. You will met prime events in the model. And you will notice compound actions (events) are built up from your prime actions. Then you are on your way This is proof theory. The prime action soften will give rise to new axioms. No problem with that as soon as you know what you can do then. And you use the normal formal logic. And soon some light will shine. I can make it formal. I tell this in short. But this is the center of formalization. The theory hangs above the model and has it's own more exact language. R=This language is no more that physical language. We call it the meta language. It's a more formal and parallel language to what happens in the model. Important is you make it first very true in the model. An rephrasing what you have, Just putting in other word. Splitting it or combining you rephrase. And always you arrive at a result. The rephrasing steps can cost a day. You have also to get used to new phrasing. A day is often enough. In an information analyzing step you often start with some formulation but at the end you know it much better. I think you know that. Of course you can loose phrasings but that is not important. Phrasings are just the language we use. It concerns the content. What you really want. When I am not clear, please ask, ed === Subject: Re: Orlow cardinality question And on my question... What do you really have so you can take it as an input for a process. Give just a provisional answer. I know this is the most dificult question I can ask. But why not right away. Don't ever be afraid, we will find a solution. I will tell you already the next question. 2. Do you know of lazy programming? Do you understand what that means for you at this moment. Please don't answer this question. First a rough answer to the first question. Back to the birthday party. Jippie. ed === Subject: Re: Orlow cardinality question And Tony With set theory you can do also much. The same as with computers. They are also equavalent. For many mathematicians all consists out of sets. The is a very basic notion. And you can alo have very different types of sets. But that is another story. === Subject: Re: Orlow cardinality question > > New thread just for Tony, because some of those old ones > are getting pretty overloaded. > > Just a couple of comments - I think it really would be a much > better > idea to use a different term: it's very clear that Tony's ideas are > incompatible with cardinality as normally defined. Bigulosity was > nice idea. > Yes, I agree. I wasn't going to haggle over that. But I won't be > using the c > word, so don't worry. I will stick to set size. I don't think I want > to use > bigulosity permanently. Or, maybe I should..... > > > > In our natural view of counting, we don't think > that order matters. We just assign labels. We > arbitrarily point to the first object and > call it 1. We arbitrarily choose a second > object and call it 2. We don't expect that > shuffling a deck of cards will change its size. > > No, but in our natural view of counting (finite sets) we also > expect > other things - for example, that if one set includes another, with > some > left over, then the one with the extra bits is larger. Choosing > to > throw away this intuition and keep the ordering one is somewhat > arbitrary. (Of course I know why: because Tony's idea ain't > actually > going anywhere, but that would be an abysmal reason for deterring > him > from trying.) > you did. My > ideas have already covered much of the area that Cantor defaced. > There are only > a couple of questions left before the truly interesting results > emerge. This > has all been just a correction. > Mathematics is open - anyone can play. If you can define something > interesting, people (mathematicians) will want to know. Assuming, of > course, that you understand what mathematics is, and what distinguishes > it from Belgian lace-making or English country dancing. But once you > start off on the Cantor defaced line, you mark yourself a crank, > vastly lessening your chances of ever persuading anyone capable of > being genuinely interested in new mathematics that you really have any > ideas at all. I don't think I ever used the term defaced, but I certainly have pointed out some problems in Cantor, and some solutions. It's no coincidence that WM and I were arguing the same point regarding the size of the set of naturals vs. their values. Unfortunately I think he sees them both as finite, so we both disagree with Cantor, and with each other at the same time. What fun! > Indidentally, grandiose predictions about the great results coming once > minor questions are resolved are another crank warning sign. Perhaps, but you don't see what I see. It's hard to put my diagrams here for you. I am working on it and have high hopes. I don't think I'm a crank. I usually end up being right in such cases and people shake their heads, and I move on. Such is life. > Meanwhile, of the many real mathematicians who have developed > non-standard models (Conway's surreal numbers, Robinson's nonstandard > analysis) none found it necessary to be unable to understand Cantor's > definitions of cardinality and so on. They could all do real set > theory: the sort in which considering the set of all x such that P(x) > only includes elements x for which P(x) is true. They were professional mathematicians, and obligated to study this wonder of mathematics and be an expert. I am not a professional, and therefore am free to express my distaste for whatever I want. I don't have to please the Oscar committee. I learned the basics in school, enough to ace the test, but the conclusions all seemed iffy at best, and downright incorrect at worst. Besides, I really see all the logical loopholes and shell games in it as contributing to the deliquency of mathematics. Conway and Robinson probably never tread on the same ground as Cantor, so as not to contradict his established mathematics. My territory apparently overlaps with Cantor's significantly, and in an incompatible way. Oh well. Sorry. > Brian Chandler > http://imaginatorium.org > > Brian Chandler > http://imaginatorium.org > > > > -- > Smiles, > > Tony -- Smiles, Tony === Subject: Re: Orlow cardinality question > > New thread just for Tony, because some of those old ones > are getting pretty overloaded. > > Just a couple of comments - I think it really would be a much > better > idea to use a different term: it's very clear that Tony's ideas are > incompatible with cardinality as normally defined. Bigulosity was > a > nice idea. > Yes, I agree. I wasn't going to haggle over that. But I won't be > using the c > word, so don't worry. I will stick to set size. I don't think I want > to use > bigulosity permanently. Or, maybe I should..... > > > > In our natural view of counting, we don't think > that order matters. We just assign labels. We > arbitrarily point to the first object and > call it 1. We arbitrarily choose a second > object and call it 2. We don't expect that > shuffling a deck of cards will change its size. > > No, but in our natural view of counting (finite sets) we also > expect > other things - for example, that if one set includes another, with > some > left over, then the one with the extra bits is larger. Choosing > to > throw away this intuition and keep the ordering one is somewhat > arbitrary. (Of course I know why: because Tony's idea ain't > actually > going anywhere, but that would be an abysmal reason for deterring > him > from trying.) extent > you did. My ******************************************************************** > ideas have already covered much of the area that Cantor defaced. ******************************************************************** > There are only > a couple of questions left before the truly interesting results > emerge. This > has all been just a correction. > Mathematics is open - anyone can play. If you can define something > interesting, people (mathematicians) will want to know. Assuming, of > course, that you understand what mathematics is, and what distinguishes > it from Belgian lace-making or English country dancing. But once you > start off on the Cantor defaced line, you mark yourself a crank, > vastly lessening your chances of ever persuading anyone capable of > being genuinely interested in new mathematics that you really have any > ideas at all. > I don't think I ever used the term defaced, ... OK: Look up to the line separated by asterisks. That's you. > ... but I certainly have pointed out > some problems in Cantor, and some solutions. Well, yes. Perceived problems. But many people here have seen exactly these same problems pointed out over and over again, with exactly the same flaws (basically, there's *always* a quantifier swap) in the logic. There was a chap called Phil a year or three back, who approached things almost exactly as you have. Couldn't accept that (our, finite) natural numbers are not a finite set, and drew uncannily similar numbers consisting of digits and dots in ever-increasingly baroque (and ill-defined) combinations. So if you think you have already done something original, you are plainly mistaken. It's no coincidence that WM and I > were arguing the same point regarding the size of the set of naturals vs. their > values. Unfortunately I think he sees them both as finite, so we both disagree > with Cantor, and with each other at the same time. What fun! > Indidentally, grandiose predictions about the great results coming once > minor questions are resolved are another crank warning sign. > Perhaps, but you don't see what I see. It's hard to put my diagrams here for > you. I am working on it and have high hopes. I don't think I'm a crank. I > usually end up being right in such cases and people shake their heads, and I > move on. Such is life. Pity if you get so keyed up about it that you don't learn something out of the debris. You've got as far as having boundless sets. Why don't you try to formalise the difference between a boundless set and an infinite set, instead of responding like you did to my making this suggestion the other day. > Meanwhile, of the many real mathematicians who have developed > non-standard models (Conway's surreal numbers, Robinson's nonstandard > analysis) none found it necessary to be unable to understand Cantor's > definitions of cardinality and so on. They could all do real set > theory: the sort in which considering the set of all x such that P(x) > only includes elements x for which P(x) is true. > They were professional mathematicians, and obligated to study this wonder of > mathematics and be an expert. I am not a professional, and therefore am free to > express my distaste for whatever I want. I don't have to please the Oscar > committee. I learned the basics in school, enough to ace the test, but the > conclusions all seemed iffy at best, and downright incorrect at worst. Hmm, you might well get crank points for that - acing the test while knowing it's all 'wrong'. Look, Conway's numbers are a different system from the conventional real numbers of set theory (they are a proper class, for a start), so of course lots of things are true in surreals and not in Cantor set theory, and vice versa. That's different from claiming that propositions with a clear and simple proof in some system are wrong. > Besides, > I really see all the logical loopholes and shell games in it as contributing to > the deliquency of mathematics. Conway and Robinson probably never tread on the > same ground as Cantor, so as not to contradict his established mathematics. My > territory apparently overlaps with Cantor's significantly, and in an > incompatible way. Oh well. Sorry. So I think your claimed distinction here just isn't. Brian Chandler http://imaginatorium.org === Subject: Re: Orlow cardinality question > > > New thread just for Tony, because some of those old ones > are getting pretty overloaded. > > Just a couple of comments - I think it really would be a much > better > idea to use a different term: it's very clear that Tony's ideas > are > incompatible with cardinality as normally defined. Bigulosity > was > a > nice idea. > Yes, I agree. I wasn't going to haggle over that. But I won't be > using the c > word, so don't worry. I will stick to set size. I don't think I > want > to use > bigulosity permanently. Or, maybe I should..... > > > > In our natural view of counting, we don't think > that order matters. We just assign labels. We > arbitrarily point to the first object and > call it 1. We arbitrarily choose a second > object and call it 2. We don't expect that > shuffling a deck of cards will change its size. > > No, but in our natural view of counting (finite sets) we also > expect > other things - for example, that if one set includes another, > with > some > left over, then the one with the extra bits is larger. > Choosing > to > throw away this intuition and keep the ordering one is somewhat > arbitrary. (Of course I know why: because Tony's idea ain't > actually > going anywhere, but that would be an abysmal reason for > deterring > him > from trying.) > extent > you did. My > ******************************************************************** > ideas have already covered much of the area that Cantor defaced. > ******************************************************************** > There are only > a couple of questions left before the truly interesting results > emerge. This > has all been just a correction. > > Mathematics is open - anyone can play. If you can define something > interesting, people (mathematicians) will want to know. Assuming, > of > course, that you understand what mathematics is, and what > distinguishes > it from Belgian lace-making or English country dancing. But once > you > start off on the Cantor defaced line, you mark yourself a crank, > vastly lessening your chances of ever persuading anyone capable of > being genuinely interested in new mathematics that you really have > any > ideas at all. > I don't think I ever used the term defaced, > ... OK: Look up to the line separated by asterisks. That's you. Oh, my bad. When you put it alone like that, I thought of Cantor is defaced, and thought I never claimed to deface him. Yes, I think he created a problematic system that many feel is a blemish, but is very difficult to undo. Sorry. > ... but I certainly have pointed out > some problems in Cantor, and some solutions. > Well, yes. Perceived problems. But many people here have seen exactly > these same problems pointed out over and over again, with exactly the > same flaws (basically, there's *always* a quantifier swap) I remember WM being accused of that. I don;t think that was in reference to me. If it was, I wouldn't mind seeing specifics. > in the > logic. There was a chap called Phil a year or three back, who > approached things almost exactly as you have. Couldn't accept that > (our, finite) natural numbers are not a finite set, and drew uncannily > similar numbers consisting of digits and dots in ever-increasingly > baroque (and ill-defined) combinations. So if you think you have > already done something original, you are plainly mistaken. I know people have been trying to make this work for a long time. I think it does. I know now what a tenacious bunch you all are, and I can see how the training in this system has adversely affected the way people think about these things. it's time for the axiomatics to get a little picture training, and stop thinking exclusively in words. That's not how the structure of benzene was discovered. > It's no coincidence that WM and I > were arguing the same point regarding the size of the set of naturals > vs. their > values. Unfortunately I think he sees them both as finite, so we both > disagree > with Cantor, and with each other at the same time. What fun! > > Indidentally, grandiose predictions about the great results coming > once > minor questions are resolved are another crank warning sign. > Perhaps, but you don't see what I see. It's hard to put my diagrams > here for > you. I am working on it and have high hopes. I don't think I'm a > crank. I > usually end up being right in such cases and people shake their > heads, and I > move on. Such is life. > Pity if you get so keyed up about it that you don't learn something out > of the debris. You've got as far as having boundless sets. Why don't > you try to formalise the difference between a boundless set and an > infinite set, instead of responding like you did to my making this > suggestion the other day. Because boundless sets are not part of my theory. That term was a tongue-in- cheek response to the need to have infinite sets of distinct finite natural numbers, which is mathematically impossible outside of set theory. My point was this: if you want to violate the rules of numbers this way, please make up a name that doesn't have a meaning outside of your theory, and resolve your anomaly so we don't fight about this nonsense. > Meanwhile, of the many real mathematicians who have developed > non-standard models (Conway's surreal numbers, Robinson's > nonstandard > analysis) none found it necessary to be unable to understand > Cantor's > definitions of cardinality and so on. They could all do real set > theory: the sort in which considering the set of all x such that > P(x) > only includes elements x for which P(x) is true. > They were professional mathematicians, and obligated to study this > wonder of > mathematics and be an expert. I am not a professional, and therefore > am free to > express my distaste for whatever I want. I don't have to please the > Oscar > committee. I learned the basics in school, enough to ace the test, > but the > conclusions all seemed iffy at best, and downright incorrect at > worst. > Hmm, you might well get crank points for that - acing the test while > knowing it's all 'wrong'. I am capable of memorizing and performing nonsensical operations. I can follow rules. I can play chess. But, chess isn't supposed to be an exact description of anything. If something is supposed to be true, then I evaluate it against other truths, bith for similarity and difference, and keep my eyes out for contradictions in my understanding. > Look, Conway's numbers are a different system > from the conventional real numbers of set theory (they are a proper > class, for a start), so of course lots of things are true in surreals > and not in Cantor set theory, and vice versa. That's different from > claiming that propositions with a clear and simple proof in some system > are wrong. Mostly I take this stance because whenever I present my ideas, Cantorians tell me I am wrong. It has happened many times here. You don't want me to attack your system? Don't call it correct, expecially when it is at odds with so many other areas of math, not to mention mathematical intuition, as well as the rules cocnerning finite sets, like proper subsets are smaller. If you want to claim your axioms are correct without justifying them, then I have the right to show that ina larger context this little system is totally confused. I mean, I am surprised you folks don't take it all the way into town, and prove that there are MORE naturals than rationals. What a triumph that would be! > Besides, > I really see all the logical loopholes and shell games in it as > contributing to > the deliquency of mathematics. Conway and Robinson probably never > tread on the > same ground as Cantor, so as not to contradict his established > mathematics. My > territory apparently overlaps with Cantor's significantly, and in an > incompatible way. Oh well. Sorry. > So I think your claimed distinction here just isn't. That sentence no objective clause. What means? > Brian Chandler > http://imaginatorium.org -- Smiles, Tony === Subject: Re: Orlow cardinality question > ... but I certainly have pointed out > some problems in Cantor, and some solutions. > Well, yes. Perceived problems. But many people here have seen exactly > these same problems pointed out over and over again, with exactly the > same flaws (basically, there's *always* a quantifier swap) > I remember WM being accused of that. I don;t think that was in reference to me. > If it was, I wouldn't mind seeing specifics. OK. I hope you really do understand that for all x (written Ax) really means for absolutely any individual x you care to give me, and not for the set of xes. Urm, ok, I'll write for any, because I don't like using non-inverted 'A'. For any (finite) set P of natnums, exists natnum m s.t. m > n for all n in P. TRUE Reverse the order of quantification: Exists natnum m s.t. for any (finite) set P of natnums, m > n for all n in P. FALSE > logic. There was a chap called Phil a year or three back, who > approached things almost exactly as you have. Couldn't accept that > (our, finite) natural numbers are not a finite set, and drew uncannily > similar numbers consisting of digits and dots in ever-increasingly > baroque (and ill-defined) combinations. So if you think you have > already done something original, you are plainly mistaken. > I know people have been trying to make this work for a long time. I think it > does. I know now what a tenacious bunch you all are, and I can see how the > training in this system has adversely affected the way people think about these > things. Then how did John Conway invent surreals? Your claims of brainwashing are silly, and at least mildly offensive. > ... it's time for the axiomatics to get a little picture training, and stop > thinking exclusively in words. That's not how the structure of benzene was > discovered. Ironically, the inspiration here was realising that the benzene molecule has no end! > It's no coincidence that WM and I > were arguing the same point regarding the size of the set of naturals > vs. their > values. Unfortunately I think he sees them both as finite, so we both > disagree > with Cantor, and with each other at the same time. What fun! You're not disagreeing with Cantor, you're claiming that a huge body of mathematics that has been thoroughly studied by tens(?) of thousands of clever people just happens to be utterly wrong, because none of these people noticed something that you have. Anyway, let's leave that aside. > Pity if you get so keyed up about it that you don't learn something out > of the debris. You've got as far as having boundless sets. Why don't > you try to formalise the difference between a boundless set and an > infinite set, instead of responding like you did to my making this > suggestion the other day. > Because boundless sets are not part of my theory. That term was a tongue-in- > cheek response to the need to have infinite sets of distinct finite natural > numbers, which is mathematically impossible outside of set theory. So does your version of whatever corresponds to set theory (bagulosity, perhaps?) allow one to consider the collection of all objects x having property P for any well-formed property P? Such as being either zero or a natnum plus one? In what way does this break down, to force this collection to include some things that don't have property P? > Look, Conway's numbers are a different system > from the conventional real numbers of set theory (they are a proper > class, for a start), so of course lots of things are true in surreals > and not in Cantor set theory, and vice versa. That's different from > claiming that propositions with a clear and simple proof in some system > are wrong. > Mostly I take this stance because whenever I present my ideas, Cantorians tell > me I am wrong. What is wrong is not whatever you have created (bigulosity, or the beginnings thereof), but your posting of plainly unsupported claims that this or that bit of standard theory is wrong. I promise not to call anything you *do* wrong, if you agree to stop ranting about Cantor being wrong. That way we might get somewhere. > Besides, > I really see all the logical loopholes and shell games in it as > contributing to > the deliquency of mathematics. Conway and Robinson probably never > tread on the > same ground as Cantor, so as not to contradict his established > mathematics. My > territory apparently overlaps with Cantor's significantly, and in an > incompatible way. Oh well. Sorry. > So I think your claimed distinction here just isn't. > That sentence no objective clause. What means? What you say is true isn't. I think that's grammatical (not commenting on truth value for the moment). I meant: you appear to claim that whereas Conway (for example) creates something disjoint from Cantorian set theory, you are going to create something that somehow contradicts it. This is not a valid distinction, IMO, because lots of statements about surreal numbers are not true about reals; you may establish facts about bigulosity that are not true of cardinality - great, but this does not in itself make cardinality wrong. The very best thing that could ever happen for you is that universities abandon the teaching of cardinality and start teaching bigulosity; they will never give courses called Why Cantor was wrong. (And not just because he wasn't.) Brian Chandler http://imaginatorium.org === Subject: Re: Orlow cardinality question > ... but I certainly have pointed out > some problems in Cantor, and some solutions. > > Well, yes. Perceived problems. But many people here have seen > exactly > these same problems pointed out over and over again, with exactly > the > same flaws (basically, there's *always* a quantifier swap) > I remember WM being accused of that. I don;t think that was in > reference to me. > If it was, I wouldn't mind seeing specifics. > OK. I hope you really do understand that for all x (written Ax) > really means for absolutely any individual x you care to give me, and > not for the set of xes. Urm, ok, I'll write for any, because I > don't like using non-inverted 'A'. Okay, I remeber now. I was accused of that mistake when doing my inductive proof that the naturals as an infinite set must contain infinite values, but that was because people got confused about what I was doing. I was using a set defined by each successive natural, and using induction to show that if the set generated by one natural had a property, thet the set generated by the successive natural had that property. The property of defining a set that has a certain property, IS a property of the natural number that defines the set. So, that property would be true of any set defined in that same way on any natural number. It wasn't that I was generalizing properties of element to sets, though I am not sure that's always inappropriate. > For any (finite) set P of natnums, exists natnum m s.t. m > n for all n > in P. TRUE Sure, you can always identify the largest in a finite set, at least in the conventional sense, and find one larger than that one. > Reverse the order of quantification: > Exists natnum m s.t. for any (finite) set P of natnums, m > n for all n > in P. FALSE No, you cannot identify a largest natural number in the conventional sense. > logic. There was a chap called Phil a year or three back, who > approached things almost exactly as you have. Couldn't accept that > (our, finite) natural numbers are not a finite set, and drew > uncannily > similar numbers consisting of digits and dots in > ever-increasingly > baroque (and ill-defined) combinations. So if you think you have > already done something original, you are plainly mistaken. > I know people have been trying to make this work for a long time. I > think it > does. I know now what a tenacious bunch you all are, and I can see > how the > training in this system has adversely affected the way people think > about these > things. > Then how did John Conway invent surreals? Your claims of brainwashing > are silly, and at least mildly offensive. Okay, I am being too generally harsh here, Set theory has actually led to excellent ways of thinking about things, notably including Conway's work on surreals. It's a great alternative way to look at things, and worthwhile. I guess my particular beef is with cardinality measures. That's the only part that really bothers me. I don't mean to put down set theory in general, and I'm sorry if I implied that. Cardinality of infinite sets is totally mishandled, however, in my opinion, and has a few inconsistencies to blame for it's poor conclusions. > ... it's time for the axiomatics to get a little picture training, > and stop > thinking exclusively in words. That's not how the structure of > benzene was > discovered. > Ironically, the inspiration here was realising that the benzene > molecule has no end! And yet endless as it may be, it can be conceived of and understood. The snake bit its tail in a dream, and benzene was understood. The snake biting its tail is a sign of infinity. Why is this? The number line is only a straight line when locally viewed. When you step back infinitely far away, it is a circle. Positive and negative infinity are one. Irony indeed. Or symbolism for the moment. > It's no coincidence that WM and I > were arguing the same point regarding the size of the set of > naturals > vs. their > values. Unfortunately I think he sees them both as finite, so we > both > disagree > with Cantor, and with each other at the same time. What fun! > You're not disagreeing with Cantor, you're claiming that a huge body > of mathematics that has been thoroughly studied by tens(?) of thousands > of clever people just happens to be utterly wrong, because none of > these people noticed something that you have. Anyway, let's leave that > aside. As I said above, i am not against set theory by any means and apologize if it sounded like that. I disagree with cardinality measures as they applied to infinite sets. > Pity if you get so keyed up about it that you don't learn something > out > of the debris. You've got as far as having boundless sets. Why > don't > you try to formalise the difference between a boundless set and > an > infinite set, instead of responding like you did to my making > this > suggestion the other day. > Because boundless sets are not part of my theory. That term was a > tongue-in- > cheek response to the need to have infinite sets of distinct finite > natural > numbers, which is mathematically impossible outside of set theory. > So does your version of whatever corresponds to set theory to cardinality, really.... > (bagulosity, perhaps?) Ha ha Good one! > allow one to consider the collection of all > objects x having property P for any well-formed property P? Such as > being either zero or a natnum plus one? In what way does this break > down, to force this collection to include some things that don't have > property P? I am not sure what you are asking here. > Look, Conway's numbers are a different system > from the conventional real numbers of set theory (they are a proper > class, for a start), so of course lots of things are true in > surreals > and not in Cantor set theory, and vice versa. That's different from > claiming that propositions with a clear and simple proof in some > system > are wrong. > Mostly I take this stance because whenever I present my ideas, > Cantorians tell > me I am wrong. > What is wrong is not whatever you have created (bigulosity, or the > beginnings thereof), but your posting of plainly unsupported claims > that this or that bit of standard theory is wrong. I promise not to > call anything you *do* wrong, if you agree to stop ranting about > Cantor being wrong. That way we might get somewhere. Perhaps. It seems extremely hard for those who have accepted cardinality to consider any other approach to set size measures for infinite sets, because they feel they have proved their resluts, and I am simply wrong, a moron, a crackpot or crank, retarded, uneducated, or a troll. I disagree with those methods and those results, and think they really need to be uprooted, and that won't happen until the flaws are understood and better method developed that avoids them. I really feel that Cantor is dividing by zero and hiding it. But, I'll try to ignore Cantor, as long as no one tries to prove to me that I am wrong using those confounded bijections. Dagnabbit! > Besides, > I really see all the logical loopholes and shell games in it as > contributing to > the deliquency of mathematics. Conway and Robinson probably never > tread on the > same ground as Cantor, so as not to contradict his established > mathematics. My > territory apparently overlaps with Cantor's significantly, and in > an > incompatible way. Oh well. Sorry. > > So I think your claimed distinction here just isn't. > That sentence no objective clause. What means? > What you say is true isn't. I think that's grammatical (not > commenting on truth value for the moment). I meant: you appear to claim > that whereas Conway (for example) creates something disjoint from > Cantorian set theory, you are going to create something that somehow > contradicts it. This is not a valid distinction, IMO, because lots of > statements about surreal numbers are not true about reals; you may > establish facts about bigulosity that are not true of cardinality - > great, but this does not in itself make cardinality wrong. The very > best thing that could ever happen for you is that universities abandon > the teaching of cardinality and start teaching bigulosity; they will > never give courses called Why Cantor was wrong. (And not just because > he wasn't.) Point well taken. It's like hate speech in political campaigns. People get sick of hearing it. I feel sorry for Cantor and the way things went down for him, and he shouldn't be forgotten and more than Freud should. Set theory is robust, but in this area I see a problem. I'll try to calm down, as long as I don't have to go through the same arguments over and over, as if cardinality is the last word on the subject. I mean, when I learned it 25 years ago, I kind of went, That doesn't sound right. I don't think I need to keep that in my infinity bag. I think it goes with the trisected angles and squared circles, and other curiosities and left it at that. The infinity bag has a lot of other stuff in it, but people don't seem to like my snakes and straight circles, or multiplying zero by infinity. They all like the bag that's like the one's everybody else has. That's human nature. I'm the kid in 5th grade with the green mohawk. Oh no, actually, that's my son. The apple doesn't fall far from the tree. > Brian Chandler > http://imaginatorium.org -- Smiles, Tony === Subject: Re: Orlow cardinality question >> No, but in our natural view of counting (finite sets) we also expect >> other things - for example, that if one set includes another, with >some >> left over, then the one with the extra bits is larger. Choosing to >> throw away this intuition and keep the ordering one is somewhat >> arbitrary. (Of course I know why: because Tony's idea ain't actually >> going anywhere, but that would be an abysmal reason for deterring him >> from trying.) >OK, I'll stick with bigulosity. >Now, I think it might actually be interesting to develop >a self-consistent bigulosity, but I don't think what >he's got is self-consistent, and I think it violates >other intuitions. Can you compare the bigulousity of two sets that have no elements in common? I'm thinking of the odd numbers and the even numbers here. Or, even better, the positive odd numbers and the negative odd numbers. >He says that the bigulosity of the rationals is |N|^2. OK, >fine. But is that bigger than |N|? Since the rationals contain the integers doesn't that mean that the bigulousity of the rationals is greater than the bigulousity of the integers by definition? >Well, it certainly looks >bigger because it's the square of |N|, right? But he >hasn't worked out his infinite arithmetic yet. As >a more trivial example, he wants to say that oo >is bigger than oo-1 and smaller than oo+1. But >for many informal proofs, we rely on an intuition >that oo+1 is equal to oo. For instance: > x = 0.3333.... > 10x = 3.3333... = 3 + x >Doesn't that rely on the notion that the decimal >part of 10x, which has an Orlow bigulosity of >oo-1, is equal to the decimal part of x? So we have cardinal arithmetic, ordinal arithmetic, and bigulous arithmetic. Just make sure you know which one you are doing. Alan -- Defendit numerus === Subject: Re: Orlow cardinality question Alan Morgan said: >> No, but in our natural view of counting (finite sets) we also expect >> other things - for example, that if one set includes another, with >some >> left over, then the one with the extra bits is larger. Choosing to >> throw away this intuition and keep the ordering one is somewhat >> arbitrary. (Of course I know why: because Tony's idea ain't actually >> going anywhere, but that would be an abysmal reason for deterring him >> from trying.) >OK, I'll stick with bigulosity. >Now, I think it might actually be interesting to develop >a self-consistent bigulosity, but I don't think what >he's got is self-consistent, and I think it violates >other intuitions. > Can you compare the bigulousity of two sets that have no > elements in common? I'm thinking of the odd numbers and > the even numbers here. Or, even better, the positive odd > numbers and the negative odd numbers. Sure. They are all measured using functional mappings of the naturals, and the functions that define them are compared. Any two sets that are both defined as functions on the naturals, or both as functions on the reals, can be compared by comparing the inverses of their mapping functions. All the sets you mentioned have size N/2, and are therefore equal in size. >He says that the bigulosity of the rationals is |N|^2. OK, >fine. But is that bigger than |N|? > Since the rationals contain the integers doesn't that mean > that the bigulousity of the rationals is greater than the > bigulousity of the integers by definition? Yes, it certainly satisfies that intuition for me, Alan. The naturals are an infinitesimal fraction of the naturals, a tiny proper subset, and obviously not equivalent in size, in my eyes. >Well, it certainly looks >bigger because it's the square of |N|, right? But he >hasn't worked out his infinite arithmetic yet. As >a more trivial example, he wants to say that oo >is bigger than oo-1 and smaller than oo+1. But >for many informal proofs, we rely on an intuition >that oo+1 is equal to oo. For instance: > x = 0.3333.... > 10x = 3.3333... = 3 + x >Doesn't that rely on the notion that the decimal >part of 10x, which has an Orlow bigulosity of >oo-1, is equal to the decimal part of x? > So we have cardinal arithmetic, ordinal arithmetic, and bigulous > arithmetic. Just make sure you know which one you are doing. comparing things to that as one would do with finites. After all, don't we do that on the finite level already, choose a unit and work with it consistently? > Alan -- Smiles, Tony === Subject: Re: Orlow cardinality question > Alan Morgan said: > Can you compare the bigulousity of two sets that have no > elements in common? I'm thinking of the odd numbers and > the even numbers here. Or, even better, the positive odd > numbers and the negative odd numbers. > Sure. They are all measured using functional mappings of the naturals, and the > functions that define them are compared. Any two sets that are both defined as > functions on the naturals, or both as functions on the reals, can be compared > by comparing the inverses of their mapping functions. All the sets you > mentioned have size N/2, and are therefore equal in size. Are you sure? If the naturals (from 1) are N, then the integers must be 2N+1. I guess this means there are N+1 even integers and N odd ones? (I think 'number' usually includes negative.) >He says that the bigulosity of the rationals is |N|^2. OK, >fine. But is that bigger than |N|? Is it OK? The rationals include negative values and zero. So I calculate b(Q) as 2N^2+1, writing b() for bigulosity, and using Tony's convention that N is the number of naturals. > So we have cardinal arithmetic, ordinal arithmetic, and bigulous > arithmetic. Just make sure you know which one you are doing. infinity and > comparing things to that as one would do with finites. After all, don't we do > that on the finite level already, choose a unit and work with it consistently? Yes, assuming [!!] that one can do these comparisons as one would with finites. Many of us think that's precisely what you _can't_ do, but don't give up yet. Brian Chandler http://imaginatorium.org === Subject: Re: Orlow cardinality question > Alan Morgan said: > > Can you compare the bigulousity of two sets that have no > elements in common? I'm thinking of the odd numbers and > the even numbers here. Or, even better, the positive odd > numbers and the negative odd numbers. > Sure. They are all measured using functional mappings of the > naturals, and the > functions that define them are compared. Any two sets that are both > defined as > functions on the naturals, or both as functions on the reals, can be > compared > by comparing the inverses of their mapping functions. All the sets > you > mentioned have size N/2, and are therefore equal in size. > Are you sure? If the naturals (from 1) are N, then the integers must be > 2N+1. I guess this means there are N+1 even integers and N odd ones? (I > think 'number' usually includes negative.) See my response to Alan on this. > >He says that the bigulosity of the rationals is |N|^2. OK, >fine. But is that bigger than |N|? > Is it OK? The rationals include negative values and zero. So I > calculate b(Q) as 2N^2+1, writing b() for bigulosity, and using Tony's > convention that N is the number of naturals. See my response to Alan. > So we have cardinal arithmetic, ordinal arithmetic, and bigulous > arithmetic. Just make sure you know which one you are doing. > infinity and > comparing things to that as one would do with finites. After all, > don't we do > that on the finite level already, choose a unit and work with it > consistently? > Yes, assuming [!!] that one can do these comparisons as one would with > finites. Many of us think that's precisely what you _can't_ do, but > don't give up yet. Whyever do you THINK that? > Brian Chandler > http://imaginatorium.org -- Smiles, Tony === Subject: Re: Orlow cardinality question >Alan Morgan said: >> >>> No, but in our natural view of counting (finite sets) we also expect >>> other things - for example, that if one set includes another, with >>some >>> left over, then the one with the extra bits is larger. Choosing to >>> throw away this intuition and keep the ordering one is somewhat >>> arbitrary. (Of course I know why: because Tony's idea ain't actually >>> going anywhere, but that would be an abysmal reason for deterring him >>> from trying.) >> >>OK, I'll stick with bigulosity. >> >>Now, I think it might actually be interesting to develop >>a self-consistent bigulosity, but I don't think what >>he's got is self-consistent, and I think it violates >>other intuitions. >> Can you compare the bigulousity of two sets that have no >> elements in common? I'm thinking of the odd numbers and >> the even numbers here. Or, even better, the positive odd >> numbers and the negative odd numbers. >Sure. They are all measured using functional mappings of the naturals, and the >functions that define them are compared. Any two sets that are both defined as >functions on the naturals, or both as functions on the reals, can be compared >by comparing the inverses of their mapping functions. Could you explain how this is done? The rational numbers can be defined as a function on the naturals, but you don't think that the rational numbers have the same bigulousity (whew! I'm really enjoying this word) as the integers. >All the sets you >mentioned have size N/2, and are therefore equal in size. What about the positive numbers? Negative numbers? I'm assuming that both of them are size |N|/2. The postive numbers, negative numbers, and 0 make up N, but that implies that |N|/2 + |N|/2 + 1 = |N|, which doesn't make a lot of sense to me. The logical (?) conclusion is that the bigulousity of the positive numbers is actually |N|/2 - 1/2. I'd be interested in knowing what the bigulousity of the prime numbers is. |N|/k won't work for any finite k. All in all, the traditional definition of cardinality seems to work much better. Alan -- Defendit numerus === Subject: Re: Orlow cardinality question Alan Morgan said: >Alan Morgan said: >> >>> No, but in our natural view of counting (finite sets) we also expect >>> other things - for example, that if one set includes another, with >>some >>> left over, then the one with the extra bits is larger. Choosing to >>> throw away this intuition and keep the ordering one is somewhat >>> arbitrary. (Of course I know why: because Tony's idea ain't actually >>> going anywhere, but that would be an abysmal reason for deterring him >>> from trying.) >> >>OK, I'll stick with bigulosity. >> >>Now, I think it might actually be interesting to develop >>a self-consistent bigulosity, but I don't think what >>he's got is self-consistent, and I think it violates >>other intuitions. >> >> Can you compare the bigulousity of two sets that have no >> elements in common? I'm thinking of the odd numbers and >> the even numbers here. Or, even better, the positive odd >> numbers and the negative odd numbers. >Sure. They are all measured using functional mappings of the naturals, and the >functions that define them are compared. Any two sets that are both defined as >functions on the naturals, or both as functions on the reals, can be compared >by comparing the inverses of their mapping functions. > Could you explain how this is done? The rational numbers can be defined as > a function on the naturals, but you don't think that the rational numbers > have the same bigulousity (whew! I'm really enjoying this word) as the > integers. The bigulosity would be defined by the inverse of the function. I can't offhand think of how to define the rationals as a proper function on a single natural variable. They are represented by two naturals with a notation, which is why I consider them to be a 2D natural number system. The grid of rationals is NxN, so i calculate the area of that square as N^2 (N=aleph_0). Can you think of an invertible function on a single natural that generates the rationals completely in a linear order. I am not saying it can't be done. I susp[ect perhaps it can. I'll think about it. One would hope that if one generated such a function, and applied the inverse function method, that one would get N^2, but that would make the function equivalent to sqrt(n), which doesn't seem to fit at all, at first glance. So maybe it can't be done. If one can define the rationals as a set defined by a function of a single natural variable, and the method doesn't work on that function, then I would consider that the first genuine objection >All the sets you >mentioned have size N/2, and are therefore equal in size. > What about the positive numbers? Negative numbers? I'm assuming that > both of them are size |N|/2. The postive numbers, negative numbers, > and 0 make up N, but that implies that |N|/2 + |N|/2 + 1 = |N|, which > doesn't make a lot of sense to me. The logical (?) conclusion is that > the bigulousity of the positive numbers is actually |N|/2 - 1/2. Well, I almost mentioned this previously, but decided to keep things consistent. But, since you brought it up, I really think the unit infinity is represented by a line, not a ray, and the whole number set (W) should be everything from negative infinity ro positive infinity, like the reals. In this case the positive whole numbers would be W/2. Really I am talking about the integers, except that I don't exclude infinite whole numbers. Now, again, this will look to you like we have an odd number of numbers, with 0 in the middle, but remember that +0=-0, when I tell you that +oo=-oo, and the whole line can be inverted so that -0 and +0 are the endpoints, and +-oo is the midpoint. The number line is a circle. 0 and oo are both positive and negative, but guess what? oo's odd, and almost surely prime. :D D: Now I really won the crank contest. :D > I'd be interested in knowing what the bigulousity of the prime numbers > is. |N|/k won't work for any finite k. Since the primes can't be formulated exactly, the functional approach won't work. But, since we know the number of primes less than x is asymptotic to x/log(x), our work is done for us. As x goes to |N|, the function goes to N/log (N). I would say this function describes the size of the set relative to N. Log (N) is of course infinite, so this is an infinite number, which is infinitely smaller than N. An interesting question I have. Is there an inverse function g(x) for f(x) =x/log(x), such that g(f(x))=x? What is this g(x), and can it be used to approximate primes? > All in all, the traditional definition of cardinality seems to work much > better. For what? Take your time. > Alan -- Smiles, Tony === Subject: Re: Orlow cardinality question Bob. Give people some freedom. Everybody has his or her own interest. And you know a lot of me, that I don't know. Please let us stay friends. Nothing in mathematics will go away. It will only grow. Do you know of discrete math? Look in Physics are the laws of Newton. But also the quantum theories. Certainly not so easy. Axiomatized in the meta theory above it. the so called MM. meta-model. So formalized. A very big job. But what do you think they use in the life sciences? Just Newton. Where are you afraid of. For analysing jobs is our good math very needed. For instance the tools that come from decrete math are for only simulations and teaching and that kind of things. But our traditinal math will be needed for sure. Do you think we can throw away the analizing job? I think everybody defends his here his or her interest. === Subject: Re: Orlow cardinality question another reply And all those goods that mathematicians have produced will stay. No one can take that away. They can only add things. So Bob I am also with you. === Subject: Re: expected value for sampling without replacement <42882A57.5060205@netscape.net> OK, I now agree that what must be meant by sampling without replacement is choosing a group of n unknown elements at a time, thus we do NOT allow the known elements to change the probability distribution of the unknown ones. Thus it is unconditional probabilities, or conditional probabilities in the sense that we sum over all possibilities. --Jeff === Subject: Re: expected value for sampling without replacement <42882A57.5060205@netscape.net> I don't follow the first 2 paragraphs, and I'm not going to think about them anymore. The 3rd paragraph is a clever argument, and I appreciate it. This whole problem for me seems to turn on what exactly is meant by sampling without replacement. Evidently it means choose a group of n unknown items at a time; if you look at the items as they are being chosed sequentially, then the probabilities of the remaing items change and it is impossible to get the result that E(X bar) = μ --Jeff === Subject: Re: expected value for sampling without replacement <42882A57.5060205@netscape.net> <42899F5F.8030804@netscape.net> The probability that the 2nd card is a 3 is either 4/51 or 3/51, depending on the first choice. Clearly, the probabilities are different depending on whether or not we allow the knowledge of previous draws to change the pdf. For my problem, stated at top of thread, somehow it is implicit that we draw an unknown group of n elements at a time. I would naively think that we would draw elements and then look at them, but evidently not, since the calculations don't give the right answers that way. --Jeff === Subject: Re: expected value for sampling without replacement > The probability that the 2nd card is a 3 is either 4/51 or 3/51, > depending on the first choice. px = P(1st card is not a 3) = 48/52 = 12/13 p3 = P(1st card is a 3) = 4/52 = 1/13 px3 = P(2nd card is a 3 | 1st card is not a 3) = 4/51 p33 = P(2nd card is a 3 | 1st card is a 3) = 3/51 p_3 = P(2nd card is a 3) = px * px3 + p3 * p33 = (12/13 * 4/51) + (1/13 * 3/51) = (48 + 3) / (13 * 51) = 51 / (13 * 51) = 1 / 13 Stephen wasn't asking for px3 or p33, Stephen was asking for p_3. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: expected value for sampling without replacement <42882A57.5060205@netscape.net> <42899F5F.8030804@netscape.net> > The probability that the 2nd card is a 3 is either 4/51 or 3/51, > depending on the first choice. Those are the conditional probabilities. The unconditional probability that the 2nd card is a 3 is P(1st card is not a 3)* (4/51) + P(1st card is a 3)*(3/51) = (48/52)*(4/51) + (4/52)*(3/51) = 4/52 My point is that the calculation above is unnecessary, because the symmetry makes it obvious that the 2nd card is as likely as the first card to be a 3. Mike === Subject: Re: Complex Prime Integers Hello Michael Orion, You should get to read a good book Fermat Last Theoremby Prof.Harold Edwards.It talks about cyclotemics integers and the eforts made by Kummer to answer your question Best whishes george Ghiata === Subject: Re: Complex Prime Integers George, You should get a life. Your postings are fraught with errors that you readily admit and yet you have the arrogance to say you have found a simple proof of FLT that has alluded thousands of great minds for centuries. Take a dose of reality man. And please, refrain from giving me any of your advice. It is not welcomed. === Subject: Re: Complex Prime Integers MO, Others have answered your questions directly. I will provide more background. First of all, the set you are talking about is generally known as the Gaussian integers. In this context, primes in the usual sense of the word are known as rational primes rather than as real primes. The Gaussian integers are a ring (which is not important for what follows) whose units, as you say, are 1, -1, i, -1, and which has unique factorization into Gaussian primes. If a rational prime p can be written as a sum of two squares, p = a^2 + b^2, then evidently it factors in the Gaussians as p = (a + bi)(a-bi) and so is not a prime. As is well known, primes congruent to 1 mod 4 can be written as sums of two squares and so do not remain prime as Gaussians. Actually the usual proof that such primes can be written this way is to use the unique factorization of Gaussians and a couple of small observations. Conversely, if a rational prime does not remain prime as a Gaussian, then it can be written as (a + bi)(a - bi), and so we must have that p = a^2 + b^2. Hence rational primes congruent to 3 mod 4 cannot factor as Gaussian integers and so must remain prime. The prime 2 is unique, because it factors, up to units, as 2 = unit * (1 + i)^2. It is said to be ramified because its prime factorization involves a repeated factor. The word ramify derives from a word in some ancient language that means branched, and I believe that the use here ultimately derives from considerations in algebraic geometry. A very useful property of the Gaussian integers is called the norm function, represented as N, and defined by N(a + bi) = a^2 + b^2 What makes this function particularly useful is that the norm of a product is equal to the product of the norms (easy computation, made even easier if you use complex conjugation). It is also true that a Gaussian is a unit if and only if it has a norm of 1. It is now easily seen that a Gaussian whose norm is a rational prime must be a Gaussian prime. The converse is false, as 3 is a Gaussian prime but N(3) = 9. There is a huge volume of work that derives from early researches into Gaussian integers. It is called algebraic number theory. There are many books on elementary number theory that include an introduction to this subject by discussing the adjunction of certain non-integers to the collection of integers and seeing what kind of arithmetic structure results. This is a field well worth looking into. Achava === Subject: Re: Complex Prime Integers Achava and Nobody, === Subject: Re: prove it measurable > He said R^n, now you say [n,n+1], a bit confusing perhaps... Oops, quite right. I did not notice that. Of course, the method that I described also applies to R^n. Jose Carlos Santos === Subject: Re: Question on Chaitin > Dik T. Winter says... > If it's a finite program, then it must a) halt, or b) enter a loop, or > inductive expansion, because we have no way to generate randomness. > > If it enters a loop, then that is basically the same thing as halting. >Eh? No, not in this area. > Well, looping is as good as halting from a certain point of view: > If a program halts, then you can prove that it halts. If > a program loops, then you can prove that it loops. (Where to loop > means to repeat exactly the same state twice.) However, if > a program neither loops nor halts, it may not be possible to > prove that it neither loops nor halts. Easy to find or not, a halting predicate is a halting predicate. -- Eray === Subject: Re: Question on Chaitin >> I didn't say anything about any heuristic principle. > But Chaitin does. It is the heuristic principle that a theory > of complexity N cannot prove theorems of complexity greater than N. Sigh. (Imagine a real Scandinavian sigh.) So what do you think Chaitin meant by theorems of complexity greater than N? I realise it is a principle of yours not to answer questions but please, just for once, indulge me. And please don't tell me I can deduce the answer by some bizarre chain of reasoning - just tell me what you think he means. Or if you have no idea, tell me that. -- 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: Question on Chaitin > So what do you think Chaitin meant by > theorems of complexity greater than N? He was of course using the explicit definition he has given in a number of papers. And as Chaitin explains, any set of axioms will have theorems with arbitrarily high complexity. Just what is it you find unclear or objectionable about this? === Subject: Re: Question on Chaitin >> So what do you think Chaitin meant by >> theorems of complexity greater than N? > He was of course using the explicit definition he has given in a > number of papers. I knew you wouldn't answer the question. For the 6th time, WHAT DO YOU THINK HE MEANT BY THE PHRASE? If you think he defined it explicitly elsewhere, just repeat the explicit definition. Please. -- 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: Question on Chaitin > For the 6th time, WHAT DO YOU THINK HE MEANT BY THE PHRASE? As I have said several times, he is of course using the definition of information content of X given in the paper. === Subject: Re: Question on Chaitin Discussion, linux) >> So what do you think Chaitin meant by >> theorems of complexity greater than N? > He was of course using the explicit definition he has given in a > number of papers. Could you be a bit more explicit? Do you mean: the complexity of a theorem is the complexity of the string representing the proposition proved? > And as Chaitin explains, any set of axioms will > have theorems with arbitrarily high complexity. Just what is it you > find unclear or objectionable about this? -- Jesse F. Hughes All information is subject to change without notice. -- California Alternative High School === Subject: Re: Question on Chaitin > Do you mean: the complexity of a theorem is the complexity of the > string representing the proposition proved? Chaitin doesn't say anything about strings representing propositions, which is typical philosophical jargon. Why don't you simply look up Chaitin's explicit definition of The algorithmic information content of an object X in the paper at issue? Apparently some people, like Keith Ramsay, seem inclined to think that in speaking of the information content of a theorem, Chaitin cannot have this definition in mind, but is instead presupposing some other, unstated definition. This idea is particularly puzzling since Chaitin himself, in the 1992 reference, very explicitly states that it is not literally true that a set of axioms of complexity N cannot yield a theorem of complexity greater than N, referring to this statement as a heuristic principle, one that should perhaps be rephrased so as to eliminate the false implication. === Subject: Re: Question on Chaitin >> Do you mean: the complexity of a theorem is the complexity of the >> string representing the proposition proved? > Chaitin doesn't say anything about strings representing > propositions, which is typical philosophical jargon. Why don't you > simply look up Chaitin's explicit definition of The algorithmic > information content of an object X in the paper at issue? Apparently ttp://www.cs.auckland.ac.nz/CDMTCS/chaitin/iti.pdf Abstract: We propose an improved definition of the complexity of a formal axiomatic system: this is now taken to be the minimum size of a self-delimiting program for enumerating the set of theorems of the formal system. Using this new definition, we show (a) that no formal system of complexity n can exhibit a specific object with complexity greater than n + c, and (b) that a formal system of complexity n can determine at most n+ c scattered bits of the halting probability. We also present a short, self-contained proof of (b). The new idea is to measure the complexity of a formal system in terms of the program-size complexity of enumerating its infinite set of theorems, not in terms of the program-size complexity of the finite string of the axioms. This new approach combines in a single number the complexity of the axioms and the rules of inference, and the new complexity K' is never more than c greater and can sometimes be up to ~ log_2 K less than the old complexity K. Thus the incompleteness results given here are never weaker and are sometimes somewhat stronger than the incompleteness results in [16]. [16] G. J. Chaitin, Algorithmic Information Theory, 3rd Printing, Cambridge: Cambridge University Press (1990). See Chaitin, G. J., G.9adel's Theorem and Information, International Journal of Theoretical Physics 22 (1982), pp 941-954. Chaitin, G. J., Information-Theoretic Incompleteness, Applied Mathematics and Computation 52 (1992), pp. 83-101. > some people, like Keith Ramsay, seem inclined to think that in > speaking of the information content of a theorem, Chaitin cannot have > this definition in mind, but is instead presupposing some other, > unstated definition. This idea is particularly puzzling since Chaitin > himself, in the 1992 reference, very explicitly states that it is not > literally true that a set of axioms of complexity N cannot yield a > theorem of complexity greater than N, referring to this statement as a > heuristic principle, one that should perhaps be rephrased so as to > eliminate the false implication. === Subject: Re: Question on Chaitin <87hdgzf0jm.fsf@phiwumbda.org> >> Do you mean: the complexity of a theorem is the complexity of the >> string representing the proposition proved? > Chaitin doesn't say anything about strings representing > propositions, which is typical philosophical jargon. Why don't you > simply look up Chaitin's explicit definition of The algorithmic > information content of an object X in the paper at issue? Apparently > ttp://www.cs.auckland.ac.nz/CDMTCS/chaitin/iti.pdf > Abstract: > We propose an improved definition of the complexity of a formal > axiomatic system: this is now taken to be the minimum size of a > self-delimiting program for enumerating the set of theorems of > the formal system. Using this new definition, we show (a) that no > formal system of complexity n can exhibit a specific object with > complexity greater than n + c, and (b) that a formal system of > complexity n can determine at most n+ c scattered bits of the > halting probability. We also present a short, self-contained proof of (b). > The new idea is to measure the complexity of a formal system in > terms of the program-size complexity of enumerating its infinite > set of theorems, not in terms of the program-size complexity of > the finite string of the axioms. > This new approach combines in a single number the complexity of > the axioms and the rules of inference, and the new complexity K' > is never more than c greater and can sometimes be up to ~ log_2 K > less than the old complexity K. Thus the incompleteness results > given here are never weaker and are sometimes somewhat stronger > than the incompleteness results in [16]. > [16] G. J. Chaitin, Algorithmic Information Theory, 3rd Printing, > Cambridge: Cambridge University Press (1990). Right, this is what Chaitin talks of, the complexity is of the enumeration of T obviously, but since Torkel DIDN'T READ Chaitin's work he CANNOT KNOW, can he? -- Eray === Subject: Re: Question on Chaitin > Right, this is what Chaitin talks of, the complexity is of the > enumeration of T obviously, That is a reasonable definition of the complexity of a theory, sure, and it is one given by Chaitin. The complexity of a theorem is a different matter. You seem strangely reluctant to actually read anything written by Chaitin. === Subject: Re: Question on Chaitin <87hdgzf0jm.fsf@phiwumbda.org> > Right, this is what Chaitin talks of, the complexity is of the > enumeration of T obviously, > That is a reasonable definition of the complexity of a theory, sure, > and it is one given by Chaitin. The complexity of a theorem is a > different matter. You seem strangely reluctant to actually read > anything written by Chaitin. Nah, you still don't understand what the theorem proving power is supposed to be. It's the number of bits of Omega that you can prove. That's what Stephen's quote is meant to indicate. I think he got the right piece here. -- Eray === Subject: Re: Question on Chaitin > Nah, you still don't understand what the theorem proving power > is supposed to be. It's the number of bits of Omega that you can prove. How many true statements of the form the n-th bit of Omega is i a theory proves depends on just how we choose to define Omega. Chaitin's heuristic principle has nothing to do with this, as he himself explains. However, it is of course open to you to ignore Chaitin's comments. === Subject: Re: Question on Chaitin <87hdgzf0jm.fsf@phiwumbda.org> > Nah, you still don't understand what the theorem proving power > is supposed to be. It's the number of bits of Omega that you can prove. > How many true statements of the form the n-th bit of Omega is i a > theory proves depends on just how we choose to define Omega. False. > Chaitin's heuristic principle has nothing to do with this, as he > himself explains. However, it is of course open to you to ignore > Chaitin's comments. Irrelevant. -- Eray === Subject: Re: Question on Chaitin > False. Come now. This is a trivial technical point. > Irrelevant. Why is that, since the Question on Chaitin specifically concerned his heuristic principle? === Subject: Re: On normed vector spaces >>>Hi all, >>>Let V and W be two closed vector subspaces of some normed vector space >>>such that their intersection is equal to {0}. If (v_n)_n is a sequence >>>of elements of V, (w_n)_n is a sequence of elements of W and >>> lim_n ||v_n + w_n|| = 0, >>>does it follow that >>> lim_n ||v_n|| = lim_n ||w_n|| = 0? >>>My guess is that the answer is negative but, so far, I've been unable >>>to find a counter-example. >>>Any ideas? >>>Jose Carlos Santos >>Considering the continuous projection P: V (+) W --> V, (v,w) -> v, the >>sequence v_n = P(v_n +w_n) converges to 0. So I guess the assertion is >>right. Or am I missing anything - since this seems to be very simple. > _exactly_ why is that projection continuous? And with regard > to what norm on V + W? (The projection is continuous if we > give V + W the product norm ||v + w|| = ||v|| + ||w||, but > that norm is not relevant here. If V + W were a Banach > space one could show that that norm was equivalent to the > norm inherited from the underlying space, but even if the > underlying space is complete V + W need not be closed.) >>J. > ************************ > David C. Ullrich *complete* as a subspace. Robert did so in his reply, too, including an example when V (+) W fails to be closed in a Banach space. J. === Subject: OUTLINE OF ELEMENTARY PROOF OF FERMAT LAST THEOREM Here is Archimedes Proof of Fermat Last theorem george Ghiata To Alf and Andrew, X^n+Y^n=Z^n is Imposible if Z,X,YAre Integers and n=prime>2 We take Z=even number X+Y-Z=B X=B+Q Y=B+P 2*B+Q+P=s*u^n=W where s=1 or n^(n-1) if Z is divisible by n B+Q+P=r*z*u wher r is 1 or n if Z is divisible by n Therefore Q+P=2*u*k *r where k =odd number m=n^(n-2) if Z is divisible by n Let's take (B+Q)^n+(B+P)^n=(B+Q+P)^n where Z is not divisible by n We multiply the Eq by (2*u)^n and get: (2*B+Q+P+Q-P)^n+(2*B+Q+P+P-Q)^n=(2*u*z)^n After we develop the parantheses and move everything to the left of Eq. and divide by (2*u)^n we get : EQG: (W^2)*A +(W)*V*(Q-P)^(n-3) +(n)*(Q-u*k)^(n-1)-z^n=0 where u=2*V Let's say T=[n*(Q-u*k)^(n-1](b+k) If we divide EQ.G by (b+k) we get that T must be divisible by 2 as many times as (u^2)*(Q-P)^(n-3) Now we take X^n+Y^n=Z^n and after division by (X+Y)=u^n we change it to: EQJ: [(X+Y)*E +n*X^(n-1)]=z^n We change this to EQ.S: [(n-1)*(u*2)*E+n*(n-1)*(b+k)*u*A+n*(Q-u*k)^(n-1)-z^n=0 where A is a odd number. We see from EQ.S that T={ [n*(Q-u*k)^(n-1)]-z^n}/(b+k) is divisible by 2 as many times as (n-1)*u is divisible by 2 Therefore case Z not divisible by n is impossible The same proof is used to prove the case for Z divisible by n Created by George Ghiata-AA === Subject: Re: OUTLINE OF ELEMENTARY PROOF OF FERMAT LAST THEOREM > Here is Archimedes Proof of Fermat Last theorem > george Ghiata > To Alf and Andrew, > X^n+Y^n=Z^n is Imposible if Z,X,YAre Integers and > n=prime>2 > We take Z=even number > X+Y-Z=B > X=B+Q > Y=B+P > 2*B+Q+P=s*u^n=W where s=1 or n^(n-1) if Z is > divisible by n > B+Q+P=r*z*u wher r is 1 or n if Z is divisible by n > Therefore Q+P=2*u*k *r where k =odd number > m=n^(n-2) if Z is divisible by n > Let's take (B+Q)^n+(B+P)^n=(B+Q+P)^n where Z is not > t divisible by n > We multiply the Eq by (2*u)^n and get: > (2*B+Q+P+Q-P)^n+(2*B+Q+P+P-Q)^n=(2*u*z)^n > After we develop the parantheses and move everything > g to > the left of Eq. and divide by (2*u)^n we get : > EQG: (W^2)*A +(W)*V*(Q-P)^(n-3) > 3) +(n)*(Q-u*k)^(n-1)-z^n=0 > where u=2*V > Let's say T=[n*(Q-u*k)^(n-1](b+k) > If we divide EQ.G by (b+k) we get that T must be > divisible by 2 as many times as (u^2)*(Q-P)^(n-3) > Now we take X^n+Y^n=Z^n and after division by > y (X+Y)=u^n we change it > to: EQJ: [(X+Y)*E +n*X^(n-1)]=z^n > We change this to > EQ.S: > [(n-1)*(u*2)*E+n*(n-1)*(b+k)*u*A+n*(Q-u*k)^(n-1)-z^n=0 > where A is a odd number. > We see from EQ.S that T={ > [n*(Q-u*k)^(n-1)]-z^n}/(b+k) is divisible by 2 as > many times as (n-1)*u is divisible by 2 > Therefore case Z not divisible by n is impossible > The same proof is used to prove the case for Z > divisible by n > Created by George Ghiata-AA Now I find this OUTLINE wrong I stick with the one posted at to Dik Winter about FLT which I tryed to symplify more and I failed. Let's remember from my What the modern mathematicien are shy to talk aboutpostings that my intension is to show that ther is a elementary proof of FLT for n=4*L+1 at Least and has to do with the topic about modern mathematicien george ghiata === Subject: Re: OUTLINE OF ELEMENTARY PROOF OF FERMAT LAST THEOREM Here it is: X,Y ,Z relative Prime numbers X+Y=W X^(n-1)-(X^(n-2))*Y +(X^(n-3))*Y^2.........+Y^(n-1)= =R= W^(n-1)-(Cn1)*[W^(n-2)]*Y+(Cn2)*[W^(n-3]*Y^2- -(Cn3)*[W^(n-4)]*Y^3+ +(Cn4)*[W^(n-5)]*Y^4-..............-(Cn2)*[W^1)*Y^(n-2) + +(Cn1)*Y^(n-1) where Cnk=[n*(n-1)*(n-2)*(n-3)*.........*(n-k+1)]/2*3*4*............*k If (X+Y) is divisible by n, you can see that R is divisible by n only. But (X+Y)*R=Z^n=(u*z*n)^n Since R is divisible only by n then (X+Y) =[n^(n-1)]]*u^n and Z=n*u*z When Z is not divisible by n we see that (X+Y)=W and R do not have any common divisor .Therfore they are relative prime and must be: R=z^n and W=X+Y=u^n and Z=u*z Is this clear?The editing on computer is not easy. george ghiata > -- > Here is Archimedes Proof of Fermat Last theorem > george Ghiata > To Alf and Andrew, > X^n+Y^n=Z^n is Imposible if Z,X,YAre Integers and > n=prime>2 > We take Z=even number > X+Y-Z=B > X=B+Q > Y=B+P > 2*B+Q+P=s*u^n=W where s=1 or n^(n-1) if Z is > divisible by n > B+Q+P=r*z*u wher r is 1 or n if Z is divisible by n > Therefore Q+P=2*u*k *r where k =odd number > m=n^(n-2) if Z is divisible by n > Let's take (B+Q)^n+(B+P)^n=(B+Q+P)^n where Z is not > t divisible by n > We multiply the Eq by (2*u)^n and get: > (2*B+Q+P+Q-P)^n+(2*B+Q+P+P-Q)^n=(2*u*z)^n > After we develop the parantheses and move everything > g to > the left of Eq. and divide by (2*u)^n we get : > EQG: (W^2)*A +(W)*V*(Q-P)^(n-3) > 3) +(n)*(Q-u*k)^(n-1)-z^n=0 > where u=2*V > Let's say T=[n*(Q-u*k)^(n-1](b+k) > If we divide EQ.G by (b+k) we get that T must be > divisible by 2 as many times as (u^2)*(Q-P)^(n-3) > Now we take X^n+Y^n=Z^n and after division by > y (X+Y)=u^n we change it > to: EQJ: [(X+Y)*E +n*X^(n-1)]=z^n > We change this to > EQ.S: > [(n-1)*(u*2)*E+n*(n-1)*(b+k)*u*A+n*(Q-u*k)^(n-1)-z^n=0 > where A is a odd number. > We see from EQ.S that T={ > [n*(Q-u*k)^(n-1)]-z^n}/(b+k) is divisible by 2 as > many times as (n-1)*u is divisible by 2 > Therefore case Z not divisible by n is impossible > The same proof is used to prove the case for Z > divisible by n > Created by George Ghiata-AA === Subject: Re: OUTLINE OF ELEMENTARY PROOF OF FERMAT LAST THEOREM > Here it is: > Suppose: X+Y=W > Now: R = X^(n-1) - (X^(n-2))*Y + (X^(n-3))*Y^2.........+Y^(n-1) = > = W^(n-1) - (Cn1)*[W^(n-2)]*Y + (Cn2)*[W^(n-3]*Y^2 - > - (Cn3)*[W^(n-4)]*Y^3 + Cn4)*[W^(n-5)]*Y^4 - > - ... - (Cn2)*[W^1)*Y^(n-2) + Cn1*Y^(n-1) > where > Cnk = [n*(n-1)*(n-2)*(n-3)*.........*(n-k+1)]/2*3*4*............*k This is the kind of editing I mean. > If (X+Y) is divisible by n, you can see that R is divisible by n only. Reformulation: R has only a single factor n (because Y is co-prime to W). You may also note that here you use that n is not even. > But (X+Y)*R=Z^n=(u*z*n)^n formulation I surmise that u = gcd(Z, X + Y). What z is I have not yet any idea. I think you mean that because R is divisible by n, and n is prime, so Z must also be divisible by n. And so z = Z / (u * n). > Since R is divisible only by n then (X+Y) =[n^(n-1)]]*u^n > and Z=n*u*z Since only a single factor n is accounted for (when X + Y is divisible by n), so... (X + Y) = [n^(n-1)] * u^n. > When Z is not divisible by n we see that (X+Y)=W and R do not have any > common divisor .Therfore they are relative prime and must be: > R=z^n and W=X+Y=u^n and Z=u*z > Is this clear?The editing on computer is not easy. Fairly, the editing above took me less time (on a computer) than trying to find what everything meant. Conclusion: If (X + Y) is divisible by n, Z can be written as u*z*n and (X + Y) = [n^(n-1)]*u^n, and R = n * z^n If Z is not divisible by n, Z can be written as u * z and (X + Y) = u^n, and R = z^n. In both cases u and z are co-prime. I am missing the case that Z is divisible by n but (X + Y) is not. But that is repaired easily enough. (Z divisible by n and (X + Y) not, means that R is divisible by n, but it starts with W^(n-1) followed by terms divisible by n, ...) > X+Y-Z=B > X=B+Q > Y=B+P > 2*B+Q+P=s*u^n=W where s=1 or n^(n-1) if Z is > divisible by n What if Z is divisible by n, but X + Y is not? > B+Q+P=r*z*u wher r is 1 or n if Z is divisible by n Now you lost me again. Ah, I see, B + Q + P = Z. So this is stating the obvious. Let's see whether I can reformulate this. We have: (X + Y) divisible by n: (X + Y) = [n^(n-1)]*u^n, Z = u*z*n, and r = n B = X + Y - Z = u * n * (n^{n-2)*u^(n-1) - z). (X + Y) not divisible by n: (X + Y) = u^n, Z = u * z, and r = 1 B = X + Y - Z = u * (u^(n-1) - z). So B = b * u * r with b some complex expression. Where do you find that u is even? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: a question for the anti-Cantorians At first I thought that the whole Cantor-discussion was just plain nonsense, but I have changed my mind. Actually I now find it quite interesting to see how various people react to such anti-intuitive notions as transfinite arithmetic or even inaccessible cardinals. Now my question for the anti-Cantorians is: does anyone have a better theory on which mathematics should be based, rather than traditional set theory (naive or axiomatic)? Or do you at least have an idea of what such a theory should look like? === Subject: Re: a question for the anti-Cantorians Kim said: > At first I thought that the whole Cantor-discussion was just plain nonsense, but I have changed my mind. Actually I now find it quite interesting to see how various people react to such anti-intuitive notions as transfinite arithmetic or even inaccessible cardinals. Now my question for the anti-Cantorians is: does anyone have a better theory on which mathematics should be based, rather than traditional set theory (naive or axiomatic)? Or do you at least have an idea of what such a theory should look like? Hi Kim - Glad you changed your mind.:D You ask a good question. Personally, I think a lot of set theory is probably alright. My beef has to do with conclusions about infinite sets and infinities in general, when they use sets of numbers and draw conclusions about their values and about equivalences between infinite sets, ignoring the functions they use to form the bijections that supposedly prove equivalence. It's Cardinality that I disagree with as a tool for saying anythign about infinity. It works for finite sets, but not for infinite sets, since it is essentially based on counting, and therefore can never really reach infinity and is inappropriate for drawing conclusions about infinity. So, I have put forth the Theory of Bigulosity, which is a work in progress, but addresses size comparisons between sets defined as functions on the integers or real numbers using the mapping functions, in a way that is intuitively satisfying, robust and precise. It addresses the relationship between what you call aleph_0 and aleph_1, the discrete and continuous infinities, and addresses dimensionality, or compound infinities. I'll probably start a thread sometime next week with a web link, so keep your eyes peeled. Not everyone who disagrees with the crowd is wrong, although that alone can drive some people nuts. Luckily, I was innoculated against that when I grew up in Manhattan. :D -- Smiles, Tony === Subject: Re: a question for the anti-Cantorians > Personally, I think a lot of set theory is probably alright. My beef > has to do with conclusions about infinite sets and infinities in > general, when they use sets of numbers and draw conclusions about > their values and about equivalences between infinite sets, ignoring > the functions they use to form the bijections that supposedly prove > equivalence. It's Cardinality that I disagree with as a tool for > saying anythign about infinity. It works for finite sets, but not for > infinite sets, since it is essentially based on counting, and > therefore can never really reach infinity and is inappropriate for > drawing conclusions about infinity. If there can be functions between sets, there can be bijections and injections and surjections among them. So either TO rejects functions between infinite sets entirely or allows the basis for a standard cardinality theory. He can't have t both ways! === Subject: Re: a question for the anti-Cantorians Virgil said: > Personally, I think a lot of set theory is probably alright. My beef > has to do with conclusions about infinite sets and infinities in > general, when they use sets of numbers and draw conclusions about > their values and about equivalences between infinite sets, ignoring > the functions they use to form the bijections that supposedly prove > equivalence. It's Cardinality that I disagree with as a tool for > saying anythign about infinity. It works for finite sets, but not for > infinite sets, since it is essentially based on counting, and > therefore can never really reach infinity and is inappropriate for > drawing conclusions about infinity. > If there can be functions between sets, there can be bijections and > injections and surjections among them. > So either TO rejects functions between infinite sets entirely or allows > the basis for a standard cardinality theory. He can't have t both ways! An invertible function is equivalent to a bijection. I say the functions be treated differently than the way bijections are treated, and that assumptions are made about the meaning of bijections which are unfounded, as I have made clear. If that's not clear to you by now there is nothing I can do for you. -- Smiles, Tony === Subject: Re: a question for the anti-Cantorians > Actually I now find it quite interesting to see how >various people react to such anti-intuitive notions as >transfinite arithmetic or even inaccessible cardinals. To say anti-intuitive is to miss the point. The point is that such ideas have no connection to reality. All of the mathematics which stands a chance of being applied to the task of understanding the world we live in, must make predictions about the results of computational experiments. Thus, real mathematics could be defined as the science of phenomena observable in the world of computation. >Now my question for the anti-Cantorians is: does anyone >have a better theory on which mathematics should be based, PA with restrictions on the use of the existential quantifier will be almost enough for discrete mathematics. The continuum requires the introduction of probabilistic logic. === Subject: Re: a question for the anti-Cantorians david petry said: > Actually I now find it quite interesting to see how >various people react to such anti-intuitive notions as >transfinite arithmetic or even inaccessible cardinals. > To say anti-intuitive is to miss the point. The point > is that such ideas have no connection to reality. All of > the mathematics which stands a chance of being applied to > the task of understanding the world we live in, must make > predictions about the results of computational experiments. > Thus, real mathematics could be defined as the science of > phenomena observable in the world of computation. Very well said! May I quote you on that? >Now my question for the anti-Cantorians is: does anyone >have a better theory on which mathematics should be based, > PA with restrictions on the use of the existential > quantifier will be almost enough for discrete mathematics. I'm sorry, PA? Momentary lapse perhpas, on my part. Could you expand on that? > The continuum requires the introduction of probabilistic > logic. Really? Can you explain that connection? I am very interested in probabilistic logic and the measure of the continuum, but the line connecting those two dots in my head is not well formed at all. Care to draw it for me? -- Smiles, Tony === Subject: Re: a question for the anti-Cantorians > The continuum requires the introduction of probabilistic > logic. Would you mind to explain that point to us ? Han de Bruijn === Subject: Re: a question for the anti-Cantorians >> Actually I now find it quite interesting to see how >>various people react to such anti-intuitive notions as >>transfinite arithmetic or even inaccessible cardinals. > To say anti-intuitive is to miss the point. The point > is that such ideas have no connection to reality. All of > the mathematics which stands a chance of being applied to > the task of understanding the world we live in, must make > predictions about the results of computational experiments. > Thus, real mathematics could be defined as the science of > phenomena observable in the world of computation. Just curious... do you folks accept the existence of irrational numbers like the square root of 2? Or the idea a continuous function? How about the Intermediate Value Theorem? Dan === Subject: Re: a question for the anti-Cantorians Dan Christensen said: >> Actually I now find it quite interesting to see how >>various people react to such anti-intuitive notions as >>transfinite arithmetic or even inaccessible cardinals. > To say anti-intuitive is to miss the point. The point > is that such ideas have no connection to reality. All of > the mathematics which stands a chance of being applied to > the task of understanding the world we live in, must make > predictions about the results of computational experiments. > Thus, real mathematics could be defined as the science of > phenomena observable in the world of computation. > Just curious... do you folks accept the existence of irrational numbers like > the square root of 2? Or the idea a continuous function? How about the > Intermediate Value Theorem? > Dan I have no problem with any of those concepts. Does cardinality's countability depend on any of them, or vice versa? -- Smiles, Tony === Subject: Re: a question for the anti-Cantorians > Dan Christensen said: >> >>> Actually I now find it quite interesting to see how >>>various people react to such anti-intuitive notions as >>>transfinite arithmetic or even inaccessible cardinals. >> >> To say anti-intuitive is to miss the point. The point >> is that such ideas have no connection to reality. All of >> the mathematics which stands a chance of being applied to >> the task of understanding the world we live in, must make >> predictions about the results of computational experiments. >> Thus, real mathematics could be defined as the science of >> phenomena observable in the world of computation. >> >> Just curious... do you folks accept the existence of irrational numbers >> like >> the square root of 2? Or the idea a continuous function? How about the >> Intermediate Value Theorem? >> Dan > I have no problem with any of those concepts. Does cardinality's > countability > depend on any of them, or vice versa? Just a hunch, but your problem with Cantor's theory -- and I not quite sure what it is exactly -- may make such concepts exceedingly difficult to formalize. Has anyone, in fact, been able to do so in your system? Dan === Subject: Re: a question for the anti-Cantorians Dan Christensen said: > Dan Christensen said: >> >> >>> Actually I now find it quite interesting to see how >>>various people react to such anti-intuitive notions as >>>transfinite arithmetic or even inaccessible cardinals. >> >> To say anti-intuitive is to miss the point. The point >> is that such ideas have no connection to reality. All of >> the mathematics which stands a chance of being applied to >> the task of understanding the world we live in, must make >> predictions about the results of computational experiments. >> Thus, real mathematics could be defined as the science of >> phenomena observable in the world of computation. >> >> >> Just curious... do you folks accept the existence of irrational numbers >> like >> the square root of 2? Or the idea a continuous function? How about the >> Intermediate Value Theorem? >> >> Dan >> > I have no problem with any of those concepts. Does cardinality's > countability > depend on any of them, or vice versa? > Just a hunch, but your problem with Cantor's theory -- and I not quite sure > what it is exactly -- may make such concepts exceedingly difficult to > formalize. Has anyone, in fact, been able to do so in your system? > Dan If you read my first post, I specifically stated what my issue is. Cardinality based on counting fails for infinite sets, because, as the Cantorians point out ad nauseum, one cannot count to infinity. Differences between sets such as the naturals and rationals, the naturals and evens, the naturals and their squares or square roots, are entirely ignored when those differences seem to exist intuitively. It violates the rule that a proper subset is smaller than its superset, a rule which should be preserved, in my opinion. Nothing I have done has broken any of the concepts you mentioned, and in fact, it relies on them. -- Smiles, Tony === Subject: Re: a question for the anti-Cantorians >> Actually I now find it quite interesting to see how >>various people react to such anti-intuitive notions as >>transfinite arithmetic or even inaccessible cardinals. > To say anti-intuitive is to miss the point. The point > is that such ideas have no connection to reality. All of > the mathematics which stands a chance of being applied to > the task of understanding the world we live in, must make > predictions about the results of computational experiments. > Thus, real mathematics could be defined as the science of > phenomena observable in the world of computation. Just curious... do you folks accept the existence irrational numbers like the square root of 2? Or the idea a continuous function? How about the Intermediate Value Theorem? Dan === Subject: Re: a question for the anti-Cantorians > Actually I now find it quite interesting to see how >various people react to such anti-intuitive notions as >transfinite arithmetic or even inaccessible cardinals. > To say anti-intuitive is to miss the point. The point > is that such ideas have no connection to reality. All of > the mathematics which stands a chance of being applied to > the task of understanding the world we live in, must make > predictions about the results of computational experiments. > Thus, real mathematics could be defined as the science of > phenomena observable in the world of computation. Up until the invention of electronic computers, number theory had so little connection with reality that G.H. Hardy (of A Mathematician's Apology) could confidently, but wrongly, predict that his work would never be debased by being put to any practical use. When he died in 1947, no one though him wrong, but within half a century, much of his work became of incredible practical value, and the modern financial world would be crippled without it. Mathematics abounds with creations for which there was originally no practical or theoretical application in the physical world, but which have since become essentials. For anyone to predict that Cantor's mathematics can never be of any practical use defies the lessons of history. === Subject: Re: a question for the anti-Cantorians Virgil said: > Actually I now find it quite interesting to see how >various people react to such anti-intuitive notions as >transfinite arithmetic or even inaccessible cardinals. > > To say anti-intuitive is to miss the point. The point > is that such ideas have no connection to reality. All of > the mathematics which stands a chance of being applied to > the task of understanding the world we live in, must make > predictions about the results of computational experiments. > Thus, real mathematics could be defined as the science of > phenomena observable in the world of computation. > Up until the invention of electronic computers, number theory had so > little connection with reality that G.H. Hardy (of A Mathematician's > Apology) could confidently, but wrongly, predict that his work would > never be debased by being put to any practical use. When he died in > 1947, no one though him wrong, but within half a century, much of his > work became of incredible practical value, and the modern financial > world would be crippled without it. > Mathematics abounds with creations for which there was originally no > practical or theoretical application in the physical world, but which > have since become essentials. > For anyone to predict that Cantor's mathematics can never be of any > practical use defies the lessons of history. Not if they can identify distinct ways in which its conclusions differ from reality, and the reasons for those discrepancies. Then, it is safe to say they will always have trouble being applied to reality. -- Smiles, Tony === Subject: Re: a question for the anti-Cantorians > Not if they can identify distinct ways in which its conclusions differ from > reality, and the reasons for those discrepancies. Then, it is safe to say they > will always have trouble being applied to reality. What makes you think that mathematics is necessarily about (physical) reality? Bob Kolker === Subject: Re: a question for the anti-Cantorians Robert Kolker said: > Not if they can identify distinct ways in which its conclusions differ from > reality, and the reasons for those discrepancies. Then, it is safe to say they > will always have trouble being applied to reality. > What makes you think that mathematics is necessarily about (physical) > reality? The comment was made that Cantor's cardinality may yet be applied to reality, by someone else. > Bob Kolker Bob, as one who always talks about the triumphs of science, you shouldn't be arguing that science isn't the fruit of mathematics. Mathematics grew out of a need to describe the universe. There is no math that doesn't have its roots in that, though it may extend far into the abstract realms and dwell there forever without being de-abstracted, applied, to reality. As long as math resides in that ether and is not tested against reality in any way, the new untested ways it applies rules to the concepts it is built from remain postulated, and not confirmed. That is not to say it is wrong, but that there is the chance that the intuitions that led to it were off. -- Smiles, Tony === Subject: Re: a question for the anti-Cantorians > Bob, as one who always talks about the triumphs of science, you shouldn't be > arguing that science isn't the fruit of mathematics. It often is. Also the reverse is true. But as a generally statement, one cannot assert mathematics is about physical reality. > Mathematics grew out of a > need to describe the universe. There is no math that doesn't have its roots in > that, though it may extend far into the abstract realms and dwell there forever > without being de-abstracted, applied, to reality. That is a stretch. If a mathematical system is grounded in abstractions from the git-go it is not physically based. However even a system built on air (in a manner of speaking) may yet find an application in reality. > As long as math resides in > that ether and is not tested against reality in any way, the new untested ways > it applies rules to the concepts it is built from remain postulated, and not > confirmed. That is not to say it is wrong, but that there is the chance that > the intuitions that led to it were off. Always a possibility. Bob Kolker === Subject: Re: a question for the anti-Cantorians Robert Kolker said: > > Bob, as one who always talks about the triumphs of science, you shouldn't be > arguing that science isn't the fruit of mathematics. > It often is. Also the reverse is true. But as a generally statement, one > cannot assert mathematics is about physical reality. > Mathematics grew out of a > need to describe the universe. There is no math that doesn't have its roots in > that, though it may extend far into the abstract realms and dwell there forever > without being de-abstracted, applied, to reality. > That is a stretch. If a mathematical system is grounded in abstractions > from the git-go it is not physically based. However even a system built > on air (in a manner of speaking) may yet find an application in reality. Well, it seems to me that when you abstract something, that is the process of making some kind of symbolic representation of something real or perceived to be real. it has to be an abstraction of some thing. > As long as math resides in > that ether and is not tested against reality in any way, the new untested ways > it applies rules to the concepts it is built from remain postulated, and not > confirmed. That is not to say it is wrong, but that there is the chance that > the intuitions that led to it were off. > Always a possibility. And how do we tell, if not by trying to apply the concepts we've derived to, at least, the computational environment in which they reside? > Bob Kolker -- Smiles, Tony === Subject: Re: a question for the anti-Cantorians It's Post-Cantorian. In set theory, people generally use a set of axioms, axiomatic set theory with axioms. One group of those is called Zermelo-Fraenkel, or ZF, set theory, the list of what are called non-logical or proper axioms that clarify what a set is, or can be. One of those axioms is called regularity or sometimes foundation. Now, one thing that was noticed by Cantor, which he called the Domain Principle, is that quantification over sets implies a universal set. The universal set would contain all sets as member elements, including itself. The axiom of regularity precludes from being sets those things that contain themselves, or that their transitive closure contains itself as an inductive lemma, where any set containing as an element null or the empty set vacuously fullfils regularity, it's a well-founded set. Anyways, the axiom of regularity was added by Zermelo and Fraenkel after B. Russell, basically, for a variety of reasons including that a universal set would be its own powerset and thus as it's not greater than itself, or that that's non-contradictory, which is a separate, nullable option, that that's a counterexample to that any set is of lesser cardinality than its own powerset, because trivially the universal, or any set, maps 1:1 to itself. One implication of accepting the Domain Principle, which I understand is an extension of some theories of Immanuel Kant, a Western technical philosopher, which I just heard of but already said, is that ZF is inconsistent because of regularity. There are a variety of modern theories that are anti-foundational, where that means without the axiom of regularity/foundation, with perhaps some other construction, or anti-foundation axiom. My theory, the null axiom theory, a set theory, basically has no proper/non-logical axioms. Instead, truth is determined from among that which is not not true. Ross F. === Subject: Re: a question for the anti-Cantorians Ross A. Finlayson said: > It's Post-Cantorian. > In set theory, people generally use a set of axioms, axiomatic set > theory with axioms. One group of those is called Zermelo-Fraenkel, or > ZF, set theory, the list of what are called non-logical or proper > axioms that clarify what a set is, or can be. > One of those axioms is called regularity or sometimes foundation. Now, > one thing that was noticed by Cantor, which he called the Domain > Principle, is that quantification over sets implies a universal set. > The universal set would contain all sets as member elements, including > itself. The axiom of regularity precludes from being sets those things > that contain themselves, or that their transitive closure contains > itself as an inductive lemma, where any set containing as an element > null or the empty set vacuously fullfils regularity, it's a > well-founded set. Anyways, the axiom of regularity was added by > Zermelo and Fraenkel after B. Russell, basically, for a variety of > reasons including that a universal set would be its own powerset and > thus as it's not greater than itself, or that that's non-contradictory, > which is a separate, nullable option, that that's a counterexample to > that any set is of lesser cardinality than its own powerset, because > trivially the universal, or any set, maps 1:1 to itself. > One implication of accepting the Domain Principle, which I understand > is an extension of some theories of Immanuel Kant, a Western technical > philosopher, which I just heard of but already said, is that ZF is > inconsistent because of regularity. > There are a variety of modern theories that are anti-foundational, > where that means without the axiom of regularity/foundation, with > perhaps some other construction, or anti-foundation axiom. > My theory, the null axiom theory, a set theory, basically has no > proper/non-logical axioms. Instead, truth is determined from among > that which is not not true. > Ross F. Hi Ross - Do you have a link or document that fuly describes your non-axiomatic approach? -- Smiles, Tony === Subject: Re: a question for the anti-Cantorians Hi Tony, It's Post-Goedelian. It's not explained fully, but it is, comprehensively, over several years, particularly the last couple years, shared via sci.math, sci.logic, and to some extent comp.theory, comp.ai.philosophy, sci.philosophy.meta, and sci.physics. I started reading sci.math, as a mathematical interest forum, in around 1998 or so. At the time I was somewhat more naive, I thought Archimedes Plutonium was entertaining. I had never heard of transfinite cardinality, through fifteen years of mathematical education up to advanced calculus, and, after reading about it, and thinking about it, I said Infinite sets are equivalent, in one of these extended discussions about the equivalency of infinite sets, Equivalency of Infinite Sets, and now I recognize pretty much every regular poster to sci.math in the last several years. Except for 2002, basically, where I stopped posting to usenet, I have the same arguments and they hold and I've been developing them for those years. They're not necessarily neatly packaged, instead, pieces of the puzzle are scattered far and wide. I ask myself sometimes, Why do I continue to post to usenet? I guess the answer is that I read usenet. You'll notice that Google, a premier search engine, has four or more primary search interfaces, including: Web, Images, Groups (that's usenet, USEnet, USENET or USenet), and the News. So, if you research my writings to usenet, you can find hundreds and thousands of posts, short essays, and discussion threads about, say, infinite sets, and they have been reviewed by other readers of sci.math, in the unmoderated vicious fishbowl. Tony, one thing about arguing against the status quo, in terms of forensic debate, in the policy debate is that in general you show why the status quo is wrong, and then explain how it can be made right. Osher Doctorow asked a similar question a few weeks or months ago, and I hope to at some point submit a paper detailing this abstract to a refereed journal. An Axiom-Free Set Theory proof strength and fundamental considerations of the primary ur-element. The ur-element is described as indeterminate yet expressible as 0 or U. The theory is shown to be representible in first-order logic. Several theorems of an axiom-free theory are presented. Relations of an axiom-free theory to number systems with numbers as primary objects are considered. The sets of the theory are described as ordinals. Zermelo-Fraenkel set theory is compared and contrasted. Issues of Goedelian completeness in an axiom-free theory are considered. A set theory generally is comprised of a set or collection of logical and non-logical axioms. The logical axioms are representative of tautology and the binary truth table. Tautology is an expression of sameness or identity. The non-logical axioms assert restrictions on set membership and construction. An ur-element or primary object of the set theory is asserted. By assert, the existence of a primary object is claimed to exist by fiat, and in the sequel shown to possess certain characteristics and that its existence is a consequence of tautology. In the meantime, I have explained most of those things already on sci.logic and sci.math. Ross F. === Subject: =?iso-8859-1?q?Show_=B3v(7+5v2)_in_the_form_x_+_yv2?= Can anyone explain to me the process of showing ñv(7+5v2) in the form x + yv2 I know the answer but don't understand the process of reaching it. === Subject: Re: Show ñv(7+5v2) in the form x + yv2 > Can anyone explain to me the process of showing > ñv(7+5v2) > in the form > x + yv2 > I know the answer but don't understand the process of reaching it. If I use 'curt' and 'sqrt' respecively for real cube root of a real and non-negative square root on a non-negative real respecively , you seem to be asking how to solve x + y^sqrt(2) = curt(7 + 5*sqrt(2)) To start with, cubing both sides gives x^3 + 3*x^2*y*sqrt(2) + 6*x*y^2 + 2*y^3*sqrt(2) = 7 + 5*sqrt(2) Assuming integral, or even rational rational, values for x and y allows us to separate this into two equations, rational part versus irrational part: x^3 + 6*x*y^2 = 7 and 3*x^2*y + 2*y^3 = 5, respectively. Supposing small integral values for x and y and noting coefficients 1 + 6 = 7 in the first equation and 2 + 3 = 5 in the second suggests the lucky guess x = y = 1. Alternately, one can eliminate, say. y by solving for y in the first equation and substituting one of the solutions for y in the second equation. The resulting mess can eventually, term shifitngs and squarings and such, be made into a polynomial equation in x with integer coefficients. The potential rational roots of such an equation can be tested one at a time to find x = 1, and so on. === Subject: =?iso-8859-1?q?Re:_Show_=B3v(7+5v2)_in_the_form_x_+_yv2?= Sorry, the v's above should be roots. === Subject: Re: Show ñv(7+5v2) in the form x + yv2 mrcrowl@gmail.com escribi.97: > Sorry, the v's above should be roots. You meant rc(7 + 5*rq(2)) = x + y*rq(2) where I use 'rc' for cubic root and 'rq' for square root. I suppose that x and y must be rationals. Then, let raise it to cube, 7 + 5*rq(2) = x^3 + 3x^2*y*rq(2) + 6x*y^2 + 2y^3*rq(2) = (x^3 + 6x*y^2) + (3x^2*y + 2y^3)rq(2) Then, as x and y are rationals, must be 7 = x^3 + 6x*y^2 5 = 3x^2*y + 2y^3 By simple inspection, it's obvius that x = y = 1 is a solution. The injectivity of f(x) = x^3, in the reals, says us that this solution is unique. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: =?ISO-8859-1?Q?Re:_Show_=B3v?=(7+5v2) in the form x + yv2 ETAtAhUAhq99B21Z/B49+CfM9p7tt4wO93sCFA+LRaHfGejnOO5uNyYnumAGF93C It's a sort of alternate casus irreducibilis, in the sense that you an't get the required form form purely analytic techniques. You're forced to resort to an experimental trial. But you can at least make it an *educated* guess. Let us assume that a and b are rational numbers such that (a+bu)^3 = 7+5u; u = sqrt(2), u^2 = 2 Then we also have the conjugate relation (a-bu)^3 = 7-5u and upon multiplication: (a^2-2b^2)^3 = 7^2-5^2u^2 = -1 The RHS is a perfect cube, so the cube root can be taken: a^2-2b^2 = -1, b^2 = (a^2+1)/2 *** Now go back to your original equaiton and expand the LHS: a^3 + 3ua^2b + 6ab^2 + 2ua^3 = 7+5u where u^2 = 2 has been applied. Equation of the rational and irational parts gives a^3 + 6ab^2 = 7 3a^2b + 2b^3 = 5 and upon substituting foir b^2 in the rational part, see the *** equation above: a^3 + 3a^3 + 3a = 7, 4a^3 + 3a - 7 = 0 This is where the casus becomes irreducibilis. True, the cubic quaiton bove has nly one real root, and upon applying the Cardano formula that root is obtained in terms of real algebraic functions only. BUT -- the Cardano formula gives no indication that the root is rational. The only way to PROVE that is to try out the rational roots that apre possible from the standard rational root theorem. That is how the root a = 1 is obtained. Now we can easily obtain ^2 = (1^2+1)/2 = 1, b = (+/-) 1. All that remains is to choose the proper sign of b. Put a = 1 into the irrational part equation noted above: 3a^2b + 2b^3 = 5, 3b + 2b^3 = 5 b = +1 gives the proper sign, so (7+5u)^(1/3) = 1+u. b = -1 gives the conjugate result (7-5u)^(1/3) = 1-u. --OL === Subject: Re: $$REWARD If You prove my argument wrong(FLT) > Here is presented the ABSTRACT of Archymedes Proof Of FLT. > If you think you found something wrong, No, no, no! There is NOTHING WRONG with your abstract! I find [64.4.51.220] thingies thrown all over the place and the phrase: e-mails containing various versions of his 'proof' and that most of the lines of his 'proof' are redundant, it is noteworthy that, strictly speaking, there is nothing wrong with Mr Ghiata's more recent arguments, until their last line. Mind you, that may be because there is only the one line and that line is just the bald assertion that Fermat's last theorem is true. That is so, by virtue of Wiles' proof, but of course does not follow ipso facto from the author's statement. It summarizes your unpublished proof very well. In fact, it inspired me to publish my own Abstract of my proof that 2+2 equals to 5 on wednesdays and thursdays and to 6 on fridays. Here it goes: x+a+r is a number. therefore, 15*x = e+m+v where w=1 or n if t=y therefore 2+2 = 5 on thursdays and wednesdays >and you have a question > and my answer or argument is wrong then you will be rewarded with > $50 to buy yourself math books > george ghiata > Sent : Saturday, May 14, 2005 6:57 PM > To : alf@math.mq.edu.au, andrew@dms.umontreal.ca, george ghiata@hotmail.com, leeming@math.uvic.ca > Subject : FW: RE: Archimedes's Proof of Fermat Last theorem > | | | Inbox > X-Originating-IP: [64.4.51.220] > X-Originating-Email: [george ghiata@hotmail.com] > Received: from 64.4.51.220 by by107fd.bay107.hotmail.msn.com with > View E-mail Message Source andrew@dms.umontreal.ca, > leeming@math.uvic.ca === >Subject: FW: RE: Archimedes's Proof of Fermat Last theorem >X-Originating-IP: [64.4.51.220] >X-Originating-Email: [george ghiata@hotmail.com] >Received: from 64.4.51.220 by by107fd.bay107.hotmail.msn.com with andrew@dms.umontreal.ca, > leeming@math.uvic.ca === >Subject: RE: Archimedes's Proof of Fermat Last theorem >X-Originating-IP: [64.4.51.220] >X-Originating-Email: [george ghiata@hotmail.com] >Received: from 64.4.51.220 by by107fd.bay107.hotmail.msn.com with HTTP;Fri, > >Here is Archimedes Proof of Fermat Last theorem. >george Ghiata > andrew@dms.umontreal.ca === >Subject: RE: Archimedes's Proof of FLT for n=4*L+1 >X-Originating-IP: [64.4.51.205] >X-Originating-Email: [george ghiata@hotmail.com] >Received: from 64.4.51.205 by by107fd.bay107.hotmail.msn.com with HTTP;Wed, > > >To Alf and Andrew, >A simplfied Fermat last theorm proof:Archimedes FLT proof when n=4*L+1 >X^n+Y^n=Z^n is Imposible if Z,X,YAre Integers and n=prime>2 >We take Z=even number >X+Y-Z=B >X=B+Q >Y=B+P >2*B+Q+P=s*u^n wher s=1 or n^(n-1) if Z is divisible by n >B+Q+P=r*z*u wher r is 1 or n if Z is divisible by n where k >is =odd number. > Therefore too we get that z= - b+v*u^(n-1).Therfore b=odd numbre and > v=n^(n-2) > (B+Q+)^n+(B+P)^n-(B+Q+P)^n=0 >can be written as: > EQ A: B^n+M + G*H*(u^n)=0 where H is 2^t which comes from > 2*(n-1)=2^t*d where d is =odd and G=odd number and >M is comming from following: >Now:EQ.C : (B+Q)^n+(B+P)^n - (Q+P)^n= B^n+ n*B*(Q+P)+B^n+M many times > as > (2^t)*u^n is divisible by 2 B^n-[B^(n-1)]*n*(Q+P) > is divisible by 2 as many times as > (2^t)*u^n is divisible by 2 > Therefore F=Z^n- B^n - (Q+P)^n-n*[B^(n-1)]*(Q+P) =(z*u)^n- > [b*u]^n-(2*u*k)^n >- n*(b^n)*2*(u^n) > can be written as : > F= (u^n)*{ z^n - b^n - (2*k)^n - n*2*(b^n)} >Since z=- b+v*u^(n-1) we get that F =(u^n)*{2*(b^n)*(n*k+1) -d*u^(n-1) > -2*k)^n} >But F is divisible by 2 as many times as (2^t)*u^n is divisible by 2 >Therfore We get that (k+1)=2^m where m>(t-1) and (n-1)=j*2^(t-1) wher j > is odd number >We know that b+2*k=z and 2*b+2*k=h*u*(n-1)where h=1 or n^(n-2) when Z is > divisible by n >Therfore z-k=b+k. >Therfore z+1 is divisible by2 as many times as k+1 is divisible by 2 >But when the EQuation (B+Q+P)^n=(B+Q)^n+(B+P)^n=(z*u)^n is divided by > (2*B+Q+P) >we get : > [(n-1)*u*V +n*Q^(n-1)}=r*z^n >If r=1 then we get: > [(n-1)*u*V +(n-1)*Q^(n-1)+Q^(n-1) +1=z^n+1 >We know that (z+1)=(2^m)*c where m>(t-1) and [2^(t-1)]*a =(n-1) >Therfore n=4*g+3 >If r=n then we get : > [(n-1)*u*V+(n-1)*Q^(n-1) > +Q^(n-1)+1=(n-1)*z^n+z^n+1 > We see that the right side of EQ. is divisible by 2^(t-1). >If (n-1) is divisible by 2 only then the left side is divisible by 4 >If (n-1) is divisible by 4 then The left side is divisible by 2 only and > the right side by 4 >Conclusion: > If n=4*L+1 then X^n +Y^n =Z^n is impossible >created by george ghiata-may 11 -05 > Second Part: n=4*L+3 then Z is > not divisible by n. > We got too that (z+1) is divisible by 4 at least and (k+1) is > divisibil by 4 at least. > Now we multiply the EQF: (B+Q)^n+(B+P)^n=(B+Q+P)^n by 2^n and > get: > > EQF1:(2*B+Q+P+Q-P)^n+(2*B+Q+P+P-Q)^n=(2*B+Q+P+Q+P)^n=(2*z*u)^n > But ( 2*B+Q+P)=u^n=W > EQG1: F(W)=(2*z*u)^n and > EG2: G(W)=(Q+P)^n=(2*u*k)^n > Now we divide EQG1 and EQG2 by (2*u)^n and get: > EQG3: F1(W)=[F(W)]/(2*u)^n=z^n and > EQG4: G1(W)/(2*u)^n=k^n > Now we write EQG5: [G1(W)/(2*u)^n]-z^n=k^n-z^n > Keeping in mind the EQG3 we divide EQG5 by (z-k)=[u^(n-1)]/2 and > get the EQG6. divisible by 2 > Therfore we got a contradiction which shows that Fermat Last theorem > is true. > Created by gheorghe Ghiata- Ann Arbor -5/14/5 > OBSERVATION: > In Fact if Zis divisible by 2*n the Second part proof still works for > every n=Prime>2 > george ghiata- 5/14/5 > >Received: from mail016.syd.optusnet.com.au ([211.29.132.167]) by > 16:23:17 -0800 >Received: from [10.0.1.3] (c211-30-90-94.belrs3.nsw.optusnet.com.au > [211.30.90.94])(authenticated)by mail016.syd.optusnet.com.au (8.11.6p2/8.11.6) > 11:23:12 +1100 >X-Message-Info: JGTYoYF78jE7SgqAtq6kf2kNvUfN8uk3 > FILETIME=[727E0890:01C3D4B4] > >The author appears to be believe in spontaneous creation; that one can > get >something for nothing. Worse, he appears to imagine that the nothing he > does >somehow has not been done many times before. > >Ghiata's technique is the familiar one of obfuscation by many variables. > In such arguments one introduces an alphabet of symbols $B$, $P$, $Q$, $z 1$, > $z 2$, $m$, $K$ --- with several incompatible meanings, $s$, $ldots,$, I > presume that the purpose of these symbols is to create an illusion that > something has been input. > >However, with the quite unnecessary intermediate argument removed, Mr > Ghiata says the following > >Suppose $X^n+Y^n=Z^n$. > >Then one obtains Case~$m$: >begin{equation}label{so what} >Z^m(X^n+Y^n)=Z^{n+m},. >tag{$F m$} >end{equation} > >Mr Ghiata now announces that eqref{so what}, for different $m$ if needs > be, yields a contradiction. > forty > e-mails containing various versions of his 'proof' and that most of the lines of > his 'proof' are redundant, it is noteworthy that, strictly speaking, there is > nothing wrong with Mr Ghiata's more recent arguments, until their last line. > Mind you, that may be because there is only the one line and that line is just > the bald assertion that Fermat's last theorem is true. That is so, by virtue of > Wiles' proof, but of course does not follow ipso facto from the author's > statement. > > > > Express yourself instantly with MSN Messenger! Download today - it's FREE! > | | | | | Inbox > Get the latest updates from MSN > MSN Home | My MSN | Hotmail | Search | Shopping | Money | People & Chat > Feedback | Help > © 2005 Microsoft TERMS OF USE Advertise TRUSTe Approved Privacy Statement Anti-Spam Policy === Subject: Minimization of an Integral with an integral constraint Hi Can someone suggest a method to solve this problem integral((x(t)+a_1(t))/(x(t)- a_2(t))*b(t) dt) subject to integral(x(t) dt) <= K where x(t), a_1(t),a_2(t),b(t), K are all positive for all t. Bhushan === Subject: truncated icosahedron angle problem Consider a truncated icosahedron (soccer ball) in its unfolded form: Consider any pentagon and then any two attached hexagons. What is the angle that is formed with respect to the pentagon-plane and the hexagon-plane, when the hexagons are rotated until their edges touch? Any suggestions? === Subject: Re: truncated icosahedron angle problem > Consider a truncated icosahedron (soccer ball) in its unfolded form: > Consider any pentagon and then any two attached hexagons. What is the > angle that is formed with respect to the pentagon-plane and the > hexagon-plane, when the hexagons are rotated until their edges touch? > Any suggestions? I think you are asking for the dihedral angles of the truncated icosahedron. If so, I don't see any point in unfolding it. The 12 vertices of a regular icosahedron can be taken as (+-3,+-3*t,0), (0,+-3,+-3*t), (+-3*t,0,+-3), where t=(1+sqrt(5))/2. The vertices of the truncated icosahedron are the trisection points find exact coordinates of two adjacent hexagons and a pentagon which touches both. Now using vectors (say) you can calculate the dihedral angles. You should get: cos(angle between touching hexagons) = -sqrt(5)/3 so dihedral angle hex to hex is about 138.18968510422140193 degrees. cos(angle between touching hex and penta) = -1/15*sqrt(75+30*sqrt(5)) so dihedral angle hex to penta is about 142.62263185935030436 degrees. -- Jim Buddenhagen http://home.earthlink.net/~jbuddenh/ === Subject: Re: truncated icosahedron angle problem >cos(angle between touching hex and penta) = -1/15*sqrt(75+30*sqrt(5)) >so dihedral angle hex to penta is about 142.62263185935030436 degrees. This is exactly what I have been looking for. Can you point me to any references for calculating these dihedral angles? MEP === Subject: Re: truncated icosahedron angle problem the hexagona *do* touch. --The Kyoto Protocol and the Permian Basin Gang! http://tarpley.net/bush7htm http://larouchepub.com http://members.tripod.com/~ame[CapitalEth]rican almanac http://www.21stcenturysciencetech.com/ === Subject: Re: Gelfand's lemma > X is a Banach space. p is map from X to R (reals), which satisfies > (1) p(x) >= 0 for all x in X > (2) p(ax)=ap(x) for all a>0 and x in X > (3) p(x+y)<=p(x)+p(y) for all x,y in X > (4) If x_n -> x then lim inf p(x_n) >= p(x). > Show that there exists M, such that p(x) <= M||x|| for all x in X. Set N(x):=norm(x)+p(x) for x e X. Then N is a norm on X such that (X,N) is complete. By Banach's theorem, the norms 'norm' and N are equivalent which implies the stated inequality. J. === Subject: Re: Gelfand's lemma > X is a Banach space. p is map from X to R (reals), > which satisfies > (1) p(x) >= 0 for all x in X > (2) p(ax)=ap(x) for all a>0 and x in X > (3) p(x+y)<=p(x)+p(y) for all x,y in X > (4) If x_n -> x then lim inf p(x_n) >= p(x). > > Show that there exists M, such that p(x) <= M||x|| > for all x in X. > > Set N(x):=norm(x)+p(x) for x e X. Then N is a norm on > X such that (X,N) > is complete. By Banach's theorem, the norms 'norm' > and N are equivalent > which implies the stated inequality. > J. N is not necesarily a norm (it is not absolutely homogeneous); in the real case one may use p(x)+p(-x). The usual proof uses the fact that the lower semicontinuity (4) implies that p is sigma-subadditive and then apply a Baire category argument. V. Anisiu === Subject: Re: Gelfand's lemma >>>X is a Banach space. p is map from X to R (reals), >>which satisfies >>>(1) p(x) >= 0 for all x in X >>>(2) p(ax)=ap(x) for all a>0 and x in X >>>(3) p(x+y)<=p(x)+p(y) for all x,y in X >>>(4) If x_n -> x then lim inf p(x_n) >= p(x). >>>Show that there exists M, such that p(x) <= M||x|| >>for all x in X. >>Set N(x):=norm(x)+p(x) for x e X. Then N is a norm on >>X such that (X,N) >>is complete. By Banach's theorem, the norms 'norm' >>and N are equivalent >>which implies the stated inequality. >>J. > N is not necesarily a norm (it is not absolutely homogeneous); in the real case one may use p(x)+p(-x). > The usual proof uses the fact that the lower semicontinuity (4) implies that p is sigma-subadditive and then apply a Baire category argument. > V. Anisiu Valeriu, indeed I implicitly assumed w.l.o.g. that p is absolutely homogeneous. J. === Subject: Re: Gelfand's lemma <14966708.1116547275737.JavaMail.jakarta@nitrogen.mathforum.org> how do you prove that p(x) = 0 implies x = 0? > Set N(x):=norm(x)+p(x) for x e X. Then N is a norm on > X such that (X,N) > is complete. By Banach's theorem, the norms 'norm' > and N are equivalent > which implies the stated inequality. > J. > N is not necesarily a norm (it is not absolutely homogeneous); in the real case one may use p(x)+p(-x). > The usual proof uses the fact that the lower semicontinuity (4) implies that p is sigma-subadditive and then apply a Baire category argument. > V. Anisiu === Subject: Re: Gelfand's lemma > how do you prove that p(x) = 0 implies x = 0? >>>Set N(x):=norm(x)+p(x) for x e X. Then N is a norm on >>>X such that (X,N) >>>is complete. By Banach's theorem, the norms 'norm' >>>and N are equivalent >>>which implies the stated inequality. >>>J. >>N is not necesarily a norm (it is not absolutely homogeneous); in the > real case one may use p(x)+p(-x). >>The usual proof uses the fact that the lower semicontinuity (4) > implies that p is sigma-subadditive and then apply a Baire category > argument. >>V. Anisiu A posteriori your claim is true. But where do you need it in the course of the proof? Perhaps the property p(x) >= 0 is sufficient in your step? J. === Subject: Re: Gelfand's lemma <14966708.1116547275737.JavaMail.jakarta@nitrogen.mathforum.org> <428D83BF.6020604@web.de> Of course we need. In order to prove it is a norm... The condition of a norm includes: > how do you prove that p(x) = 0 implies x = 0? > A posteriori your claim is true. But where do you need it in the course > of the proof? Perhaps the property p(x) >= 0 is sufficient in your step? > J. === Subject: Re: Gelfand's lemma > Of course we need. > In order to prove it is a norm... > The condition of a norm includes: >>>how do you prove that p(x) = 0 implies x = 0? >>A posteriori your claim is true. But where do you need it in the > course >>of the proof? Perhaps the property p(x) >= 0 is sufficient in your > step? >>J. What do you want to show: that p is a norm or that N is a norm? === Subject: Re: Gelfand's lemma <14966708.1116547275737.JavaMail.jakarta@nitrogen.mathforum.org> <428D83BF.6020604@web.de> <428D89CF.7040804@web.de> Oh...I made a mistake. Now I want to prove that (X, N(x)) is complete. If N(x_m - x_n) < epsilon then ||x_m - x_n || < epsilon, hence x_n -> x in X. Then N(x_n - x) = ||x_n - x|| + p(x_n - x) + p(x - x_n)... How to prove that p(x_n - x) -> 0? > Of course we need. > In order to prove it is a norm... > The condition of a norm includes: > > What do you want to show: that p is a norm or that N is a norm? === Subject: Re: Gelfand's lemma > Oh...I made a mistake. > Now I want to prove that (X, N(x)) is complete. > If N(x_m - x_n) < epsilon then ||x_m - x_n || < epsilon, hence x_n -> x > in X. > Then N(x_n - x) = ||x_n - x|| + p(x_n - x) + p(x - x_n)... > How to prove that p(x_n - x) -> 0? >>>Of course we need. >>>In order to prove it is a norm... >>>The condition of a norm includes: >>What do you want to show: that p is a norm or that N is a norm? Have a look at your last condition of p ... === Subject: Re: Gelfand's lemma <14966708.1116547275737.JavaMail.jakarta@nitrogen.mathforum.org> <428D83BF.6020604@web.de> <428D89CF.7040804@web.de> <428D9618.6020207@web.de> Well, I don't get it. lim inf p(x_n - x) = 0 but it does not mean that p(x_n - x) -> 0 > How to prove that p(x_n - x) -> 0? > Have a look at your last condition of p ... === Subject: Re: Gelfand's lemma >Well, I don't get it. >lim inf p(x_n - x) = 0 >but it does not mean that >p(x_n - x) -> 0 >> How to prove that p(x_n - x) -> 0? It took me a minute to see this, but it's true, and not that hard. Hint: It uses the last condition on p, and it _also_ uses the triangle inequality (as well as the fact that lim inf p(x_n - x) = 0.) >> Have a look at your last condition of p ... ************************ David C. Ullrich === Subject: Re: Gelfand's lemma Cc: Li Yi >>Of course we need. >>In order to prove it is a norm... >>The condition of a norm includes: >>>>how do you prove that p(x) = 0 implies x = 0? >>>A posteriori your claim is true. But where do you need it in the >>course >>>of the proof? Perhaps the property p(x) >= 0 is sufficient in your >>step? >>>J. > What do you want to show: that p is a norm or that N is a norm? More precisely: that p(x)=0 implies x = 0 or N(x)=0 implies x=0? === Subject: Re: Gelfand's lemma <14966708.1116547275737.JavaMail.jakarta@nitrogen.mathforum.org> If we know N(x) = ||x|| + p(x) + p(-x) is a complete norm, then how can we get the conclusion? N(x) is stronger than ||x||, they are defined on a Banach space, hence they are equivalent. There exists M such that N(x) <= M||x||...how there is p(x)+p(-x) on the left side. > Set N(x):=norm(x)+p(x) for x e X. Then N is a norm on > X such that (X,N) > is complete. By Banach's theorem, the norms 'norm' > and N are equivalent > which implies the stated inequality. > J. > N is not necesarily a norm (it is not absolutely homogeneous); in the real case one may use p(x)+p(-x). > The usual proof uses the fact that the lower semicontinuity (4) implies that p is sigma-subadditive and then apply a Baire category argument. > V. Anisiu === Subject: Re: Gelfand's lemma > If we know N(x) = ||x|| + p(x) + p(-x) is a complete norm, > then how can we get the conclusion? > N(x) is stronger than ||x||, they are defined on a Banach space, hence > they are equivalent. > There exists M such that N(x) <= M||x||...how there is p(x)+p(-x) on > the left side. >>>Set N(x):=norm(x)+p(x) for x e X. Then N is a norm on >>>X such that (X,N) >>>is complete. By Banach's theorem, the norms 'norm' >>>and N are equivalent >>>which implies the stated inequality. >>>J. >>N is not necesarily a norm (it is not absolutely homogeneous); in the > real case one may use p(x)+p(-x). >>The usual proof uses the fact that the lower semicontinuity (4) > implies that p is sigma-subadditive and then apply a Baire category > argument. >>V. Anisiu You could use p(x) <= p(x) + p(-x) by axiom (1). Then the main result follwos. J. === Subject: Re: None of you are smart enough to get this right ! On 18 May 2005 15:10:14 -0700, Klassixx >Two clock towers, (Tower A and Tower B), facing North East and North >North East respectively, stand almost one mile apart. > Both towers are taller than each other by roughly 54 feet but less wide >than one another by a similar value. These two pieces of data, as written, are useless. Do you mean to say that ONE tower is roughly 54 feet higher than the other? And does the roughly 54 feet? I'm not too good at reading minds, sorry.. === Subject: Re: None of you are smart enough to get this right ! >> What the hell is the point to this bull????? > Two things aren't taller than each other, period. Therefore, the wording > is intentionally misleading and vague. > It should read: the towers are taller than each other, *in each others' > horizontal line of sight*. On what planet do you live that has a concave surface? Or is it that the atmosphere has such a high density gradient that it only _looks_ concave? Rich === Subject: Re: None of you are smart enough to get this right ! >>> What the hell is the point to this bull????? >> Two things aren't taller than each other, period. Therefore, the wording >> is intentionally misleading and vague. >> It should read: the towers are taller than each other, *in each others' >> horizontal line of sight*. >On what planet do you live that has a concave surface? Or is it that >the atmosphere has such a high density gradient that it only _looks_ >concave? Earth has many concave surfaces, often called basins. -- There's no such thing as a free lunch, but certain accounting practices can result in a fully-depreciated one. === Subject: Re: None of you are smart enough to get this right ! ... > Bull. Arrogance explains it. > http://bbcamerica.com/genre/comedy_games/look_around_you/look_around_you_mat h _quiz.jsp Answer 4: -BRAINS Rich === Subject: Re: None of you are smart enough to get this right ! I find this thread quite funny, on two different levels. lol. Anyway, in case anyone cares, here is a little puzzle I just thought up myself: A 1*1*1 meter box is pushed against a wall. A 5 meter long ladder is then leaned against the wall, such that its weight is taken by the floor, the wall, and an edge of the box. What are the two possible distances from where the edge of the ladder touches the floor, to where the floor and the wall meet? PS This one actually makes sense. === Subject: Re: None of you are smart enough to get this right ! On 19 May 2005 11:09:18 -0700, Klassixx >I find this thread quite funny, on two different levels. lol. >Anyway, in case anyone cares, here is a little puzzle I just thought up >myself: >A 1*1*1 meter box is pushed against a wall. >A 5 meter long ladder is then leaned against the wall, such that its >weight is taken by the floor, the wall, and an edge of the box. >What are the two possible distances from where the edge of the ladder >touches the floor, to where the floor and the wall meet? >PS This one actually makes sense. I have a nearest possibility of 1.481 meters to a farthest of 4.776 meters (approximate). === Subject: Re: None of you are smart enough to get this right ! <4s6q815fbba6cthnflqqa8kslk9b4ote8m@4ax.com> > On 19 May 2005 11:09:18 -0700, Klassixx >I find this thread quite funny, on two different levels. lol. >Anyway, in case anyone cares, here is a little puzzle I just thought up >myself: >A 1*1*1 meter box is pushed against a wall. >A 5 meter long ladder is then leaned against the wall, such that its >weight is taken by the floor, the wall, and an edge of the box. >What are the two possible distances from where the edge of the ladder >touches the floor, to where the floor and the wall meet? >PS This one actually makes sense. > I have a nearest possibility of 1.481 meters to a farthest of 4.776 > meters (approximate). Not very good approximations. Instead, try 1.260518 and 4.838501. For more accuracy add 1 to the positive roots of the polynomial equation x^4 + 2x^3 - 23x^2 + 2x + 1 = 0 MzF === Subject: Re: None of you are smart enough to get this right ! Reckon the two possible distances are either zero or 1 metre for the ladder to touch floor. And floor and wall to meet must be zero, unless it's ty victorian rat hole'd incase it's about 4 covered by skirting board. === Subject: Re: None of you are smart enough to get this right ! > I find this thread quite funny, on two different levels. lol. > Anyway, in case anyone cares, here is a little puzzle I just thought up > myself: > A 1*1*1 meter box is pushed against a wall. > A 5 meter long ladder is then leaned against the wall, such that its > weight is taken by the floor, the wall, and an edge of the box. > What are the two possible distances from where the edge of the ladder > touches the floor, to where the floor and the wall meet? > PS This one actually makes sense. That's a very old trig workbook assignment. Rich === Subject: Re: None of you are smart enough to get this right ! > I find this thread quite funny, on two different levels. lol. > Anyway, in case anyone cares, here is a little puzzle I just thought up > myself: > A 1*1*1 meter box is pushed against a wall. > A 5 meter long ladder is then leaned against the wall, such that its > weight is taken by the floor, the wall, and an edge of the box. > What are the two possible distances from where the edge of the ladder > touches the floor, to where the floor and the wall meet? > PS This one actually makes sense. > That's a very old trig workbook assignment. I'm sure something like it probably is. There's only so many variations possible on that theme. It was only intended to be a harmless minute's worth of entertainment anyway. If you think that was too easy (or maybe not - you didn't actually give any answer), then try this: Can you figure out if 321,600,000,001 is, or is not a prime, using no computer or calculator? It's extremely do-able - there's a nice little trick envolved. === Subject: Re: None of you are smart enough to get this right ! >>>I find this thread quite funny, on two different levels. lol. >>>Anyway, in case anyone cares, here is a little puzzle I just > thought up >>>myself: >>>A 1*1*1 meter box is pushed against a wall. >>>A 5 meter long ladder is then leaned against the wall, such that > its >>>weight is taken by the floor, the wall, and an edge of the box. >>>What are the two possible distances from where the edge of the > ladder >>>touches the floor, to where the floor and the wall meet? >>>PS This one actually makes sense. >>That's a very old trig workbook assignment. > I'm sure something like it probably is. There's only so many variations > possible on that theme. It was only intended to be a harmless minute's > worth of entertainment anyway. > If you think that was too easy (or maybe not - you didn't actually give > any answer), then try this: > Can you figure out if 321,600,000,001 is, or is not a prime, using no > computer or calculator? > It's extremely do-able - there's a nice little trick envolved. The number is clearly divisible by 7, since 321-600+000-001 = -280 is divisible by 7. -- Mark Thornquist === Subject: Re: None of you are smart enough to get this right ! > Can you figure out if 321,600,000,001 is, or is not a prime, using no > computer or calculator? > It's extremely do-able - there's a nice little trick envolved. > The number is clearly divisible by 7, since 321-600+000-001 = > -280 is divisible by 7. Yes, but can you do: 2,438,100,000,001 It's a little harder and there's a different trick envolved. === Subject: Re: None of you are smart enough to get this right ! >> >> Can you figure out if 321,600,000,001 is, or is not a prime, using >> computer or calculator? >> >> It's extremely do-able - there's a nice little trick envolved. >> The number is clearly divisible by 7, since 321-600+000-001 = >> -280 is divisible by 7. >Yes, but can you do: 2,438,100,000,001 >It's a little harder and there's a different trick envolved. spoiler . . . . . . . . . . . . . . . . . . . . 1 + x^4 + x^5 = (1 + x + x^2)(1 - x + x^3) x=200, 300 give factorisations for your 2 numbers (though it's the same trick each time) -- J.E.H.Shaw [Ewart Shaw] strgh@uk.ac.warwick TEL: +44 2476 523069 Department of Statistics, University of Warwick, Coventry CV4 7AL, UK http://www.warwick.ac.uk/statsdept http://www.ewartshaw.co.uk 3 ((4&({*.(=+/))++/=3:)@([:,/0&,^:(i.3)@|:2^:2))&.>@]^:(i.@[) <#:3 6 2 === Subject: Re: None of you are smart enough to get this right ! I imagine you. Classixx, as me on one of my super high days... Foaming at the mouth, laughing at everything because everythings funny right now and harbouring a massive erection that won't be stroked unless the ego is too. You and me have a lot of common ground. === Subject: Re: None of you are smart enough to get this right ! > I find this thread quite funny, on two different levels. lol. > Anyway, in case anyone cares, here is a little puzzle I just thought up > myself: > A 1*1*1 meter box is pushed against a wall. > A 5 meter long ladder is then leaned against the wall, such that its > weight is taken by the floor, the wall, and an edge of the box. > What are the two possible distances from where the edge of the ladder > touches the floor, to where the floor and the wall meet? > PS This one actually makes sense. http://www.drgrammar.org/faqs/#116 None is or none are? According to Merriam Webster's Dictionary of English Usage, Clearly, none has been both singular and plural since Old English and still is. The notion that it is singular only is a myth of unknown origin that appears to have arisen in the 19th century. If in context it seems like a singular to you, use a singular verb; if it seems like a plural, use a plural verb. Both are acceptable beyond serious criticism (664). Proceed... PLaToPeS === Subject: Re: None of you are smart enough to get this right ! >> I find this thread quite funny, on two different levels. lol. >> Anyway, in case anyone cares, here is a little puzzle I just thought > up >> myself: >> A 1*1*1 meter box is pushed against a wall. >> A 5 meter long ladder is then leaned against the wall, such that its >> weight is taken by the floor, the wall, and an edge of the box. >> What are the two possible distances from where the edge of the ladder >> touches the floor, to where the floor and the wall meet? >> PS This one actually makes sense. > http://www.drgrammar.org/faqs/#116 > None is or none are? > According to Merriam Webster's Dictionary of English Usage, Clearly, none > has been both singular and plural since Old English and still is. The > notion that it is singular only is a myth of unknown origin that appears > to have arisen in the 19th century. If in context it seems like a singular > to you, use a singular verb; if it seems like a plural, use a plural verb. > Both are acceptable beyond serious criticism (664). > Proceed... You have a set of n items. Three of them are blue, two of them are white, one of them is black, but none of them ___ green? A) is B) are Rich === Subject: Re: None of you are smart enough to get this right ! > http://www.drgrammar.org/faqs/#116 > None is or none are? > According to Merriam Webster's Dictionary of English Usage, Clearly, none > has been both singular and plural since Old English and still is. The > notion that it is singular only is a myth of unknown origin that appears > to have arisen in the 19th century. If in context it seems like a singular > to you, use a singular verb; if it seems like a plural, use a plural verb. > Both are acceptable beyond serious criticism (664). > Proceed... > You have a set of n items. Three of them are blue, two of them are white, > one of them is black, but none of them ___ green? > A) is > B) are > Rich I like is, as in not one of them is, but, not any of them are is apparently correct, too (damnit!) Wait a minute! How about *this* - You have a set of n items. Three of them are blue, two of them are white, one of them is black, but none of them be green? How be that? ~:?) PLaToPeS [plate of peas] === Subject: Re: Trisecting and angle after infinite steps > I praise you for struggling valiantly. I will give you a hint. Use > formal power series. > Bob Kolker what formal means), so I'm sure you're right. This problem was at the end of a section on power series, where the sum of an infinite series of the form ar^n equals a/(1-r) if |r|<1. I was apparently supposed to use that to solve it, but I couldn't figure out how to express anything I saw in the form ar^n, though. The denominators were 2^n, but the numerators contained additions, and I couldn't find a way to arrange them in the form of repeated multiplication as required by r^n. === Subject: Re: Trisecting and angle after infinite steps >> I praise you for struggling valiantly. I will give you a hint. Use >> formal power series. >> Bob Kolker >what formal means) It means you consider them only as strings of coefficients. You're not thinking of them as representing functions in any way. I mean, 1 + x + x^2 + ... is a power series; you can think of it as being a representation of the function f(x) = 1/(1-x) since the series and the function have the same values for every x between -1 and 1. But this series doesn't converge for other x's, and some series are even worse, e.g. 1 + x + 2 x^2 + 6 x^3 + ... + n! x^n + ... will only converge if x = 0 ! Nonetheless, it's a useful way of combining all the factorials into a single thing. dave === Subject: Re: Trisecting and angle after infinite steps <3f2r7jF5n0hoU1@individual.net> > I assume that the way to prove that A(n) approaches (2^n)/3 as n > approaches infinity is to write a closed form expression of A(n) and > take its limit as n->inf, but I can't see how to write the closed form. > If you are trying to prove that A(n) approaches (2^n)/3, it might be > worthwhile trying to prove that 3*A(n) approaches 2^n. Take a look at the > sequence of numbers produced by 3*A(n). > -- > Clive Tooth > http://www.clivetooth.dk Ah, yes. Just one tiny step farther and I would have seen it. 3*A(n) = 2^n + (-1)^n so if I multiply both the numerator and denominator by 3, X(n) = [2^n + (-1)^n] / [3*2^n] which clearly approaches 1/3 as n approaches infinity. === Subject: Re: Trisecting and angle after infinite steps <428c3700$0$236$edfadb0f@dread12.news.tele.dk> > Well, this is a standard problem. You start by substituting X(n) = a^n into > the above equation. That gives you (after rearranging): > 2 a^2 - a - 1 = 0. > This quadratic equation has the two solutions a1 and a2, where a1=1 and > a2=-1/2. > Then the complete solution to your problem above is: > X(n) = c1 * a1^n + c2 * a2^n, > where c1 and c2 are constants independent of n. They are determined from the > first two terms. > You'll find c1 = 1/3 and c2 = -1/3. So the solution may be written as > X(n) = 1/3 * (1 - (-1/2)^n). > -Michael. before. It was not covered in this book, so apparently I was meant to use another, simpler approach. But, since my purpose for doing this is to improve my skills, I'd like to learn more about this more general approach. Where would I find this described? In a discrete math text? === Subject: Re: Trisecting and angle after infinite steps > It was not covered in this book, so apparently I was meant to > use another, simpler approach. Is the problem, perhaps, in the section with the sum of a geometric series? === Subject: Re: Trisecting and angle after infinite steps <428c3700$0$236$edfadb0f@dread12.news.tele.dk> <200520050735191724%anniel@nym.alias.net.invalid> Yes, it is, so it's not as if it never occurred to me to look for such a solution. The question doesn't ask about the sum of a series, it asks about a single term X(n) in a sequence, as n approaches infinity. My problem is that I couldn't find any way to express a single term as the sum of a series of terms where each term in the series was of the form ar^n. === Subject: equivalence of l^p norms on R^n (simple) slightly stuck on something, but I just can't make it work. It's known that the l^p norms on R^n are equivalent, in the the usual sense: Let 1 < r < s. Then constants C, D > 0 can be found so that C ||a||_s >= ||a||_r >= D ||a||_s for all vectors a. I'd like to know what is the best (i.e. smallest) possible choice for C. I'm fairly sure it is n^( 1/r - 1/s ) but I can't prove it. Could anyone help me out with this? I think I've successfully proven that the best (i.e. largest) possible choice for D is 1. Dave. === Subject: Re: equivalence of l^p norms on R^n (simple) > slightly stuck on something, but I just can't make it work. > It's known that the l^p norms on R^n are equivalent, in the > the usual sense: > Let 1 < r < s. Then constants C, D > 0 can be found so that > C ||a||_s >= ||a||_r >= D ||a||_s > for all vectors a. > I'd like to know what is the best (i.e. smallest) possible > choice for C. I'm fairly sure it is > n^( 1/r - 1/s ) > but I can't prove it. Could anyone help me out with this? n^(1/r - 1/s) works in the above by Holder. To show it's the smallest such C, try a_k = 1 for all k. > I think I've successfully proven that the best (i.e. largest) > possible choice for D is 1. The standard approach here is to choose a such that ||a||_r = 1. Then ||a||_s <= 1 is immediate; ||a||_r >= ||a||_s for all a follows by homogeneity. To see that 1 is the best constant, look at a = (1, 0, ..., 0). === Subject: Re: equivalence of l^p norms on R^n (simple) To WWW, I did have everything except the proof that n^(1/r - 1/s) works, but I've got it now. I gather the method is to write sum |a_k|^r = sum ( 1 * |a_k|^r ) and use that in the inner product side of the Holder inequality, so sum |a_k|^r <= ( sum |a_k|^rp )^{1/p} * ( sum 1 )^{1/p'} holds for all p >1, then make the particular choice of p = s/r. Neat! I doubt I would have thought of it without the hint. Dave. === Subject: Revision of the Product Rule The Product Rule of calculus says that, given the function k(x) such that k(x) = f(x)g(x) the derivative k'(x) of k(x) is k'(x) = f(x)g'(x) + g(x)f'(x) A simple example shows why I think this is incorrect. Consider a rectangle with area A = wh, where w is its width and h its height. We wish to vary the width by dw and the height by dh, and determine the resulting change in area. Here's a drawing to make it more clear. ____________________ dh | | | |________________|___| | | | | | | h | | | | | | | | | |________________|___| w dw The change in the area dA is dA = (w + dw)(h + dh) - wh = wh + wdh + hdw + dwdh - wh = wdh + hdw + dwdh (1) This differs from the usual Product Rule in this case dA = wdh + hdw by the term dwdh. Substituting k, f, and g for A, w, and h in (1), we have dk = fdg + gdf + dfdg Dividing by dx and multiplying the last term by dx/dx, we have dk/dx = f(dg/dx) + g(df/dx) + (df/dx)(dg/dx)dx or k'(x) = f(x)g'(x) + g(x)f'(x) + f'(x)g'(x)dx (2) I will try now to present a mathematical proof of (2). Please excuse the ascii. Proof: Let k(x) = f(x)g(x) and h = delta x. We wish to show that k'(x) = f(x)g'(x) + g(x)f'(x) + f'(x)g'(x)dx If k'(x) exists, then k'(x) = lim (k(x + h) - k(x))/h h->0 = lim (f(x + h)g(x + h) - f(x)g(x))/h h->0 To change the form of the quotient so that the limit may be evaluated, we subtract and add the expressions f(x)g(x + h), g(x)f(x + h), and f(x)g(x) in the numerator. Thus k'(x) = lim (f(x + h)g(x + h) - f(x)g(x) + f(x)g(x + h) - f(x)g(x + h) h->0 + g(x)f(x + h) - g(x)f(x + h) + f(x)g(x) - f(x)g(x))/h which may be written k'(x) = lim (f(x)(g(x + h) - g(x))/h + g(x)(f(x + h) - f(x))/h h->0 + (f(x + h) - f(x))(g(x + h) - g(x))/h) = [lim f(x)][lim (g(x + h) - g(x))/h] [h->0 ][h->0 ] + [lim g(x)][lim (f(x + h) - f(x))/h] [h->0 ][h->0 ] + [lim (f(x + h) - f(x))/h][lim (g(x + h) - g(x))/h] h [h->0 ][h->0 ] Since lim f(x) = f(x) h->0 and lim g(x) = g(x) h->0 we can write k'(x) = f(x)[lim (g(x + h) - g(x))/h] [h->0 ] + g(x)[lim (f(x + h) - f(x))/h] [h->0 ] + [lim (f(x + h) - f(x))/h][lim (g(x + h) - g(x))/h] h [h->0 ][h->0 ] Applying the definition of the derivative to f(x) and g(x), and noting that delta x = dx in the definition of the differential, we get k'(x) = f(x)g'(x) + g(x)f'(x) + f'(x)g'(x)dx This completes my proof. Please comment on its validity. If this proof is valid, we would also need to make the appropriate changes to the Quotient Rule and Integration by Parts Rule. -- Dave Rutherford New Transformation Equations and the Electric Field Four-vector http://www.softcom.net/users/der555/newtransform.pdf Applications: 4/3 Problem Resolution http://www.softcom.net/users/der555/elecmass.pdf Action-reaction Paradox Resolution http://www.softcom.net/users/der555/actreact.pdf Energy Density Correction http://www.softcom.net/users/der555/enerdens.pdf Proposed Quantum Mechanical Connection http://www.softcom.net/users/der555/quantum.pdf Biot-Savart's Companion http://www.softcom.net/users/der555/biotcomp.pdf === Subject: Re: Revision of the Product Rule : the derivative k'(x) of k(x) is : k'(x) = f(x)g'(x) + g(x)f'(x) Right, that's true. : This completes my proof. Please comment on its validity. It's invalid. Please learn some basic calculus. Justin === Subject: Re: Revision of the Product Rule > The Product Rule of calculus says that, given the function k(x) such that > k(x) = f(x)g(x) > the derivative k'(x) of k(x) is > k'(x) = f(x)g'(x) + g(x)f'(x) > A simple example shows why I think this is incorrect. Consider a > rectangle with area A = wh, where w is its width and h its height. We > wish to vary the width by dw and the height by dh, and determine the > resulting change in area. > Here's a drawing to make it more clear. > ____________________ > dh | | | > |________________|___| > | | | > | | | > h | | | > | | | > | | | > |________________|___| > w dw > The change in the area dA is > dA = (w + dw)(h + dh) - wh > = wh + wdh + hdw + dwdh - wh > = wdh + hdw + dwdh (1) > This differs from the usual Product Rule in this case > dA = wdh + hdw > by the term dwdh. This is a second-order approximation to the change in area - it's not wrong, but in the limit the first derivative only takes account of the first-order approximation. > Substituting k, f, and g for A, w, and h in (1), we have > dk = fdg + gdf + dfdg > Dividing by dx and multiplying the last term by dx/dx, we have > dk/dx = f(dg/dx) + g(df/dx) + (df/dx)(dg/dx)dx > or > k'(x) = f(x)g'(x) + g(x)f'(x) + f'(x)g'(x)dx (2) > I will try now to present a mathematical proof of (2). Please excuse the > ascii. > Proof: > Let k(x) = f(x)g(x) and h = delta x. We wish to show that > k'(x) = f(x)g'(x) + g(x)f'(x) + f'(x)g'(x)dx > If k'(x) exists, then > k'(x) = lim (k(x + h) - k(x))/h > h->0 > = lim (f(x + h)g(x + h) - f(x)g(x))/h > h->0 > To change the form of the quotient so that the limit may be evaluated, > we subtract and add the expressions f(x)g(x + h), g(x)f(x + h), and > f(x)g(x) in the numerator. Thus > k'(x) = lim (f(x + h)g(x + h) - f(x)g(x) + f(x)g(x + h) - f(x)g(x + h) > h->0 + g(x)f(x + h) - g(x)f(x + h) + f(x)g(x) - f(x)g(x))/h > which may be written > k'(x) = lim (f(x)(g(x + h) - g(x))/h + g(x)(f(x + h) - f(x))/h > h->0 + (f(x + h) - f(x))(g(x + h) - g(x))/h) > = [lim f(x)][lim (g(x + h) - g(x))/h] > [h->0 ][h->0 ] > + [lim g(x)][lim (f(x + h) - f(x))/h] > [h->0 ][h->0 ] > + [lim (f(x + h) - f(x))/h][lim (g(x + h) - g(x))/h] h > [h->0 ][h->0 ] > Since > lim f(x) = f(x) > h->0 > and > lim g(x) = g(x) > h->0 > we can write > k'(x) = f(x)[lim (g(x + h) - g(x))/h] > [h->0 ] > + g(x)[lim (f(x + h) - f(x))/h] > [h->0 ] > + [lim (f(x + h) - f(x))/h][lim (g(x + h) - g(x))/h] h > [h->0 ][h->0 ] > Applying the definition of the derivative to f(x) and g(x), and noting > that delta x = dx in the definition of the differential, we get > k'(x) = f(x)g'(x) + g(x)f'(x) + f'(x)g'(x)dx It's not exactly wrong. However, what is the utility of the dx on the right side? If you think of it as an infinitesimal, for any computational purposes you may as well replace it by 0 and you then end up with the usual product rule. If you evaluate your limits rigorously, you must include lim (h) [h->0] which is obviously 0, not dx, and again, you end up with the usual product rule. Nora B. > This completes my proof. Please comment on its validity. > If this proof is valid, we would also need to make the appropriate > changes to the Quotient Rule and Integration by Parts Rule. > -- > Dave Rutherford > New Transformation Equations and the Electric Field Four-vector > http://www.softcom.net/users/der555/newtransform.pdf > Applications: > 4/3 Problem Resolution > http://www.softcom.net/users/der555/elecmass.pdf > Action-reaction Paradox Resolution > http://www.softcom.net/users/der555/actreact.pdf > Energy Density Correction > http://www.softcom.net/users/der555/enerdens.pdf > Proposed Quantum Mechanical Connection > http://www.softcom.net/users/der555/quantum.pdf > Biot-Savart's Companion > http://www.softcom.net/users/der555/biotcomp.pdf === Subject: Re: Revision of the Product Rule >>The change in the area dA is >>dA = (w + dw)(h + dh) - wh >> = wh + wdh + hdw + dwdh - wh >> = wdh + hdw + dwdh (1) >>This differs from the usual Product Rule in this case >>dA = wdh + hdw >>by the term dwdh. > This is a second-order approximation to the change in area - > it's not wrong, but in the limit the first derivative only takes > account of the first-order approximation. It's only an approximation if the term dwdh is not included. With the term dwdh included, it is the _exact_ change in area, not an approximation. Without the term dwdh, dA = wdh + hdw is _not_ an equation, since the RHS is not equal to the LHS. If you want to be correct, you should say dA ~~ wdh + hdw. >>k'(x) = f(x)g'(x) + g(x)f'(x) + f'(x)g'(x)dx > It's not exactly wrong. However, what is the utility of the > dx on the right side? If you think of it as an infinitesimal, > for any computational purposes you may as well replace it by 0 > and you then end up with the usual product rule. I don't know what equations people use to model the weather, but if they contain derivatives of products, the inclusion of the extra term may help the accuracy of their predictions. I'm sure there could be other important applications, too. > If you evaluate your limits rigorously, you must include > lim (h) > [h->0] > which is obviously 0, not dx, and again, you end up with the usual > product rule. If lim(h->0) h = 0, then you have zero in the denominator in the definition of the derivative, as I noted in a previous message. Thus you have an invalid definition. It must be lim(h->0) h = dx. -- Dave Rutherford New Transformation Equations and the Electric Field Four-vector http://www.softcom.net/users/der555/newtransform.pdf Applications: 4/3 Problem Resolution http://www.softcom.net/users/der555/elecmass.pdf Action-reaction Paradox Resolution http://www.softcom.net/users/der555/actreact.pdf Energy Density Correction http://www.softcom.net/users/der555/enerdens.pdf Proposed Quantum Mechanical Connection http://www.softcom.net/users/der555/quantum.pdf Biot-Savart's Companion http://www.softcom.net/users/der555/biotcomp.pdf === Subject: Re: Revision of the Product Rule >The Product Rule of calculus says that, given the function k(x) such that >k(x) = f(x)g(x) >the derivative k'(x) of k(x) is >k'(x) = f(x)g'(x) + g(x)f'(x) >A simple example shows why I think this is incorrect. Consider a chop ln k = ln f + ln g dk/k = df/f + dg/g dk/dx = kdf/dx/f + kdg/dx/g dk/dx = gdf/dx + fdg/dx since k/f = g and k/g = f Mr. Dual Space If you have something to say, write an equation. If you have nothing to say, write an essay === Subject: Re: Revision of the Product Rule > The Product Rule of calculus says that, given the function k(x) such that > k(x) = f(x)g(x) > the derivative k'(x) of k(x) is > k'(x) = f(x)g'(x) + g(x)f'(x) > A simple example shows why I think this is incorrect. I preferred your revolutionary metric ds^2 = c dt^2 + dx^2 + dy^2 + dz^2 live shorter than their laboratory counterparts. Dirk Vdm === Subject: Re: Revision of the Product Rule Your objection to the product rule is not totally new. Have a look at: D.E. Smith: A Source Book in Mathematics, Dover, 1959, there: pp. 627, Berkeley: The Analyst: A Discourse Adressed to an Infidel Mathematician, see P. 630 and also the footnote next page. To solve the here appearing problems with differentials the limit Ciao Karl === Subject: Re: Revision of the Product Rule > Your objection to the product rule is not totally new. Have a look at: > D.E. Smith: A Source Book in Mathematics, Dover, 1959, there: pp. 627, > Berkeley: The Analyst: A Discourse Adressed to an Infidel Mathematician, > see P. 630 and also the footnote next page. > To solve the here appearing problems with differentials the limit -- Dave Rutherford New Transformation Equations and the Electric Field Four-vector http://www.softcom.net/users/der555/newtransform.pdf Applications: 4/3 Problem Resolution http://www.softcom.net/users/der555/elecmass.pdf Action-reaction Paradox Resolution http://www.softcom.net/users/der555/actreact.pdf Energy Density Correction http://www.softcom.net/users/der555/enerdens.pdf Proposed Quantum Mechanical Connection http://www.softcom.net/users/der555/quantum.pdf Biot-Savart's Companion http://www.softcom.net/users/der555/biotcomp.pdf === Subject: Re: Revision of the Product Rule > Let k(x) = f(x)g(x) and h = delta x. We wish to show that > k'(x) = f(x)g'(x) + g(x)f'(x) + f'(x)g'(x)dx > If k'(x) exists, then > k'(x) = lim (k(x + h) - k(x))/h > h->0 > = lim (f(x + h)g(x + h) - f(x)g(x))/h > h->0 > k'(x) = lim (f(x + h)g(x + h) - f(x)g(x) + f(x)g(x + h) - f(x)g(x + h) > h->0 + g(x)f(x + h) - g(x)f(x + h) + f(x)g(x) - f(x)g(x))/h > which may be written > k'(x) = lim (f(x)(g(x + h) - g(x))/h + g(x)(f(x + h) - f(x))/h > h->0 + (f(x + h) - f(x))(g(x + h) - g(x))/h) > = [lim f(x)][lim (g(x + h) - g(x))/h] > [h->0 ][h->0 ] > + [lim g(x)][lim (f(x + h) - f(x))/h] > [h->0 ][h->0 ] > + [lim (f(x + h) - f(x))/h][lim (g(x + h) - g(x))/h] h > [h->0 ][h->0 ] You forgot lim(h->0) h. > Since > lim f(x) = f(x) > h->0 > lim g(x) = g(x) > h->0 > we can write > k'(x) = f(x)[lim (g(x + h) - g(x))/h] > [h->0 ] > + g(x)[lim (f(x + h) - f(x))/h] > [h->0 ] > + [lim (f(x + h) - f(x))/h][lim (g(x + h) - g(x))/h] h > [h->0 ][h->0 ] You've not included lim(h->0) h = 0 > Applying the definition of the derivative to f(x) and g(x), and noting > that delta x = dx in the definition of the differential, we get > k'(x) = f(x)g'(x) + g(x)f'(x) + f'(x)g'(x)dx By what fantasy did you change h, ie lim(h->0) h to dx and what does dx mean in an equation that's not balanced left and right infinitesimally? > This completes my proof. Please comment on its validity. k'(x) = f(x)g'(x) + g(x)f'(x) + f'(x)g'(x) * lim(h->0) h = f(x)g'(x) + g(x)f'(x) + f'(x)g'(x) * 0 = f(x)g'(x) + g(x)f'(x) + 0 = f(x)g'(x) + g(x)f'(x) === Subject: Re: Revision of the Product Rule >>Let k(x) = f(x)g(x) and h = delta x. We wish to show that >>k'(x) = f(x)g'(x) + g(x)f'(x) + f'(x)g'(x)dx >>If k'(x) exists, then >>k'(x) = lim (k(x + h) - k(x))/h >> h->0 >> = lim (f(x + h)g(x + h) - f(x)g(x))/h >> h->0 >>k'(x) = lim (f(x + h)g(x + h) - f(x)g(x) + f(x)g(x + h) - f(x)g(x + h) >> h->0 + g(x)f(x + h) - g(x)f(x + h) + f(x)g(x) - f(x)g(x))/h >>which may be written >>k'(x) = lim (f(x)(g(x + h) - g(x))/h + g(x)(f(x + h) - f(x))/h >> h->0 + (f(x + h) - f(x))(g(x + h) - g(x))/h) >> = [lim f(x)][lim (g(x + h) - g(x))/h] >> [h->0 ][h->0 ] >> + [lim g(x)][lim (f(x + h) - f(x))/h] >> [h->0 ][h->0 ] >> + [lim (f(x + h) - f(x))/h][lim (g(x + h) - g(x))/h] h >> [h->0 ][h->0 ] > You forgot lim(h->0) h. lim(h->0) instead of my more messy one, k'(x) = [lim(h->0)f(x)][lim(h->0)(g(x+h)-g(x))/h] + [lim(h->0)g(x)][lim(h->0)(f(x+h)-f(x))/h] + [lim(h->0)(f(x+h)-f(x))/h][lim(h->0)(g(x+h)-g(x))/h][lim(h->0)h] and using h = delta x, and my definition of lim(h->0) h lim(h->0) h = lim(delta x->0) delta x = dx this becomes k'(x) = f(x)g'(x) + g(x)f'(x) + f'(x)g'(x)dx > and what does dx mean in an equation that's not balanced > left and right infinitesimally? I don't know what you mean by balanced left and right infinitesimally. And why does it need to be balanced that way? -- Dave Rutherford New Transformation Equations and the Electric Field Four-vector http://www.softcom.net/users/der555/newtransform.pdf Applications: 4/3 Problem Resolution http://www.softcom.net/users/der555/elecmass.pdf Action-reaction Paradox Resolution http://www.softcom.net/users/der555/actreact.pdf Energy Density Correction http://www.softcom.net/users/der555/enerdens.pdf Proposed Quantum Mechanical Connection http://www.softcom.net/users/der555/quantum.pdf Biot-Savart's Companion http://www.softcom.net/users/der555/biotcomp.pdf === Subject: Re: Revision of the Product Rule >>Let k(x) = f(x)g(x) and h = delta x. We wish to show that >> >>k'(x) = f(x)g'(x) + g(x)f'(x) + f'(x)g'(x)dx >> >>If k'(x) exists, then >> >>k'(x) = lim (k(x + h) - k(x))/h >> h->0 >> >> = lim (f(x + h)g(x + h) - f(x)g(x))/h >> h->0 >> >>k'(x) = lim (f(x + h)g(x + h) - f(x)g(x) + f(x)g(x + h) - f(x)g(x + h) >> h->0 + g(x)f(x + h) - g(x)f(x + h) + f(x)g(x) - f(x)g(x))/h >> >>which may be written >> >>k'(x) = lim (f(x)(g(x + h) - g(x))/h + g(x)(f(x + h) - f(x))/h >> h->0 + (f(x + h) - f(x))(g(x + h) - g(x))/h) >> >> = [lim f(x)][lim (g(x + h) - g(x))/h] >> [h->0 ][h->0 ] >> >> + [lim g(x)][lim (f(x + h) - f(x))/h] >> [h->0 ][h->0 ] >> >> + [lim (f(x + h) - f(x))/h][lim (g(x + h) - g(x))/h] h >> [h->0 ][h->0 ] >> > You forgot lim(h->0) h. > lim(h->0) instead of my more messy one, > k'(x) = [lim(h->0)f(x)][lim(h->0)(g(x+h)-g(x))/h] > + [lim(h->0)g(x)][lim(h->0)(f(x+h)-f(x))/h] > + [lim(h->0)(f(x+h)-f(x))/h][lim(h->0)(g(x+h)-g(x))/h][lim(h->0)h] Somespacesinthoselongequationswouldsaveeyestrain. Sotosaveeyestrain,I'llskiptothenextline. > and using h = delta x, and my definition of lim(h->0) h > lim(h->0) h = lim(delta x->0) delta x = dx Note then by your definition dx = lim(h->0) h = 0 > this becomes > k'(x) = f(x)g'(x) + g(x)f'(x) + f'(x)g'(x)dx k'(x) = f(x)g'(x) + g(x)f'(x) + f'(x)g'(x) * 0 k'(x) = f(x)g'(x) + g(x)f'(x) === Subject: Re: Revision of the Product Rule >>k'(x) = [lim(h->0)f(x)][lim(h->0)(g(x+h)-g(x))/h] >> + [lim(h->0)g(x)][lim(h->0)(f(x+h)-f(x))/h] >> + [lim(h->0)(f(x+h)-f(x))/h][lim(h->0)(g(x+h)-g(x))/h][lim(h->0)h] > Somespacesinthoselongequationswouldsaveeyestrain. > Sotosaveeyestrain,I'llskiptothenextline. I had to do it that wa y to get the whole ter m on one line. >>and using h = delta x, and my definition of lim(h->0) h >>lim(h->0) h = lim(delta x->0) delta x = dx > Note then by your definition > dx = lim(h->0) h = 0 No. You have your definitions, I have mine. In mine, lim(h->0) h = dx =/= 0 -- Dave Rutherford New Transformation Equations and the Electric Field Four-vector http://www.softcom.net/users/der555/newtransform.pdf Applications: 4/3 Problem Resolution http://www.softcom.net/users/der555/elecmass.pdf Action-reaction Paradox Resolution http://www.softcom.net/users/der555/actreact.pdf Energy Density Correction http://www.softcom.net/users/der555/enerdens.pdf Proposed Quantum Mechanical Connection http://www.softcom.net/users/der555/quantum.pdf Biot-Savart's Companion http://www.softcom.net/users/der555/biotcomp.pdf === Subject: Re: Revision of the Product Rule > lim(h->0) instead of my more messy one, > k'(x) = [lim(h->0)f(x)][lim(h->0)(g(x+h)-g(x))/h] > + [lim(h->0)g(x)][lim(h->0)(f(x+h)-f(x))/h] > + [lim(h->0)(f(x+h)-f(x))/h][lim(h->0)(g(x+h)-g(x))/h][lim(h->0)h] > and using h = delta x, and my definition of lim(h->0) h > lim(h->0) h = lim(delta x->0) delta x = dx Lim(h -> 0) h = 0, and lim(delta x -> 0) elta x = 0 > this becomes > k'(x) = f(x)g'(x) + g(x)f'(x) + f'(x)g'(x)dx k'(x) = f(x)g'(x) + g(x)f'(x) + f'(x)g'(x)*0 = f(x)g'(x) + g(x)f'(x) just as the old fashioned product rule says it should! > and what does dx mean in an equation that's not balanced > left and right infinitesimally? > I don't know what you mean by balanced left and right infinitesimally. If any term in an equation contains an infinitesimal, all terms must. > And why does it need to be balanced that way? Because it is like multipling both sides of an equation by a non-zero quantity, the alleged infinitesimal. If you don't do it right the new equation is so much garbage. === Subject: Re: Revision of the Product Rule >>lim(h->0) instead of my more messy one, >>k'(x) = [lim(h->0)f(x)][lim(h->0)(g(x+h)-g(x))/h] >> + [lim(h->0)g(x)][lim(h->0)(f(x+h)-f(x))/h] >> + [lim(h->0)(f(x+h)-f(x))/h][lim(h->0)(g(x+h)-g(x))/h][lim(h->0)h] >>and using h = delta x, and my definition of lim(h->0) h >>lim(h->0) h = lim(delta x->0) delta x = dx > Lim(h -> 0) h = 0, and lim(delta x -> 0) elta x = 0 By a theorem of calculus lim(x->a) (f(x)/g(x)) = (lim(x->a) f(x))/(lim(x->a) g(x)) if lim(x->a) g(x) =/= 0. But, according to you, lim(h->0) [(f(x + h) - f(x))/h] = (lim(h->0) (f(x + h) - f(x)))/(lim(h->0) h) = (lim(h->0) (f(x + h) - f(x)))/0 This is dividing by zero, which is not allowed. >>>and what does dx mean in an equation that's not balanced >>>left and right infinitesimally? >>I don't know what you mean by balanced left and right infinitesimally. > If any term in an equation contains an infinitesimal, all terms must. The equation dA = wdh + hdw + dwdh in my original post is a valid equation. It contains two 'infinitesimals' in the last term, but only one in all the others. I guess you could say that's unbalanced, however, it doesn't affect the validity of the equation (see the drawing in my original post). >>And why does it need to be balanced that way? > Because it is like multipling both sides of an equation by a non-zero > quantity, the alleged infinitesimal. How so? -- Dave Rutherford New Transformation Equations and the Electric Field Four-vector http://www.softcom.net/users/der555/newtransform.pdf Applications: 4/3 Problem Resolution http://www.softcom.net/users/der555/elecmass.pdf Action-reaction Paradox Resolution http://www.softcom.net/users/der555/actreact.pdf Energy Density Correction http://www.softcom.net/users/der555/enerdens.pdf Proposed Quantum Mechanical Connection http://www.softcom.net/users/der555/quantum.pdf Biot-Savart's Companion http://www.softcom.net/users/der555/biotcomp.pdf === Subject: Re: Revision of the Product Rule Dave, Find any book on elementary real analysis (Rudin's Principles of any proof of the product rule. That should show you where you went wrong. Try calculating the derivative of x^6 as a product, say x^3 * x^3 or x^2 * x^4 and you will see that the traditional product rule gives the correct answer both ways. I doubt that your product rule gives the same answer both ways, but I don't have time to try these things at the moment. The basic problem with your original statement is that the the product dw*dh goes to 0 so rapidly that it does not show up in the derivative. Achava === Subject: Re: Revision of the Product Rule ETAsAhRinDL0UR1eO6C7Wrwki3hU5qZZagIUP88rwRvbJwR0eTm8J7aN5E8z77I= The problem is that dw dh is not the same as w'h' dx It's (w' dx)(h' dx). You lost an order of differentiation. It's better to work with deltas and use the limit definition of the derivative. When you do, upon taing the limit you will find that the delta w*delta h term does not contribute to the limit. --OL === Subject: Re: Revision of the Product Rule > The Product Rule of calculus says that, given the function k(x) such that > k(x) = f(x)g(x) > the derivative k'(x) of k(x) is > k'(x) = f(x)g'(x) + g(x)f'(x) > A simple example shows why I think this is incorrect. You would be laughable if you weren't pathetic. Why didn't you psot in LaTeX? Meds kicking in? -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: CANTOR's theorem > ... > > > Nope. But convince yourself after all by the mapping > > > N --> P(N) M > > > which I claim is bijective, unless you prove it is not. > > > > That is too easy. Say your mapping is f. Define the following > > mapping g from N to P(N): > > g(1) = M > > g(n) = f(n - 1) for n > 1. > > If your mapping is surjective, this one is also surjective. > > But we know > > it is not surjective, so your map is also not surjective. > > > > But we know ... is not permitted, because that is taken from the > > proof in question. As I said, this proof is not a proof of > > non-surjectivity but of non-existence of a special self-reflexive set > > {M, m, f'}. Therefore I challenge you to prove your assertion without > > reference to that dubios proof. > To prove that the map is non-surjective, I have only to display a set > that is not in the image of the map. So I have to define a set, and > next show that it is not f'(n) for some n in N. Right? And to define > a set, I need to display the properties of the elements of that set. > Right? I have no idea what you mean with the set {M, m, f'}. In > Cantor's the set M is defined depending on f. That is step 1. > Next it is shown that the set is not in the map, by proving that it > is impossible to find the pre-image. So that proves non-surjectivity. > Now we start again with the map f: N -> P(N). Define a range of sets > of natural numbers, depending on f, and indexed by the integers 0, 1,..., > as follows: > M(i) = {x in N, x + i not in f(x)} Sorry, this attempt was wrong. So a second attempt now, and I think it works. Let's have a map f: N -> P(N) (as given above). Take any injection g: N -> N. Define a set H = {x in N, g(x) not in f(x)}. Define M(g) = g(H) (i.e. the set of all elements in H under the injection g). Now suppose that for some i, i maps on M(g). Suppose that i in H, then g(i) not in f(i); this means that when g(i) in M(g) we have g(i) not in f(i), so f(i) can not be M(g). Conclusion: i not in H. But in that case g(i) in f(i), so g(i) not in M(g) -> g(i) in f(i) so again f(i) can not be equal to M(g). Hence, M(g) is not in the image of f for any injection g. The standard M we get if g is the indentity mapping. Now can the restriction f': N -> P{M} be a surjection? Suppose it is. We know from above that for any injection g, M(g) is not in the map, so for f' to be surjective, all M(g) must be identical to M. Now I will define a sequence of sets M, parametrised by a non-negative integer. M(i) is set M(g) for g the injection N -> N, with g(n) = n + i. The traditional M we have is M(0). But we know, from the definition, that all elements of M(i) are larger than i. So if all those M(i) must be equal to M, M = {}, and so *all* M(i) are equal to {}. Now let's see whether we can find what 1 maps to. 1 is not in M(0), so 1 is in f(1). In general, 1 is not in M(i), so 1 + i is in f(1). Hence 1 maps to N. In the same way we find that any natural n maps to a subset of N that contains all numbers larger or equal to N (and if f' is an injection it *is* the subset that contains all those numbers and no others). So clearly there is *no* pre-image of the set of even numbers, because for any n there are both even and non-even numbers larger than that number. . -- 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: Naked Singularity Electron or ... ? That clears that up. So... are you for subsidizing biofuels? tesseract === Subject: Re: Naked Singularity Electron or ... ? the very first equation reminds me *suspiciously* of HSJ's pair of trisecting hyperbolae (or what ever they're supposed to do to each-other). > That clears that up. --The Kyoto Protocol and the Permian Basin Gang! http://tarpley.net/bush7htm http://larouchepub.com http://members.tripod.com/~ame[CapitalEth]rican almanac http://www.21stcenturysciencetech.com/ === Subject: Re: To Dik Winter about FLT To Dik Winter The Second Part has a mistake So Ignore the Second Part. In fact my first Intention in the What the modern mathematicien are shy to talk about postings was to show only that exist a elementary proof of FLT for n=4*+1 to make uneasy the modern mathematicien when they have to answer the questionsIs there a general elementary proof of FLT?and as I did with the question ?Why Euler did not find a proof of Pell's equation? Now I answer your question: Here it is: X,Y ,Z relative Prime numbers X+Y=W X^(n-1)-(X^(n-2))*Y +(X^(n-3))*Y^2.........+Y^(n-1)= =R= W^(n-1)-(Cn1)*[W^(n-2)]*Y+(Cn2)*[W^(n-3]*Y^2- -(Cn3)*[W^(n-4)]*Y^3+ +(Cn4)*[W^(n-5)]*Y^4-..............-(Cn2)*[W^1)*Y^(n-2) + +(Cn1)*Y^(n-1) where Cnk=[n*(n-1)*(n-2)*(n-3)*.........*(n-k+1)]/2*3*4*............*k If (X+Y) is divisible by n, you can see that R is divisible by n only. But (X+Y)*R=Z^n=(u*z*n)^n Since R is divisible only by n then (X+Y) =[n^(n-1)]]*u^n and Z=n*u*z When Z is not divisible by n we see that (X+Y)=W and R do not have any common divisor .Therfore they are relative prime and must be: R=z^n and W=X+Y=u^n and Z=u*z Is this clear?The editing on computer is not easy. george ghiata > Hello Dik Winter > Here it is a better editing of the proof of FLT > > Now we going to see the elementary genaral proof of > f FLT. > george ghiata > Here is another exemple to suport my point what this > message > and title is about besides the exemple given wit > t Pell's equation > The text below shows the proof of FLT for n=4*L+1 > I have not seen or heard about this kind the proof. > Since this proof exists, we have to think twice > before we say the great mathematicien could not > t had > missed a elementary proof for long time. > That is because now it looks possible that it might > exist a similar one for=n*L+3. > And guess what? Yes ,there is one. > Next time I will present it. > george ghiata > Here is Archimedes Proof of Fermat Last theorem for > n=4*L+1 > george Ghiata > To Alf and Andrew, > A simplfied Fermat last theorm proof:Archimedes FLT > proof when n=4*L+1 > X^n+Y^n=Z^n is Imposible if Z,X,YAre Integers and > n=prime>2 > We take Z=even number > X+Y-Z=B > X=B+Q > Y=B+P > 2*B+Q+P=s*u^n wher s=1 or n^(n-1) if Z is divisible > by n > B+Q+P=r*z*u wher r is 1 or n if Z is divisible by n > Therefore Q+P=2*u*k *r where k =odd number > Therefore too we get that z= - b+v*u^(n-1).Therfore > e b=odd number and > v=1 or n^(n-2) when Z is divisible by n > equation : > (B+Q+)^n+(B+P)^n-(B+Q+P)^n=0 > can be written as: > EQ A: B^n+M + G*H*(u^n)=0 where H is 2^t > which comes from 2*(n-1)=2^t*d where d =odd > and G=odd number and M is comming from following: > Now:EQ.C : (B+Q)^n+(B+P)^n - (Q+P)^n= B^n+ > n*B*(Q+P)+B^n+M > by 2 > as many times as (2^t)*u^n is divisible by 2 > B^n-[B^(n-1)]*n*(Q+P) > is divisible by 2 as many times (2^t)*u^n is > s divisible by 2 > Therefore F=Z^n- B^n - (Q+P)^n-n*[B^(n-1)]*(Q+P) > ) =(z*u)^n- [b*u]^n-(2*u*k*r)^n- > - n*(b^n)*r*k*2*(u^n) can be written as : > F= (u^n)*{ z^n - b^n - (2*k*r)^n - n*2*k*r*(b^n)} > Since z=- b+v*u^(n-1) we get that > F=(u^n)*{2*(b^n)*(n*k*r+1) -d*u^(n-1) -(2*k*r)^n} > But F is divisible by 2 as many times as (2^t)*u^n > is divisible by 2 > Therfore We get that (k+1)=2^m where m>(t-1) > and (n-1)=j*2^(t-1) wher j is odd number > We know that b+2*k=z and 2*b+2*k=h*u*(n-1)where h=1 > or n^(n-2) when Z > is divisible by n > Therfore z-k=b+k. > Therfore z+1 is divisible by2 as many times as k+1 is > divisible by 2 > But when the EQuation > (B+Q+P)^n=(B+Q)^n+(B+P)^n=(z*u)^n is divided > by (2*B+Q+P)we get : > H= [(n-1)*u*V +n*Q^(n-1)}=r*z^n > If r=1 then we get: > [(n-1)*u*V +(n-1)*Q^(n-1)+Q^(n-1) +1=z^n+1 > We know that (z+1)=(2^m)*c where m>(t-1 and > [2^(t-1)]*a =(n-1) > Therfore n=4*g+3 > If r=n then we get : > [(n-1)*u*V+(n-1)*Q^(n-1)+ > + [(n-1)*u*V+(n-1)*Q^(n-1)+Q^(n-1)+1=(n-1)*z^n+z^n+1 > We see that the right side of EQ is divisible by > y 2^(t-1). > If (n-1) is divisible by 2 only then the left side is > divisible by 4 > If (n-1) is divisible by 4 then Theleft side is > divisible by 2 only and the right side by 4 > Conclusion: If n=4*L+1 then X^n +Y^n =Z^ is > impossible > Created by george ghiata > Second Part of The general proof of FLT.:n=4*L+3 > We use the information from the case n=4*L+1 > is divisible by 2 > then Z is not divisible by n > Let's take (B+Q)^n+(B+P)^n=(B+Q+P)^n > We multiply the Eq by (2*u)^n and get: > (2*B+Q+P+Q-P)^n+(2*B+Q+P+P-Q)^n=(2*B+Q+P+Q+P)^n > After we develop the parantheses and move everything > g to > the left of Eq. and divide by (2*u)^n we get : > EQ.S1= [2^(n-3)]*(u^n)*C +(t^n)*[2^(n-1)]*E > E +n*(Q-u*k)^(n-1)-n*[(k*u)^(n-1)]/2-k^n=0 > Now we take X^n+Y^n=Z^n and after division by > y (X+Y)=u^n we change it > to: EQJ: [(X+Y)*E +n*X^(n-1)]=z^n > We change this to > EQ.S: > : [(X+Y)*E+n*(n-1)*(b+k)*u*A+n*(Q-u*k)^(n-1)-z^n=0 > where A is a odd number. > We see thatT={ [n*(Q-u*k)^(n-1)]-z^n}/(b+k) is > divisible by 2 as > many times as (n-1)*u is divisible by 2 > Now we substract and add up in the EQ.S1 the the > e value z^n and then > l n*k^(n-1)and get EQ.S2. > Since n=4*L+3 we get that the EQ.S2 is divisible by > y 2 only as many times as (u) is divisible > by 2 [That is because T= {n*(Q-u*k)^(n-1)-z^n}/(b+k) > ) is > divisible by 2*u conform to the EQ.S] > Therefore EQ.S2 is not possible in integers. > Therefore Fermat last theorem is impossible for > r n=4*L+3 too. > Therefore Fermat Last theorem is true. > Created by George Ghiata-AA > OBSERVATION: In fact we can prove the general case Of > FLT by > using only the proof of Second Part ! GOOD Luck > george ghiata === Subject: Re: To Dik Winter about FLT Sorry, I do not follow this. I just have spent over 60 minutes to decipher what you meant, and have included comments, and now you come with the message that I should not have done so. I quit. -- 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: To Dik Winter about FLT I am sorry.I was talking about Second part. I re-written the first part( and modofy it) such to be very easy to follow It is the same as that posted at GG'Ansewr to......... > Here it is the proof of FLT for n=prime>2by using > n=5 as an exemple > The proof is written to the level which you asked me > to do. > In fact my first Intention in the What the modern > mathematicien are shy > to talk about postings was to show only that > exist > a elementary proof of > FLT for n=4*+1 to make uneasy the modern > mathematicien > when they have to answer the questionsIs there a > general > elementary proof of FLT?and as I did with the > e question > ?Why Euler did not find a proof of Pell's > equation? > I am going to show proof of FLT for n=5 which is the > same as the proof > for n=prime>3 > X^5+Y^5=Z^5 wher X,Y,Z are integers and Z=even > X+Y-Z=B > X=B+Q > Y=B+P > Z=B+Q+P=r*u*z where r =1 or r=5 if Z is divisible by > X+Y=2*B+Q+P=s*u^5 where s =1 or s=5^4 if Z is > is divisible by n= 5 > When we do X+Y-Z we get that B=r*b*u > Therfore (Q+P)=2*r*u*k > Therfore we see that z=-b+v*u^(n-1) where v=1 or > r v=n^(n-2) > when Z is divisible by n=5 > > OBservation: > X,Y ,Z relative Prime numbers > X+Y=W > X^(n-1)-(X^(n-2))*Y > +(X^(n-3))*Y^2.........+Y^(n-1)= > =R= W^(n-1)-(Cn1)*[W^(n-2)]*Y+(Cn2)*[W^(n-3]*Y^2- > -(Cn3)*[W^(n-4)]*Y^3+ > +(Cn4)*[W^(n-5)]*Y^4-..............-(Cn2)*[W^1)*Y^(n-2 > ) + > +(Cn1)*Y^(n-1) > where > Cnk=[n*(n-1)*(n-2)*(n-3)*.........*(n-k+1)]/2*3*4*.... > ........*k > If (X+Y) is divisible by n, you can see that R is > divisible by n only. > But (X+Y)*R=Z^n=(u*z*n)^n > Since R is divisible only by n then (X+Y) > =[n^(n-1)]]*u^n > and Z=n*u*z > When Z is not divisible by n we see that (X+Y)=W > and > R do not have any common divisor . > Therfore they are relative prime and must be: > R=z^n and W=X+Y=u^n and Z=u*z > The end of Observation. > Let'choose Z not divisible by 5 > Let's take F=(B+Q)^5+(B+P)^5-(B+Q+P)^5=0 > and develope the parantheses: > F=B^5 +M+D -(Q+P)^5=0 where > M=(C52)*(B^3)*(Q^2+P^2) > +(C53)*(B^2)*(Q^3+P^3)+(C54)*B*(Q^4+P^4) . > -D=(C52)*(B^3)*(Q+P)^2+(C53)*(B^2)*(Q+P)^3+(C54)*B*(Q+ > P)^4 > times as > T= (B^3)* (C52)*(Q+P)^2 is divisible by 2 > Therefore G=B^5+M is divisible by 2 as many times T > is divisible by 2. > Now T can be written as : > T=[(b)^3]*[(k)^2]*(u^5)*5*(n-1)*2 where n=5 > Therfore G/u^5 is divisible by 2 as many times > 2*(n-1) is > divisible by 2 > 1): Let's take H=(B+Q)^5+(B+P)^5 > 2).H can be written as :H=(B^5+M) +B^5 > +(C51)*(B^4)*(Q+P+2*B-2*B) > 3). H=(B^5+M) > 3). H=(B^5+M) > 3). H=(B^5+M) +[(b)^5]*u^5-n*2*[(b)^5]*(u^5)+V > > where V=(C51)*[(b)^4]*(u^4)*(u^5) > If we divide H in 3). by (2*B+Q+P) we get: > H1={[(X+Y)*A > *A +n*X^(n-1)}=[B^5+M]/(X+Y)+b^5-2*n*b^5+V/(u^5 > where n=5 [SEE Observation) > H2={X+Y)*A +n*X^(n-1) > n-1) +(b^5)*(2*n-1)=[B^5+M}/(X+Y) +V/(u^5). > n=5 > We see that V/(u^5) is divisible by 2 as many times > u^4 is > divisible by 2. > We see that [B^5+M]/(u^5 is divisible by 2 as many > times as 2*(n-1) > is divisible by 2. We know n=5 > Therefore H2 must be divisible by 2 as many times > 2*(n-1) > is divisible by 2 . We know n=5 > L=n*X^(n-1)+(b^5)*(2*n-1) is divisible by 2 > as many times as 2*n-1 is divisible by 2 . (n=5) > 4) Therfore from L we get that D=n*X^(n-1) +b^5 is > divisible by 2 > as many times as 2*(n-1) is divisible by 2. > But H1=(X^n+Y^n)/(X+Y=z^n=[-b+u^(n-1)]^n where n=5 > Therfore H1=(X+Y)*A +n*X^(n-1)+b^5=C*u^(n-1) where > C=odd > Therfore D =n*X^(n-1) +b^5 must be divisible by 2 as > many times > as u^(n-1) is divisible by 2 > Therfore 2*(n-1) is divisible by 2 as many times as > u^(n-1); n=5 > That is impossible.(possible only for n=1) > The same proof can be used to prove the case Z > divisible by n > Therfore FLT for n=5 is impossible > Created by George Ghiata === Subject: Re: To Dik Winter about FLT This is a corection of a error: M must include Q^n+P^n where n=5 The omission does not affect the proof george ghiata I am sorry.I was talking about Second part. > I re-written the first part( and modofy it) such to > be very easy to follow > It is the same as that posted at GG'Ansewr > to......... > Here it is the proof of FLT for n=prime>2by > using > n=5 as an exemple > The proof is written to the level which you asked > me > to do. > In fact my first Intention in the What the > modern > mathematicien are shy > to talk about postings was to show only that > exist > a elementary proof of > FLT for n=4*+1 to make uneasy the modern > mathematicien > when they have to answer the questionsIs there > general > elementary proof of FLT?and as I did with the > e question > ?Why Euler did not find a proof of Pell's > equation? > > I am going to show proof of FLT for n=5 which is > the > same as the proof > for n=prime>3 > X^5+Y^5=Z^5 wher X,Y,Z are integers and Z=even > X+Y-Z=B > X=B+Q > Y=B+P > Z=B+Q+P=r*u*z where r =1 or r=5 if Z is divisible > by > 5 > X+Y=2*B+Q+P=s*u^5 where s =1 or s=5^4 if Z is > is divisible by n= 5 > When we do X+Y-Z we get that B=r*b*u > Therfore (Q+P)=2*r*u*k > Therfore we see that z=-b+v*u^(n-1) where v=1 or > r v=n^(n-2) > when Z is divisible by n=5 > > OBservation: > X,Y ,Z relative Prime numbers > X+Y=W > X^(n-1)-(X^(n-2))*Y > +(X^(n-3))*Y^2.........+Y^(n-1)= > =R= > W^(n-1)-(Cn1)*[W^(n-2)]*Y+(Cn2)*[W^(n-3]*Y^2- > -(Cn3)*[W^(n-4)]*Y^3+ > > +(Cn4)*[W^(n-5)]*Y^4-..............-(Cn2)*[W^1)*Y^(n-2 > ) + > +(Cn1)*Y^(n-1) > where > > Cnk=[n*(n-1)*(n-2)*(n-3)*.........*(n-k+1)]/2*3*4*.... > > ........*k > If (X+Y) is divisible by n, you can see that R > is > divisible by n only. > But (X+Y)*R=Z^n=(u*z*n)^n > Since R is divisible only by n then (X+Y) > =[n^(n-1)]]*u^n > and Z=n*u*z > When Z is not divisible by n we see that (X+Y)=W > and > R do not have any common divisor . > Therfore they are relative prime and must be: > R=z^n and W=X+Y=u^n and Z=u*z > The end of Observation. > Let'choose Z not divisible by 5 > Let's take F=(B+Q)^5+(B+P)^5-(B+Q+P)^5=0 > and develope the parantheses: > F=B^5 +M+D -(Q+P)^5=0 where > M=(C52)*(B^3)*(Q^2+P^2) > +(C53)*(B^2)*(Q^3+P^3)+(C54)*B*(Q^4+P^4) . > -D=(C52)*(B^3)*(Q+P)^2+(C53)*(B^2)*(Q+P)^3+(C54)*B*(Q+ > P)^4 > times as > T= (B^3)* (C52)*(Q+P)^2 is divisible by 2 > Therefore G=B^5+M is divisible by 2 as many times > is divisible by 2. > Now T can be written as : > T=[(b)^3]*[(k)^2]*(u^5)*5*(n-1)*2 where n=5 > Therfore G/u^5 is divisible by 2 as many times > 2*(n-1) is > divisible by 2 > > 1): Let's take H=(B+Q)^5+(B+P)^5 > > 2).H can be written as :H=(B^5+M) +B^5 > +(C51)*(B^4)*(Q+P+2*B-2*B) > 3). H=(B^5+M) > 3). H=(B^5+M) > 3). H=(B^5+M) +[(b)^5]*u^5-n*2*[(b)^5]*(u^5)+V > > where V=(C51)*[(b)^4]*(u^4)*(u^5) > If we divide H in 3). by (2*B+Q+P) we get: > H1={[(X+Y)*A > *A +n*X^(n-1)}=[B^5+M]/(X+Y)+b^5-2*n*b^5+V/(u^5 > where n=5 [SEE Observation) > H2={X+Y)*A +n*X^(n-1) > n-1) +(b^5)*(2*n-1)=[B^5+M}/(X+Y) +V/(u^5). > n=5 > We see that V/(u^5) is divisible by 2 as many > times > u^4 is > divisible by 2. > We see that [B^5+M]/(u^5 is divisible by 2 as many > times as 2*(n-1) > is divisible by 2. We know n=5 > Therefore H2 must be divisible by 2 as many times > 2*(n-1) > is divisible by 2 . We know n=5 > L=n*X^(n-1)+(b^5)*(2*n-1) is divisible by 2 > as many times as 2*n-1 is divisible by 2 . (n=5) > 4) Therfore from L we get that D=n*X^(n-1) +b^5 is > divisible by 2 > as many times as 2*(n-1) is divisible by 2. > > But H1=(X^n+Y^n)/(X+Y=z^n=[-b+u^(n-1)]^n where > n=5 > Therfore H1=(X+Y)*A +n*X^(n-1)+b^5=C*u^(n-1) where > C=odd > Therfore D =n*X^(n-1) +b^5 must be divisible by 2 > as > many times > as u^(n-1) is divisible by 2 > Therfore 2*(n-1) is divisible by 2 as many times > as > u^(n-1); n=5 > That is impossible.(possible only for n=1) > The same proof can be used to prove the case Z > divisible by n > Therfore FLT for n=5 is impossible > Created by George Ghiata === Subject: Re: To Dik Winter about FLT > This is a corection of a error: > M must include Q^n+P^n where n=5 > The omission does not affect the proof > george ghiata > I am sorry.I was talking about Second part. > I re-written the first part( and modofy it) such > to > be very easy to follow > It is the same as that posted at GG'Ansewr > to......... > Here it is the proof of FLT for n=prime>2by > using > n=5 as an exemple > The proof is written to the level which you > asked > me > to do. > In fact my first Intention in the What the > modern > mathematicien are shy > to talk about postings was to show only > that > exist > a elementary proof of > FLT for n=4*+1 to make uneasy the modern > mathematicien > when they have to answer the questionsIs > there > a > general > elementary proof of FLT?and as I did with the > e question > ?Why Euler did not find a proof of Pell's > equation? > > I am going to show proof of FLT for n=5 which is > the > same as the proof > for n=prime>3 > X^5+Y^5=Z^5 wher X,Y,Z are integers and > Z=even > X+Y-Z=B > X=B+Q > Y=B+P > Z=B+Q+P=r*u*z where r =1 or r=5 if Z is > divisible > by > 5 > X+Y=2*B+Q+P=s*u^5 where s =1 or s=5^4 if Z is > is divisible by n= 5 > When we do X+Y-Z we get that B=r*b*u > Therfore (Q+P)=2*r*u*k > Therfore we see that z=-b+v*u^(n-1) where v=1 > or > r v=n^(n-2) > when Z is divisible by n=5 > > OBservation: > X,Y ,Z relative Prime numbers > X+Y=W > X^(n-1)-(X^(n-2))*Y > +(X^(n-3))*Y^2.........+Y^(n-1)= > =R= > W^(n-1)-(Cn1)*[W^(n-2)]*Y+(Cn2)*[W^(n-3]*Y^2- > -(Cn3)*[W^(n-4)]*Y^3+ > > > +(Cn4)*[W^(n-5)]*Y^4-..............-(Cn2)*[W^1)*Y^(n-2 > > ) + > +(Cn1)*Y^(n-1) > where > > > Cnk=[n*(n-1)*(n-2)*(n-3)*.........*(n-k+1)]/2*3*4*.... > > > ........*k > If (X+Y) is divisible by n, you can see that R > is > divisible by n only. > But (X+Y)*R=Z^n=(u*z*n)^n > Since R is divisible only by n then (X+Y) > =[n^(n-1)]]*u^n > and Z=n*u*z > When Z is not divisible by n we see that > (X+Y)=W > and > R do not have any common divisor . > Therfore they are relative prime and must be: > R=z^n and W=X+Y=u^n and Z=u*z > The end of Observation. > Let'choose Z not divisible by 5 > Let's take F=(B+Q)^5+(B+P)^5-(B+Q+P)^5=0 > and develope the parantheses: > F=B^5 +M+D -(Q+P)^5=0 where > M=(C52)*(B^3)*(Q^2+P^2) > +(C53)*(B^2)*(Q^3+P^3)+(C54)*B*(Q^4+P^4) . > > -D=(C52)*(B^3)*(Q+P)^2+(C53)*(B^2)*(Q+P)^3+(C54)*B*(Q+ > > P)^4 > many > times as > T= (B^3)* (C52)*(Q+P)^2 is divisible by 2 > Therefore G=B^5+M is divisible by 2 as many > times > T > is divisible by 2. > Now T can be written as : > T=[(b)^3]*[(k)^2]*(u^5)*5*(n-1)*2 where n=5 > Therfore G/u^5 is divisible by 2 as many times > 2*(n-1) is > divisible by 2 > > 1): Let's take H=(B+Q)^5+(B+P)^5 > > 2).H can be written as :H=(B^5+M) +B^5 > +(C51)*(B^4)*(Q+P+2*B-2*B) > 3). H=(B^5+M) > 3). H=(B^5+M) > 3). H=(B^5+M) +[(b)^5]*u^5-n*2*[(b)^5]*(u^5)+V > > where V=(C51)*[(b)^4]*(u^4)*(u^5) > If we divide H in 3). by (2*B+Q+P) we get: > H1={[(X+Y)*A > *A +n*X^(n-1)}=[B^5+M]/(X+Y)+b^5-2*n*b^5+V/(u^5 > where n=5 [SEE Observation) > H2={X+Y)*A +n*X^(n-1) > n-1) +(b^5)*(2*n-1)=[B^5+M}/(X+Y) +V/(u^5). > n=5 > We see that V/(u^5) is divisible by 2 as many > times > u^4 is > divisible by 2. > We see that [B^5+M]/(u^5 is divisible by 2 as > many > times as 2*(n-1) > is divisible by 2. We know n=5 > Therefore H2 must be divisible by 2 as many > times > 2*(n-1) > is divisible by 2 . We know n=5 > L=n*X^(n-1)+(b^5)*(2*n-1) is divisible by 2 > as many times as 2*n-1 is divisible by 2 . > (n=5) > 4) Therfore from L we get that D=n*X^(n-1) +b^5 > is > divisible by 2 > as many times as 2*(n-1) is divisible by 2. > > But H1=(X^n+Y^n)/(X+Y=z^n=[-b+u^(n-1)]^n where > n=5 > Therfore H1=(X+Y)*A +n*X^(n-1)+b^5=C*u^(n-1) > where > C=odd > Therfore D =n*X^(n-1) +b^5 must be divisible by > as > many times > as u^(n-1) is divisible by 2 > Therfore 2*(n-1) is divisible by 2 as many times > as > u^(n-1); n=5 > That is impossible.(possible only for n=1) > The same proof can be used to prove the case Z > divisible by n > Therfore FLT for n=5 is impossible > Created by George Ghiata === Subject: Re: To Dik Winter about FLT As I said:This proof,present it here, is the same as the proof of Fermat Last theorem for n=prime>3 Somebody is mixing up my text. > This is a corection of a error: > M must include Q^n+P^n where n=5 > The omission does not affect the proof > george ghiata > > I am sorry.I was talking about Second part. > I re-written the first part( and modofy it) such > to > be very easy to follow > It is the same as that posted at GG'Ansewr > to......... > Here it is the proof of FLT for n=prime>2by > using > n=5 as an exemple > The proof is written to the level which you > asked > me > to do. > In fact my first Intention in the What the > modern > mathematicien are shy > to talk about postings was to show only > that > exist > a elementary proof of > FLT for n=4*+1 to make uneasy the modern > mathematicien > when they have to answer the questionsIs > there > a > general > elementary proof of FLT?and as I did with > the > e question > ?Why Euler did not find a proof of Pell's > equation? > > I am going to show proof of FLT for n=5 which > is > the > same as the proof > for n=prime>3 > X^5+Y^5=Z^5 wher X,Y,Z are integers and > Z=even > X+Y-Z=B > X=B+Q > Y=B+P > Z=B+Q+P=r*u*z where r =1 or r=5 if Z is > divisible > by > 5 > X+Y=2*B+Q+P=s*u^5 where s =1 or s=5^4 if Z is > is divisible by n= 5 > When we do X+Y-Z we get that B=r*b*u > Therfore (Q+P)=2*r*u*k > Therfore we see that z=-b+v*u^(n-1) where > v=1 > or > r v=n^(n-2) > when Z is divisible by n=5 > > OBservation: > X,Y ,Z relative Prime numbers > X+Y=W > X^(n-1)-(X^(n-2))*Y > +(X^(n-3))*Y^2.........+Y^(n-1)= > =R= > W^(n-1)-(Cn1)*[W^(n-2)]*Y+(Cn2)*[W^(n-3]*Y^2- > -(Cn3)*[W^(n-4)]*Y^3+ > > > > +(Cn4)*[W^(n-5)]*Y^4-..............-(Cn2)*[W^1)*Y^(n-2 > > > ) + > +(Cn1)*Y^(n-1) > where > > > > Cnk=[n*(n-1)*(n-2)*(n-3)*.........*(n-k+1)]/2*3*4*.... > > > > ........*k > If (X+Y) is divisible by n, you can see that > is > divisible by n only. > But (X+Y)*R=Z^n=(u*z*n)^n > Since R is divisible only by n then (X+Y) > =[n^(n-1)]]*u^n > and Z=n*u*z > When Z is not divisible by n we see that > (X+Y)=W > and > R do not have any common divisor . > Therfore they are relative prime and must be: > R=z^n and W=X+Y=u^n and Z=u*z > The end of Observation. > Let'choose Z not divisible by 5 > Let's take F=(B+Q)^5+(B+P)^5-(B+Q+P)^5=0 > and develope the parantheses: > F=B^5 +M+D -(Q+P)^5=0 where > M=(C52)*(B^3)*(Q^2+P^2) > +(C53)*(B^2)*(Q^3+P^3)+(C54)*B*(Q^4+P^4) . > > > -D=(C52)*(B^3)*(Q+P)^2+(C53)*(B^2)*(Q+P)^3+(C54)*B*(Q+ > > > P)^4 > many > times as > T= (B^3)* (C52)*(Q+P)^2 is divisible by 2 > Therefore G=B^5+M is divisible by 2 as many > times > T > is divisible by 2. > Now T can be written as : > T=[(b)^3]*[(k)^2]*(u^5)*5*(n-1)*2 where n=5 > Therfore G/u^5 is divisible by 2 as many times > 2*(n-1) is > divisible by 2 > > 1): Let's take H=(B+Q)^5+(B+P)^5 > > 2).H can be written as :H=(B^5+M) +B^5 > +(C51)*(B^4)*(Q+P+2*B-2*B) > 3). H=(B^5+M) > 3). H=(B^5+M) > 3). H=(B^5+M) > +[(b)^5]*u^5-n*2*[(b)^5]*(u^5)+V > > where V=(C51)*[(b)^4]*(u^4)*(u^5) > If we divide H in 3). by (2*B+Q+P) we get: > H1={[(X+Y)*A > *A > +n*X^(n-1)}=[B^5+M]/(X+Y)+b^5-2*n*b^5+V/(u^5 > where n=5 [SEE Observation) > H2={X+Y)*A +n*X^(n-1) > n-1) +(b^5)*(2*n-1)=[B^5+M}/(X+Y) +V/(u^5). > n=5 > We see that V/(u^5) is divisible by 2 as many > times > u^4 is > divisible by 2. > We see that [B^5+M]/(u^5 is divisible by 2 as > many > times as 2*(n-1) > is divisible by 2. We know n=5 > Therefore H2 must be divisible by 2 as many > times > 2*(n-1) > is divisible by 2 . We know n=5 > L=n*X^(n-1)+(b^5)*(2*n-1) is divisible by 2 > as many times as 2*n-1 is divisible by 2 . > (n=5) > 4) Therfore from L we get that D=n*X^(n-1) > +b^5 > is > divisible by 2 > as many times as 2*(n-1) is divisible by 2. > > But H1=(X^n+Y^n)/(X+Y=z^n=[-b+u^(n-1)]^n > where > n=5 > Therfore H1=(X+Y)*A +n*X^(n-1)+b^5=C*u^(n-1) > where > C=odd > Therfore D =n*X^(n-1) +b^5 must be divisible > by > 2 > as > many times > as u^(n-1) is divisible by 2 > Therfore 2*(n-1) is divisible by 2 as many > times > as > u^(n-1); n=5 > That is impossible.(possible only for n=1) > The same proof can be used to prove the case Z > divisible by n > Therfore FLT for n=5 is impossible > Created by George Ghiata === Subject: Re: To Dik Winter about FLT <25141896.1116560017990.JavaMail.jakarta@nitrogen.mathforum.org> he says he quits, so i'm afraid he is no longer interested in your proof Mr. Ghiata. === Subject: Re: To Dik Winter about FLT > he says he quits, so i'm afraid he is no longer > interested in your > proof Mr. Ghiata. HellO Jay, They look at the postings GG' ansewr to Anzaurres about Fermat last therem proofwhere a symplified Fermat Last therem proof has been posted, and in fact, has been sent to mr.Professor David Leeming and Mr.Professor Andrew Granville too. Best wishes George ghiata === Subject: What is the measure of a body's mass? The measure of a body's mass is a measure of it's inertia: A unit of mass is the _Constant_ ratio of the net force (f) exerted on and/or by it, divided by the acceleration (a) that it causes (f/a); which is equal to the _Constant_ ratio of the weight (w) of the mass, divided by the acceleration (g) at which it will free fall; at the location where it is weighed. The unit of mass in the metric system is ONE kilogram, and is constant, in that the net force (f) varies in proportion to the acceleration due to gravity; which varies at various locations. The unit of mass in the customary system is ONE slug, and is constant, in that the net force (f) varies in proportion to the acceleration due to gravity; which varies at various locations. Don === Subject: Re: What is the measure of a body's mass? The numerical value of the mass of an object is: The times the standard unit of mass conserved in Paris, it is necessary to place in the left pan of an equal arms balance, when the object is placed in the right pan. That's the definition of the International Bureau of Poids et Mesures. Ludovicus === Subject: Re: What is the measure of a body's mass? > [...trivial stuff snipped...] That's hardly rocket science, Don. I already proved you're just quoting the obvious. So what NEW insights do you have to offer? A. Friend === Subject: Re: What is the measure of a body's mass? [snip crap] Hey Dumb Donny HEAD, do you know what momomania is? Do you know how ing stooopid you are? Can you hope for one out of two? -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: What is the measure of a body's mass? > [snip crap] I think you mean _monomania_; idiot(;^) Don === Subject: Re: What is the measure of a body's mass? > The measure of a body's mass is a measure of it's inertia: > A unit of mass is the _Constant_ ratio of the net force (f) > exerted on and/or by it, divided by the acceleration (a) > that it causes (f/a); which is equal to the _Constant_ ratio > of the weight (w) of the mass, divided by the acceleration > (g) at which it will free fall; at the location where it is > weighed. > The unit of mass in the metric system is ONE kilogram, and > is constant, in that the net force (f) varies in proportion > to the acceleration due to gravity; which varies at various > locations. > The unit of mass in the customary system is ONE slug, and is > constant, in that the net force (f) varies in proportion to > the acceleration due to gravity; which varies at various > locations. Please let us know why we should give you any credence at all, when you are capable of writing nonsense such as the following return to Aristotolean theory: One pound of force will accelerate any body weighing one pound at one foot per sec, per second; 32# of force will accelerate any body weighing 32# at 32 feet per second, per second, and/or 32.174# of force will accelerate one slug of matter at 32.174 feet per second, per second. Jerry === Subject: Re: What is the measure of a body's mass? > The measure of a body's mass is a measure of it's inertia: > A unit of mass is the _Constant_ ratio of the net force (f) > exerted on and/or by it, divided by the acceleration (a) > that it causes (f/a); which is equal to the _Constant_ ratio > of the weight (w) of the mass, divided by the acceleration > (g) at which it will free fall; at the location where it is > weighed. > The unit of mass in the metric system is ONE kilogram, and > is constant, in that the net force (f) varies in proportion > to the acceleration due to gravity; which varies at various > locations. > The unit of mass in the customary system is ONE slug, and is > constant, in that the net force (f) varies in proportion to > the acceleration due to gravity; which varies at various > locations. > Please let us know why we should give you any credence at all, > when you are capable of writing nonsense such as the following > return to Aristotolean theory: > One pound of force will accelerate any body weighing one pound > at one foot per sec, per second; 32# of force will accelerate > any body weighing 32# at 32 feet per second, per second, and/or > 32.174# of force will accelerate one slug of matter at 32.174 > feet per second, per second. > Jerry Webster's says: 7 : the gravitational unit of mass in the foot-pound-second system to which a pound force can impart an acceleration of one foot per second per second and which is equal to the mass of an object weighing 32 pounds . Now I ask you, if this is the case, that a pound force can impart a slug with an acceleration of one foot per second per second, what is the reason that a slug is equal to the mass of an object weighing 32 pounds? Don === Subject: Re: What is the measure of a body's mass? > Webster's says: 7 : the gravitational unit of mass in the > foot-pound-second system to which a pound force can impart an > acceleration of one foot per second per second and which is equal to > the mass of an object weighing 32 pounds . You use Webster's as a physics textbook? [...] -Mark Martin === Subject: Re: What is the measure of a body's mass? > Webster's says: 7 : the gravitational unit of mass in the > foot-pound-second system to which a pound force can impart an > acceleration of one foot per second per second and which is equal to > the mass of an object weighing 32 pounds . > You use Webster's as a physics textbook? [...] > -Mark Martin Sometimes; especially when ordinary physics texts don't have the answers. What other text is reviewed by my peers? Don === Subject: Re: What is the measure of a body's mass? > Sometimes; especially when ordinary physics texts don't have the > answers. What other text is reviewed by my peers? So the way you see it, Webster's physics words & definitions are peer reviewed by people like you, and therefore it becomes a repository of accurate physical knowledge and insight. Yes or no? -Mark Martin === Subject: Re: What is the measure of a body's mass? > Sometimes; especially when ordinary physics texts don't have the > answers. What other text is reviewed by my peers? > So the way you see it, Webster's physics words & definitions are > peer reviewed by people like you, and therefore it becomes a > repository of accurate physical knowledge and insight. Yes or no? No Mark, like my peers and me, Webster's too is fraught with inconsistencies, and errors so it isn't always right: A slug isn't the same weight as a body with a mass equal to 32 pounds. A slug is the mathematical ratio of the force exerted on and/or by a body, divided by the acceleration that it causes; that is the only way a slug carries _any_ weight. The measure of mass is inertia! Don > -Mark Martin === Subject: Math help I have to give a counterexample to prove the following statement false... for all integers a and n, if a|n (n is squared) example a = 4 n=5 which is 4|25 | means divides.. like 4|20 = 5.. 4 divides 20 5 times. also a < or equal to n.. then a|n I tried a=2 and n=4 i get 2|16 but 2|4 works so its not a counterexample i have tried forever and cant find any counterexample!! this is hurting my brain.. any of you people smart enough to come up with 1 counterexample? === Subject: Re: Math help > I have to give a counterexample to prove the following statement false... > for all integers a and n, if a|n (n is squared) example > a = 4 n=5 which is 4|25 > | means divides.. like 4|20 = 5.. 4 divides 20 5 times. > also a < or equal to n.. then a|n > I tried a=2 and n=4 i get 2|16 > but 2|4 works so its not a counterexample How do you propose that a number would not be divisible by something its square is divisible by? if a1...an are the prime factors of n then: n = a1*a2*a3*...*an n^2 = (a1*a2*a3*...*an)*(a1*a2*a3*...*an) just pick a=a_i*a_j so that n^2 is divisible by a but n is not. i.e. n^2 = (2*3*7)*(2*3*7) = 1764 Then you might choose any combination of these factors for n except 2, 3, 7, 2*3, 2*7, 3*2, 3*7, 7*2, 7*3, or 2*3*7 to form your counterexample -Mysid === Subject: Re: Math help Thanx Tomek.. that is a counterexample and I just cant understand why I didnt think of that one... but Mysidia how do 2 and 7 work? === Subject: Re: Math help <20487666.1116557313968.JavaMail.jakarta@nitrogen.mathforum.org> > Thanx Tomek.. that is a counterexample and I just cant understand why I didnt think of that one... but Mysidia how do 2 and 7 work? 2*2 = 4 3*3 = 9 So look at... 4|(42^2) = 4|1764 4|42 9|1764 9|42 -Mysid === Subject: Re: Math help k thanx for ur help guys! === Subject: Re: Math help > I have to give a counterexample to prove the following statement false... > for all integers a and n, if a|n (n is squared) example > a = 4 n=5 which is 4|25 > | means divides.. like 4|20 = 5.. 4 divides 20 5 times. > also a < or equal to n.. then a|n > I tried a=2 and n=4 i get 2|16 > but 2|4 works so its not a counterexample > i have tried forever and cant find any counterexample!! this is hurting my brain.. any of you people smart enough to come up with 1 counterexample? 4|100 but not 4|10. === Subject: Re: The newton is the metric unit of force > The newton (N) is the metric unit of force, but most people find it too > small, and prefer to use kilos and kilograms. Please let us know why we should give you any credence at all, when you are capable of writing nonsense such as the following: No Jerry. One pound of force will accelerate any body weighing one pound at one foot per sec, per second; 32# of force will accelerate any body weighing 32# at 32 feet per second, per second, and/or 32.174# of force will accelerate one slug of matter at 32.174 feet per second, per second. === Subject: Re: The newton is the metric unit of force newton is metric unit for . force is potential distance. as long as the body has force it'll distance away from the equilibrium point. the metric unit for force is more likely to be meter as it is the case with the distance. www.geocities.com/dedanoe the physics they won't tell you about === Subject: Re: The newton is the metric unit of force <428bb7c3$1_1@spool9-west.superfeed.net> >newton is metric unit for . force is potential distance. as long >as the body has force it'll distance away from the equilibrium point. >the metric unit for force is more likely to be meter as it is the case >with the distance. In the English language a sentence always begins with an UPPERCASE letter (as do proper nouns). This is the txt generation gone mad. You've obviously got a U.S.A spell-checker. How about installing a grammar checker? Are you trying to create a content-free post? -- Jeremy Boden === Subject: Re: The newton is the metric unit of force <428bb7c3$1_1@spool9-west.superfeed.net> > Are you trying to create a content-free post? Not just a content-free post, but also a content-free website. Side Note: He's accusing Newton of plagiarism, but the logo at the end of this website (The Cross of DedaNoe's Dynamic Lever) is clearly a rip-off from the movie Triple X and err, well, dunno, but this looks pretty close to a swastika to me. Good luck, A. Friend === Subject: Re: A CLASS OF NUMBERS THAT YOU CAN SHUFFLE THE DIAGONAL! <7uckl2-1cd.ln1@sirius.athghost7038suus.net> <4287de8d$0$550$ed2619ec@ptn-nntp-reader03.plus.net> <428be3fb$0$26123$ed2619ec@ptn-nntp-reader02.plus.net> what happens when you run the algorithm on a real computer? (say its multitasking and you can examine B while it runs). Herc === Subject: Re: A CLASS OF NUMBERS THAT YOU CAN SHUFFLE THE DIAGONAL! <7uckl2-1cd.ln1@sirius.athghost7038suus.net> <4287de8d$0$550$ed2619ec@ptn-nntp-reader03.plus.net> <428be3fb$0$26123$ed2619ec@ptn-nntp-reader02.plus.net> What about these 2? A = < 1, 2, 3, 4, 5, 6, 7, 8, 9 > B = <> START B = B U INT(RND*9)+1 IF SIZE(B)<9 GOTO START Herc === Subject: Re: A CLASS OF NUMBERS THAT YOU CAN SHUFFLE THE DIAGONAL! > Mostly right, I had done the same analysis of finite swaps but didn't > pursue it. > You missed the case with oo swaps AND L = L'. Your example relied on a > diagonal dependant to the list. > scrambled with random shuffling while the set of elements is the > same. > QED > Its a trivial result, you just gave the same rebuttal to the 1st > shuffling algorithm on the L last week, again to this shuffle but on > the diagonal itself. > I agree the addendum is an oversimplification which your point shows, > but you're really just relying on the fact > 3) you haven't proved anything about random shuffling *yet* > YET! > All you've said here is 'Complete the Proof!. YES YES. Go on - COMPLETE THE PROOF!... :-) > All I'm saying is ANSWER THE RELEVANT POST > RE Are these 2 sets equal. > A = < 1, 3, 5, 7, 9, 11, ...> > B = <> > n = 1 > START > If (rnd<0.5) > B = B U A[n] > n++ > endif > GOTO START I don't consider this THE RELEVANT POST. This is not COMPLETING THE INSTEAD OF COMPLETING THE PROOF! I.e. for me, the relevent post will be the one in which you fill in the missing bits of your proof... I'm curious as to whether you know the answers to the questions you ask - if not then I don't see how you have any hope of completing your proof... If you do know the answers to your questions, then presumably you must be in a position to complete your proof right now, and you should do exactly this - then I'll comment at that point. Maybe all these questions of yours help you towards your final proof, which is fine I guess, but I think I'll just wait until a definite proof comes along from you... (Or maybe one of your questions will intrigue me and I'll jump in despite myself ;-) Mike. === Subject: Re: A CLASS OF NUMBERS THAT YOU CAN SHUFFLE THE DIAGONAL! <7uckl2-1cd.ln1@sirius.athghost7038suus.net> <4287de8d$0$550$ed2619ec@ptn-nntp-reader03.plus.net> <428be3fb$0$26123$ed2619ec@ptn-nntp-reader02.plus.net> <428d2278$0$39063$ed2e19e4@ptn-nntp-reader04.plus.net> but its EXACTLY the same algorithm I'm using to shuffle. I'll outline the transformation between the two so you can see it but there's no point if you can't even analyse an isolated algorithm. I think you're relying on excuses from here on, I'm aware that you won't understand the proof at any point in future so I see no point in continuing if you're the only one who can follow it this far and you have no knowledge of complexity analysis. Herc : to complete the proof someone must answer this question. Trev : complain complain excuse excuse. Really, if people didn't abort threads and ignore points that are put forth every time it looked like the proof was going against them, I'd have proven God way back on 02 02 2002. Herc === Subject: GG 'ANSEWR to ANZAURRES About Fermat Last theorem Proof Here it is the proof of FLT for n=prime>2by using n=5 as an exemple The proof is written to the level which you asked me to do. > In fact my first Intention in the What the modern > mathematicien are shy > to talk about postings was to show only that exist > a elementary proof of > FLT for n=4*+1 to make uneasy the modern > mathematicien > when they have to answer the questionsIs there a > general > elementary proof of FLT?and as I did with the > e question > ?Why Euler did not find a proof of Pell's equation? I am going to show proof of FLT for n=5 which is the same as the proof for n=prime>3 X^5+Y^5=Z^5 wher X,Y,Z are integers and Z=even X+Y-Z=B X=B+Q Y=B+P Z=B+Q+P=r*u*z where r =1 or r=5 if Z is divisible by 5 X+Y=2*B+Q+P=s*u^5 where s =1 or s=5^4 if Z is divisible by n= 5 When we do X+Y-Z we get that B=r*b*u Therfore (Q+P)=2*r*u*k Therfore we see that z=-b+v*u^(n-1) where v=1 or v=n^(n-2) when Z is divisible by n=5 OBservation: > X,Y ,Z relative Prime numbers > X+Y=W > X^(n-1)-(X^(n-2))*Y +(X^(n-3))*Y^2.........+Y^(n-1)= > =R= W^(n-1)-(Cn1)*[W^(n-2)]*Y+(Cn2)*[W^(n-3]*Y^2- > -(Cn3)*[W^(n-4)]*Y^3+ > +(Cn4)*[W^(n-5)]*Y^4-..............-(Cn2)*[W^1)*Y^(n-2 ) + > +(Cn1)*Y^(n-1) > where > Cnk=[n*(n-1)*(n-2)*(n-3)*.........*(n-k+1)]/2*3*4*.... > ........*k > If (X+Y) is divisible by n, you can see that R is > divisible by n only. > But (X+Y)*R=Z^n=(u*z*n)^n > Since R is divisible only by n then (X+Y) =[n^(n-1)]]*u^n > and Z=n*u*z > When Z is not divisible by n we see that (X+Y)=W and > R do not have any common divisor . Therfore they are relative prime and must be: > R=z^n and W=X+Y=u^n and Z=u*z The end of Observation. Let'choose Z not divisible by 5 Let's take F=(B+Q)^5+(B+P)^5-(B+Q+P)^5=0 and develope the parantheses: F=B^5 +M+D -(Q+P)^5=0 where M=(C52)*(B^3)*(Q^2+P^2) +(C53)*(B^2)*(Q^3+P^3)+(C54)*B*(Q^4+P^4) . -D=(C52)*(B^3)*(Q+P)^2+(C53)*(B^2)*(Q+P)^3+(C54)*B*(Q+P)^4 T= (B^3)* (C52)*(Q+P)^2 is divisible by 2 Therefore G=B^5+M is divisible by 2 as many times T is divisible by 2. Now T can be written as : T=[(b)^3]*[(k)^2]*(u^5)*5*(n-1)*2 where n=5 Therfore G/u^5 is divisible by 2 as many times 2*(n-1) is divisible by 2 1): Let's take H=(B+Q)^5+(B+P)^5 2).H can be written as :H=(B^5+M) +B^5 +(C51)*(B^4)*(Q+P+2*B-2*B) 3). H=(B^5+M) +[(b)^5]*u^5-n*2*[(b)^5]*(u^5)+V where V=(C51)*[(b)^4]*(u^4)*(u^5) If we divide H in 3). by (2*B+Q+P) we get: H1={[(X+Y)*A +n*X^(n-1)}=[B^5+M]/(X+Y)+b^5-2*n*b^5+V/(u^5 where n=5 [SEE Observation) H2={X+Y)*A +n*X^(n-1) +(b^5)*(2*n-1)=[B^5+M}/(X+Y) +V/(u^5). n=5 We see that V/(u^5) is divisible by 2 as many times u^4 is divisible by 2. We see that [B^5+M]/(u^5 is divisible by 2 as many times as 2*(n-1) is divisible by 2. We know n=5 Therefore H2 must be divisible by 2 as many times 2*(n-1) is divisible by 2 . We know n=5 as many times as 2*n-1 is divisible by 2 . (n=5) 4) Therfore from L we get that D=n*X^(n-1) +b^5 is divisible by 2 as many times as 2*(n-1) is divisible by 2. But H1=(X^n+Y^n)/(X+Y=z^n=[-b+u^(n-1)]^n where n=5 Therfore H1=(X+Y)*A +n*X^(n-1)+b^5=C*u^(n-1) where C=odd Therfore D =n*X^(n-1) +b^5 must be divisible by 2 as many times as u^(n-1) is divisible by 2 Therfore 2*(n-1) is divisible by 2 as many times as u^(n-1); n=5 That is impossible.(possible only for n=1) The same proof can be used to prove the case Z divisible by n Therfore FLT for n=5 is impossible Created by George Ghiata === Subject: Re: GG 'ANSEWR to ANZAURRES About Fermat Last theorem Proof correction: In fact from the proof we can get that is it good for n=3 too besides n=prime>3 george ghiata > Here it is the proof of FLT for n=prime>2by using > n=5 as an exemple > The proof is written to the level which you asked me > to do. > In fact my first Intention in the What the modern > mathematicien are shy > to talk about postings was to show only that > exist > a elementary proof of > FLT for n=4*+1 to make uneasy the modern > mathematicien > when they have to answer the questionsIs there a > general > elementary proof of FLT?and as I did with the > e question > ?Why Euler did not find a proof of Pell's > equation? > I am going to show proof of FLT for n=5 which is the > same as the proof > for n=prime>3 > X^5+Y^5=Z^5 wher X,Y,Z are integers and Z=even > X+Y-Z=B > X=B+Q > Y=B+P > Z=B+Q+P=r*u*z where r =1 or r=5 if Z is divisible by > X+Y=2*B+Q+P=s*u^5 where s =1 or s=5^4 if Z is > is divisible by n= 5 > When we do X+Y-Z we get that B=r*b*u > Therfore (Q+P)=2*r*u*k > Therfore we see that z=-b+v*u^(n-1) where v=1 or > r v=n^(n-2) > when Z is divisible by n=5 > > OBservation: > X,Y ,Z relative Prime numbers > X+Y=W > X^(n-1)-(X^(n-2))*Y > +(X^(n-3))*Y^2.........+Y^(n-1)= > =R= W^(n-1)-(Cn1)*[W^(n-2)]*Y+(Cn2)*[W^(n-3]*Y^2- > -(Cn3)*[W^(n-4)]*Y^3+ > +(Cn4)*[W^(n-5)]*Y^4-..............-(Cn2)*[W^1)*Y^(n-2 > ) + > +(Cn1)*Y^(n-1) > where > Cnk=[n*(n-1)*(n-2)*(n-3)*.........*(n-k+1)]/2*3*4*.... > ........*k > If (X+Y) is divisible by n, you can see that R is > divisible by n only. > But (X+Y)*R=Z^n=(u*z*n)^n > Since R is divisible only by n then (X+Y) > =[n^(n-1)]]*u^n > and Z=n*u*z > When Z is not divisible by n we see that (X+Y)=W > and > R do not have any common divisor . > Therfore they are relative prime and must be: > R=z^n and W=X+Y=u^n and Z=u*z > The end of Observation. > Let'choose Z not divisible by 5 > Let's take F=(B+Q)^5+(B+P)^5-(B+Q+P)^5=0 > and develope the parantheses: > F=B^5 +M+D -(Q+P)^5=0 where > M=(C52)*(B^3)*(Q^2+P^2) > +(C53)*(B^2)*(Q^3+P^3)+(C54)*B*(Q^4+P^4) . > -D=(C52)*(B^3)*(Q+P)^2+(C53)*(B^2)*(Q+P)^3+(C54)*B*(Q+ > P)^4 > times as > T= (B^3)* (C52)*(Q+P)^2 is divisible by 2 > Therefore G=B^5+M is divisible by 2 as many times T > is divisible by 2. > Now T can be written as : > T=[(b)^3]*[(k)^2]*(u^5)*5*(n-1)*2 where n=5 > Therfore G/u^5 is divisible by 2 as many times > 2*(n-1) is > divisible by 2 > 1): Let's take H=(B+Q)^5+(B+P)^5 > 2).H can be written as :H=(B^5+M) +B^5 > +(C51)*(B^4)*(Q+P+2*B-2*B) > 3). H=(B^5+M) > 3). H=(B^5+M) > 3). H=(B^5+M) +[(b)^5]*u^5-n*2*[(b)^5]*(u^5)+V > > where V=(C51)*[(b)^4]*(u^4)*(u^5) > If we divide H in 3). by (2*B+Q+P) we get: > H1={[(X+Y)*A > *A +n*X^(n-1)}=[B^5+M]/(X+Y)+b^5-2*n*b^5+V/(u^5 > where n=5 [SEE Observation) > H2={X+Y)*A +n*X^(n-1) > n-1) +(b^5)*(2*n-1)=[B^5+M}/(X+Y) +V/(u^5). > n=5 > We see that V/(u^5) is divisible by 2 as many times > u^4 is > divisible by 2. > We see that [B^5+M]/(u^5 is divisible by 2 as many > times as 2*(n-1) > is divisible by 2. We know n=5 > Therefore H2 must be divisible by 2 as many times > 2*(n-1) > is divisible by 2 . We know n=5 > L=n*X^(n-1)+(b^5)*(2*n-1) is divisible by 2 > as many times as 2*n-1 is divisible by 2 . (n=5) > 4) Therfore from L we get that D=n*X^(n-1) +b^5 is > divisible by 2 > as many times as 2*(n-1) is divisible by 2. > But H1=(X^n+Y^n)/(X+Y=z^n=[-b+u^(n-1)]^n where n=5 > Therfore H1=(X+Y)*A +n*X^(n-1)+b^5=C*u^(n-1) where > C=odd > Therfore D =n*X^(n-1) +b^5 must be divisible by 2 as > many times > as u^(n-1) is divisible by 2 > Therfore 2*(n-1) is divisible by 2 as many times as > u^(n-1); n=5 > That is impossible.(possible only for n=1) > The same proof can be used to prove the case Z > divisible by n > Therfore FLT for n=5 is impossible > Created by George Ghiata === Subject: Re: GG 'ANSEWR to ANZAURRES About Fermat Last theorem Proof Corection: 1).M includes Q^n+P^n TOO. (n=5) 2).The proof is good for n=prime >3 > Here it is the proof of FLT for n=prime>2by using > n=5 as an exemple > The proof is written to the level which you asked me > to do. > In fact my first Intention in the What the modern > mathematicien are shy > to talk about postings was to show only that > exist > a elementary proof of > FLT for n=4*+1 to make uneasy the modern > mathematicien > when they have to answer the questionsIs there a > general > elementary proof of FLT?and as I did with the > e question > ?Why Euler did not find a proof of Pell's > equation? > I am going to show proof of FLT for n=5 which is the > same as the proof > for n=prime>3 > X^5+Y^5=Z^5 wher X,Y,Z are integers and Z=even > X+Y-Z=B > X=B+Q > Y=B+P > Z=B+Q+P=r*u*z where r =1 or r=5 if Z is divisible by > X+Y=2*B+Q+P=s*u^5 where s =1 or s=5^4 if Z is > is divisible by n= 5 > When we do X+Y-Z we get that B=r*b*u > Therfore (Q+P)=2*r*u*k > Therfore we see that z=-b+v*u^(n-1) where v=1 or > r v=n^(n-2) > when Z is divisible by n=5 > > OBservation: > X,Y ,Z relative Prime numbers > X+Y=W > X^(n-1)-(X^(n-2))*Y > +(X^(n-3))*Y^2.........+Y^(n-1)= > =R= W^(n-1)-(Cn1)*[W^(n-2)]*Y+(Cn2)*[W^(n-3]*Y^2- > -(Cn3)*[W^(n-4)]*Y^3+ > +(Cn4)*[W^(n-5)]*Y^4-..............-(Cn2)*[W^1)*Y^(n-2 > ) + > +(Cn1)*Y^(n-1) > where > Cnk=[n*(n-1)*(n-2)*(n-3)*.........*(n-k+1)]/2*3*4*.... > ........*k > If (X+Y) is divisible by n, you can see that R is > divisible by n only. > But (X+Y)*R=Z^n=(u*z*n)^n > Since R is divisible only by n then (X+Y) > =[n^(n-1)]]*u^n > and Z=n*u*z > When Z is not divisible by n we see that (X+Y)=W > and > R do not have any common divisor . > Therfore they are relative prime and must be: > R=z^n and W=X+Y=u^n and Z=u*z > The end of Observation. > Let'choose Z not divisible by 5 > Let's take F=(B+Q)^5+(B+P)^5-(B+Q+P)^5=0 > and develope the parantheses: > F=B^5 +M+D -(Q+P)^5=0 where > M=(C52)*(B^3)*(Q^2+P^2) > +(C53)*(B^2)*(Q^3+P^3)+(C54)*B*(Q^4+P^4) . > -D=(C52)*(B^3)*(Q+P)^2+(C53)*(B^2)*(Q+P)^3+(C54)*B*(Q+ > P)^4 > times as > T= (B^3)* (C52)*(Q+P)^2 is divisible by 2 > Therefore G=B^5+M is divisible by 2 as many times T > is divisible by 2. > Now T can be written as : > T=[(b)^3]*[(k)^2]*(u^5)*5*(n-1)*2 where n=5 > Therfore G/u^5 is divisible by 2 as many times > 2*(n-1) is > divisible by 2 > 1): Let's take H=(B+Q)^5+(B+P)^5 > 2).H can be written as :H=(B^5+M) +B^5 > +(C51)*(B^4)*(Q+P+2*B-2*B) > 3). H=(B^5+M) > 3). H=(B^5+M) > 3). H=(B^5+M) +[(b)^5]*u^5-n*2*[(b)^5]*(u^5)+V > > where V=(C51)*[(b)^4]*(u^4)*(u^5) > If we divide H in 3). by (2*B+Q+P) we get: > H1={[(X+Y)*A > *A +n*X^(n-1)}=[B^5+M]/(X+Y)+b^5-2*n*b^5+V/(u^5 > where n=5 [SEE Observation) > H2={X+Y)*A +n*X^(n-1) > n-1) +(b^5)*(2*n-1)=[B^5+M}/(X+Y) +V/(u^5). > n=5 > We see that V/(u^5) is divisible by 2 as many times > u^4 is > divisible by 2. > We see that [B^5+M]/(u^5 is divisible by 2 as many > times as 2*(n-1) > is divisible by 2. We know n=5 > Therefore H2 must be divisible by 2 as many times > 2*(n-1) > is divisible by 2 . We know n=5 > L=n*X^(n-1)+(b^5)*(2*n-1) is divisible by 2 > as many times as 2*n-1 is divisible by 2 . (n=5) > 4) Therfore from L we get that D=n*X^(n-1) +b^5 is > divisible by 2 > as many times as 2*(n-1) is divisible by 2. > But H1=(X^n+Y^n)/(X+Y=z^n=[-b+u^(n-1)]^n where n=5 > Therfore H1=(X+Y)*A +n*X^(n-1)+b^5=C*u^(n-1) where > C=odd > Therfore D =n*X^(n-1) +b^5 must be divisible by 2 as > many times > as u^(n-1) is divisible by 2 > Therfore 2*(n-1) is divisible by 2 as many times as > u^(n-1); n=5 > That is impossible.(possible only for n=1) > The same proof can be used to prove the case Z > divisible by n > Therfore FLT for n=5 is impossible > Created by George Ghiata === Subject: Re: GG 'ANSEWR to ANZAURRES About Fermat Last theorem Proof Hello to ANZAURREs, ARe you happy About this new presentation?What about the Content? Please I want to hear some positive thinking from you. Do not fall Silent. Best wishes george ghiata > Here it is the proof of FLT for n=prime>2by using > n=5 as an exemple > The proof is written to the level which you asked me > to do. > In fact my first Intention in the What the modern > mathematicien are shy > to talk about postings was to show only that > exist > a elementary proof of > FLT for n=4*+1 to make uneasy the modern > mathematicien > when they have to answer the questionsIs there a > general > elementary proof of FLT?and as I did with the > e question > ?Why Euler did not find a proof of Pell's > equation? > I am going to show proof of FLT for n=5 which is the > same as the proof > for n=prime>3 > X^5+Y^5=Z^5 wher X,Y,Z are integers and Z=even > X+Y-Z=B > X=B+Q > Y=B+P > Z=B+Q+P=r*u*z where r =1 or r=5 if Z is divisible by > X+Y=2*B+Q+P=s*u^5 where s =1 or s=5^4 if Z is > is divisible by n= 5 > When we do X+Y-Z we get that B=r*b*u > Therfore (Q+P)=2*r*u*k > Therfore we see that z=-b+v*u^(n-1) where v=1 or > r v=n^(n-2) > when Z is divisible by n=5 > > OBservation: > X,Y ,Z relative Prime numbers > X+Y=W > X^(n-1)-(X^(n-2))*Y > +(X^(n-3))*Y^2.........+Y^(n-1)= > =R= W^(n-1)-(Cn1)*[W^(n-2)]*Y+(Cn2)*[W^(n-3]*Y^2- > -(Cn3)*[W^(n-4)]*Y^3+ > +(Cn4)*[W^(n-5)]*Y^4-..............-(Cn2)*[W^1)*Y^(n-2 > ) + > +(Cn1)*Y^(n-1) > where > Cnk=[n*(n-1)*(n-2)*(n-3)*.........*(n-k+1)]/2*3*4*.... > ........*k > If (X+Y) is divisible by n, you can see that R is > divisible by n only. > But (X+Y)*R=Z^n=(u*z*n)^n > Since R is divisible only by n then (X+Y) > =[n^(n-1)]]*u^n > and Z=n*u*z > When Z is not divisible by n we see that (X+Y)=W > and > R do not have any common divisor . > Therfore they are relative prime and must be: > R=z^n and W=X+Y=u^n and Z=u*z > The end of Observation. > Let'choose Z not divisible by 5 > Let's take F=(B+Q)^5+(B+P)^5-(B+Q+P)^5=0 > and develope the parantheses: > F=B^5 +M+D -(Q+P)^5=0 where > M=(C52)*(B^3)*(Q^2+P^2) > +(C53)*(B^2)*(Q^3+P^3)+(C54)*B*(Q^4+P^4) . > -D=(C52)*(B^3)*(Q+P)^2+(C53)*(B^2)*(Q+P)^3+(C54)*B*(Q+ > P)^4 > times as > T= (B^3)* (C52)*(Q+P)^2 is divisible by 2 > Therefore G=B^5+M is divisible by 2 as many times T > is divisible by 2. > Now T can be written as : > T=[(b)^3]*[(k)^2]*(u^5)*5*(n-1)*2 where n=5 > Therfore G/u^5 is divisible by 2 as many times > 2*(n-1) is > divisible by 2 > 1): Let's take H=(B+Q)^5+(B+P)^5 > 2).H can be written as :H=(B^5+M) +B^5 > +(C51)*(B^4)*(Q+P+2*B-2*B) > 3). H=(B^5+M) > 3). H=(B^5+M) > 3). H=(B^5+M) +[(b)^5]*u^5-n*2*[(b)^5]*(u^5)+V > > where V=(C51)*[(b)^4]*(u^4)*(u^5) > If we divide H in 3). by (2*B+Q+P) we get: > H1={[(X+Y)*A > *A +n*X^(n-1)}=[B^5+M]/(X+Y)+b^5-2*n*b^5+V/(u^5 > where n=5 [SEE Observation) > H2={X+Y)*A +n*X^(n-1) > n-1) +(b^5)*(2*n-1)=[B^5+M}/(X+Y) +V/(u^5). > n=5 > We see that V/(u^5) is divisible by 2 as many times > u^4 is > divisible by 2. > We see that [B^5+M]/(u^5 is divisible by 2 as many > times as 2*(n-1) > is divisible by 2. We know n=5 > Therefore H2 must be divisible by 2 as many times > 2*(n-1) > is divisible by 2 . We know n=5 > L=n*X^(n-1)+(b^5)*(2*n-1) is divisible by 2 > as many times as 2*n-1 is divisible by 2 . (n=5) > 4) Therfore from L we get that D=n*X^(n-1) +b^5 is > divisible by 2 > as many times as 2*(n-1) is divisible by 2. > But H1=(X^n+Y^n)/(X+Y=z^n=[-b+u^(n-1)]^n where n=5 > Therfore H1=(X+Y)*A +n*X^(n-1)+b^5=C*u^(n-1) where > C=odd > Therfore D =n*X^(n-1) +b^5 must be divisible by 2 as > many times > as u^(n-1) is divisible by 2 > Therfore 2*(n-1) is divisible by 2 as many times as > u^(n-1); n=5 > That is impossible.(possible only for n=1) > The same proof can be used to prove the case Z > divisible by n > Therfore FLT for n=5 is impossible > Created by George Ghiata