mm-3889 === Subject: Parallel_execution_of_an_integral_approach_of_the_passage_to_ t he_security_derivative_that_it_fixes_the_price_Ltd_world_of_supercomputers_o f _the_qu=E1drico_we_we_use_an_integral_approach_of_the_passage?= www.Sworde.com Parallel execution of an integral approach of the passage to the security derivative that it fixes the price Ltd world of supercomputers of the qu.87drico we we use an integral approach of the passage deciding the underlying random equations shape financier to fix the price derivative of the security. The integral methodology of the passage is known well in the field of mechanics of quantum and defines values of the expectation as integral exceeding functionaries or histories of the additions (passages) of quantum or the random dynamic system. Such integrals are a limit of the sequence of the finite-dimensional multiple integrals gotten discretizing the interval of the time under the consideration, that corresponds to the time to the expiration in the box of derivatives financial. Essentially, a continuous random process is specified by its density of the functional probability better that for its law of the evolution (a random distinguishing equation, SDE). The formula of Feynman-Kac guarantees the equivalence of the two formulations. Known mount numerical Carlo well (MC) and methods quasi of Monte Carlo (QMC) for the calculation of conditional values of the expectation in random processes is seen as devices normally generating appropriate random passages of the excess of the averages. In the structure of the integral approach of the passage discretized of the density of the probability is needed to generate appropriate functions multivariate. When MC and QMC will be essentially the only existing numerical methods for the raised dimensional problems (that they correspond to derivatives with many seguran.8das underlying), suffer from the slow properties of the convergence. In the example of low dimensional problems many techniques are available, based in integrating for differencies finite partial the distinguishing equation (PDE) that it corresponds to the SDE in the hand or a simplified dr.88stica assumption in the probabilities of the transistion (that is methods binomial). We use an alternative method for the numerical computation of formulation integral of the passage of the random financial problem that is elegant, easy to extend for options passage-dependents and Americans, amenable to a parallel execution and with good properties of the convergence and the stability. The numerical method deterministic of the green function (GFDNM) trusts approaching the probabilities of the transistion for the stages discretized of the time, and computing the integrals for the numerical quadrature standard on a grating discretized. In the fact, the probability of the transistion represents the green function of the PDE that corresponds to the underlying SDE the evolution of the financial security. The conditional values of the expectation are simply products of the vector of payoff for one determined number of matrices of the transistion. www.Sworde.com === Subject: giving applied math a bad name http://www.sundaytimes.news.com.au/common/story_page/0,7034,18752304%5E95 0,00.html FEW women would claim to have the perfect bottom. But for those in need of reassurance that it is within reach, a scientist has come to the rescue by working out a mathematical formula they believe adds up to the perfect posterior. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: giving applied math a bad name > http://www.sundaytimes.news.com.au/common/story_page/0,7034,18752304%5E95 > 0,00.html > FEW women would claim to have the perfect bottom. But for those in need > of reassurance that it is within reach, a scientist has come to the > rescue by working out a mathematical formula they believe adds up to > the perfect posterior. Laughable? I'm volunteering to conduct first hand research testing the firmness factor. I think this formula has the potential to revolutionize, er, something-or-other. More so than JSH's prime counting function at any rate. === Subject: Re: giving applied math a bad name 0,00.html > FEW women would claim to have the perfect bottom. But for those in need > of reassurance that it is within reach, a scientist has come to the > rescue by working out a mathematical formula they believe adds up to > the perfect posterior. > Laughable? I'm volunteering to conduct first hand research testing the > firmness factor. > I think this formula has the potential to revolutionize, er, > something-or-other. Revolutinize applied mathematics! Kind of like the Monty Python joke about taxing ... er, how do I say it in this forum? ... thingy ... Number twos? No. You know ... thingy. Well, it would sure make chartered accountancy more interesting. --- Christopher Heckman (Going by memory here.) === Subject: Re: giving applied math a bad name > FEW women would claim to have the perfect bottom. But for those in need > of reassurance that it is within reach, a scientist has come to the > rescue by working out a mathematical formula they believe adds up to > the perfect posterior. Note: The above is not actually a quotation from Gerry! This should fix it: http://tinyurl.com/j47lv === Subject: Re: Pythagoras described himself as a philosopher and Scientist..... <3524g.8500$j7.302503@news.indigo.ie Pythagoras described himself as a philosopher. He wanted to set up as a > teacher of philosophy, astronomy and mathematics, so began his society. > He founded an ethico-mathematical school following ethical beliefs > and pursuits and worked on mathematics. Many persons were attracted by > the society and joined it. Pythagoras was the head of the society. It > was in fact Pythagorean brotherhood observing some rules and following > a particular way of life. Pythagorean School regarded men and women > equally. They enjoyed a common way of life. The society had an inner > circle and an outer circle of followers. > This sounds quite a sensible religion. > Are there any followers today? > It would be nice to go to a church > where the sermons might be about quadratic number fields. It would be nice to go to a church where no one has been eating beans. > -- > Timothy Murphy > e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie > tel: +353-86-2336090, +353-1-2842366 > s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: Pythagoras described himself as a philosopher and Scientist..... <3524g.8500$j7.302503@news.indigo.ie Pythagoras described himself as a philosopher. He wanted to set up as a > teacher of philosophy, astronomy and mathematics, so began his society. > He founded an ethico-mathematical school following ethical beliefs > and pursuits and worked on mathematics. Many persons were attracted by > the society and joined it. Pythagoras was the head of the society. It > was in fact Pythagorean brotherhood observing some rules and following > a particular way of life. Pythagorean School regarded men and women > equally. They enjoyed a common way of life. The society had an inner > circle and an outer circle of followers. > This sounds quite a sensible religion. > Are there any followers today? > It would be nice to go to a church > where the sermons might be about quadratic number fields. I suspect you didn't read the second paragraph. **from second paragraph** They had to observe top secrecy about every activity and works of the society. They were bound by an oath not to speak of the doctrines and discoveries of their school. Loyalty towards Pythagoreans and their leader was the basic tenet of the Pythagoreans. **end quote** It was a cult. === Subject: Re: Degree of so/unrtedness - please help a programmer > The problem with experimental approach is that it is highly dependent > on a behavior of particular algorithm. [...] Right, but I wouldn't call it a problem with the experimental approach. The problem is whether there *is* a way to measure unsortedness that is appropriate for all algorithms. The approach I described is one way to discover that. === Subject: Re: Degree of so/unrtedness - please help a programmer comes up trumps. Respect, Al. === Subject: Re: Random Matrix Product Expectation > I'm trying to compute E [Z] in the following equation: > E [ Z ] - E [ A Z A' ] = E [ f(A) ] > Hre Z and A are INDEPENDENT random matrices. How can I factor out the > matrix A on the left hand side of the above equation, so i can get an > analytical expression for E [Z}? > Any help or directions is appreciated. For example, if A is a diagonal matrix whose entries are plus or minus 1 with probability 1/2, and Z is a non-random matrix, then E[AZA'] gives you the matrix with the diagonal entries of Z. You can assume without loss of generality that Z is non-random matrix because E[AZA']=E[A(E[Z])A']. So you get sum_{j,k} E[a_{ij}a_{lk}] E[z_{jk}] It depends only upon the means of the entries of Z, but depends on the variances and covariances as well as the means of the entries of A. I guess you could write it as (sum_{j,k} Cov[a_{ij},a_{lk}] E[z_{jk}]) - (E(A) E(Z) E(A'))_{il} === Subject: Re: JSH: Now you can see it happening <445118b9$0$16470$892e7fe2@authen.yellow.readfreenews.net> : There was this stunned pause on the newsgroups for a while, as some >> : posters realized I was right, and a few even admitted it was this neat >> : thing. >> You should understand here that what you consider a stunned pause might >> just be the fact that people here have lives and only get around to >> addressing your posts in their spare time, for a bit of a laugh. >> Justin > James doesn't understand what a life is. He's wasted all of his on useless > crap that won't get him any of the things he thinks he'll get. I can tell > you that Oprah, for one, certainly won't give a crap about his research > whether or not it's correct. And secondly, women won't be impressed with > being famous for being a crank. > Dave > He has figured out the right statements to post to always get a response. > And that is something And readers can look in this very thread to see how it works. I make a major discovery, and then posters start trivializing it. MOST people reasonably suppose that some mathematicians somewhere would seize upon a major discovery, so they start believing that if what they mostly see are negative replies, it must NOT be a major discovery. And then intriguing facts--like my having had a paper published and then sci.math'ers got it yanked with an email campaign--don't matter to them, as people probably keep thinking that some one, somewhere would step in to save the day if I am the one who is right, but no one does. YEARS go by, and I get another result, and the cycle continues. The reason my posts get so much interest is that posters have to keep up the negativity out of fear that if they do not, some people might actually start thinking that what I say is true and important. So my posts usually generate a response. That's why. Years ago, back when I had a lot of failed arguments, I could post, and get not a single response in reply. But these people tracking things today don't have that luxury, so now they reply obsessively to create a hostile atmosphere to my ideas, where naive people trusting that mathematicians as a group could not behave in this way, are pulled to the conclusion that my research is not important. But my research tends to connect in one way or another to factoring. That is significant. James Harris === Subject: Re: JSH: Now you can see it happening >> : There was this stunned pause on the newsgroups for a while, as some > : posters realized I was right, and a few even admitted it was this neat > : thing. >> You should understand here that what you consider a stunned pause might > just be the fact that people here have lives and only get around to > addressing your posts in their spare time, for a bit of a laugh. >> Justin > James doesn't understand what a life is. He's wasted all of his on useless >> crap that won't get him any of the things he thinks he'll get. I can tell >> you that Oprah, for one, certainly won't give a crap about his research >> whether or not it's correct. And secondly, women won't be impressed with >> being famous for being a crank. >> Dave >> He has figured out the right statements to post to always get a response. >> And that is something >And readers can look in this very thread to see how it works. >I make a major discovery, and then posters start trivializing it. >MOST people reasonably suppose that some mathematicians somewhere would >seize upon a major discovery, so they start believing that if what they >mostly see are negative replies, it must NOT be a major discovery. Um. Most people reasonably suppose that when you announce a major discovery and people post proofs more or less immediately then the discovery is probably not so major. >And then intriguing facts--like my having had a paper published and >then sci.math'ers got it yanked with an email campaign--don't matter to >them, as people probably keep thinking that some one, somewhere would >step in to save the day if I am the one who is right, but no one does. >YEARS go by, and I get another result, and the cycle continues. >The reason my posts get so much interest is that posters have to keep >up the negativity out of fear that if they do not, some people might >actually start thinking that what I say is true and important. >So my posts usually generate a response. >That's why. Years ago, back when I had a lot of failed arguments, I >could post, and get not a single response in reply. >But these people tracking things today don't have that luxury, Because you no longer have a lot of failed arguments? That's very funny. Like you're living in a totally different world. >so now >they reply obsessively to create a hostile atmosphere to my ideas, >where naive people trusting that mathematicians as a group could not >behave in this way, are pulled to the conclusion that my research is >not important. >But my research tends to connect in one way or another to factoring. >That is significant. >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Now you can see it happening > I make a major discovery, and then posters start trivializing it. Which discovery are you claiming? === Subject: Re: Fibonacci and the roullette wheel <4277905.1145886917367.JavaMail.jakarta@nitrogen.mathforum.org Using the fibonacci sequence 1,1,2,3,5,8,13,21,34......n, at any given point in the sequence the total bets placed for that sequence is equal to 2 interations along the sequence minus 1. > Therefore in the example above if you had got to betting 8 units when the sequence finally ends you would have bet 20 units in total. If the winning bet pays 2 to 1 plus return of stake then the return is 24 units. > In fact the further along the sequence you go the higher is the total profit when the sequence ends. > Most people though seem to think that the casino always wins because of the 0 or double 0. However these are irrelevant to the sequence since they merely represent a losing bet and you move further along the sequence. > Since the bets would be placed on each of the block bets, thus covering every number other then the 0 or double 0, you would require a long sequence of results landing on 0 or double 0 to prevent a profit. Or alternatively for one of the block bets to run for a very long time before finishing. > As stated in my original question, other than a long losing sequence on one of the block bets, are there any other reasons why this would not work. Lets say that on some particular spinl, you bet 2, 3 & 21 units The odds of any bet winning is 33.33% The expected value is .333*6 +.333*9 + .333*63 = 2 + 3 + 21 which is exactly what you bet! If 0 or 00 never comes , in the long run you will break even! But 0 and 00 always happens! === Subject: Re: Fibonacci and the roullette wheel odds - a block of 12 does pay 2 to 1 but you also get your stake back which makes it the equivaqlent of 3 to 1. > Using the fibonacci sequence 1,1,2,3,5,8,13,21,34......n, at any given point in the sequence the total bets placed for that sequence is equal to 2 interations along the sequence minus 1. > Therefore in the example above if you had got to betting 8 units when the sequence finally ends you would have bet 20 units in total. If the winning bet pays 2 to 1 plus return of stake then the return is 24 units. > In fact the further along the sequence you go the higher is the total profit when the sequence ends. > Most people though seem to think that the casino always wins because of the 0 or double 0. However these are irrelevant to the sequence since they merely represent a losing bet and you move further along the sequence. > Since the bets would be placed on each of the block bets, thus covering every number other then the 0 or double 0, you would require a long sequence of results landing on 0 or double 0 to prevent a profit. Or alternatively for one of the block bets to run for a very long time before finishing. > As stated in my original question, other than a long losing sequence on one of the block bets, are there any other reasons why this would not work. > Lets say that on some particular spinl, you bet 2, 3 & 21 units > The odds of any bet winning is 33.33% The expected value is > .333*6 +.333*9 + .333*63 = 2 + 3 + 21 which is exactly what you > bet! > If 0 or 00 never comes , in the long run you will break even! But 0 > and 00 always > happens! On average, on a fair wheel, 0 and 00 don't always happen. They happen just often enough. === Subject: integrating bessel functions Hi All, I'm running in to an integral which I'm not sure if it can be done analytically viz. int_0^b dx I_0(x) Is it possible to do this integral analytically? If not, does anyone have an idea on how to proceed numerically? (Abramovitz and Stegun has a procedure for computing this using tables, but I was hoping for a quicker and more accurate method) === Subject: Re: integrating bessel functions >Hi All, >I'm running in to an integral which I'm not sure if it can be done >analytically viz. >int_0^b dx I_0(x) >Is it possible to do this integral analytically? If not, does anyone >have an idea on how to proceed numerically? (Abramovitz and Stegun has >a procedure for computing this using tables, but I was hoping for a >quicker and more accurate method) Maple says: b*BesselI(0,b)-1/2*Pi*b*StruveL(0,b)*BesselI(1,b)+1/2*Pi*b*StruveL(1,b)*Bess elI(0,b) Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: passing a dish around problem <9rk642d482np9pglutrcm79i8a180ugm5r@4ax.com> <02pd42dqaq9jlgdhrovcddvug7ek8jdj2l@4ax.com> <29fzHZ6Ym3TEFwH8@212648.invalid> .... >>It would help if I had a better understanding of expected value. To >>me, >>it is the value which gives at least 50& chance of success? In this >>case; E(n) = 1 for all n. Is it possible that your definition of EV is >>significantly different? Because, I cannot see the purpose of E(n)? >> It is, roughly speaking, what you would expect to be the average of the >> values obtained over many repeats of the process in question. Here E(n) >> is the average number of moves for the dish to get from person 0 to >> person 1. You get it by summing the possible values weighted by their >> probabilities. See http://en.wikipedia.org/wiki/Expected_value >I> David Hartley >Let P(k) = probability that p_1 is reached in exactly 'k' flips. Then >for large n, > k P(k) P(k)* k > 1 0.500000 0.5000 > 2 0 > 3 0.125000 0.3750 > 4 0 > 5 0.062500 0.3125 > 6 0 > 7 0.039063 0.2734 > 8 0 > 9 0.019531 0.1758 should be 0.027344 (it's (2^-9)(8!/4!5!) > Sum P(k)=0.7465 > Sum P(k)* k=1.6367 >Suppose that P(k) remains constant for m flips starting with k = 11. But it doesn't! (I assume you mean the next m values of k with non-zero P(k), i.e. 11,13,..., 11 + 2(m-1).) P(k) is non-zero for all odd k so, since sum(k=1 to oo) P(k) = 1, P(k) must tend to 0 as k tends to oo. P(2i+3) = P(2i+1)*(2i+1)/(2i+4) so P(k) (for odd k) is strictly decreasing *until* k gets as large as n. Then paths the long way around the circle become possible. I guess that cuts off more of the paths that would possible on a larger circle than it allows new ones, so p(k) decreases more quickly for k > n. (Enough to make the expectation finite.) >The average P(k) would be 0.2535/m The median k = (m+10). >Then, Sum{P(k)*k} from k=11 to 11+2*(m-1) = (0.2535)*(m+10)/(m) >The limit of (0.2535)*(m+10)/(m) approaches 0.2535 as m increases. >Therefore E(n) = 2 or 3 for n > 11+2*(m-1), m = 3, 4, 5, 6, ... >Note that E(n) decreases as m increases. Extend your table above for a few more values of k and then repeat this calculation, you'll get a different answer. You can't get expected vslues this way. -- David Hartley === Subject: Quadratic residues and factoring Wouldn't you know I go to factoring, yet again, as, for one thing, just about all of my research relates to factoring one way or another, and for another, I think that the factoring problem has a big enough economic importance that I focus on it as a way to break the impasse. This result might be a way to prove Goldbach's Conjecture, but forget that, I prove that and mathematicians can just lie about that result like everything else. So I don't see value in working to prove Goldbach's Conjecture, as it would really depress me to prove it, and still be arguing with sci.math people, like with all my other research results. Mathematicians can ignore a proof of Goldbach's Conjecture, but I get a factoring result, and well... So what is this thing? The result is that given odd naturals summed to get a composite: n_1 + n_2 = C it must be true that, if k is an odd difference of factors of 2*C that 8n_2 + k^2 must be a quadratic residue of p, where p is a prime factor of n_1, or (8n_2 + k^2) mod p, where p is a prime factor of n_1, must be quadratic residue of n_1, or to state it the way I've been told is correct terminology, 8n_2 + k^2 must be a square modulo p, where p is a prime factor of n_1, or (8n_2 + k^2) mod p must be a square modulo p. So how can that be used to factor? Well, if n_1 is a composite with two primes as factors, then maybe you can find their individual quadratic residues. For example, using n_1 = 55, I will let n_2 = 17, so I have 2*C = 144, and picking the factorization 16(9), I have k = 7. Then 8*17 + 49 = 185 and 185 mod 55 = 20. And 20 is 0 mod 5 and 9 mod 11, so they are squares, but if you didn't know that, what would you know? I don't know. You could check the 0 square modulo p possibility by factoring 20, which would give you 5. But hey, you can use another k, so let's use the factorization 48(3), which gives k = 45, then 8*17 + 45^2 = 2161, and 2161 mod 55 = 16. And I don't know what that tells me, if anything. Oh well, I'll try n_2 = 7, which gives C = 62, and 2*C = 124, and I'll use the factorization 31(4), so k = 27, and I have 8*7 + 27^2 = 785, and 785 mod 55 = 15. And I don't know what that gives me either. But, kowing that 11 and 5 are factors, I do see the 0 square modulo 5, and 4 is, of course a square. Maybe I'm using primes that are too small, and I wonder how often the 0 case pops up? Maybe about 2/(p + 1) where p is the smallest prime, so that's no help. So far with these I have residues with C that are bigger than either prime factor, but if you get a residue less than sqrt(C) then maybe that could tell you something? Lots of speculation here, or is there gold in there somewhere? If so, have I just given away a route to factoring composites made up of large primes? Or is this approach completely useless when it comes to practicality? James Harris === Subject: Re: Quadratic residues and factoring [...] I think you're wrong. Your proof relies on the fact that: a*(x+y) = a*x + a*y which needn't be true. Michael === Subject: Re: Quadratic residues and factoring jstevh@msn.com : ... > So I don't see value in working to prove Goldbach's Conjecture, as it > would really depress me to prove it, and still be arguing with sci.math > people, ... You seem to be getting sicker, Harris. Another example: > And 20 is 0 mod 5 and 9 mod 11, so they are squares, but if you didn't > know that, what would you know? > I don't know. You should quit making false claims about what you know and what you can do. > If so, have I just given away a route to factoring composites made up of > large primes? The answer is no, yet I expect you will again claim that you can factor anything whatever. === Subject: Re: Quadratic residues and factoring : for another, I think that the factoring problem has a big enough : economic importance that I focus on it as a way to break the impasse. What impasse? : This result might be a way to prove Goldbach's Conjecture, Okay, you have a result?! Excellent, let's see what you've got. : So what is this thing? Good question! : The result is that given odd naturals summed to get a composite: : n_1 + n_2 = C : it must be true that, if k is an odd difference of factors of 2*C that : 8n_2 + k^2 must be a quadratic residue of p, where p is a prime factor : of n_1, or : (8n_2 + k^2) mod p, where p is a prime factor of n_1, : must be quadratic residue of n_1, or to state it the way I've been told : is correct terminology, : 8n_2 + k^2 must be a square modulo p, where p is a prime factor of n_1, : or : (8n_2 + k^2) mod p must be a square modulo p. : So how can that be used to factor? I'm ready! : Well, if n_1 is a composite with two primes as factors, then maybe you : can find their individual quadratic residues. Oh...maybe? : For example, using n_1 = 55, I will let n_2 = 17, so I have 2*C = 144, : and picking the factorization 16(9), I have k = 7. (snip) : I don't know. Er, okay. That's...nice. (snip) : And I don't know what that gives me either. You don't know much, do you? : So far with these I have residues with C that are bigger than either : prime factor, but if you get a residue less than sqrt(C) then maybe : that could tell you something? It tells me you don't have a damned thing. : Lots of speculation here, or is there gold in there somewhere? Seems like crap to me. : If so, have I just given away a route to factoring composites made up : of large primes? You haven't given away anything, given that you've admitted multiple times in one post that you have no idea what your little fiddling with quadratic residues says about anything. So a prime crops up somewhere...so what? Are you suggesting that your comment makes it easier to see those primes? If so, let's have some evidence. : Or is this approach completely useless when it comes to practicality? What approach? You haven't got anything. Justin === Subject: Re: Quadratic residues and factoring : economic importance that I focus on it as a way to break the impasse. > What impasse? > : This result might be a way to prove Goldbach's Conjecture, > Okay, you have a result?! Excellent, let's see what you've got. > : So what is this thing? > Good question! > : The result is that given odd naturals summed to get a composite: > : n_1 + n_2 = C > : it must be true that, if k is an odd difference of factors of 2*C that > : 8n_2 + k^2 must be a quadratic residue of p, where p is a prime factor > : of n_1, or > : (8n_2 + k^2) mod p, where p is a prime factor of n_1, > : must be quadratic residue of n_1, or to state it the way I've been told > : is correct terminology, > : 8n_2 + k^2 must be a square modulo p, where p is a prime factor of n_1, > : or > : (8n_2 + k^2) mod p must be a square modulo p. > : So how can that be used to factor? > I'm ready! Hint, hint didn't do it for you from the rest of your reply which I deleted. It's like, 20 questions. You can take any composite, and just add it to anything you want, and factor twice the result, plug into the equations and get info about the quadratic residues of the primes that make up that composite. So then, how do you do it exactly? What's the perfect method that would allow someoone to go hog wild all over the world, and break RSA at will? I don't know. I don't know if you can. I'm just tossing out this idea that occurred to me today. But if the 20 questions can be asked right, and if there is a way, and some brilliant person finds it, I guess it'd take about 24 hours for some leakage of that info to come out. If nothing in 24 hours, then I consider the problem more carefully. If I can't do anything with it and decide it's a dead route, I may simply, finally go to my more complicated hyperbolic factoring method. Or, I'll do something else. But I have the power. More power than just about any person on this planet. I hold the world in my hands. James Harris === Subject: Re: Quadratic residues and factoring : But I have the power. More power than just about any person on this : planet. I hold the world in my hands. I figure we're all actually pretty lucky that you're doing math. If, for example, you decided that you know all there is to know about nuclear power, we might really be in for trouble as you'd probably start screwing around with enriched uranium. As it stands, the fact that you're off to the side playing around with basic algebra is really no worry. Justin === Subject: Re: Quadratic residues and factoring ... stuff deleted ... > Or, I'll do something else. > But I have the power. More power than just about any person on this > planet. > I hold the world in my hands. Does that include the itty-bitty babies? What about you and me, brother? > James Harris It's both sad and entertaining to see you acting so delusional. I suppose it's a good thing that you keep this behavior so close to the line, because if it became clear that this was an act it would just be pathetic, yet if it became clear that is was genuine, it would be even more pathetic. Dale === Subject: Re: Quadratic residues and factoring > But I have the power. More power than just about any person on this > planet. > I hold the world in my hands. Guess who forgot to take his pills. :-) Jose Carlos Santos === Subject: Re: Proving a trig identity. >> The preferred techique is to manipulate the two sides in isolation from >> each other, until they are transformed into identical forms. No moving >> items terms from one side to the other, no cross-multiplication >> allowed,,, > There are lots of ways to prove identities; there's no reason to become > dogmatic about it. > Indenties of this kind are often easy to prove by the half-angle > tangent method: > If t = tan(s/2), then sin(s) = 2t/(1+t^2), cos(s) = (1-t^2)/(1+t^2), > tan(s)=2t/(1-t^2) and etc. The rational parametrization of a circle, > showing it is birationally equivalent to a projective line. > In this particular case, the indentity is proven from sec-cos-tan*sin = > (1+t^2)/(1-t^2)-(1-t^2)/(1+t^2)-(2t/(1-t^2)(2t/(1+t^2) = 0. Now that is clever === Subject: Re: with high probability > Sorry, I just realized that my joke was less than clear. Six Sigma > already stands for +- 3 sigma, and I wanted to make fun of Motorola > (and others) using six sigma as a buzzword for doing what engineers > normally do, but on a larger scale. I took the Six Sigma course last year; it's the new thing in management where I work. As I understand it, it really means +- 6 standard deviations, http://www.isixsigma.com/sixsigma/six_sigma.asp To achieve Six Sigma, a process must not produce more than 3.4 defects per million opportunities. If you allow +-3 standard deviations, the acceptable area is .9974, allowing .0026 errors per opportunity. Multiplying by a million allows 2,600 errors per million opportunities, not 3.4. Paul === Subject: Re: with high probability <250420061012114398%anniel@nym.alias.net.invalid> <444E2FBE.9060506@hotmail.com> <8NGdnbE9NotR9MzZnZ2dnUVZ_vmdnZ2d@comcast.com> So my joke was better than I thought. Tell me one thing, though: with 6 sigma, I would get a failure probability of 1.97E-9, the 3.4E-6 would only be 4.6 sigma; so six sigma stands for +- 4.6 sigma? Why is then 3.4 defects per million opportunities used in the six sigma concept? === Subject: Re: Q[sqr p] >>On Wed, 12 Apr 2006 19:11:38 -0700, William Elliot > ... What I don't get is why the multiplicative groups >Q*[sqr 2] and Q*[sqr 3] aren't isomorphic. >>I think they _are_ isomorphic. >So suppose we have 2 algebraic number fields such that: >(1) Both fields have the same roots of 1. >(2) For both fields, the ring of algebraic integers is a UFD. >I claim that the multiplicative groups are isomorphic. >To justify the claim, let the multiplicative groups be G,H >respectively. >Since roots of 1 are the same for both fields, it's immediate that G_1 >and H_1 are isomorphic. >G_u and H_u may or may not be isomorphic but it doesn't matter. >The key is that G_u x G_n and H_u x H_n are both free abelian with a >countably infinite number of generators, hence are isomorphic. >It follows that G and H are isomorphic. >quasi > The interesting question is whether it's possible to find two fields > where the rings of algebraic integers are not UFDs, but for which the > multiplicative groups are isomorphic. > A necessary condition is that they must have the same roots of unity. > If the multiplicative groups are isomorphic, that implies that the > nature of the lack of unique factorization is in some sense the same > for both rings of integers. The in the argument above use of UFD can be replaced by weaker property: rings of algebraic integers are Dedekind rings. In particluar, there is a complete set of valuations, which means that the multiplicative group modulo units embeds into a free abelian group. Since a subgroup of a free abelian group is a free abelian group we see that G/G_u is a countably generated free abelian group. Since G_u is a product of roots of unity with a free group also G is a product of roots of unity with a free group. So the answer to original question is positive: for all positive primes p the mutiplicative groups of Q[sqr p] are isomorphic. -- Waldek Hebisch hebisch@math.uni.wroc.pl === Subject: Re: Q[sqr p] >On Wed, 12 Apr 2006 19:11:38 -0700, William Elliot > ... What I don't get is why the multiplicative groups >>Q*[sqr 2] and Q*[sqr 3] aren't isomorphic. >>I think they _are_ isomorphic. >So suppose we have 2 algebraic number fields such that: >>(1) Both fields have the same roots of 1. >>(2) For both fields, the ring of algebraic integers is a UFD. >>I claim that the multiplicative groups are isomorphic. >>To justify the claim, let the multiplicative groups be G,H >>respectively. >>Since roots of 1 are the same for both fields, it's immediate that G_1 >>and H_1 are isomorphic. >>G_u and H_u may or may not be isomorphic but it doesn't matter. >>The key is that G_u x G_n and H_u x H_n are both free abelian with a >>countably infinite number of generators, hence are isomorphic. >>It follows that G and H are isomorphic. >>quasi >> The interesting question is whether it's possible to find two fields >> where the rings of algebraic integers are not UFDs, but for which the >> multiplicative groups are isomorphic. >> A necessary condition is that they must have the same roots of unity. >> If the multiplicative groups are isomorphic, that implies that the >> nature of the lack of unique factorization is in some sense the same >> for both rings of integers. >The in the argument above use of UFD can be replaced by weaker property: >rings of algebraic integers are Dedekind rings. In particluar, there is >a complete set of valuations, which means that the multiplicative group >modulo units embeds into a free abelian group. Since a subgroup of >a free abelian group is a free abelian group we see that G/G_u is >a countably generated free abelian group. Since G_u is a product of >roots of unity with a free group also G is a product of roots of unity >with a free group. >So the answer to original question is positive: for all positive primes p >the mutiplicative groups of Q[sqr p] are isomorphic. When you write Q[sqr p], do you actually mean Q(sqrt(p)), the field generated by Q and sqrt(p)? Are you then saying that for all primes p, even those for which Z[sqrt(p)] is not a UFD, the multiplicative group of the field Q(sqrt(p)) is isomorphic to Z_2 x (Z x Z x Z x ...)? It seems to me that the lack of unique factorization should prevent G/G_u from being free abelian. quasi === Subject: Re: Numerically solving a double integration... The actual intergration is as follows: int_x_0^t int_y_0^t F(x) exp(-a|x-y|) G(y) dy dx F(x) = (3x/t); 0 < x <(t/3) = (2t-3y)/t; (t/3) < x < (2t/3) = 0; otherwise G(y) = F(y) Where a, t are real numbers. And there is absolute value of x-y, i.e., it is |x-y| in the exponent. If there is some way I can enter this in mathematica, that would also give me an insight into how the integrand is behaving. But, I donot know how to feed the functions of the form F(x), which take different values for different ranges... Regds, Ashesh > Would like to solve a double integration numerically. The function is > similar to: > f(x,y)=x(x+3y) > Can you give an example of the actual function? === Subject: Re: Numerically solving a double integration... Would like to solve a double integration numerically. The function is > similar to: > f(x,y)=x(x+3y) > Can you give an example of the actual function? > The actual intergration is as follows: > int_x_0^t int_y_0^t F(x) exp(-a|x-y|) G(y) dy dx > F(x) = (3x/t); 0 < x <(t/3) > = (2t-3y)/t; (t/3) < x < (2t/3) > = 0; otherwise > G(y) = F(y) > Where a, t are real numbers. And there is absolute value of x-y, i.e., > it is |x-y| in the exponent. > If there is some way I can enter this in mathematica, that would also > give me an insight into how the integrand is behaving. But, I donot > know how to feed the functions of the form F(x), which take different > values for different ranges... Perhaps you meant to have x instead of y in the second line, e.g.: F(x) = (3x/t); 0 < x <(t/3) = (2t-3x)/t; (t/3) < x < (2t/3) = 0; otherwise which implies that t > 0 if this definition is to make sense. You are integrating over a square [0,t]x[0,t], so with a change of variable u = x/t, v = y/t we can rewrite the (double) integral as: t^2 * INTEGRAL exp(-at |u - v|) f(u) f(v) du dv OVER [0,1]x[[0,1] where f(u) = F(x) = F(tu): f(u) = 3u when 0 < u < 1/3 = 2 - 3u when 1/3 < u < 2/3 = 0 otherwise Now since the integrand is identically 0 by this definition unless u,v < 2/3, the integral really only has to be carried out over that region, ie. [0,2/3]x[0,2/3]. The piecewise linear function f(u) on [0,2/3] is continuous, rising from 0 to 1 at the midpoint and then dropping back to 0 at u=2/3. This kind of thing is called a hat function, which sounds better in French. The absolute value in the exponential argument also introduces a piecewise continous aspect to the integrand's definition. It can be written by symmetry as twice the integral over the triangle where u >= v, like your original post's example. If we omit for now the external factor of t^2, we have left: 2 INTEGRAL INTEGRAL exp( -at(u-v) ) f(u) f(v) dv du [0,2/3] [0,u] which then breaks up into three explicitly integrable pieces: 0 < v < u < 1/3 ; 0 < v < 1/3 < u < 2/3 ; 1/3 < v < u < 2/3 Note that these are respectively, a triangle, a rectangle, and a triangle. In fact these integrals are easy enough to do by hand. === Subject: Re: Numerically solving a double integration... Hi Chip, Regs. === Subject: JSH: See what I mean? I was really excited for a while about this new result, and then the sci.math'ers started in, like they've done now for years, and you can see how depressing it all becomes. So there is this relationship between the primes, quadratic residues and factorizations of sums, so what? Mathematicians don't give a damn about the truth, they just want to maintain the status quo. Welcome to the real world. The status quo is that you get a Ph.D in mathematics, play the game, and follow certain highly specific rules to join the club and be allowed to be known for what you produce. Be an outsider like me--especially one who is a harsh critic of todays mathematical community--and they use the most powerful weapon in their arsenal--they ignore you. Oh, I have a more general result now that with odd natural composites n_1 and n_2, and where k is a difference of factors of 2*C, where C = n_1 + n_2, it must be true that for any prime factor p of n_1 8*n_2 + k^2 is a square modulo p or (8*n_2 + k^2) mod p is a square modulo p and it's just this amazing thing--pure math. And I'm doing my best to figure out a way to make it practical. Forget Goldbach's Conjecture, as I proved Fermat's Last Theorem over two years ago, and people just lie about the proof--yes it is actually a short proof of FLT--and I get in stupid arguments about the distributive property talking about it. I have my prime counting function which I've had for years, and people say stupid stuff about it, and no one can even step through the derivation on their own, and it doesn't matter. I have MAJOR results all over the place. If I can use this approach to prove Goldbach's Conjecture, I just don't think it'd make a difference. So I'm focusing on practical, and the factoring problem. It would be ironic, wouldn't it, if I am the only person who can see this all the way through to proving Goldbach's Conjecture, so it never gets proven just because I'm so sick of this crap? But my opinion of the mathematical community couldn't get much lower. They are political machines. I think they're checking the public mood whenever I get another major result, trying to figure out if they can get away with ignoring more important research. I see them checking Usenet, to see if Magidin, and Ullrich and the other people are still protecting them, still keeping all of you in check--obedient little children, trusting in academics at top universities who I believe have sold you out for their hides. My results overturn so much that for some of them, everything they've done from their doctoral thesis on gets tossed, so they may see themselves as protecting against the worst case--the world knowing correct mathematics in this area. And if the factoring problem falls as I focus on it, and you lose Wi-Fi, and your cellphone is less secure, and you can't make purchases on the Internet like before, what are you really going to do to them? Maybe they figure they lose more no matter what than if they just sit tight, and wait, depending on people like Magidin and Ullrich and the rest of the sci.math crew to hold the line, protect them from the truth, and stop the research from going forward, as humanity would then not be their concern. The future of the species would be some silly little thing, if that scenario is what's playing out, as one of the biggest issues for many of them would be, their mortgages. Got to keep that professor's salary to keep paying the mortgage, you know? James Harris === Subject: Re: JSH: See what I mean? > I was really excited for a while about this new result, and then the > sci.math'ers started in, like they've done now for years, and you can > see how depressing it all becomes. > So there is this relationship between the primes, quadratic residues > and factorizations of sums, so what? > Mathematicians don't give a damn about the truth, they just want to > maintain the status quo. > Welcome to the real world. > The status quo is that you get a Ph.D in mathematics, play the game, > and follow certain highly specific rules to join the club and be > allowed to be known for what you produce. > Be an outsider like me--especially one who is a harsh critic of todays > mathematical community--and they use the most powerful weapon in their > arsenal--they ignore you. > Oh, I have a more general result now that with odd natural composites > n_1 and n_2, and where k is a difference of factors of 2*C, where C = > n_1 + n_2, it must be true that for any prime factor p of n_1 > 8*n_2 + k^2 is a square modulo p > or > (8*n_2 + k^2) mod p is a square modulo p > and it's just this amazing thing--pure math. > And I'm doing my best to figure out a way to make it practical. > Forget Goldbach's Conjecture, as I proved Fermat's Last Theorem over > two years ago, and people just lie about the proof--yes it is actually > a short proof of FLT--and I get in stupid arguments about the > distributive property talking about it. > I have my prime counting function which I've had for years, and people > say stupid stuff about it, and no one can even step through the > derivation on their own, and it doesn't matter. > I have MAJOR results all over the place. If I can use this approach to > prove Goldbach's Conjecture, I just don't think it'd make a difference. > So I'm focusing on practical, and the factoring problem. > It would be ironic, wouldn't it, if I am the only person who can see > this all the way through to proving Goldbach's Conjecture, so it never > gets proven just because I'm so sick of this crap? > But my opinion of the mathematical community couldn't get much lower. > They are political machines. I think they're checking the public mood > whenever I get another major result, trying to figure out if they can > get away with ignoring more important research. > I see them checking Usenet, to see if Magidin, and Ullrich and the > other people are still protecting them, still keeping all of you in > check--obedient little children, trusting in academics at top > universities who I believe have sold you out for their hides. > My results overturn so much that for some of them, everything they've > done from their doctoral thesis on gets tossed, so they may see > themselves as protecting against the worst case--the world knowing > correct mathematics in this area. > And if the factoring problem falls as I focus on it, and you lose > Wi-Fi, and your cellphone is less secure, and you can't make purchases > on the Internet like before, what are you really going to do to them? > Maybe they figure they lose more no matter what than if they just sit > tight, and wait, depending on people like Magidin and Ullrich and the > rest of the sci.math crew to hold the line, protect them from the > truth, and stop the research from going forward, as humanity would then > not be their concern. > The future of the species would be some silly little thing, if that > scenario is what's playing out, as one of the biggest issues for many > of them would be, their mortgages. > Got to keep that professor's salary to keep paying the mortgage, you > know? > James Harris Another keeper! === Subject: Re: JSH: See what I mean? [jstevh@msn.com] > I was really excited for a while about this new result, Nothing wrong with that! More power to you. > and then the sci.math'ers started in, like they've done now for years, > and you can see how depressing it all becomes. It is? Why? > So there is this relationship between the primes, quadratic residues > and factorizations of sums, so what? That's a legitimate question. What follows from it? A result isn't important _just_ because it's true. You should study the Fibonacci numbers sometime: there are literally thousands of known theorems about those, many very simply stated, and wading in such an embarrassment of riches might give you an understanding for why a bare it's true! rarely merits much attention. True is easy; important is hard. > Mathematicians don't give a damn about the truth, > they just want to maintain the status quo. > Welcome to the real world. > The status quo is that you get a Ph.D in mathematics, play the game, > and follow certain highly specific rules to join the club and be > allowed to be known for what you produce. > Be an outsider like me--especially one who is a harsh critic of todays > mathematical community--and they use the most powerful weapon in their > arsenal--they ignore you. > Oh, I have a more general result now that with odd natural composites > n_1 and n_2, If you push a little harder, I believe you'll find that it doesn't matter whether they're odd, and it doesn't matter whether they're composite -- any integers >= 1 should work. > and where k is a difference of factors of 2*C, where C = > n_1 + n_2, it must be true that for any prime factor p of n_1 > 8*n_2 + k^2 is a square modulo p > or > (8*n_2 + k^2) mod p is a square modulo p You could also (and could always have done so) simplify the conclusion to just the last part: (8*n_2 + k^2) mod p is a square modulo p _subsumes_ the 8*n_2 + k^2 is a square modulo p part; the latter is just a special case of the former, so nothing is lost by dropping the first part of the conclusion, but some clarity is gained. BTW, congratulations on switching to better terminology here: using the right technical words truly helps others understand what you're saying. > and it's just this amazing thing--pure math. > And I'm doing my best to figure out a way to make it practical. That's one way to show it's important (if it is). > Forget Goldbach's Conjecture, That's another way (not forgetting it ;-), but proving it). > as I proved Fermat's Last Theorem over two years ago, > and people just lie about the proof--yes it is actually > a short proof of FLT--and I get in stupid arguments about the > distributive property talking about it. If your purported proof of FLT relies on that endlessly repeated 7 multiplies through argument, then it's surely not a proof. > I have my prime counting function which I've had for years, and people > say stupid stuff about it, and no one can even step through the > derivation on their own, Of course many can step through _a_ derivation. It can be derived quite easily from reasoning about how the Sieve of Eratosthenes works. I expect your own derivation was much more convoluted than that (but don't know for sure, because I wasn't able to follow the one derivation of yours I tried to follow -- but if you _knew_ the kind of derivation I'm speaking of, you couldn't possibly believe that it takes Harrisian mutant-brain powers to complete). > and it doesn't matter. You're saying your prime-counting formula doesn't matter? You'll get a lot of agreement on that, if so :-) > I have MAJOR results all over the place. I'm afraid you're alone in believing that -- and no, that's not because everyone else is a liar and/or scum and/or incompetent and/or evil and/or poorly educated (and if I missed one of the slurs you routinely spit at others, that was just an oversight, not a confession that you hit _that_ one on the nose ;-)). > If I can use this approach to prove Goldbach's Conjecture, I just don't > think it'd make a difference. Of course it would! You'd get some well-deserved fame if you could give a legitimate proof. > So I'm focusing on practical, and the factoring problem. > It would be ironic, wouldn't it, if I am the only person who can see > this all the way through to proving Goldbach's Conjecture, so it never > gets proven just because I'm so sick of this crap? Your choice, your responsibility. It would be good to stop blaming others for your choices. > But my opinion of the mathematical community couldn't get much lower. > They are political machines. I think they're checking the public mood > whenever I get another major result, trying to figure out if they can > get away with ignoring more important research. > I see them checking Usenet, to see if Magidin, and Ullrich and the > other people are still protecting them, still keeping all of you in > check--obedient little children, trusting in academics at top > universities who I believe have sold you out for their hides. > My results overturn so much that for some of them, everything they've > done from their doctoral thesis on gets tossed, so they may see > themselves as protecting against the worst case--the world knowing > correct mathematics in this area. > And if the factoring problem falls as I focus on it, and you lose > Wi-Fi, and your cellphone is less secure, and you can't make purchases > on the Internet like before, what are you really going to do to them? > Maybe they figure they lose more no matter what than if they just sit > tight, and wait, depending on people like Magidin and Ullrich and the > rest of the sci.math crew to hold the line, protect them from the > truth, and stop the research from going forward, as humanity would then > not be their concern. > The future of the species would be some silly little thing, if that > scenario is what's playing out, as one of the biggest issues for many > of them would be, their mortgages. > Got to keep that professor's salary to keep paying the mortgage, you > know? What do you imagine someone reading one of these rants thinks of you? Do you believe wow -- is that guy nuts? is an unlikely reaction? === Subject: Re: JSH: See what I mean? >> Be an outsider like me--especially one who is a harsh critic of todays >> mathematical community--and they use the most powerful weapon in their >> arsenal--they ignore you. >> Oh, I have a more general result now that with odd natural composites >> n_1 and n_2, > If you push a little harder, I believe you'll find that it doesn't matter > whether they're odd, and it doesn't matter whether they're composite -- any > integers >= 1 should work. The problem with pushing a little harder is that sometimes that makes you realize that your statement is not interesting at all. If someone says the last digit of every Fermat prime greater than 5 is the digit 7, this may look impressive. However, if you push a little harder then you get the statement the last digit of every number of the form 2^(2^n) with n > 1 is the digit 6, which is much less impressive. And, of course, you can go one step further and get the last digit of every number of the form 2^(4 n) is the digit 6, which is still less impressive. Jose Carlos Santos === Subject: Re: See what I mean? > Forget Goldbach's Conjecture, as I proved Fermat's Last Theorem over > two years ago, === Subject: Re: JSH: See what I mean? > Forget Goldbach's Conjecture, as I proved Fermat's Last Theorem over > two years ago, and people just lie about the proof--yes it is actually > a short proof of FLT--and I get in stupid arguments about the > distributive property talking about it. > I have my prime counting function which I've had for years, and people > say stupid stuff about it, and no one can even step through the > derivation on their own, and it doesn't matter. I see. So you're choosing your research agenda based on trying to get other people's attention. Strange behavior for someone who's only interested in the truth! If all you care about is truth, why does it matter who ignores you? === Subject: Re: JSH: See what I mean? : The status quo is that you get a Ph.D in mathematics, play the game, : and follow certain highly specific rules to join the club and be : allowed to be known for what you produce. Okay, I'll try this again, James. Here's a question for you. I'm a yearly-contracted lecturer with a PhD. who has no interest in tenure or research (I turned down one post-doc I was offered without application after I finished my degree in '00). Nobody knows who I am except my students, who really like me. I could be fired tomorrow if the budget goes south, at which point I'd go and get a job as a programmer at four times my current salary. What status quo am I supporting? Do tell. : And I'm doing my best to figure out a way to make it practical. Though you have no evidence that it is. : Forget Goldbach's Conjecture, as I proved Fermat's Last Theorem over : two years ago, and people just lie about the proof--yes it is actually : a short proof of FLT--and I get in stupid arguments about the : distributive property talking about it. This is because you have no proof and because even after all our discussions you have yet to figure out what the DP doesn't say. : I have my prime counting function which I've had for years, and people : say stupid stuff about it, and no one can even step through the : derivation on their own, and it doesn't matter. We said it's been done, and it has, and yours is slow. : I have MAJOR results all over the place. Name one. : If I can use this approach to prove Goldbach's Conjecture, I just don't : think it'd make a difference. If you can produce a valid proof of GP I'll quit my job and worship you. : Got to keep that professor's salary to keep paying the mortgage, you : know? I don't have a professor's salary. Justin === Subject: Re: JSH: See what I mean? : and follow certain highly specific rules to join the club and be > : allowed to be known for what you produce. > Okay, I'll try this again, James. Here's a question for you. I'm a > yearly-contracted lecturer with a PhD. who has no interest in tenure or > research (I turned down one post-doc I was offered without application > after I finished my degree in '00). Nobody knows who I am except my > students, who really like me. I could be fired tomorrow if the budget > goes south, at which point I'd go and get a job as a programmer at four > times my current salary. > What status quo am I supporting? Do tell. The status quo where Wiles isn't just another guy with a failed approach at proving FLT. The status quo where ideal theory is actually useful. The status quo where a hundred years of mathematical ideas in number theory don't go up in smoke, casting doubt, in the minds of mathematicians, on their very reasons for existence. If they couldn't see such a huge error, then what good are they? You're fighting to preserve your sense of rightness, your feelings of confidence in your discipline, and your abilities, and your sense of having a good grasp of your world. > : And I'm doing my best to figure out a way to make it practical. > Though you have no evidence that it is. I have a hyperbolic factoring method already. It's kind of complicated though. I've started going over an approach using this latest research which could tease out properties of factors of composites by looking over quadratic residues. It's a lot simpler. If I succeed then you won't be able to get a job doing squat, not even computer programming as you will be known, your identity revealed, and you won't have students either. Maybe you can go deliver pizzas. That's a LOT of motivation to lie. > : Forget Goldbach's Conjecture, as I proved Fermat's Last Theorem over > : two years ago, and people just lie about the proof--yes it is actually > : a short proof of FLT--and I get in stupid arguments about the > : distributive property talking about it. > This is because you have no proof and because even after all our > discussions you have yet to figure out what the DP doesn't say. The logic is trivial. But people like you ignore the logic to distract people for social reasons so you can hold on to your incorrect mathematical ideas. After all, who would doubt that given a*(f(x) + b) = a*f(x) + a*b it must be true that the value of f(x) has no impact? But, if no one doubted me on this point, why do people who claim that f(0) = 0 is a special case get away with it? You people are like so many human beings over the centuries, you sell out the truth to preserve your feeling of comfort. I have easy proofs, across much of number theory, but you deny the proofs, then deny that you deny proofs. > : I have my prime counting function which I've had for years, and people > : say stupid stuff about it, and no one can even step through the > : derivation on their own, and it doesn't matter. > We said it's been done, and it has, and yours is slow. Only the pure math version is slow. I found, with little effort, fast algorithms based from it, and can show that much of it--in sieve form--is equivalent to the fastest known. But yet again, you don't give a damn about the truth. You don't give a damn about your students, but only about what they give you, or better yet, what you take from them, or you would care about teaching them the truth. The pure math version may be slow, but unlike anything else known for counting primes, it relies on a partial difference equation. The partial difference equation leads directly to a partial differential equation, which can be integrated, to check the Riemann Hypothesis. So many answers available, but you people don't want those answers, as the Riemann Hypothesis is likely false. > : I have MAJOR results all over the place. > Name one. I found a proof of Fermat's Last Theorem, to start. Arguing about that proof I brainstormed my way to my prime counting function. Then I went back to arguments over the proof of FLT and abstracted out one piece where posters argued the most, and expanded it with some changes into a paper. That paper was published. People on sci.math heard about it, and some of them emailed the journal and convinced the editor the paper was false, and he abruptly yanked it. A few months later the entire freaking journal keeled over and died. I kept doing research and worked out, slowly, with a lot of mistakes, a new way to factor. And just now, I brainstormed a fundamental result relating primes, their quadratic residues, and the factorization of an important sum. ANY one of those results is massive. > : If I can use this approach to prove Goldbach's Conjecture, I just don't > : think it'd make a difference. > If you can produce a valid proof of GP I'll quit my job and worship you. Bull. You lie. > : Got to keep that professor's salary to keep paying the mortgage, you > : know? > I don't have a professor's salary. > Justin Others do. Wiles does. So does Ribet, and Taylor. They have a lot to protect, from the truth. Yes, you may be some minor player, but hey, you're on Usenet, so that's not a surprise. I'm only here because here is one of my outlets to the world, to let people know what's going on. But you're here because you want to be. James Harris === Subject: Re: JSH: See what I mean? : You're fighting to preserve your sense of rightness, your feelings of : confidence in your discipline, and your abilities, and your sense of : having a good grasp of your world. Actually I don't care much. Look at it this way, if you were right and if still be employed teaching calculus to engineering students but people would know me because I stood by the great James Harris as he gutted modern ideal theory. : I have a hyperbolic factoring method already. It's kind of complicated : though. Yay. : If I succeed then you won't be able to get a job doing squat, not even : computer programming as you will be known, your identity revealed, and : you won't have students either. If you succeed in factoring, I'll lose my job as a teacher? Er, okay. : After all, who would doubt that given : a*(f(x) + b) = a*f(x) + a*b : it must be true that the value of f(x) has no impact? No impact on what? Nobody denies that the above statement is true no matter what f(x) is, but that's not the problem with your attempted proof. : You don't give a damn about your students, but only about what they : give you, or better yet, what you take from them, or you would care : about teaching them the truth. Today I'm teaching, in my three classes: (1) Trigonometry of right triangles. (2) Solutions of systems of nonlinear equations. (3) The ElGamal Cryptosystem. So what am I teaching that's not true? Unless perhaps you've also solved the discrete logarithm problem... : I found a proof of Fermat's Last Theorem, to start. Arguing about that : proof I brainstormed my way to my prime counting function. But you're mistaken about your FLT proof. : That paper was published. It was published with an error. Don't feel bad. My PhD. thesis has a small error in a small lemma. Luckily the lemma is still true but my proof is wrong and nobody caught it. : Yes, you may be some minor player, but hey, you're on Usenet, so that's : not a surprise. Yes, yes I am a minor player, which brings me full-circle to my original point. Why should I not gain fame and fortune by supporting you? Justin === Subject: Re: JSH: See what I mean? : and follow certain highly specific rules to join the club and be > : allowed to be known for what you produce. > Okay, I'll try this again, James. Here's a question for you. I'm a > yearly-contracted lecturer with a PhD. who has no interest in tenure or > research (I turned down one post-doc I was offered without application > after I finished my degree in '00). Nobody knows who I am except my > students, who really like me. I could be fired tomorrow if the budget > goes south, at which point I'd go and get a job as a programmer at four > times my current salary. > What status quo am I supporting? Do tell. > The status quo where Wiles isn't just another guy with a failed > approach at proving FLT. > The status quo where ideal theory is actually useful. The term ideal(s) occurs 134 times in Matthew Briggs's M.Sc. thesis An Introduction to the General Number Field Sieve. If we look at http://www.crypto-world.com/FactorRecords.html we see that RSA-200 was factored in 2005 by Bahr, Boehm, Franke and Kleinjung using the general number field sieve. What's wrong with ideals anyway? David Bernier > The status quo where a hundred years of mathematical ideas in number > theory don't go up in smoke, casting doubt, in the minds of > mathematicians, on their very reasons for existence. > If they couldn't see such a huge error, then what good are they? > You're fighting to preserve your sense of rightness, your feelings of > confidence in your discipline, and your abilities, and your sense of > having a good grasp of your world. > : And I'm doing my best to figure out a way to make it practical. > Though you have no evidence that it is. > I have a hyperbolic factoring method already. It's kind of complicated > though. > I've started going over an approach using this latest research which > could tease out properties of factors of composites by looking over > quadratic residues. > It's a lot simpler. > If I succeed then you won't be able to get a job doing squat, not even > computer programming as you will be known, your identity revealed, and > you won't have students either. > Maybe you can go deliver pizzas. > That's a LOT of motivation to lie. > : Forget Goldbach's Conjecture, as I proved Fermat's Last Theorem over > : two years ago, and people just lie about the proof--yes it is actually > : a short proof of FLT--and I get in stupid arguments about the > : distributive property talking about it. > This is because you have no proof and because even after all our > discussions you have yet to figure out what the DP doesn't say. > The logic is trivial. But people like you ignore the logic to distract > people for social reasons so you can hold on to your incorrect > mathematical ideas. > After all, who would doubt that given > a*(f(x) + b) = a*f(x) + a*b > it must be true that the value of f(x) has no impact? > But, if no one doubted me on this point, why do people who claim that > f(0) = 0 is a special case get away with it? > You people are like so many human beings over the centuries, you sell > out the truth to preserve your feeling of comfort. > I have easy proofs, across much of number theory, but you deny the > proofs, then deny that you deny proofs. > : I have my prime counting function which I've had for years, and people > : say stupid stuff about it, and no one can even step through the > : derivation on their own, and it doesn't matter. > We said it's been done, and it has, and yours is slow. > Only the pure math version is slow. I found, with little effort, fast > algorithms based from it, and can show that much of it--in sieve > form--is equivalent to the fastest known. > But yet again, you don't give a damn about the truth. > You don't give a damn about your students, but only about what they > give you, or better yet, what you take from them, or you would care > about teaching them the truth. > The pure math version may be slow, but unlike anything else known for > counting primes, it relies on a partial difference equation. > The partial difference equation leads directly to a partial > differential equation, which can be integrated, to check the Riemann > Hypothesis. > So many answers available, but you people don't want those answers, as > the Riemann Hypothesis is likely false. > : I have MAJOR results all over the place. > Name one. > I found a proof of Fermat's Last Theorem, to start. Arguing about that > proof I brainstormed my way to my prime counting function. > Then I went back to arguments over the proof of FLT and abstracted out > one piece where posters argued the most, and expanded it with some > changes into a paper. > That paper was published. > People on sci.math heard about it, and some of them emailed the journal > and convinced the editor the paper was false, and he abruptly yanked > it. > A few months later the entire freaking journal keeled over and died. > I kept doing research and worked out, slowly, with a lot of mistakes, a > new way to factor. > And just now, I brainstormed a fundamental result relating primes, > their quadratic residues, and the factorization of an important sum. > ANY one of those results is massive. > : If I can use this approach to prove Goldbach's Conjecture, I just don't > : think it'd make a difference. > If you can produce a valid proof of GP I'll quit my job and worship you. > Bull. > You lie. > : Got to keep that professor's salary to keep paying the mortgage, you > : know? > I don't have a professor's salary. > Justin > Others do. Wiles does. So does Ribet, and Taylor. > They have a lot to protect, from the truth. > Yes, you may be some minor player, but hey, you're on Usenet, so that's > not a surprise. > I'm only here because here is one of my outlets to the world, to let > people know what's going on. > But you're here because you want to be. > James Harris === Subject: Re: JSH: See what I mean? James, YOU SERIOUSLY, AND I ASKED FOR A REVISED VERSION OF YOUR PAPER, AND I HAVE YET TO SEE ANYTHING. IF YOU ARE A REAL MATHEMATICIAN, YOU SHOULD NOT CARE WHAT PEOPLE ON HERE THINK. YOU SHOULD STOP POSTING HERE OVER AND OVER, AND START WRITING LOTS AND LOTS OF STUFF!! WRITE A FOOL-PROOF PAPER ON YOUR PROOF OF FERMAT'S LAST THEOREM IF YOU WANT!!! JUST MAKE IT WRITTEN WELL!!! BUT BE WILLING TO SEE THE MISTAKES IN IT THAT ARE THERE, CORRECT THEM, AND MOVE ON. THAT IS THE MARK OF A REAL MATHEMATICIAN. Einstein's image has gradually become an icon which sums up the sometimes eccentric brilliance of the physicist. When an outraged Hitler produced a paper entitled 100 Scientists Against Einstein he simply replied, If I were wrong, one would have been enough. Rob === Subject: Re: JSH: See what I mean? YOU SERIOUSLY, AND I ASKED FOR A REVISED VERSION OF YOUR PAPER, AND I > HAVE YET TO SEE ANYTHING. I wouldn't hold my breath, if I were you. I made a similar request and haven't received any replies. > IF YOU ARE A REAL MATHEMATICIAN, YOU SHOULD NOT CARE WHAT PEOPLE ON > HERE THINK. YOU SHOULD STOP POSTING HERE OVER AND OVER, AND START > WRITING LOTS AND LOTS OF STUFF!! WRITE A FOOL-PROOF PAPER ON YOUR > PROOF OF FERMAT'S LAST THEOREM IF YOU WANT!!! JUST MAKE IT WRITTEN > WELL!!! BUT BE WILLING TO SEE THE MISTAKES IN IT THAT ARE THERE, > CORRECT THEM, AND MOVE ON. THAT IS THE MARK OF A REAL MATHEMATICIAN. > Einstein's image has gradually become an icon which sums up the > sometimes eccentric brilliance of the physicist. When an outraged > Hitler produced a paper entitled 100 Scientists Against Einstein he > simply replied, If I were wrong, one would have been enough. Or, as Carl Sagan once said: They laughed at Galileo. They laughed at Newton. But they also laughed at Bozo the Clown. JSH will never get it. --- Christopher Heckman === Subject: Re: JSH: Power of myth, discoverer reality >>And eventually, there would have to be a result that resonated? > Why would there *have* to be one such? > Does anyone think you have anything both original and non-trivial? > Since this *has* to happen 'eventually', maybe you could give some > estimate of when this will be - some kind of *testable* prediction > perhaps... Ever hear of a monkey who lives forever and bangs away constantly at a typewriter? Eventually said monkey will write all the works of Shakespeare, plus any other written works that you can think of! Bob Terwilliger === Subject: Re: JSH: Power of myth, discoverer reality > And eventually, there would have to be a result that resonated? > Why would there *have* to be one such? > Does anyone think you have anything both original and non-trivial? I've had a paper published, but then sci.math'ers mounted that email campaign against it. I've had people come on and post in support of me, and sci.math'ers went after them. Just to test, I asked if it would matter if I had someone from NASA post in support, and posters attacked NASA. > Since this *has* to happen 'eventually', maybe you could give some > estimate of when this will be - some kind of *testable* prediction > perhaps... As long as mathematicians as a group maintain the respect of the world so that people cannot imagine them doing what they're doing, then it's a process of attrition. Maybe the word goes out and some money doesn't get appropriated. Maybe a mathematician at a party gets a few odd glances, or hears an odd question here or there, so you never know what the full impact of it is. And, maybe enrollments of undergraduates drop, and graduate student enrollment drops, or you have some undergrads who suddenly and inexplicably change majors, but it's like this slight thing for a long time. And then one day the impact really gets felt as the effect slowly snowballs. James Harris === Subject: Re: JSH: Power of myth, discoverer reality > [...] > I've had people come on and post in support of me, and sci.math'ers > went after them. [...] That's true for some people who support you. No one ever attacked me, though, when I supported you, in the beginning. --- Christopher Heckman === Subject: Re: JSH: Power of myth, discoverer reality > And eventually, there would have to be a result that resonated? > Why would there *have* to be one such? > Does anyone think you have anything both original and non-trivial? > I've had a paper published, but then sci.math'ers mounted that email > campaign against it. > I've had people come on and post in support of me, and sci.math'ers > went after them. > Just to test, I asked if it would matter if I had someone from NASA > post in support, and posters attacked NASA. NASA ain't exactly the sharpest knife in the drawer. They blew up a space shuttle. Twice. They made a math error and lost a Mars probe. One of several that were lost. Why don't you apply for a job there? > Since this *has* to happen 'eventually', maybe you could give some > estimate of when this will be - some kind of *testable* prediction > perhaps... > As long as mathematicians as a group maintain the respect of the world > so that people cannot imagine them doing what they're doing, then it's > a process of attrition. > Maybe the word goes out and some money doesn't get appropriated. > Maybe a mathematician at a party gets a few odd glances, or hears an > odd question here or there, so you never know what the full impact of > it is. > And, maybe enrollments of undergraduates drop, and graduate student > enrollment drops, or you have some undergrads who suddenly and > inexplicably change majors, but it's like this slight thing for a long > time. > And then one day the impact really gets felt as the effect slowly > snowballs. > James Harris === Subject: Re: JSH: Power of myth, discoverer reality >> And eventually, there would have to be a result that resonated? >> Why would there *have* to be one such? >> Does anyone think you have anything both original and non-trivial? > I've had a paper published, but then sci.math'ers mounted that email > campaign against it. > I've had people come on and post in support of me, and sci.math'ers > went after them. > Just to test, I asked if it would matter if I had someone from NASA > post in support, and posters attacked NASA. >> Since this *has* to happen 'eventually', maybe you could give some >> estimate of when this will be - some kind of *testable* prediction >> perhaps... > As long as mathematicians as a group maintain the respect of the world > so that people cannot imagine them doing what they're doing, then it's > a process of attrition. > Maybe the word goes out and some money doesn't get appropriated. > Maybe a mathematician at a party gets a few odd glances, or hears an > odd question here or there, so you never know what the full impact of > it is. Onward Victim Psudo-mathematicians.......... My poor puppy got sick because he was rejected in algebra class. I am failing because all you sci.math people will not worship or recognize me or my research. It is Your fault that I cannot do math My Prime-Counting Algorithm drives our Nations GDP === Subject: Jensen - pronounce How do you pronounce Jensen ? He was danish, should we pronounce it as the J were like in yet ? === Subject: Re: Jensen - pronounce > How do you pronounce Jensen ? He was danish, should we pronounce it as the > J were like in yet ? Jes. === Subject: Integrating the tail of a distribution Okay, for long I have accepted that integrating the tail of a distribution yields the expectation of a random variable. How does one prove this? === Subject: Re: Integrating the tail of a distribution > Okay, for long I have accepted that integrating the tail of a > distribution yields the expectation of a random variable. How does one > prove this? You might begin with non-negative integer-valued random variables, then regard a more general case as a limit. If p_i, i=0,1,2, ... are the probabilities at x_i = 0,1,2, ..., then (assuming a finite EX) we have EX = 1p_1 + 2p_2 + 3p_3 + ... = (p_1 + p_2 + ...) + (p_2 + p_3 + ...) + (p_3 + p_4 + ...) + ... = P{X>0} + P{X>1} + P{X>2} + ... . (Note that you can re-arrange the series because it converges and its terms are >= 0.) R.G. Vickson === Subject: Re: Integrating the tail of a distribution >Okay, for long I have accepted that integrating the tail of a >distribution yields the expectation of a random variable. How does one >prove this? This is true only for a nonnegative random variable; let's call it X. Let I(x) be the indicator variable for the event {X > x}. Then X = integral(x=0..infty, I(x)). (Draw the picture.) Thus, EX = E integral(x=0..infty, I(x)) = integral(x=0..infty, EI(x)) by Tonelli = integral(x=0..infty, P{X > x}). -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Letters to a Young Mathematician review A review of the book Letters to a Young Mathematician by Ian Stewart FRS of Warwick: http://www.csmonitor.com/2006/0425/p14s02-bogn.html === Subject: Re: Now you can see it happening <44503bc0$0$79632$892e7fe2@authen.yellow.readfreenews.net> I'm reminding now with this quadratic residue result of when I came up >> with my prime counting function. >> Why don't you post any Mathematics ? >> This is a Mathematics group. > Hey, I made a mistake as I may have started out wanting to talk of > reminding now with the quadratic residue result and then shifted to > saying reminded, but did not go back to change from reminding to > reminded. > As for math, why am I on freaking Usenet now anyway? > I am because the math establishment has succeeded at ignoring my > results for years, so I have few ways to get the word out. > My point to you all is that they have to know what they are doing. > As you watch now with yet another result I want to emphasize up front > that here we go again, so you can watch these people as I have watched > them over the years with major result after major result where they > just explain away, rationalize, and just plain lie about my > research--and people believe them. > This is at least my fourth major result. > Notice though you will not see headlines tomorrow about this major > research find which could lead to a proof of one of the most famous > still unsolved problems in mathematics. > But if Wiles, or Ribet or Taylor, or any of a number of other members > of the in-crowd of mathematics had the same exact results, math people > would be falling all over themselves getting excited and talking about > the possibilities surrouding this extraordinary link between primes, > their quadratic residues and factorization. > The point here is that I am not a crackpot. > I do make mistakes--using modern problem solving techniques, like > brainstorming. > Despite my explanation of what I do, and the demonstrations, > repeatedly, of how powerful modern problem solving techniques are, I > get ridiculed for using them. > My results are usually huge, and still people ignore them. > The issues here are political. > Here we go again, but I hope that some of you will start acting like > educated adults, and this time I will see less support for the people > who make it their business to attack my research using any distraction, > lie, or whatever they can think of, to convince you. > I still hope--wonder how that is possible giving what I have seen over > the YEARS--that some of you, will maybe, this time, use critical > thinking. > James Harris > You are a crackpot. You refuse to learn anything, which if you learned > something about math, you'd see you're wrong. [...] Either that, or that what JSH is doing is only an exercise in the textbook. I had a nervous breakdown and stayed out of grad school for a year. During that year, I proved the classical result about the number of integral solutions to the equation 1/a + 1/b = 1/M, where M is a fixed positive integer. When I went back, I bothered several of the faculty about this new result, until I mentioned the problem to me, or whether I got a glimpse of it; my memories around that time are a bit hazy.) Now, if I had been JSH, I would have kept on insisting that it was new and tried to get it published. Instead, I acted like a professional. --- Christopher Heckman === Subject: Re: A physics question about infinity >I think the largest number in the universe is on the order of 10^67, or >something like that. >There is no largest number. >>The largest number to have a physical attribute, to represent something. > Distance: > The Planck scale is about 10^-38 m, and the visible universe is > something like 50 billion light-years in radius, or about 5 x 10^23 m. > Assuming a perfect sphere for the shape, this gives roughly > 5 x 10^185 as the total number of cubic quantum distance units. > Time: > The Planck time scale is about 10^-43 sec, and the age of the visible > universe is about 14 billion years. This gives a total range of > roughly 4.4 x 10^60 quantum time ticks. Yeah, sure. Now do the same thing for the Planck mass, which is in the order of 20-50 micrograms: http://scienceworld.wolfram.com/physics/PlanckMass.html Han de Bruijn === Subject: Re: A physics question about infinity > I think the largest number in the universe is on the order of 10^67, or > something like that. > There is no largest number. >> The largest number to have a physical attribute, to represent something. > Distance: > The Planck scale is about 10^-38 m, and the visible universe is > something like 50 billion light-years in radius, or about 5 x 10^23 m. > Assuming a perfect sphere for the shape, this gives roughly > 5 x 10^185 as the total number of cubic quantum distance units. > Time: > The Planck time scale is about 10^-43 sec, and the age of the visible > universe is about 14 billion years. This gives a total range of > roughly 4.4 x 10^60 quantum time ticks. I think he was remembering that 10^85 (or thereabouts) is the estimated number of subatomic particals in the known universe. === Subject: Re: Question Just because you don't like the answer does not mean it is not an >> answer. Your question was the logical equivalent of what kind of duck >> is a fox terrier? It would seem you would consider the answer a fox >> terrier is not a duck to be an escape rather than an answer. >> A sequence is, BY DEFINITION, a map whose domain is the natural >> numbers. >Fair enough, this mean that the biggest number of terms in a sequence >would be Omega. > You really need to stop using standard mathematical words with your > own private meaning. > Omega is not a quantity; you do not speak about there being omega > things. Omega is an ordinal. It corresponds to a well-ordering type. There should be some expression which can describe the quantity of terms in a sequance Since a sequence is a function who's domain is the natural numbers. Then it should have Aleph-0 of terms in it, because the multiplicity of terms in the natural numbers set is descriped by Aleph-0. So 0.9999......... if it is to represent a real number then it should be a cauchy sequence and accordingly it should have Aleph-0 of nines in it. Is that Correct? > And, no. It means EXACTLY that a sequence is a function whose domain > is the natural numbers. There are ->exactly<- one term for each > element of omega. Not the biggest number of terms. There is exactly > one term for each element of omega. >Or in other words a sequence either contains n terms when n is finite, >or Omega of terms if the number of terms in it is infinite. > Will you ever learn to read? > A sequence is a function whose DOMAIN is the natural numbers. Not > whose domain is a subset of the natural numbers. There is always one > term for each natural number. > If you want to talk about finite sequences, whose domains are the > sets I called n (where the set 0 is the empty set, and the set n+1 > is the set {0, 1, ..., n}), then call them finite sequences. >Therefore the decimal expansion of 0.99999......... has Omega of 9 in >it. > No. Omega is not a quantity. omega is an ordinal. >and I repeat it would be lesser than 1 by 1/10^Omega. > And it does not matter how many times you repeat it, it is still > nonsense. The decimal expansion represents a sequence. The sequence is > equivalent to the constant sequence 1, and therefore, since 1 and > 0.999... are two distinct representations of the same equivalence >Perhaps this is non sense. I don't know. > Yes, you do know. Because you have been told this many times. You just > ignore it and repeat it over and over and over. No matter how often > you repeat it, it is still nonsense. The symbol 10^omega does not > represent a real number. You can define addition and multiplication of > ordinals, as well as exponentiation; under the standard definitions, > 10^omega is the ordinal omega. And there is no standard definition of > division of ordinals, so no matter how many times you repeat the > nonsense of writing 1/10^omega, it is still a symbol that lack any > meaning. It is nonsense. > Just because you can write down a symbol it does not, ipso facto, > grant it sense. >However you succeeded in illustrating to me that the symbole >0.9999.......9 is not a sequence >simply because the last 9 though a finite number but it is at a >transfinite position that is Omega+1 > Actually, no. It would be in position omega, not position omega+1. The > order type of the digits is Omega+1; that is, your symbol represents a > function with DOMAIN omega+1, which has value 9 at each natural number > and value 9 at omega (recall that, BY DEFINITION, > omega+1 = omega / {omega} = {0, 1, 2, ..., n, ...} / {omega} >and therefore it is not a sequence. >well according to what you are saying then 1/10^Omega =0 > No. It still has no meaning whatsoever. >according to >standard mathematics. > No. According to standard mathematics, the symbol you insist on > repeating is still complete and utter nonsense. >which something very strange as I see. > That is quite plain. > -- === Subject: Re: Question days. My association with the Department is that of an alumnus. Learn to edit your replies and remove that which you are no longer addressing. You followed up on 16 lines of text, and left 90 lines which are no longer relevant. You are wasting space and resources. > Just because you don't like the answer does not mean it is not an > answer. Your question was the logical equivalent of what kind of duck > is a fox terrier? It would seem you would consider the answer a fox > terrier is not a duck to be an escape rather than an answer. >> A sequence is, BY DEFINITION, a map whose domain is the natural > numbers. >>Fair enough, this mean that the biggest number of terms in a sequence >>would be Omega. >> You really need to stop using standard mathematical words with your >> own private meaning. >> Omega is not a quantity; you do not speak about there being omega >> things. Omega is an ordinal. It corresponds to a well-ordering type. >There should be some expression which can describe the quantity of >terms in a sequance There is. But it is not the ordinal numbers. Ordinal numbers are about ->order<-, hence the name. What is used to describe quantities are the cardinal numbers. Of course, the fact that you do not know what either of them are (or what the real numbers are, for that matter) goes a long way towards explaining why you continue to spout nonsense. >Since a sequence is a function who's domain is the natural numbers. >Then it should have >Aleph-0 of terms in it, Yes; because the cardinality of the natural numbers is aleph_0. We use cardinals when speaking about cardinality, ordinals when we are dealing with well orders. >So 0.9999......... if it is to represent a real number No if about it. This is short hand for a Cauchy sequence of rationals, and therefore ->does<- represent a (unique) real number. Namely, 1, because the Cauchy sequence converges to 1. >then it should be a cauchy sequence It ->represents<- a Cauchy sequence of rationals. Decimal notation is shorthand for a sequence. The sequence has as its d-th term (i.e., the image of the natural number d) the rational number obtained by truncating the decimal expression at the d-th decimal position. In this case, the terms of the sequence are (0, 0.9, 0.99, 0.999, 0.9999, ....) or (0, (10-1)/10, (10^2-1)/10^2, (10^3-1)/10^3, ..., (10^n-1)/10^n, ...) >and accordingly it should have Aleph-0 of nines in it. It has aleph_0 nines in it. A countably infinite number of nines in it. -- === Subject: Re: Question > Just because you don't like the answer does not mean it is not an >> answer. Your question was the logical equivalent of what kind of duck >> is a fox terrier? It would seem you would consider the answer a fox >> terrier is not a duck to be an escape rather than an answer. >> A sequence is, BY DEFINITION, a map whose domain is the natural >> numbers. >Fair enough, this mean that the biggest number of terms in a sequence >would be Omega. > You really need to stop using standard mathematical words with your > own private meaning. > Omega is not a quantity; you do not speak about there being omega > things. Omega is an ordinal. It corresponds to a well-ordering type. > There should be some expression which can describe the quantity of > terms in a sequance > Since a sequence is a function who's domain is the natural numbers. > Then it should have > Aleph-0 of terms in it, because the multiplicity of terms in the > natural numbers set is descriped > by Aleph-0. > So 0.9999......... if it is to represent a real number then it should > be a cauchy sequence > and accordingly it should have Aleph-0 of nines in it. > Is that Correct? > And, no. It means EXACTLY that a sequence is a function whose domain > is the natural numbers. There are ->exactly<- one term for each > element of omega. Not the biggest number of terms. There is exactly > one term for each element of omega. >Or in other words a sequence either contains n terms when n is finite, >or Omega of terms if the number of terms in it is infinite. > Will you ever learn to read? > A sequence is a function whose DOMAIN is the natural numbers. Not > whose domain is a subset of the natural numbers. There is always one > term for each natural number. > If you want to talk about finite sequences, whose domains are the > sets I called n (where the set 0 is the empty set, and the set n+1 > is the set {0, 1, ..., n}), then call them finite sequences. >Therefore the decimal expansion of 0.99999......... has Omega of 9 in >it. > No. Omega is not a quantity. omega is an ordinal. >and I repeat it would be lesser than 1 by 1/10^Omega. > And it does not matter how many times you repeat it, it is still > nonsense. The decimal expansion represents a sequence. The sequence is > equivalent to the constant sequence 1, and therefore, since 1 and > 0.999... are two distinct representations of the same equivalence >Perhaps this is non sense. I don't know. > Yes, you do know. Because you have been told this many times. You just > ignore it and repeat it over and over and over. No matter how often > you repeat it, it is still nonsense. The symbol 10^omega does not > represent a real number. You can define addition and multiplication of > ordinals, as well as exponentiation; under the standard definitions, > 10^omega is the ordinal omega. And there is no standard definition of > division of ordinals, so no matter how many times you repeat the > nonsense of writing 1/10^omega, it is still a symbol that lack any > meaning. It is nonsense. > Just because you can write down a symbol it does not, ipso facto, > grant it sense. >However you succeeded in illustrating to me that the symbole >0.9999.......9 is not a sequence >simply because the last 9 though a finite number but it is at a >transfinite position that is Omega+1 > Actually, no. It would be in position omega, not position omega+1. The > order type of the digits is Omega+1; that is, your symbol represents a > function with DOMAIN omega+1, which has value 9 at each natural number > and value 9 at omega (recall that, BY DEFINITION, > omega+1 = omega / {omega} = {0, 1, 2, ..., n, ...} / {omega} >and therefore it is not a sequence. >well according to what you are saying then 1/10^Omega =0 > No. It still has no meaning whatsoever. >according to >standard mathematics. > No. According to standard mathematics, the symbol you insist on > repeating is still complete and utter nonsense. >which something very strange as I see. > That is quite plain. > -- I wonder why subtraction and division are not defined for Transfinite Ordinal numbers? For example : n + Omega = Omega but Omega + n > Omega Now we can define two operators of subtraction , on is left subtraction denoted as L - and the other is right subtraction denoted as -R as followes Define: Omega L- n = Omega Omega -R n < Omega n is a finite number Also about division we can also define two division operators in a similar way L / and /R as below Omega L / n= x is x: n.x = Omega Omega /R n = z is z : z.n = Omega For example Omega L / 3 = ( 1L / 3 ) Omega < Omega, because 3.Omega>Omega While Omega / R 3 = Omega. 3 = 3+3+3+3+.......... = Omega Now 1 L / Omega < Omega = infentismal = Left one omegath but 1 / R Omega = ??? Any how if these are valid both right and left division operators cannot be zero. because even if the right division operator is not defined yet it cannot be zero because Omega of zeros is a zero and not one. It is the left divion operator that might be interesting 1 L / Omega = 0 means 0.Omega =1 means Omega L - Omega = 1 now Omega L - Omega would be and indeterminate number like 0/0 . While Omega -R Omega = Zero. However all the above might seem to be loonatic to professional mathematicians. Anyhow I am still conviniced that we can form numbers like 1/2^Omega or 1/2^Aleph-x And they cannot be zeros, but of coarse we should specify the division operator. Zuhair === Subject: Re: Question days. My association with the Department is that of an alumnus. You quote over a hundred lines just so you can ignore them later. In addition to your ignorance about mathematics, you also exhibit an appalling ignorance about usenet ettiquette. That one is easier to learn; why don't you? >I wonder why subtraction and division are not defined for Transfinite >Ordinal numbers? I told you why: subtraction is really addition of additive inverses. There are no additive inverses for ordinal numbers. In other words: the symbol -x is the (unique) object which, when ->added<- to x, is equal to 0. By definition, y-x is short hand for y + (-x), i.e., adding -x to y. Alternatively, you can say that x-y = z if and only if z is the UNIQUE object such that x = z+y, provided this exists. The symbol x^{-1} is, by definition, the (unique) object which, when multiplied with x, is equal to 1, when it exists; by definition, y/x is shorthand for y*(x^{-1}). Alternatively you can say that y/x = z if and only if z is the UNIQUE object such that y = z*x, provided such a unique element exists. Ordinal addition is non-commutative. So is ordinal multiplication. That, by itself is no obstacle. But ordinal addition does NOT have additive inverses. And the equation x = z+y does not always have a unique solution z. So for example, writing omega - n = alpha would mean that alpha is the unique object such that omega = alpha+n. But there is no such element at all, unless n = omega, in which case the alpha is any natural number, and therefore is not unique. No matter how you cut it, you don't have a well defined subtraction for ordinals. Even if you go to cardinals, where addition and multiplication is commutative, you still run into problems. For infinite cardinals, kappa + lambda = kappa*lambda = max{kappa, lambda}, so again there are no unique solutions to the equations, so no well defined addition nor multiplication. Just because you can write down a symbol does not mean this makes sense. >For example : n + Omega = Omega >but Omega + n > Omega >Now we can define two operators of subtraction , No, you cannot. Because your equations do not represent unique solutions. >on is left subtraction >denoted as L - and the other is right subtraction denoted as -R as >followes >Define: >Omega L- n = Omega You can make any definitions you want, but this objects do NOT satisfy the rules that relate addition with subtraction, and this subtraction does not have the usual properties of subtraction. As such, it is foolish to call it subtraction, because it is nothing like subtraction. With usual subtraction, if a -x = a - y, then x=y. Here, no such thing. Just one example of why it is foolish to try to call it subtraction. >Omega -R n < Omega >n is a finite number The ONLY ordinals strictly less than omega are the natural numbers. So here you would be saying that right subtracting a finite number from omega yields a finite number. Which makes this subtraction something that has absolutely nothing to do with addition or subtraction. Which is why we do not use it. The rest of your nonsense suffers form the same problems. You define, but you never bother to think about the consequences of such definitions. >Now 1 L / Omega < Omega = infentismal = Left one omegath There are no infinitesimals among the reals or among the ordinals. This is all nonsense in that context. Even in the context of nonstandard analysis, where infinitesimals exist, it is still false that 0.9999... is different from 1. -- === Subject: Re: Question addition to your ignorance about mathematics, you also exhibit an > appalling ignorance about usenet ettiquette. That one is easier to > learn; why don't you? >I wonder why subtraction and division are not defined for Transfinite >Ordinal numbers? > I told you why: subtraction is really addition of additive > inverses. There are no additive inverses for ordinal numbers. > In other words: the symbol -x is the (unique) object which, when > ->added<- to x, is equal to 0. By definition, y-x is short hand for > y + (-x), i.e., adding -x to y. > Alternatively, you can say that x-y = z if and only if z is the UNIQUE > object such that x = z+y, provided this exists. > The symbol x^{-1} is, by definition, the (unique) object which, when > multiplied with x, is equal to 1, when it exists; by definition, y/x > is shorthand for y*(x^{-1}). > Alternatively you can say that y/x = z if and only if z is the > UNIQUE object such that y = z*x, provided such a unique element exists. > Ordinal addition is non-commutative. So is ordinal > multiplication. That, by itself is no obstacle. But ordinal addition > does NOT have additive inverses. And the equation x = z+y does not > always have a unique solution z. So for example, writing omega - n = > alpha would mean that alpha is the unique object such that omega = > alpha+n. But there is no such element at all, unless n = omega, in > which case the alpha is any natural number, and therefore is not > unique. No matter how you cut it, you don't have a well defined > subtraction for ordinals. > Even if you go to cardinals, where addition and multiplication is > commutative, you still run into problems. For infinite cardinals, > kappa + lambda = kappa*lambda = max{kappa, lambda}, so again there are > no unique solutions to the equations, so no well defined addition nor > multiplication. > Just because you can write down a symbol does not mean this makes sense. >For example : n + Omega = Omega >but Omega + n > Omega >Now we can define two operators of subtraction , > No, you cannot. Because your equations do not represent unique solutions. >on is left subtraction >denoted as L - and the other is right subtraction denoted as -R as >followes >Define: >Omega L- n = Omega > You can make any definitions you want, but this objects do NOT satisfy > the rules that relate addition with subtraction, and this > subtraction does not have the usual properties of subtraction. As > such, it is foolish to call it subtraction, because it is nothing like > subtraction. > With usual subtraction, if a -x = a - y, then x=y. Here, no such > thing. Just one example of why it is foolish to try to call it subtraction. Hmmm........... With usual addition if a+x = b+y , a<>b , then x <> y While a + Omega = b + Omega , a <> b and it is not foolish to try to call it addition! >Omega -R n < Omega >n is a finite number > The ONLY ordinals strictly less than omega are the natural numbers. So > here you would be saying that right subtracting a finite number from > omega yields a finite number. Which makes this subtraction > something that has absolutely nothing to do with addition or > subtraction. Which is why we do not use it. > The rest of your nonsense suffers form the same problems. You > define, but you never bother to think about the consequences of such > definitions. >Now 1 L / Omega < Omega = infentismal = Left one omegath > There are no infinitesimals among the reals or among the > ordinals. This is all nonsense in that context. > Even in the context of nonstandard analysis, where infinitesimals > exist, it is still false that 0.9999... is different from 1. > -- Let me see: If we visualise Omega as a raw or stars *****....... and the finite natural number n as ****....n* Now I will attempt to visualize ordinal summation and then to derive ordinal subraction from it. ***...n* + *****....... = *****...n*****......... = ***...... While ****.... + ***...n* = ***....... ***..n* > ****..... ( - ***...n* ) + *****....... = 0000...n0 ****....... = ****.......... a unique result While ****...... + (-***...n*) = ****.......000...n0 < ****...... a unique result Now multiplication is of two types n.Omega = Omega + Omega+Omega +.......+ nth Omega > Omega and Omega .n = n+n+n+n+............. = Omega So it follows that we should have two types of division of ordinals. 1) Omega/ n = z , z: n.z=Omega here z< Omega and it equal one nth of Omega and 2) Omegan = z , z: z.n = Omega here it is obvious that z= Omega In a similar way n/Omega = z , z: Omega.z= n here z should be one Omegath of n. and n Omega = z , z: z.Omega= n here the only chance for z is when it is zero. because Omega-Omega= n can be true. I don't see these definitions all together non sense. Zuhair === Subject: Re: Question days. My association with the Department is that of an alumnus. I am tired of you not bothering to clean up your replies; of you repeating the same thing; and of you ignoring what I write to continue insisting on the same assertions even when I have explained what the problems are. And this does not even touch on how tired I am of your continued misuse of standard mathematical terminology. I'll make it simple: GO READ A MATH BOOK ON THIS TOPIC. I'll make it even simpler: read the following book: _Naive Set Theory_ by Paul R. Halmos. Undergraduate Texts in Mathematics Springer-Verlag It is a small book: a mere 104 pages (including two in the index), plus two of introduction, and one for a table of contents. It will cover the basics of set theory, and the basics of natural numbers, ordinals, and cardinals. Then, and only then, return to these musings of yours. Until you provide some evidence that you have bothered to attempt to acquire the minimum of knowledge necessary to continue these discussions, and the minimum of usenet ettiquette to do so without wasting time and space, I will beg off. -- === Subject: Re: Question repeating the same thing; and of you ignoring what I write to continue > insisting on the same assertions even when I have explained what the > problems are. And this does not even touch on how tired I am of your > continued misuse of standard mathematical terminology. > I'll make it simple: GO READ A MATH BOOK ON THIS TOPIC. I'll make it > even simpler: read the following book: > _Naive Set Theory_ > by Paul R. Halmos. > Undergraduate Texts in Mathematics > Springer-Verlag > It is a small book: a mere 104 pages (including two in the index), > plus two of introduction, and one for a table of contents. It will > cover the basics of set theory, and the basics of natural numbers, > ordinals, and cardinals. > Then, and only then, return to these musings of yours. Until you > provide some evidence that you have bothered to attempt to acquire the > minimum of knowledge necessary to continue these discussions, and the > minimum of usenet ettiquette to do so without wasting time and space, > I will beg off. > -- You are very stuborn, you only wasted my time. I was looking forward to discuss these matters with someone who is more enlightend than you, some body like Tony. Go and dwell in the darkness of your standard math. Stay their like a rock in a mountine. Zuhair === Subject: Re: Question > I am tired of you not bothering to clean up your replies; of you > repeating the same thing; and of you ignoring what I write to continue > insisting on the same assertions even when I have explained what the > problems are. And this does not even touch on how tired I am of your > continued misuse of standard mathematical terminology. > I'll make it simple: GO READ A MATH BOOK ON THIS TOPIC. I'll make it > even simpler: read the following book: > _Naive Set Theory_ > by Paul R. Halmos. > Undergraduate Texts in Mathematics > Springer-Verlag > It is a small book: a mere 104 pages (including two in the index), > plus two of introduction, and one for a table of contents. It will > cover the basics of set theory, and the basics of natural numbers, > ordinals, and cardinals. > Then, and only then, return to these musings of yours. Until you > provide some evidence that you have bothered to attempt to acquire the > minimum of knowledge necessary to continue these discussions, and the > minimum of usenet ettiquette to do so without wasting time and space, > I will beg off. > -- > You are very stuborn, you only wasted my time. I was looking forward to > discuss these matters with someone who is more enlightend than you, > some body like Tony. > Go and dwell in the darkness of your standard math. Stay their like a > rock in a mountine. > Zuhair If you are taking TO as your ideal, whatever you are interested in, it is not mathematics and it is not knowledge. === Subject: Re: Question days. My association with the Department is that of an alumnus. >> I am tired of you not bothering to clean up your replies; of you >> repeating the same thing; and of you ignoring what I write to continue >> insisting on the same assertions even when I have explained what the >> problems are. And this does not even touch on how tired I am of your >> continued misuse of standard mathematical terminology. >> I'll make it simple: GO READ A MATH BOOK ON THIS TOPIC. I'll make it >> even simpler: read the following book: >> _Naive Set Theory_ >> by Paul R. Halmos. >> Undergraduate Texts in Mathematics >> Springer-Verlag >> It is a small book: a mere 104 pages (including two in the index), >> plus two of introduction, and one for a table of contents. It will >> cover the basics of set theory, and the basics of natural numbers, >> ordinals, and cardinals. >> Then, and only then, return to these musings of yours. Until you >> provide some evidence that you have bothered to attempt to acquire the >> minimum of knowledge necessary to continue these discussions, and the >> minimum of usenet ettiquette to do so without wasting time and space, >> I will beg off. >You are very stuborn, Right. Because I'm the one who repeated the same thing even after agreeing it was nonsense as written, over and over again... > you only wasted my time. You wasted your own time, and you wasted mine by ignoring the replies > I was looking forward to >discuss these matters with someone who is more enlightend than you, You mean, you were hoping someone would say you were right. Too bad you were not. >some body like Tony. >Go and dwell in the darkness of your standard math. Stay their like a >rock in a mountine. Go buy a spellchecker. -- === Subject: Re: Question days. My association with the Department is that of an alumnus. You still seem incapable of adhering to usenet ettiquette and edit your replies. More evidence of laziness, I gather. >> You can make any definitions you want, but this objects do NOT satisfy >> the rules that relate addition with subtraction, and this >> subtraction does not have the usual properties of subtraction. As >> such, it is foolish to call it subtraction, because it is nothing like >> subtraction. >> With usual subtraction, if a -x = a - y, then x=y. Here, no such >> thing. Just one example of why it is foolish to try to call it subtraction. >Hmmm........... >With usual addition if a+x = b+y , a<>b , then x <> y >While a + Omega = b + Omega , a <> b >and it is not foolish to try to call it addition! We do not call it addition, we call it ordinal addition. [.snip a lot of stuff that you should have removed yourself, for instance my sig; physical laziness to complement your mental one.] >Let me see: >If we visualise Omega as a raw or stars *****....... >and the finite natural number n as ****....n* >Now I will attempt to visualize ordinal summation and then to derive >ordinal subraction from it. You are not deriving ordinal subtraction; you are, at best, finding possible answers for specific expressions involving ->specific<- ordinals (to wit, omega and finite ordinals). You are still mssing a lot of ordinals. >***...n* + *****....... = *****...n*****......... = ***...... >While ****.... + ***...n* = ***....... ***..n* > ****..... This is just a complicated way of saying: for ordinal addition, n+omega = omega, and omega+n = omega + n <> omega. > ( - ***...n* ) + *****....... = 0000...n0 ****....... = ****.......... > a unique result Nonsense in this notation. There are no 0s. All you have are *s. The finite ordinals were finite sequences of *s, the ordinal omega was a (countably infinite) sequence of stars, AND THAT IS ALL YOU DEFINED. Where did this idiocy of 0s come from? Sloppy thinking (if any thinking went into it). And, the problem is not the unique answer here. Once again, you demonstrate that you DID NOT bother to read what was written, and prefer instead to babble on. You can define it any way you want and have a unique answer. I could define it as (-n) + omega = apple tree and it would be a unique answer. But n + appletree is not equal to omega. and n is not the unique thing to which you can add apple tree to get omega. >While ****...... + (-***...n*) = ****.......000...n0 < ****...... a >unique result Your expression ***... 0000 is nonsense in this notation. All you have are *s, where did this nonsense 0 come from? >Now multiplication is of two types No. You have lots of types. The ordinals do not end at omega. >n.Omega = Omega + Omega+Omega +.......+ nth Omega > Omega >and >Omega .n = n+n+n+n+............. = Omega This is at odds with the standard definition of ordinal multiplication. Standard ordinal multiplication in these cases would be that Omega * 0 = 0 Omega * (n+1) = Omega*n + Omega and in general alpha * 0 = 0 alpha * (beta + 1) = (alpha*beta) + alpha and for limit ordinals gamma, alpha * gamma = union (alpha * beta) with the union ranging over all betaI don't see these definitions all together non sense. Nobody said you could not give definitions. You can define every ordinal addition and every ordinal multiplication to be anything you want. You can define them to be a pony; you can define them to be Santa Claus; you can define them to be all equal to 3. But the definitions simply do not satisfy the properties you want. The definitions will not necessarily satisfy such properties as alpha - beta < alpha for all beta>0. They will not necessarily satisfy that if alpha < beta then alpha - gamma < beta - gamma. They will not necessarily satisfy that if alpha and beta are ordinals then alpha - beta will be an ordinal. Just giving a definition is not sufficient. Just writing down that you will declare a string of symbols to be equal to another string of symbols does not impart upon them the properties that the symbols have in other contexts. And you insist not only in wanting to define the symbols for subtraction, you also want them to satisfy the properties of subtraction such as alpha - beta < alpha. It is simply assuming that such is true that is nonsense, and it is simply assuming that there is an obvious definition which is also nonsense. -- === Subject: Re: Question your replies. More evidence of laziness, I gather. >> You can make any definitions you want, but this objects do NOT satisfy >> the rules that relate addition with subtraction, and this >> subtraction does not have the usual properties of subtraction. As >> such, it is foolish to call it subtraction, because it is nothing like >> subtraction. >> With usual subtraction, if a -x = a - y, then x=y. Here, no such >> thing. Just one example of why it is foolish to try to call it subtraction. >Hmmm........... >With usual addition if a+x = b+y , a<>b , then x <> y >While a + Omega = b + Omega , a <> b >and it is not foolish to try to call it addition! > We do not call it addition, we call it ordinal addition. > [.snip a lot of stuff that you should have removed yourself, > for instance my sig; physical laziness to complement your > mental one.] >Let me see: >If we visualise Omega as a raw or stars *****....... >and the finite natural number n as ****....n* >Now I will attempt to visualize ordinal summation and then to derive >ordinal subraction from it. > You are not deriving ordinal subtraction; you are, at best, finding > possible answers for specific expressions involving ->specific<- > ordinals (to wit, omega and finite ordinals). You are still mssing a > lot of ordinals. >***...n* + *****....... = *****...n*****......... = ***...... >While ****.... + ***...n* = ***....... ***..n* > ****..... > This is just a complicated way of saying: for ordinal addition, > n+omega = omega, and omega+n = omega + n <> omega. > ( - ***...n* ) + *****....... = 0000...n0 ****....... = ****.......... > a unique result > Nonsense in this notation. There are no 0s. All you have are > *s. The finite ordinals were finite sequences of *s, the ordinal > omega was a (countably infinite) sequence of stars, AND THAT IS ALL > YOU DEFINED. Where did this idiocy of 0s come from? * - * = 0 got it, is it that difficult. Zuhair > Sloppy thinking (if any thinking went into it). > And, the problem is not the unique answer here. Once again, you > demonstrate that you DID NOT bother to read what was written, and > prefer instead to babble on. You can define it any way you want and > have a unique answer. I could define it as > (-n) + omega = apple tree > and it would be a unique answer. > But n + appletree is not equal to omega. and n is not the unique > thing to which you can add apple tree to get omega. >While ****...... + (-***...n*) = ****.......000...n0 < ****...... a >unique result > Your expression ***... 0000 is nonsense in this notation. All > you have are *s, where did this nonsense 0 come from? >Now multiplication is of two types > No. You have lots of types. The ordinals do not end at omega. >n.Omega = Omega + Omega+Omega +.......+ nth Omega > Omega >and >Omega .n = n+n+n+n+............. = Omega > This is at odds with the standard definition of ordinal > multiplication. Standard ordinal multiplication in these cases would > be that > Omega * 0 = 0 > Omega * (n+1) = Omega*n + Omega > and in general > alpha * 0 = 0 > alpha * (beta + 1) = (alpha*beta) + alpha > and for limit ordinals gamma, > alpha * gamma = union (alpha * beta) with the union ranging over all beta And still you confuse being able to define an expression with the > expression making sense in the context you wish to use it. You are > able to define anything any way you want. But your definitions do NOT > interact well with ordinal addition, ordinal multiplication (whether > the standard or your opposite convention), ordinal inequalities, and > the like. > Your major error consists of thinking that just because you can write > down an expression omega - n, and just because you can write down > omega - n = k for some k, that automatically implies all the usual > theorems about inequalities and the like. They do not; those theorems > come from properties of these operations which your definitions simply > do NOT have. >I don't see these definitions all together non sense. > Nobody said you could not give definitions. You can define every > ordinal addition and every ordinal multiplication to be anything you > want. You can define them to be a pony; you can define them to be > Santa Claus; you can define them to be all equal to 3. > But the definitions simply do not satisfy the properties you want. The > definitions will not necessarily satisfy such properties as alpha - > beta < alpha for all beta>0. They will not necessarily satisfy that > if alpha < beta then alpha - gamma < beta - gamma. They will not > necessarily satisfy that if alpha and beta are ordinals then alpha - > beta will be an ordinal. > Just giving a definition is not sufficient. Just writing down that you > will declare a string of symbols to be equal to another string of > symbols does not impart upon them the properties that the symbols have > in other contexts. And you insist not only in wanting to define the > symbols for subtraction, you also want them to satisfy the > properties of subtraction such as alpha - beta < alpha. It is simply > assuming that such is true that is nonsense, and it is simply assuming > that there is an obvious definition which is also nonsense. > -- === Subject: Re: Question > ( - ***...n* ) + *****....... = 0000...n0 ****....... = ****.......... > a unique result > Nonsense in this notation. There are no 0s. All you have are > *s. The finite ordinals were finite sequences of *s, the ordinal > omega was a (countably infinite) sequence of stars, AND THAT IS ALL > YOU DEFINED. Where did this idiocy of 0s come from? > * - * = 0 > got it, is it that difficult. One might make a case for * - * = OR * - * = {}, (THE EMPTY SET} , but unless * represents some number, there is no case to be made for * - * = 0 > Zuhair Arturo's only idiocy is continuing to try to teach one so invincibly ignorant as Zuhair anything about mathematics. === Subject: Re: Question days. My association with the Department is that of an alumnus. >>***...n* + *****....... = *****...n*****......... = ***...... >>While ****.... + ***...n* = ***....... ***..n* > ****..... >> This is just a complicated way of saying: for ordinal addition, >> n+omega = omega, and omega+n = omega + n <> omega. >> ( - ***...n* ) + *****....... = 0000...n0 ****....... = ****.......... >> a unique result >> Nonsense in this notation. There are no 0s. All you have are >> *s. The finite ordinals were finite sequences of *s, the ordinal >> omega was a (countably infinite) sequence of stars, AND THAT IS ALL >> YOU DEFINED. Where did this idiocy of 0s come from? Can't even write without messing up the attributions, can you? >* - * = 0 >got it, is it that difficult. Nonsense. This is simply NOT contemplated in the notation YOU introduced. The notation YOU introduced contained NOTHING but *s. Then, suddenly, you introduce a new symbol which has NO meaning in the notation YOU created. That makes it nonense. No, it is clear that ->YOU<- are a lazy idiot. Either when you introduced the notation you failed to give it COMPLETELY, or else you changed notation mid-stream. Either way, you spoke nonsense. Were ordinals expressed ONLY as lists of *s? That's what YOU DEFINED THEM TO BE. Then putting 0s makes it nonsense. Duh. -- === Subject: Re: Question ***...n* + *****....... = *****...n*****......... = ***...... >>While ****.... + ***...n* = ***....... ***..n* > ****..... >> This is just a complicated way of saying: for ordinal addition, >> n+omega = omega, and omega+n = omega + n <> omega. >> ( - ***...n* ) + *****....... = 0000...n0 ****....... = ****.......... >> a unique result >> Nonsense in this notation. There are no 0s. All you have are >> *s. The finite ordinals were finite sequences of *s, the ordinal >> omega was a (countably infinite) sequence of stars, AND THAT IS ALL >> YOU DEFINED. Where did this idiocy of 0s come from? > Can't even write without messing up the attributions, can you? >* - * = 0 >got it, is it that difficult. > Nonsense. This is simply NOT contemplated in the notation YOU > introduced. The notation YOU introduced contained NOTHING but > *s. Then, suddenly, you introduce a new symbol which has NO meaning > in the notation YOU created. That makes it nonense. > No, it is clear that ->YOU<- are a lazy idiot. > Either when you introduced the notation you failed to give it > COMPLETELY, or else you changed notation mid-stream. Either way, you > spoke nonsense. > Were ordinals expressed ONLY as lists of *s? That's what YOU DEFINED > THEM TO BE. Then putting 0s makes it nonsense. Duh. > -- -***...n* + ****......... = 0000...n0***.......... The zeros denote the removed stars . yet since 000...n0***..... can be bijected to ****........ then they are equal But ****........ + (-***...n* ) = ****....000...n0 = Omega-n and this is a transfinite ordinal. this number cannot be bijected to ****..... , in exactly the same why ****..... cannot be bijected to ****..... ***...n* ( bijection her is respecting order ) The smallest transfinite ordinal that do not contain zeros in it after the natural numbers is Omega but ordinally speaking Omega is the smallest ordinal after the natural numbers as Cantor thought right finite subtracties of Omega are of smaller ordinal value than omega but yet bigger ordinal value than any finite ordinal. All of the above in absolutelly nonsense Smiles Just to wast you time Arturo Zuhair === Subject: Re: Question >>***...n* + *****....... = *****...n*****......... = ***...... >>While ****.... + ***...n* = ***....... ***..n* > ****..... >> This is just a complicated way of saying: for ordinal addition, >> n+omega = omega, and omega+n = omega + n <> omega. >> ( - ***...n* ) + *****....... = 0000...n0 ****....... = ****.......... >> a unique result >> Nonsense in this notation. There are no 0s. All you have are >> *s. The finite ordinals were finite sequences of *s, the ordinal >> omega was a (countably infinite) sequence of stars, AND THAT IS ALL >> YOU DEFINED. Where did this idiocy of 0s come from? > Can't even write without messing up the attributions, can you? > >* - * = 0 >got it, is it that difficult. > Nonsense. This is simply NOT contemplated in the notation YOU > introduced. The notation YOU introduced contained NOTHING but > *s. Then, suddenly, you introduce a new symbol which has NO meaning > in the notation YOU created. That makes it nonense. > No, it is clear that ->YOU<- are a lazy idiot. > Either when you introduced the notation you failed to give it > COMPLETELY, or else you changed notation mid-stream. Either way, you > spoke nonsense. > Were ordinals expressed ONLY as lists of *s? That's what YOU DEFINED > THEM TO BE. Then putting 0s makes it nonsense. Duh. > -- > -***...n* + ****......... = 0000...n0***.......... > The zeros denote the removed stars . Then you are essentially using a binary coding system based on two symbols . * and either 0, or possible a space, rather than a unary one based only on one symmbol. > Just to wast you time Arturo > Zuhair Waste mostly snipped. === Subject: Re: Question days. My association with the Department is that of an alumnus. >> The definition states that if the difference converges into zero then >> the two Cauchy sequences are equivalent, I disagree with that. >One cannot disagree with a definition in mathematics. One either >accepts it or rejects it. >> They should be equivalent only when the difference REACHES zero. >Using that as your equivalence relation, what you would get is only >another copy of the rational numbers, not the usual real numbers at all. I don't think you get the rational numbers. Cauchy sequences that converged to a rational number but were not eventually constant would not be equivalent to the constant sequence corresponding to that rational; and any such sequence would be non-equivalent to any shift on that sequence. So the sequences (1, 1, 1, ...) (1/2, 3/4, 7/8, ..., (2^n-1)/2^n, ...) (0, 1/2, 3/4, 7/7, ..., 2^{n-1}-1/2^{n-1}, ... ) would lie in distinct equivalence classes. Your equivalence classes are just the classes of all sequence that are eventually equal, so I think you would get essentially just the set of all Cauchy sequences back (essentially, mind you). -- === Subject: Re: Question >> The definition states that if the difference converges into zero then >> the two Cauchy sequences are equivalent, I disagree with that. >One cannot disagree with a definition in mathematics. One either >accepts it or rejects it. >> >> They should be equivalent only when the difference REACHES zero. >Using that as your equivalence relation, what you would get is only >another copy of the rational numbers, not the usual real numbers at all. > I don't think you get the rational numbers. Cauchy sequences that > converged to a rational number but were not eventually constant would > not be equivalent to the constant sequence corresponding to that > rational; and any such sequence would be non-equivalent to any shift > on that sequence. So the sequences > (1, 1, 1, ...) > (1/2, 3/4, 7/8, ..., (2^n-1)/2^n, ...) > (0, 1/2, 3/4, 7/7, ..., 2^{n-1}-1/2^{n-1}, ... ) > would lie in distinct equivalence classes. Your equivalence classes > are just the classes of all sequence that are eventually equal, so I > think you would get essentially just the set of all Cauchy sequences > back (essentially, mind you). I spoke too quickly. The zero equivalence class would be the set of sequences with finite carrier (zero for all but finitely many terms). An equivalence class in general would be the set of all sequences which differed from a given sequence at only finitely many terms. The result would not even be a field, since there are obvious zero divisors. e.g., the product of a sequence zero for all odd terms times a sequence zero at all odd positions would always be the zero sequence. === Subject: Generating the alternating group The symmetric group on n letters, S_n, can be generated by two elements. What is the minimal size of a generating set for A_n, the alternating group on n letters? Thx, G-H === Subject: Re: Generating the alternating group >The symmetric group on n letters, S_n, can be generated by two >elements. What is the minimal size of a generating set for A_n, the >alternating group on n letters? Answer is: 0 for n=0,1,2 1 for n=3 2 for n>3, n finite, n for n infinite. Derek Holt. === Subject: Re: Generating the alternating group > For any n, an easy induction argument shows that A_n is generated by > the 3-cycles (1,2,3), (2,3,4), ..., (n-2,n-1,n), and it follows that > A_n = < (1,2,3,...,n), (1,2,3) > when n is odd. > Similarly, you can prove A_n = < (2,3,...,n-1,n), (1,2,3) > for n even. Argh, I should have searched the group before... === Subject: Re: Variational problem? > Suppose we have a (kernel) function y = K_s(x) which is bounded and > continuous and has the following properties: > (i) K_s(x) >= 0 > (ii) integral K_s(x) = 1 > (iii) Take a > 0 fixed, > then integral_[ |t-T| > a ] K_s(t-T) dt -> 0 as s -> 0 > (iv) K_s(x) = K_s(-x) : symmetric in y-axis > The quantity (s) is called the spread of K_s(x). > Now we calculate the second order variance: integral K_s(x) x^2 dx . > We expect the outcome to be a (form factor) times the spread squared. > The (form factor) seems to be dependent of the shape of K_s(x) only. > Question: can it be proved that, for example, the Gassian shape makes > our form factor maximal or minimal? What shape makes the form factor > extreme? Can (a modification of) this question ueberhaupt be answered? Heh, heh. First define s . As follows: s^2 = integral K_s(x) x^2 dx . So, please, forget this vicious circle question. Sorry! Han de Bruijn === Subject: Corners in metric spaces I thought about combining solids in R^3. For example an ellipsoid and a ball are both smooth in an intuitive sense. Their union (if they have common points) is also smooth except at the points that belong to the boundary of both objects. I tried to think what characterizes this smoothness - property. What is true is that when corner point is defined properly then a (finite) ellipsoid does not have corner points and a (finite) smooth torus does not have corner points. Neither has their union corner points except at the points that belong to the boundary of both objects. Also the set of the corner points of their union is cornerless. I tried to define corner in metric space so that the definition would match the intuitive picture of a corner. Let M = (X, d) be a metric space. Definition of corner point Let S be subset of X. p is a corner point of S in X iff there exists a sequence B_1, B_2, B_3,... of open balls such that: 1. the elements in the sequence are pairwise disjoint 3. for each B_i: 3.1 there exists (at least) two distinct points such that both are boundary points of the closure of S and both are boundary points of the closure of B_i 3.2 the intersection of B_i and boundary of S is empty 3.3 the intersection of the closure of B_i and the closure of B_i+1 is a singleton and not in the closure of S 2. each sequence p_1, p_2, p_3, ... where each p_i is in B_i, has p as a limit point This definition works in many cases, for instance in R^2, with the usual metric, the solution set of y = |X| has a corner point at the origin while y = x^2 does not have corner points. In R^3 the set of corner points of the intersection of any two ellipsoids (degenerate or not) is cornerless. Definition of function C: C(S) is the set of corner points of S. Definition of cornerless set: S is cornerless <-> C(S) is empty. Definition of n-cornerless set: S is 0-cornerless <-> S is cornerless, S is n+1-cornerless <-> C(S) is n-cornerless Since the empty set is cornerless, S is n-cornerless -> S is n+1-cornerless I would like to have the following sentence true: Sentence 1: Let S_1 and S_2 be any two n-cornerless sets. Then their union is n+1-cornerless and their intersection is n+1-cornerless. Definition of rough set: A set is rough iff it is not n-cornerless for any finite n. For instance the Sierpinski-triangle in R^2 is rough. I think that the following sentence should be true at least in some spaces, for instance R^2 with usual metric: Sentence 2: For any set S in M, for the functions CSeq_0(S) = S, CSeq_n+1(S) = C(CSeq_n(S)), there exists a n such that CSeq_n+1(S) = CSeq_n(S) (this is trivially true for n-cornerless (with finite n) sets) Definition of stable corner points of a set S: Stable corner points of S is CSeq_n(S) such that n is the smallest non-negative integer such that CSeq_n+1(S) = CSeq_n(S). The set of stable corner points of a set in a metric space is not neccessarily empty, for instance in R^2 with the usual metric the Sierpinski-triangle is equal to its corner points and there for equal to its stable corner points. The stable corner points of the union of a disjoint rectangle and a Sierpinski-triangle is the Sierpinski-triangle. Definition of a slice in M = (X, d): A set S in M is a slice iff for every n-cornerless subset x in X, the intersection of the S and x is n-cornerless. The following should perhaps be true under right definitions Sentence 3: The intersection of any two slices is a slice. The following sentence may be true Sentence 4: The Mandelbrot-set does not have corner-points in R^2, usual metric Also because of the Sierpinski triangle, I guess that roughness has something to do with dimensionality, for example: Sentence 5: The set of stable corner points of a rough set S in R^n, usual metric, has non-integer Hausdorff-dimension as a metric space. I thought that with the proper definition, Sentence 1 would have been easy to prove even for me. If the definition does not work, I would appreciate references or working definitions. Then I would like to learn a proof or counterexample of Sentence 1 and to get scetch of proof or counterexample of Sentence 2 We Pretty === Subject: homomorphism of module na.de> I wouldn't call 2Z a subring at all; precisely since I prefer my rings to > have identity. It is a submodule, sure. It is thus, also an ideal. But > normally, to me, not a subring. Submodule? No. Modules are a 2-sorted theory, rings a 1-sorted theory. Yes, 2Z is an idea. What about (2,2)Z = { (2n,2n) | n in Z } ? It isn't an ideal, so what is it? A ring with an idenity crises or an ideal of (1,1)Z = { (n,n) | n in Z } ? > If you care to work with rings and drop the requirement of existence of 1 > and 1!=0, then you could definitely have 2Z a subring of Z; and that's > entirely reasonable; only I normally don't. Authors do vary upon this item. Are there any expressions short of ring with identity or ring with unity that clarify in what context ring is being used? === Subject: Re: homomorphism of module > 2Z is an unidentified subring? > I wouldn't call 2Z a subring at all; precisely since I prefer my rings to > have identity. It is a submodule, sure. It is thus, also an ideal. But > normally, to me, not a subring. > Submodule? No. > Modules are a 2-sorted theory, rings a 1-sorted theory. But every ring is also a model for that two sorted theory, just write down the relevant equations. (actually, for a fixed ring R you can also recode the theory of R-modules as a one-sorted theory by considering every element of R as an operator) >[...] > If you care to work with rings and drop the requirement of existence of 1 > and 1!=0, then you could definitely have 2Z a subring of Z; and that's > entirely reasonable; only I normally don't. > Authors do vary upon this item. Are there any expressions short of > ring with identity > or > ring with unity > that clarify in what context ring is being used? It is similar to the choice between (a) compact and compact Hausdorff and (b) quasicompact and compact In short, authors will usually state at the beginning which of the two variants (a) ring and ring with identity or (b) ring without identity and ring will be used. The concept which is used more often in that context usually receives the shorter name, thereby avoiding the tedious longer version most of the time. That is why version (b) is used in many texts on algebra, especially commutative algebra (where often ring means commutative ring), whereas you may find version (a) more often in analysis. I have seen rng for ring without Identity in Rowen, and it seems to be also in Jacobsons algebra book. There is also rig for ring without Negatives which in turn is still shorter than semiring with unit and absorbing zero. Marc === Subject: Re: homomorphism of module > I wouldn't call 2Z a subring at all; precisely > since I prefer my rings to > have identity. It is a submodule, sure. It is thus, > also an ideal. But > normally, to me, not a subring. > Submodule? No. > Modules are a 2-sorted theory, rings a 1-sorted > theory. What's that? Every ideal of a commutative ring R is a submodule of the ring R as a module over itself. This is the definition of the notion of an ideal. For non-commutative rings one has to adapt this definition depending on whether right or left or two-sided ideals are meant. > Yes, 2Z is an idea. What about > (2,2)Z = { (2n,2n) | n in Z } ? > It isn't an ideal, so what is it? > A ring with an idenity crises or an ideal of > (1,1)Z = { (n,n) | n in Z } ? If in (1,1)Z addition and multiplication are defined to be component-wise, then this ring is just Z. And (2,2)Z then is just the ideal 2Z in Z. > If you care to work with rings and drop the > requirement of existence of 1 > and 1!=0, then you could definitely have 2Z a > subring of Z; and that's > entirely reasonable; only I normally don't. > Authors do vary upon this item. Are there any > expressions short of > ring with identity > or > ring with unity > that clarify in what context ring is being used? Jacobson calls rings without 1 >rngs<. Of course my son of age 1.5 years can pronounce this word better than I can ... H === Subject: Re: homomorphism of module On Fri, 28 Apr 2006 01:28:54 -0700, William Elliot >> 2Z is an unidentified subring? >> I wouldn't call 2Z a subring at all; precisely since I prefer my rings to >> have identity. It is a submodule, sure. It is thus, also an ideal. But >> normally, to me, not a subring. >Submodule? No. >Modules are a 2-sorted theory, rings a 1-sorted theory. >Yes, 2Z is an idea. What about > (2,2)Z = { (2n,2n) | n in Z } ? >It isn't an ideal, so what is it? >A ring with an idenity crises or an ideal of > (1,1)Z = { (n,n) | n in Z } ? >> If you care to work with rings and drop the requirement of existence of 1 >> and 1!=0, then you could definitely have 2Z a subring of Z; and that's >> entirely reasonable; only I normally don't. >Authors do vary upon this item. Are there any expressions short of > ring with identity > ring with unity >that clarify in what context ring is being used? Most books/courses state at the outset that all rings are assumed to have a multiplicative identity 1, with 1 not equal 0. Thus, once that's declared as the default, they just say ring. Presumably, if at some point they need to use a ring without identity, they would then have to specify that explicitly. quasi === Subject: Re: homomorphism of module have a multiplicative identity 1, with 1 not equal 0. Thus, once > that's declared as the default, they just say ring. Presumably, if > at some point they need to use a ring without identity, they would > then have to specify that explicitly. My reference John Beachy, Abstract Algebra defines communative ring and communative ring with indentity. Is it common for authors to consider all rings communative? As common as some authors consider all spaces are Hausdorff? === Subject: Re: homomorphism of module >>Most books/courses state at the outset that all rings are assumed to >>have a multiplicative identity 1, with 1 not equal 0. Thus, once >>that's declared as the default, they just say ring. Presumably, if >>at some point they need to use a ring without identity, they would >>then have to specify that explicitly. > My reference John Beachy, Abstract Algebra defines > communative ring and communative ring with indentity. Jacobson was a standard grad school algebra textbook (i.e. it was when I went to grad school), and he defines a ring to always have a multiplicative neutral element. He calls a ring without identity a rng. I think that this suggestion hasn't caught the fancy of other authors:) Certainly in algebra/number theory, it is IMVHO standard practice for all the rings to have an identity, and for IMVHO mostly historical reasons mention this in the preface or whatever. In analysis one does encounter stuctures called algebras (= something that in algebra is both a ring and a vector space) that don't have a 1. > Is it common for authors to consider all rings communative? > As common as some authors consider all spaces are Hausdorff? Well, in e.g. algebraic geometry it probably is relatively common for the reason that they won't need non-commutative rings. In group theory obviously the reverse is true. Number theory is probably in-between. My guesstimate is that it is more common to define a compact space to also be Hausdorff than to insist on rings being commutative. Jyrki === Subject: simplifying sqrt(14 -6sqrt(5))+(26-15sqrt(3))^(1/3) +(173+76sqrt(3)+60sqrt(5)+44sqrt(15))^(1/4) Which method will you propose to work out and if possible simplify this expression with radicals ? sqrt(14 -6*sqrt(5) ) + (26-15*sqrt(3) )^ (1/3) + (173+76*sqrt(3) + 60*sqrt(5)+44*sqrt(15) )^ (1/4) Alain === Subject: Re: Reference Needed > What does square free mean ? There does not exist a prime p such that p*p divides the number. Bob Kolker === Subject: Re: Reference Needed >> What does square free mean ? > There does not exist a prime p such that p*p divides the number. > Bob Kolker Or, equivalently, that the number n, when factored as a product of powers of primes, each prime has only an exponent 1. factor, and, what is more, can be factored into squarefree factors. === Subject: Re: Reference Needed >> If A,B,C,D,a,b,c,d are integers > 1, with a,b,c,d squarefree, >> (a,b) = (c,d) = 1, /x := sqrt(x) >> then A /a + B /b = C /c + D /d >> -> {a , b} = {c , d} > Now... In my prior post here, using only very simple high-school algebra, I proved {A /a , B /b} = {C /c , D /d} and hence {AA a , BB b} = {CC c , DD d} [square prior]. Therefore {a , b} = {c , d} [take squarefree part] I doubt that there exists a proof significantly simpler than that. If the whole proof isn't clear, please feel free to ask questions. --Bill Dubuque === Subject: Re: Reference Needed (1) > ab = cd (1.1) ; AB = CD (1.2) > Aa^1/2 + Bb^ 1/2 = Cc^ 1/2 + D^ d 1/2 (1) > Now the condition (a, b) = 1 is still valid. But (c, d) = 1 is not > imposed. Why not, isn't it part of the premise? > Do you still agree that a = c and b = d. Or a = d and b = c No. > I believe it is correct. Kindly give your reasonings. Without context included in your post, only sloppy answers are possible. -- To Google and MathForum users: Reply only if adequate context is included _within_ the reply. Otherwise all contexts are removed from my view, the flow of thought disrupted and chaos reigns. In particular for Google users: Instead of simply hitting the prominent Reply link, which doesn't include a copy of the post to which one is replying, click the Show Options link (toward the top of an item in the thread), which causes a shaded area of links to appear next to the top of the item, including Reply (first) that does introduce a copy of the previous text (offset by > signs in the usual fashion). ---- === Subject: sequence finite linear programming problems converges to infinite linear programming problem Hello group! I need to prove a theorem that shows that some sequence of linear programming problems converges to a infinite linear programming problem: something like If Prob_n: is max (1/n) sum_{i=1}^n y(i) s.t. 0 <= x(i) <= 2 0 <= y(i) <= 4 0 <= (1/n) sum_{j=1}^i [x(j) - y(j)] <= 1 for all i= 1, 2, ..., n then prob_n tends to prob^* Prob^* max int_0^1 y(t) dt s.t. 0 <= x(t) <= 2 0 <= y(t) <= 4 0 <= int_0^t x(t) - y(t) dt<= 1 t in [0,1] and that the discrete approximation y(i) converges to y(t) I have been looking in ANDERSON-NASH Linear programming in infinite.dimensional spaces, but they just assume that this limit exists. Any pointers on how to prove this or some reference? Diego === Subject: Integer Quadrangle I found this on MathWorld: http://mathworld.wolfram.com/RationalDistanceProblem.html - can you squeeze that 4-point constellation on a grid such that all four points have integer coordinates too? (You may multiply the distances with any surd if neccessary.) If not - is it possible at all to have four grid points with rational distances? (No rectangle, no rhombus, no circle, no 3 on a line, no cheating at all :-) -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de His-Ala-Sec-Lys-Glu Arg-Glu-Asp-Asp-Met-Ala-Asn-Asn === Subject: Re: Integer Quadrangle Edit: It isn't. So, rephrased question: Can triangles with sidelengths a,b,c;a,e,f;b,d,f;c,d,e; all be Heronian? (Only in that case they can be drawn on a grid.) http://www.mathe2.uni-bayreuth.de/axel/erdoes_diophantine_graphs.pdf Doesn't answer the question on a *non-special* configuration, though. -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de His-Ala-Sec-Lys-Glu Arg-Glu-Asp-Asp-Met-Ala-Asn-Asn === Subject: Re: Looking for a CRC algorithm > I am looking for a CRC or checksum algorithm with the property that a > small change in the content of the file results in a small change in > the checksum. Typically, most of the checksum computing algorithms such > as MD5, SHA, etc go to the other extreme i.e. a small change in the > file contents results in a large change in the checksum. Does anybody > know of any algorithm with the reverse property? count the number of 1 bits in the file lika all checksum algorthms it won't detect all possible changes but should have the property you want. Bye. Jasen === Subject: Re: Looking for a CRC algorithm > The purpose is to detect small changes in a document. Also the original > document will not be available when the information that the document > has changed is required. And I need to answer the question as to > whether the changes are small or large which precludes using a regular > hash or CRC. That's impossible. checkssums (that are smaller than the file) can't even tell you if a file hasn't been altered. Bye. Jasen === Subject: Re: Looking for a CRC algorithm (snip on detecting small or large changes in a document with a hash, checksum, or CRC.) > I do not think this is really possible. With a standard method you will > detect a small number of changes, but a large number of changes may remain > undetected. You can not have short information that allows you to detect > all changes, whether they are small or large. Whatever the method, there > will be documents that are (possibly) completely unrelated that will > generate the same checksum, unless the checksum you generate is at least > as large as the document itself. That is true, but that doesn't mean you can't find one that has a high probability of detecting small or large changes in a document. My thought for a text document is to first apply a hash function to each line. That will reduce each line to a constant small number of bits, maybe 16 or 32. If the result is still too big, maybe some function applied to those hash values. How about first hash each line to a 16 bit value, such as with a 16 bit CRC function. Next take a 65536 bit string and set the bit corresponding to each hash value. Most small changes will only change a small number of bits, most large changes a large number of bits for some definition of small and large. For larger files, it is likely that 16 bit hash functions are too small, but the idea can be extended. If 65536 bits is too many, just add up the number of ones, again it should not change too much for small changes, though it is more likely not to change much for large changes. -- glen === Subject: Re: Looking for a CRC algorithm a.k.a. Keogh's compression dissimilarity measure. > Hmm. Considering its obviousness (if you know of Kolmogorov complexity), > and the fact that it is at least 30 years old, he has done well to get > it named after him. > Nick Maclaren. Yeah, he does cite earlier applications going back to the 1970's I had the same idea when doing some postdoc research in time series anomaly detection but then I found the above paper and discovered my idea was not so new after all. -- Matt Mahoney === Subject: Re: Looking for a CRC algorithm Originator: nmm1@draco.cus.cam.ac.uk |> I had the same idea when doing some postdoc research in time series |> anomaly detection but then I found the above paper and discovered my |> idea was not so new after all. Of the few thousand techniques I have invented, a short search found that 90% were already known. Later (well-hidden) data showed that a further 9% had been. I suspect that 0.9% of the remainder have been, but I haven't found the references yet .... As this is normal, it shows why software patents (as currently proposed) are such a bad idea. Nick Maclaren. === Subject: Re: Looking for a CRC algorithm length N is very likely to be close to N*127.5, so the chance of a checksum > collision between random files of the same length is rather large. Yes, I agree with your analysis. Fortunately, most files are not the same length. In any case, the original query was not find a way to identify which files are close to each other among a large set by looking at their check value, but rather if you know the check value the file used to have, being able to answer whether has it changed a lot or a little based on the check value for that file now. Your analysis of the checksum does affect the case for no insertions or deletions, where the alteration vector (the arithmetic difference between the files) tends to be clustered around zero, even for large random changes. A simple checksum is one way to try to get a handle on whether it's changed a little or a lot. However for several reasons, including the change cancellations in a sum that you point out, as well as the arithmetic value of a change having a wide range, it's not a very good way. But it's a lot better than the usual signatures which, by design, will amplify even one small change in the file into a large change in the signature. So does someone know a better way? In case this problem doesn't have a name, I will now attempt to name it for future reference: The approximately proportional signature problem. I.e. come up with a signature so that by comparing two signatures of a changed file, you can tell approximately how much of the file has changed. The problem could be made more explicit by defining a measure of difference for alterations, insertions, and deletions, and defining approximately, but it's probably best to leave it open to interpretation for now to see what kind of answers pop up. mark === Subject: Re: Fourier coefficients of a complex borel measure On Thu, 27 Apr 2006 10:14:33 -0700, The World Wide Wade >>The problem is the following: mu is a complex borel measure on the unit >>circle T. Fourier coefficients mu' (n) are defined as integral of >>exp(-i*n*t) w.r.t. mu(t) (n= all integers) How do we prove that if >>mu'(n) goes to 0 as n goes to +infinity, mu'(n) goes to 0 as n goes to >>-infinity. >>I think I can use complex conjugate and approximate a Radon-Nikodym >>derivative by trigonometric polynomials. Is that right? >> Yes. Say nu(E) is the conjugate of mu(E), so that nu'(n) = -mu'(n). >Did you mean mu^(-n) = conj(nu^(n))? No, but of course that's what I meant to mean. >> Show that nu << mu, then approximate d nu/ dmu by a trigonometric >> polynomial (in L^1(mu)) and see what follows. ************************ David C. Ullrich === Subject: Re: New Theory about 0.999... and 1 >There is a new theory that may end the debate about 1=0.999... There is no such debate. >the whole theory can be found in this link >http://www.geocities.com/humood_theory/index.htm >here is a Quote about cracking the famouse proof: What's below is not a proof. It can't possibly be a proof because it has no connection with the _definition_ of 0.999... . But the fact that what's below is not a proof does not imply that no proof is possible. The proof is not hard, once you have the definitions straight, and in fact 0.999... = 1. Exactly. >Proof (A) > x=0.999... > 10x= 9.999... > 10x-x=9.999... - 0.999...=9 > 9x=9 > x=1 >Quote: >-------------------------------------------------------------- >I will start by cracking this proof and show it is in fact a Paradox >not a proof. >Let x = 0.999... >10x = 9x + x = [ 9* (0.999...) ] + 0.999... [ I expressed 10x as (9x >+x)] >10x - x = ( 9x + x ) - x = [ 9* (0.999...) ] + 0.999... - 0.999... >[Positive 0.999... goes with Negative 0.999...] >9x = 9* (0.999...) >x = 9* ( 0.999...) / 9 = 1 * 0.999... >x =0.999 >We did not end with x=1 >In fact what made it looks like a proof was the mistake we did in this >step: >10 x - x = 9.999... - 0.999... = 9 >This step assumes the value of infinite 9s after the decimal point in ( >10x ) equals the value of infinite 9s in (x) >The infinite decimals in x = 0.999... have different behavior than the >infinite 9s in 10x =9.999... Because 9s in x= 0.999... Go to infinity >faster than 9s in 10x=9.999... >(10 x ) is always one decimal behind (x) >When x = 0.99 then 10x=9.9 >When x = 0.9999 then 10x=9.999 >When x = 0.999999999 then 10x=9.99999999 >Value of Infinite decimals in (x) - Value of infinite decimals in >(10x) = >Lim x-> 8 9 / (10^x) >So the difference between the two values equals an infinitely small >fraction which is Lim x-> 8 9 / (10^x) >This means in Proof (A) when we say: 10 x - x = 9.999... - 0.999... = >We are already Assuming that a number minus an infinitely small >fraction is the number itself. >Which means we are already assuming that Lim x-> 8 1 - ( 1/x ) = >0.999... = 1 >So, in Proof (A) we already assumed 1= 0.999... Then no wonder we >finally ended with our assumption, which make it not a Proof at all. >You cannot prove any statement by assuming it to be true from the >beginning. >But when we explain (10 x ) in a very fundamental method no body can >argue with, we will say: >10x =0.999... + 0.999... + 0.999... + 0.999... +0.999... +0.999... >+0.999... +0.999... +0.999... +0.999... >And when we subtract x = 0.999... from it : >10x - x =0.999... + 0.999... + 0.999... + 0.999... +0.999... +0.999... >+0.999... +0.999... +0.999... + 0.999... - 0.999...= 0.999... * 9 >The infinite 9s in x after the decimal is gone with only one of those >0.999... Not all of them as we did in Proof (A). >------------------------------------------------------------------ >All the theory and its results can be seen in the link: >http://www.geocities.com/humood_theory/index.htm >For comments please write it here in this gruop or send it to the >e-mail >said_yam@yahoo.com ************************ David C. Ullrich === Subject: Re: New Theory about 0.999... and 1 > There is a new theory that may end the debate about 1=0.999... > the whole theory can be found in this link > http://www.geocities.com/humood_theory/index.htm > here is a Quote about cracking the famouse proof: > Proof (A) > x=0.999... > 10x= 9.999... > 10x-x=9.999... - 0.999...=9 > 9x=9 > x=1 > Quote: > -------------------------------------------------------------- > I will start by cracking this proof and show it is in fact a Paradox > not a proof. Well, you cracked something at any rate... > Let x = 0.999... > 10x = 9x + x = [ 9* (0.999...) ] + 0.999... [ I expressed 10x as (9x > +x)] > 10x - x = ( 9x + x ) - x = [ 9* (0.999...) ] + 0.999... - 0.999... > [Positive 0.999... goes with Negative 0.999...] > 9x = 9* (0.999...) > x = 9* ( 0.999...) / 9 = 1 * 0.999... > x =0.999 > We did not end with x=1 > In fact what made it looks like a proof was the mistake we did in this > step: > 10 x - x = 9.999... - 0.999... = 9 > This step assumes the value of infinite 9s after the decimal point in ( > 10x ) equals the value of infinite 9s in (x) > The infinite decimals in x = 0.999... have different behavior than the > infinite 9s in 10x =9.999... Because 9s in x= 0.999... Go to infinity > faster than 9s in 10x=9.999... At what velocity do the repeating 9's go to infinity? Is infinity a place where one (or 1 or 9) could go? If you take 1 and subtract 1/3 from it how fast do the repeating 6's go to infinity, fast enough to convince you that 666... is = 2/3 or only fast enough for it to be the mark of the beast? > (10 x ) is always one decimal behind (x) Yes, and as Nigel Tufnel observed 11 is 1 louder than 10. === Subject: Re: New Theory about 0.999... and 1 >> There is a new theory that may end the debate about 1=0.999... >> the whole theory can be found in this link [...] > What debate? > In the real number system (with all the standard axioms), 1 = 0.999... > In other systems (including the Surreal Numbers, or even reals plus > infinitesimals), they need not be the same. I thought that 0.999.. = 1 in Surreals as well - [{0.9, 0.99, 0.999 ....} | {1}] = [{1} | {1}] Could be wrong. Agree with your general thrust of argument, of course. > You seem to be challenging an axiom which states that 9.999... = 9 + > 0.999..., so you're no longer in the real number system. Hence you're > making a big deal out of nothing. > --- Christopher Heckman === Subject: Re: New Theory about 0.999... and 1 <44519a5a$0$4760$afc38c87@news.optusnet.com.au> There is a new theory that may end the debate about 1=0.999... >> the whole theory can be found in this link [...] > What debate? > In the real number system (with all the standard axioms), 1 = 0.999... > In other systems (including the Surreal Numbers, or even reals plus > infinitesimals), they need not be the same. > I thought that 0.999.. = 1 in Surreals as well - [{0.9, 0.99, 0.999 ....} | > {1}] = [{1} | {1}] > Could be wrong. Agree with your general thrust of argument, of course. No, 0.999... and 1 are different. And [{0.9, 0.99, 0.999, ...} | {1}] and [{1} | {1}] are two different things. In particular, [{1} | {1}] is not a (surreal) number, since there is an element in the left-hand set which is >= an element in the right-hand set. [{0.9, 0.99, 0.999, ...} | {1}] is a (surreal) number, though, since every element of the left-hand set is < every element of the right-hand set. In fact, if w = [{1,2,3,4,...} | E ] (where E is the empty set), then [{0.9, 0.99, 0.999, ...} | {1}] = 1 - 1/w. (Exercise for people familiar with surreal numbers and how to multiply and add them: Prove that [{0.9, 0.99, 0.999, ...} | {1}] * [{1,2,3,4,...} | E ] + [{0} | E] = [{1,2,3,4,...} | E ], and thus the equation [{0.9, 0.99, 0.999, ...} | {1}] = 1 - 1/w.) --- Christopher Heckman === Subject: Re: New Theory about 0.999... and 1 > There is a new theory that may end the debate about 1=0.999... > the whole theory can be found in this link > http://www.geocities.com/humood_theory/index.htm > here is a Quote about cracking the famouse proof: > Proof (A) > x=0.999... > 10x= 9.999... > 10x-x=9.999... - 0.999...=9 > 9x=9 > x=1 > Quote: > -------------------------------------------------------------- > I will start by cracking this proof and show it is in fact a Paradox > not a proof. > Let x = 0.999... > 10x = 9x + x = [ 9* (0.999...) ] + 0.999... [ I expressed 10x as (9x > +x)] > 10x - x = ( 9x + x ) - x = [ 9* (0.999...) ] + 0.999... - 0.999... > [Positive 0.999... goes with Negative 0.999...] > 9x = 9* (0.999...) > x = 9* ( 0.999...) / 9 = 1 * 0.999... > x =0.999 > We did not end with x=1 > In fact what made it looks like a proof was the mistake we did in this > step: > 10 x - x = 9.999... - 0.999... = 9 > This step assumes the value of infinite 9s after the decimal point in ( > 10x ) equals the value of infinite 9s in (x) > The infinite decimals in x = 0.999... have different behavior than the > infinite 9s in 10x =9.999... Because 9s in x= 0.999... Go to infinity > faster than 9s in 10x=9.999... > (10 x ) is always one decimal behind (x) > When x = 0.99 then 10x=9.9 > When x = 0.9999 then 10x=9.999 > When x = 0.999999999 then 10x=9.99999999 > Value of Infinite decimals in (x) - Value of infinite decimals in > (10x) = > Lim x-> 8 9 / (10^x) > So the difference between the two values equals an infinitely small > fraction which is Lim x-> 8 9 / (10^x) > This means in Proof (A) when we say: 10 x - x = 9.999... - 0.999... = > We are already Assuming that a number minus an infinitely small > fraction is the number itself. > Which means we are already assuming that Lim x-> 8 1 - ( 1/x ) = > 0.999... = 1 > So, in Proof (A) we already assumed 1= 0.999... Then no wonder we > finally ended with our assumption, which make it not a Proof at all. > You cannot prove any statement by assuming it to be true from the > beginning. > But when we explain (10 x ) in a very fundamental method no body can > argue with, we will say: > 10x =0.999... + 0.999... + 0.999... + 0.999... +0.999... +0.999... > +0.999... +0.999... +0.999... +0.999... > And when we subtract x = 0.999... from it : > 10x - x =0.999... + 0.999... + 0.999... + 0.999... +0.999... +0.999... > +0.999... +0.999... +0.999... + 0.999... - 0.999...= 0.999... * 9 > The infinite 9s in x after the decimal is gone with only one of those > 0.999... Not all of them as we did in Proof (A). > ------------------------------------------------------------------ > All the theory and its results can be seen in the link: > http://www.geocities.com/humood_theory/index.htm > For comments please write it here in this gruop or send it to the > e-mail > said_yam@yahoo.com I am glad that you brought this one up. This is a very important question. It is a matter of life and death. I say that 1 = .9999..., and you say that it is not. Let us say for a bit that 1 is not equal to .999... (infinitely many nines after decimal). Then take 1 - .999.... When you do the subtraction, and I know you are good at that, you see that the number that you get has infinitely many zeros after the decimal. I say that that number is zero, but you see that 1 after infinitely many zeros. But to write that number you have to write infinitely many zero after the decimal. Go write that number with all the infinitely many zeros. When you have finished writing that number come tell me I would agree with you that 1 is not equal to .999.... Until then long live geometric series which tells me that this number .999... is equal to 1. Now you cannot say that 1-.999... =.000...01, because I would still say that the right hand side is still 0 and to shut me up completely you would have to write all those infinitely many zeros. But the trouble is that if you have a million generations (I hope you do) and every one in would still be finitely many zeros after the decimal. So you see the 1 that you seem to see at the end of the tunnel is not there because you cannot reach it! You see my brother(s) infinite is very large. When someone says infinitely many 0's after decimal you rest assured that there cannot be anything after that because you cannot put anything after that; because you cannot get there. This is why people devise tricks to avoid having to deal with infinite sums or to deal with infinity in a round about way. Muhammad === Subject: Re: New Theory about 0.999... and 1 > There is a new theory that may end the debate about 1=0.999... > Sometimes we write, for example, pi = 3.1415926... and the ... > means and stuff. If that same meaning is used in the equation 1 = > 0.999... , then indeed, the equation is only an approximation. When > people assert that 1 = 0.999... , they're asking you to pretend that we > can talk about a infinite string of 9's beyond what's written down. If > you accept that little fantasy, then you are pretty much forced to the > conclusion that the equation is true. Were you beaten up by a geometric series as a child? === Subject: Re: New Theory about 0.999... and 1 There is a new theory that may end the debate about 1=0.999... > Sometimes we write, for example, pi = 3.1415926... and the ... > means and stuff. If that same meaning is used in the equation 1 = > 0.999... , then indeed, the equation is only an approximation. When > people assert that 1 = 0.999... , they're asking you to pretend that we > can talk about a infinite string of 9's beyond what's written down. If > you accept that little fantasy, then you are pretty much forced to the > conclusion that the equation is true. > Were you beaten up by a geometric series as a child? Has someone come up with 1 is not equal to 1.000...0001 or 1 is not equal to 1 point infinitely many 0's 1?. I am asking because I have a nice response for that: Go put the 1 after infinitely many zeros. When you succeed I would agree with you. Muhammad === Subject: Re: New Theory about 0.999... and 1 > There is a new theory that may end the debate about 1=0.999... > Sometimes we write, for example, pi = 3.1415926... and the ... > means and stuff. If that same meaning is used in the equation 1 = > 0.999... , then indeed, the equation is only an approximation. When > people assert that 1 = 0.999... , they're asking you to pretend that we > can talk about a infinite string of 9's beyond what's written down. If > you accept that little fantasy, then you are pretty much forced to the > conclusion that the equation is true. > Were you beaten up by a geometric series as a child? > Has someone come up with 1 is not equal to 1.000...0001 or 1 is not > equal to 1 point infinitely many 0's 1?. I am asking because I have a > nice response for that: Go put the 1 after infinitely many zeros. When > you succeed I would agree with you. > Muhammad Actually the people who want 0.999... to be different from 1.000... always seem to have in mind that there should be a last digit in 1.000... from which they can subtract 1 to get 0.999... === Subject: Re: New Theory about 0.999... and 1 > There is a new theory that may end the debate about 1=0.999... There never was a debate among mathematicians who use a proper definition of numbers with those dots. Many novices fail to see that without a definition of ... it makes no sense to discuss the value of 0.999... That's like discussing the value of %&9/(]$@ without defining the symbols. But that doesn't stop some definition-less people from making calculations on 0.999..., by silently assuming it satisfies some properties similar to numbers without .... If they pick different properties, they may get different results. Trying to define 0.999... as 0. followed by infinitely many 9's is meaningless if the latter has no defined value. > http://www.geocities.com/humood_theory/index.htm Your site claims 0.999...=1 is wrong because [Lim x->oo ( 1- (1/x))]^x = 1/e. That equation is false. You must be thinking about: [Lim x->oo ( 1- (1/x))^x] = 1/e. But you cannot swap the order of Lim and ^ to get your equation. x is a free variable in [..]^x, even if x is bound inside [..]. [Lim x->oo ( 1- (1/x))]^x = [1]^x =1 for any real x. -- Jens Kruse Andersen === Subject: Useful identity for ax^2+bxy+cy^2=dz^2 Hello all, Just stumbled on this identity while doing some related work. Given the equation, ax^2+bxy+cy^2 = dz^2 and initial solution (x1,y1,z1) = (m,n,p), then an infinite more can be found using, ax^2+bxy+cy^2-dz^2 = (am^2+bmn+cn^2-dp^2)(au^2+buv+cv^2)^2 x = (am+bn)u^2 + 2cnuv - cmv^2 y = -anu^2 + 2amuv + (bm+cn)v^2 z = p(au^2 + buv +cv^2) for arbitrary u,v. Example. The equation x^2+xy+y^2 = 7z^2 has small solution (x1,y1,z1) = (m,n,p) = (2,1,1) thus giving the parametrization, x = 3u^2+2uv-2v^2 y = -u^2+4uv+3v^2 z = u^2+uv+v^2 Neat, huh? Seen this identity before? (Might be in Dickson, though not sure.) --Tito === Subject: Re: Useful identity for ax^2+bxy+cy^2=dz^2 [snipped description of an identity I cannot comment on] > Example. The equation x^2+xy+y^2 = 7z^2 has small solution (x1,y1,z1) = > (m,n,p) = (2,1,1) thus giving the parametrization, > x = 3u^2+2uv-2v^2 > y = -u^2+4uv+3v^2 > z = u^2+uv+v^2 > Neat, huh? Seen this identity before? (Might be in Dickson, though not > sure.) So essentially you are looking at a projective variant of a quadratic curve, IOW a conic section. There the following well-known trick is available (explaining why these curves give rise to the rational function field): 1) Select a rational point P from the curve C (in the projective version of the curve you can always scale the coordinates to be integers) 2) Form the equation of the line L thru P that has an unknown slope t 3) Solve for the other point of intersection of L and C. As the equation of C is quadratic, and we have the trivial solution P at hand, the coordinates of the other solution will be rational functions of the unknown t such that the numerator and denominator are both at most quadratic polynomials of t. On the curve x^2+xy+y^2=7 (dehomogenized version of your curve) and point P I got (unless I made a mistake) x=(2t^2-2t-3)/(t^2+t+1) and y=(-3t^2-4t+1)/(t^2+t+1) (and being dehomogenized we automatically have z=1). Plug in t=v/u and multiplying (homogenizing) all the coordinates by u^2+uv+v^2 gives your parametrization up to certain trivial sign changes. Observe that 4) If you have another rational point Q, then the slope of the line connecting P and Q is also rational, say v/u, so you get all the rational points on C from this parametrization (when you interpret vertical lines having an infinite slope appropriately). To summarize: It seems to me that you have rediscovered a slight generalization of the perhaps even better known parametrization x=m^2-n^2, y=2mn, z=m^2+n^2 of Pythagorean triples (x,y,z) satisfying the equation x^2+y^2=z^2. The generalization to any curves of genus zero (i.e. those with a rational function field) is e.g. in an early chapter in Shafarevich's classic on algebraic geometry (forgot the exact title). The same approach works for singular cubic plane curves, as their function field is also rational. Many of us (yours truly included) first encountered this in textbooks/lecture notes on elementary calculus, as the rational parametrization of the Foil of Descartes is based on exactly the same idea. Jyrki === Subject: Labage?/Labay?/ Labayge? ... Integration My professor in a course said that he was studying it. I don't know exactly what he said it was, and I was wondering could anyone help me determine what the topic was? He said that it allows you to integrate functions like a function is one for all rational numbers and zero for all irrational numbers. And, that it is pretty dis-continuous in real analysis. Christopher Lusardi === Subject: Re: Labage?/Labay?/ Labayge? ... Integration >My professor in a course said that he was studying it. I don't know >exactly what he said it was, and I was wondering could anyone help me >determine what the topic was? >He said that it allows you to integrate functions like a function is >one for all rational numbers and zero for all irrational numbers. And, >that it is pretty dis-continuous in real analysis. Bartle, The Elements of Integration, provides an excellent introduction to Lebesgue integration. It is concise and clear. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: Labage?/Labay?/ Labayge? ... Integration > My professor in a course said that he was studying it. I don't know > exactly what he said it was, and I was wondering could anyone help me > determine what the topic was? > He said that it allows you to integrate functions like a function is > one for all rational numbers and zero for all irrational numbers. And, > that it is pretty dis-continuous in real analysis. > Christopher Lusardi That would be Lebesgue integration. http://mathworld.wolfram.com/LebesgueIntegral.html http://en.wikipedia.org/wiki/Lebesgue_integration === Subject: Re: Labage?/Labay?/ Labayge? ... Integration > My professor in a course said that he was studying it. > I don't know exactly what he said it was, and I was > wondering could anyone help me determine what the topic was? > He said that it allows you to integrate functions like > a function is one for all rational numbers and zero for > all irrational numbers. And, that it is pretty > dis-continuous in real analysis. Also, repeat the last 4 searches with an 's' at the end of both rational and irrational. Dave L. Renfro === Subject: jpeg2000 compression: window size? Hi The jpeg2000 compression format works by 1. splitting the image into disjoint rectangles and 2. applying a discrete wavelet transform to each rectangle. The size of the rectangles can be chosen -- jpeg2000 doesn't specify it. Does any one know a. What are typical rectangle sizes used? b. What are the respective advantages or disadvantages of larger or smaller rectangle sizes? The JPEG2000 still image coding system: an overview Christopoulos, C.; Skodras, A.; Ebrahimi, T.; Consumer Electronics, IEEE Transactions on Volume 46, Issue 4, Nov. 2000 Page(s):1103 - 1127 === Subject: Re: jpeg2000 compression: window size? > The jpeg2000 compression format works by > 1. splitting the image into disjoint rectangles and > 2. applying a discrete wavelet transform to each rectangle. Actually, the best way to go is to *avoid* these rectangles and encode the full image at once. They're called tiles. > The size of the rectangles can be chosen -- jpeg2000 doesn't specify > it. Does any one know > a. What are typical rectangle sizes used? One tile, as large as the image. > b. What are the respective advantages or disadvantages of larger or > smaller rectangle sizes? Since the wavelet transformation is interrupted at tile boundaries, you get artifacts there and thus see tile boundaries especially at lower bitrates. The advantage might be that - at least for a lazy coding design - you only need to buffer the image data for at most one tile. So long, Thomas === Subject: Analog of factorial for odd or even numbers only? Is there a defined function that is analogous to the factorial function, but involves only odd or even numbers? By way of illustration, let's use the arbitrary symbology n? (sort of analogous to n!) to show what I mean. The function that I'm looking for would work like this: For n even: n?=n(n-2)(n-4)(n-6)...(6)(4)(2) For n odd: n?=n(n-2)(n-4)(n-6)...(5)(3)(1) === Subject: Re: Analog of factorial for odd or even numbers only? > Is there a defined function that is analogous to the factorial > function, but involves only odd or even numbers? By way of > illustration, let's use the arbitrary symbology n? (sort of analogous > to n!) to show what I mean. The function that I'm looking for would > work like this: > For n even: n?=n(n-2)(n-4)(n-6)...(6)(4)(2) > For n odd: n?=n(n-2)(n-4)(n-6)...(5)(3)(1) That's the double factorial. See, for example, . David === Subject: Re: Analog of factorial for odd or even numbers only? <20060428083631.179$31@newsreader.com> LOL...I breezed right by that citation without slowing down. That kills me when something I was looking for was right under my nose all === Subject: please help with inequality problem I need to prove that for n>100 we have n>2*SQRT(n) * (log (n) +1) where SQRT stands for square root and log is the logarithm on basis 10. === Subject: Re: please help with inequality problem > I need to prove that for n>100 we have > n>2*SQRT(n) * (log (n) +1) > where SQRT stands for square root and log is the logarithm on basis > 10. As was suggested, rewrite as sqrt(n) > 2*(L10(n)+1). For n=100, this is 10 > 2*(2+1) = 6, which is true. To show f(n) > g(n) for n >= n0, enough to show that f(n0) > g(n0) and f'(n) > g'(n) for n >= n0. This is 1/(2*sqrt(n)) > 2/(n*L10) (where L10 = ln(10)) or sqrt(n) > 4/L10 ~ 4/2.3... which is certainly true if sqrt(n) > 2 or n > 4. (unless I made a mistake somewhere) === Subject: Re: please help with inequality problem >I need to prove that for n>100 we have >n>2*SQRT(n) * (log (n) +1) >where SQRT stands for square root and log is the logarithm on basis >10. First divide by SQRT(n). More precisely, prove the inequality with both sides divided by SQRT(n). Then justify why you can multiply your result by SQRT(n). -- VP Cheney Burr-ed his gun as a bird flew past The nation responds burr as we await bird flu shots and fight a real cold war. pmontgom@cwi.nl Microsoft Research and CWI Home: Bellevue, WA === Subject: Re: please help with inequality problem > I need to prove that for n>100 we have > n>2*SQRT(n) * (log (n) +1) > where SQRT stands for square root and log is the logarithm on basis > 10. Another hint. For n greater than some value you will have log(n) > 1. Socks === Subject: Re: please help with inequality problem > I need to prove that for n>100 we have > n>2*SQRT(n) * (log (n) +1) > where SQRT stands for square root and log is the logarithm on basis > 10. > Another hint. > For n greater than some value you will have log(n) > 1. Hmmm... People are really making a meal of this. n ? 2 sqrt(n) (log(n) + 1) Note that for n> 10, log(n) > 1 so n ? 4 sqrt(n) log(n) > 2 sqrt(n) (log(n) + 1) Hopefully it's pretty easy to show that n > 4sqrt(n)log(n) for n>100. Socks === Subject: Re: please help with inequality problem I'll give a hint for your homework problem: Consider n - 2*sqrt(n) * (log(n) + 1) Check to see if it is positive at n=100 Show (using calculus?) that it has no zeroes for n>100 === Subject: Re: please help with inequality problem LuckyOne escreveu: > I'll give a hint for your homework problem: This is not my homework. It is only a problem that i decided to try by myself. I don't know if the inequality is true. If is my problem is solved. Otherwise i have to keep trying. > Consider > n - 2*sqrt(n) * (log(n) + 1) > Check to see if it is positive at n=100 It is. > Show (using calculus?) that it has no zeroes for n>100 How? I tryed to prove that the derivative is always positive, but the problem seems as dificult as the original... === Subject: Re: please help with inequality problem > LuckyOne escreveu: > I'll give a hint for your homework problem: > This is not my homework. It is only a problem that i decided to try by > myself. I don't know if the inequality is true. If is my problem is > solved. Otherwise i have to keep trying. > Consider > n - 2*sqrt(n) * (log(n) + 1) > Check to see if it is positive at n=100 > It is. > Show (using calculus?) that it has no zeroes for n>100 > How? I tryed to prove that the derivative is always positive, but the > problem seems as dificult as the original... Okay, how about a different approach. You want to show that n > 2*sqrt(n)*(log(n) + 1) for n > 100 A little manipulation: sqrt(n) > 2((log(n) + 1) Divide through by sqrt(n) (sqrt(n)-2)/2 > log(n) Isolate log 10^((sqrt(n) - 2)/2) > n Exponentiate both sides Let u = (sqrt(n) - 2)/2 Then n = 4u^2 + 8u + 4 so that: 10^u > 4u^2 + 8u + 4 It should be clear that 10^u will be greater than 4u^2 + 8u + 4 for all u greater than some value, and that 10^u grows much more quickly than 4u^2. In fact, when n = 100 then u = 4: 10^4 > 4*(16) + 8*4 + 4 10,000 > 100 === Subject: Re: Continuum = Aleph What? I have a related and probably stupid question. I am more interested in what an ordinal corresponding to c would look like. I have read (and sort of understood) the stuff about Cantor normal form in Wikipedia, and these Ordinals all look countable to me. Even the jump to the epsilon character does not seem to make the set uncountable; it is simply a single additional limit point. So, if cardinals represent sets of equinumerous ordinals, what could an ordinal corresponding to c actually look like? === Subject: Re: Continuum = Aleph What? Peter Webb a .8ecrit : > I have a related and probably stupid question. > I am more interested in what an ordinal corresponding to c would look like. > I have read (and sort of understood) the stuff about Cantor normal form in > Wikipedia, and these Ordinals all look countable to me. Even the jump to the > epsilon character does not seem to make the set uncountable; it is simply a > single additional limit point. > So, if cardinals represent sets of equinumerous ordinals, what could an > ordinal corresponding to c actually look like? What kind of answer do you want? It should be obvious that any construction starting with w and limits will only give you a denumerable ordinal. Otoh, the ordinal (0,1,2,...,w,w+1,...,...,e_0,...,w_1,w_1+1,...,...,w_42=c) is a sort of answer... === Subject: Re: Continuum = Aleph What? Things are very interesting if the continuum is a real-valued measurable cardinal. And this contradicts CH (S. Ulam). Isn't it better when mathematics is more interesting? -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Continuum = Aleph What? Originator: grubb@lola >Things are very interesting if the continuum is a real-valued measurable >cardinal. And this contradicts CH (S. Ulam). Isn't it better >when mathematics is more interesting? This is about the only way that I'd see the falsity of CH being 'beautiful'. I usually prefer GCH simply because computations are so much easier. --Dan Grubb === Subject: Re: Continuum = Aleph What? >Things are very interesting if the continuum is a real-valued measurable >cardinal. And this contradicts CH (S. Ulam). Isn't it better >when mathematics is more interesting? > This is about the only way that I'd see the falsity of CH being > 'beautiful'. I usually prefer GCH simply because computations > are so much easier. > --Dan Grubb I disagree. Interesting mathematics is better than boring mathematics like GCH. For example: any PCA set (or is it CA set?) in R can be proved to be one of: countable, power aleph_1, or power c. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Continuum = Aleph What? [...] > for some ordinal number n. The CH states that n=1. If the CH is false, > then n must be greater than 1. I recall seeing an extension of Cohen's > work that showed that for a rather large set of values or n, C = Aleph_n > is consistent with ZFC. Out of curiosity, which Aleph do you believe is > C? In other words, for which ordinal number n is C = Aleph_n? I really don't know. But I think that ZFC+CH implies that the set of real numbers can be well-ordered in such a way that any real number has countably many predecessors. David Bernier === Subject: Re: Continuum = Aleph What? >> Out of curiosity, which Aleph do you believe is C? In other >> words, for which ordinal number n is C = Aleph_n? There are already some limitations on the value of n. For example, we know C cannot be Aleph_omega; more generally, n cannot be a limit number of cofinality omega. However, there are no reasons why it can't It's even possible for C to be greater than a weakly inaccessible cardinal (although it obviously cannot be larger than a strongly inaccessible one). It ultimately depends on your model of Set Theory. Putting on my Platonist hat for a moment, it's hard to believe in the Continuum Hypothesis, since it seems to hold true only in the most restrictive models of Set Theory. Models begin to appear more like our own Set Theory at C = Aleph_2, but even then, they seem a bit too limiting. Personally, I am inclined to think that n must be infinite to truly represent our Set Theoretical universe, but since omega is out of the running, I don't really have an aesthetic feeling what it should be. Jonathan Hoyle === Subject: Re: Continuum = Aleph What? > Out of curiosity, which Aleph do you believe is > C? In other words, for which ordinal number n is C = Aleph_n? > C = Aleph_2. Why 2? -- Chris Henrich http://www.mathinteract.com God just doesn't fit inside a single religion. === Subject: Re: Continuum = Aleph What? <270420062032211824%chenrich@monmouth.com > Out of curiosity, which Aleph do you believe is > C? In other words, for which ordinal number n is C = Aleph_n? > C = Aleph_2. > Why 2? Various more or less reasonable axioms lead to 2, and to the extent (not very great) that the experts are willing to pick anything, the lead contender seems to be 2, or at least as far as I, not in the set theory field, can tell. On the other hand GCH makes life a lot simpler; if you look at Woodin and Dales on superreal fields it's striking how much simpler it makes life. But most the top set theorists either seem to take a formalist position or think CH is false. Anyway, Bounded Martin's Maximum is one such axiom; Woodin in his survey in the Notices remarks that Subsequent investigation has shown that this theorem [2^aleph_0 = aleph_2] holds for almost any axiom which is nontrivally stronger than Martin's axiom omega_1. So, curiously, insisting that sets of size aleph_1 resemble countable sets necessitates that 2^aleph_0 = aleph_2. Woodin's Axiom (*), which generalizes Projective Determinacy, together with the assumption that there are a lot of (a proper class) of Woodin cardinals, leads to a strong form of Bounded Martin's Maximum and likewise leads to 2^aleph_0 = aleph_2. But both Projective Determinacy and large cardinal axioms are inherently plausible. === Subject: Re: Continuum = Aleph What? > The Continuum Hypothesis is probably the most accessible, and hence the > most famous unsolved problem in Mathematics today. More accessible than Goldbach? than twin primes? than odd perfect numbers? really? since my system doesn't subscribe nor allow me to post to it. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Continuum = Aleph What? most famous unsolved problem in Mathematics today. > More accessible than Goldbach? than twin primes? than odd perfect > numbers? really? You've forgotten ... Archimedes Plutonium has proven all of these. Seriously, I was going to mention the Goldbach Conjecture until I saw your post, since the 4CT seems to have been taken care of. In Graph Theory, probably the biggest open problem is the reconstruction problem (given G - v for all vertices v of G, find G). --- Christopher Heckman === Subject: Re: Continuum = Aleph What? > A related question: obviously, > C = Aleph_n > for some ordinal number n. The CH states that n=1. If the CH is false, > then n must be greater than 1. I recall seeing an extension of Cohen's > work that showed that for a rather large set of values or n, C = Aleph_n > is consistent with ZFC. Out of curiosity, which Aleph do you believe is > C? In other words, for which ordinal number n is C = Aleph_n? I believe that come the day when ZFC is found to be inconsistent, that this question will cease to have meaning in the new set theory that will arise out of the ashes of destruction. === Subject: Re: Continuum = Aleph What? > A related question: obviously, C = Aleph_n for some ordinal number n. The CH states that n=1. If the CH is false, > then n must be greater than 1. I recall seeing an extension of Cohen's > work that showed that for a rather large set of values or n, C = Aleph_n > is consistent with ZFC. Out of curiosity, which Aleph do you believe is > C? In other words, for which ordinal number n is C = Aleph_n? > I believe that come the day when ZFC is found to be inconsistent, that > this question will cease to have meaning in the new set theory that will > arise out of the ashes of destruction. This seems to imply that you think ZF is inconsistent. Why? -- Chris Henrich http://www.mathinteract.com God just doesn't fit inside a single religion. === Subject: Re: Continuum = Aleph What? >A related question: obviously, > C = Aleph_n >for some ordinal number n. The CH states that n=1. If the CH is false, >then n must be greater than 1. I recall seeing an extension of Cohen's >work that showed that for a rather large set of values or n, C = Aleph_n >is consistent with ZFC. Out of curiosity, which Aleph do you believe is >C? In other words, for which ordinal number n is C = Aleph_n? >>I believe that come the day when ZFC is found to be inconsistent, that >>this question will cease to have meaning in the new set theory that will >>arise out of the ashes of destruction. > This seems to imply that you think ZF is inconsistent. Why? Well since no inconsistency in ZF or ZFC has been found, this is only a gut feeling. But my sense is that the axiom of replacement is much too powerful, and generates sets that most mathemticians have absolutely no need of. Anyway, any theory that produces things like inaccessible ordinals has basically got to be very rocky. === Subject: Re: Continuum = Aleph What? <270420062127234501%chenrich@monmouth.com> <1ve4g.920915$x96.308032@attbi_s72 But my sense is that the axiom of replacement is much too powerful, and > generates sets that most mathemticians have absolutely no need of. You need it to show that Borel sets are determined (game theoretically); choice without replacement is not good enough. That's one excuse for it. === Subject: Re: Continuum = Aleph What? <270420062127234501%chenrich@monmouth.com> <1ve4g.920915$x96.308032@attbi_s72 But my sense is that the axiom of replacement is much too powerful, and > generates sets that most mathemticians have absolutely no need of. That hardly shows it is likely to be contradictory. It's saying the image of a set is a set, which you mostly know anyway. So you need to cook up something where it leads to a contradiction. Good luck. > Anyway, any theory that produces things like inaccessible ordinals has > basically got to be very rocky. But it *doesn't* produce them. === Subject: Re: Continuum = Aleph What? >>But my sense is that the axiom of replacement is much too powerful, and >>generates sets that most mathemticians have absolutely no need of. > That hardly shows it is likely to be contradictory. It's saying the > image of a set is a set, which you mostly know anyway. So you need to > cook up something where it leads to a contradiction. Good luck. >>Anyway, any theory that produces things like inaccessible ordinals has >>basically got to be very rocky. > But it *doesn't* produce them. Alright, any theory in which it is even seriously a consideration has to be rocky. === Subject: Re: Continuum = Aleph What? this question will cease to have meaning in the new set theory that will > arise out of the ashes of destruction. Do you think the Peano axioms will go down in flames also? === Subject: Re: Continuum = Aleph What? >>I believe that come the day when ZFC is found to be inconsistent, that >>this question will cease to have meaning in the new set theory that will >>arise out of the ashes of destruction. > Do you think the Peano axioms will go down in flames also? That's a tough call. === Subject: Re: Continuum = Aleph What? > I believe that come the day when ZFC is found to be inconsistent, that > this question will cease to have meaning in the new set theory that will > arise out of the ashes of destruction. >> Do you think the Peano axioms will go down in flames also? > That's a tough call. Do you have any idea of what the inconsistency in ZFC might be? Even a vague thought? === Subject: Re: Continuum = Aleph What? Paul Cohen showed in 1963 that the continuum hypothesis is independent of Zermelo-Fraenkel set theory even with the axiom of xhoice included. Cohen's work showed that there are many possibilities for c, the cardinality of the continuum, that are consistent with the usual axioms of set theory. For example it is consistent to have 2^(aleph-0) = aleph-2. === Subject: Integral of a factorial I just cam accross problem which I don't understand. What is the indefinite integral of a factorial: Int[ (x-1)!] dx I guess there is something with the Gamma function, but I simply don't know how. Best, stefan === Subject: Re: Integral of a factorial > I just cam accross problem which I don't understand. > What is the indefinite integral of a factorial: > Int[ (x-1)!] dx > I guess there is something with the Gamma function, but I simply don't > know how. This doesn't make sense, because (x - 1)! is defined only when _x_ is natural. And if you're after a primitive of Gamma(x), I don't think that it has a closed form. Jose CArlos Santos === Subject: Re: Integral of a factorial >> I just cam accross problem which I don't understand. >> What is the indefinite integral of a factorial: >> Int[ (x-1)!] dx >> I guess there is something with the Gamma function, but I simply don't >> know how. > This doesn't make sense, because (x - 1)! is defined only when _x_ is > natural. One can define y! for all complex numbers which are not negative integers by y! = Gamma(y+1) The Gamma function is analytic on all of the complex plane with the exception of isolated singularities at the negative integers and restricts to the factorial function on positive integers. So, the definition makes sense. > And if you're after a primitive of Gamma(x), I don't think that it has > a closed form. > Jose CArlos Santos Take care, -bobovski === Subject: Re: Integral of a factorial > I just cam accross problem which I don't understand. > What is the indefinite integral of a factorial: > Int[ (x-1)!] dx > I guess there is something with the Gamma function, but I simply don't > know how. >> This doesn't make sense, because (x - 1)! is defined only when _x_ is >> natural. > One can define y! for all complex numbers which are not negative integers > by > y! = Gamma(y+1) > The Gamma function is analytic on all of the complex plane with the > exception of isolated singularities at the negative integers* and > restricts to the factorial function on positive integers. So, the > definition makes sense. * correction: non-positive integers, as I believe the Gamma function has a pole at 0... -bobovski >> And if you're after a primitive of Gamma(x), I don't think that it has >> a closed form. >> Jose CArlos Santos > Take care, > -bobovski === Subject: Re: Integral of a factorial > I just cam accross problem which I don't understand. > What is the indefinite integral of a factorial: > Int[ (x-1)!] dx > I guess there is something with the Gamma function, but I simply don't > know how. > This doesn't make sense, because (x - 1)! is defined only when _x_ is > natural. It doesn't make sense _that way_, true. But remember, in mathematics, alas, there is no rigid standardization. I know of some important papers dealing with the Gamma function which consider Gamma[z+1] and z! to be the _same_ function (in particular, to have the same domain). > And if you're after a primitive of Gamma(x), I don't think that it has > a closed form. Right, at least none known to me. David === Subject: Re: Integral of a factorial > It doesn't make sense _that way_, true. But remember, in mathematics, alas, > there is no rigid standardization. I know of some important papers > dealing with the Gamma function which consider Gamma[z+1] and z! to be > the _same_ function (in particular, to have the same domain). In particular: in 19th century books and manuscripts you will see z! used this way. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Integral of a factorial >> It doesn't make sense that way , true. But remember, in mathematics, alas, >> there is no rigid standardization. I know of some important papers >> dealing with the Gamma function which consider Gamma[z+1] and z! to be >> the same function (in particular, to have the same domain). > In particular: in 19th century books and manuscripts you will see z! > used this way. Hmm, in fact there were /three/ notations: (1) 1730 Euler extended the factorial function n(n - 1)(n - 2) ... 3.2.1 to all real numbers greater than -1 by introducing his integral and by observing that this integral converges for noninteger values of n. (See Opera, Vol. 14, p. 1-24). (2) 1808 (Quoting http://hometown.aol.com/jeff570/stat.html ) The notation n! was introduced by Christian Kramp (1760-1826) in 1808 as a convenience to the printer. In his '.83l.8emens d'arithm.8etique Je me sers de la notation tr.8es simple n! pour d.8esigner le produit de nombres d.8ecroissans depuis n jusqu'.88 l'unit.8e, savoir n(n - 1)(n - 2) ... 3.2.1. L'emploi continuel de l'analyse combinatoire que je fais dans la plupart de mes d.8emonstrations, a rendu cette notation indispensable. (3) 1811 Legendre, Exercices de Calcul integral..., introduced the notation Gamma(s) for PI(s-1), where PI(s) is the integral introduced by Euler. His reason for doing so are obscure. (4) 1813 C.F. Gauss, in 'Circa seriem infinitam ...' introduced the notation PI(s) for Eulers integral. Thus in particular: in 19th century books and manuscripts you will see PI(s) (used by Gauss and Riemann) and Gamma(s) (used in France) for the Euler integral. === Subject: Re: Arithmetic with (only) exponentiation Why would arithematic with muliplication and addition be undecidable? Muliplication is defined by addition, afterall. Colin .9b.8d.93.b9.81F > Pressburger showed that arithmetic with just addition (i.e., the > first-order theory where addition is the only function) is a decidable > theory. Skolem showed that arithmetic with just multiplication is a > decidable theory. Does anyone know the status of arithmetic with just > exponentiation? My guess is that it would not turn out to be decidable, > but I was hoping this matter might be settled and someone could clue me > in. === Subject: Re: Arithmetic with (only) exponentiation Hi Multiplication is defined by addition and recursion, for example. Peano axioms are not: multiplication = function(addition) Peano axioms are: multiplication = function(addition,multiplication) In particular: 0*X := 0. (X+1)*Y := X*Y+Y. So multiplication is not a hierarchically defined thing, it is rather a inductively defined thing. Clearly there are recursively defined functions, which don't make it undecidable. If you take addition and multiplication in the integers you get so called diaphantine equations, which are semi decidable. BTW this paper seems to expand on diophantine equations quite extensive: Bye > Why would arithematic with muliplication and addition be undecidable? > Muliplication is defined by addition, afterall. > Colin .9b.8d.93.b9.81F >>Pressburger showed that arithmetic with just addition (i.e., the >>first-order theory where addition is the only function) is a decidable >>theory. Skolem showed that arithmetic with just multiplication is a >>decidable theory. Does anyone know the status of arithmetic with just >>exponentiation? My guess is that it would not turn out to be decidable, >>but I was hoping this matter might be settled and someone could clue me >>in. === Subject: Re: Arithmetic with (only) exponentiation > Let's try this again, with just the axioms and without the metaphysics. >> ...[among other things] >> exists (unique) x,y for all z(e(z,x)=y) > They don't have to be unique; if they are, that > will be provable as a theorem; his proposed axiom > (if I can rename the letters just to be more intuitive) > EyzAn[n^z=y] > Clearly, y will be one, z will be zero, and n will be any number. What are you trying to do here? Find an axiom scheme for a binary exp function on N so that we can prove the peano axioms (in some sence)? >> Further, developing the idea of A Niel, the set of triples x,y,z that >> satisfy >> for all w (e(e(w,x),y)=e(w,z) >> is the graph of multiplication. > So you can try to begin to define multiplication > by introducing it into the language and asserting > Axyz[ Aw[(w^x)^y=w^z] => x*y=z ] > But that's going to need cleanup and augmentation > before you can finish defining multiplication. The standard layout for this sort of approach is to consider L_1 :={x^y} L_2 :={x^y,*,+,>,0,1} Then to name L_1 - formulae phi_*,...phi_1 Then let T_1 be the L_1-theory which says that phi_* defines a binary funcion and...and phi_1 defines a single element Then any model of T_1 can be made into an L_2-structure in the obvious way. This is called an expansion by definitions, or definitial expansion. Then, it is clear that any L_2-sentence is T_1-equivalent to a L_1- sentence. The information of how to do this in the previous posts. In other words (N,*,+,0,1,x^y) is a definitional expansion (or an expansion by definitions) of (N,x^y). The resulting formulae (reformulation of peanos axioms in L_1) will look horrible though. To me, the fact that it can be done was interesting, and I am glad that A N Niel pointed it out. I am interested in the (seemingly dificult) problem of the decidability of the real exponential field. The same property should hold there (i.e. (R,*,+,-,0,1,>,exp) is definable in (R,exp)). >> Similarly, as above, one can define addition. > These are not standard eliminative/substitutive > definitions. Do the identity-elements follow as theorems > from these? === Subject: Re: Arithmetic with (only) exponentiation <220420061706206211%anniel@nym.alias.net.invalid> <79cl42pks0sltpstoghi9f9p0mqo80j66t@4ax.com> <1hd9ykbgsq9so$.15daq7q45gdcb$.dlg@40tude.net> <1nh1jtjqyt0ks$.10p5wf7vgl2kk.dlg@40tude.net EyzAn[n^z=y] > Clearly, y will be one, z will be zero, and n will be any number. > What are you trying to do here? Find an axiom scheme for a binary exp > function on N so that we can prove the peano axioms (in some sence)? More than JUST that. It can't just be ANY old binary function. It has to be the ACTUAL exponential function. It has to be a binary function that obeys what we know about the standard exp function. And, yes, (in some sense) was EXACTLY what was being debated -- in WHAT sense? The sense can't be exact because the languages are different -- nobody ever clarified whether s(.) was or wasn't in the signature (or 2 either, for that matter). But I personally do have a sense in mind, yes. > The standard layout for this sort of approach is Is irrelevant bull, because, precisely as you say, > The resulting formulae (reformulation of peanos axioms in L_1) > will look horrible though. Right, which is why NObody is even CONtemplating doing it THAT way. *I* was planning on doing it using intuitive famililar identities about exponentiation as axioms. The question then becomes whether the INTUITIVE REASONABLE axioms will be SUFFICIENT to get you (at least) PA as theorems. Obviously the resulting theory is going to be stronger than PA, simply because it has exponentiation and PA doesn't. It's going to have a much clearer statement of Fermat's Last Theorem, for example. You were also wrong to claim > Then any model of T_1 can be made into an L_2-structure in the obvious way. > This is called an expansion by definitions, or definitial expansion. > Then, it is clear that any L_2-sentence is T_1-equivalent to a L_1- > sentence. > The information of how to do this in the previous posts. That is a complete mis-characterization of the content of the previous posts. The previous posts had nothing whatsoever about naming any formulas. > I am interested in the (seemingly dificult) problem of the decidability of > the real exponential field. The same property should hold there (i.e. > (R,*,+,-,0,1,>,exp) is definable in (R,exp)). There is a first-order axiomatization of the reals that is decidable, to begin with. That is a very different state of affairs from PA. Here, we showed undecidability by showing (however informally and handwavily) that you could define the constants, functions, and axioms of PA using only ^. Undecidability then follows from the undecidability of first-order PA. If you re-state all the axioms of a first- order formalization of the reals using only ^ (and things definable from it), though, then you have merely re- formalized something that was already known to be decidable (as opposed already known to be undecidable, like PA), so much remains unclear. === Subject: Re: Absolute continuity Originator: grubb@lola >The follwoing HW problem is little bit hard for me. Any one can give >any idea for that? When we assume 1[a,b], the 1st order derivative of f, f' is L^p function, and alpha=1/q >where q is the exponent conjugate to p, how do we prove function f is >Lip alpha (That is, f satisfies a Lipschitz condition of order alpha)? >I tried to start with Mean Value Theorem, but it didn't seem to work. >Any help would be really appreciated even if one line. MVT+Holder === Subject: Re: Absolute continuity >The follwoing HW problem is little bit hard for me. Any one can give >any idea for that? When we assume 1[a,b], the 1st order derivative of f, f' is L^p function, and alpha=1/q >where q is the exponent conjugate to p, how do we prove function f is >in >Lip alpha (That is, f satisfies a Lipschitz condition of order alpha)? >I tried to start with Mean Value Theorem, but it didn't seem to work. >Any help would be really appreciated even if one line. > MVT+Holder MVT or FTC? === Subject: Re: Noosphere Academy spams our poor Noosphere - Sample # 3 On web-site http://www.geocities.com/aladjak/index1.htm you can find very interesting information about so-called leading expert on CAS V. Bondarenko. Unfortunately, that information is on Russian, but now it is being translated onto English. === Subject: The Real Bible Code - Decoding in Real Time The Real Bible Code - Decoding in Real Time Read The Last News - 28 of April 2006 - Good-bye, Darkness! Good-bye, Darkness! - http://www.atoll-geocad.com/The-Message/Diary-ENG.html Decoding Diary: http://www.atoll-geocad.com/The-Message/Diary-ENG1.html === Subject: Re: compact 3-manifold cohomology problem prime decomposition. === Subject: Re: compact 3-manifold cohomology problem >>Then use the universal coefficients theorem again to >>show that the Euler characteristic being zero with Z/2Z coefficients >>implies that the usual Euler characteristic with Z coefficients is also >>zero. H_n(X;Z2)=H_n(X;Z)@Z2 + Tor(H_(n-1)(X;Z),Z); looking at this and Sum_(-1)^i * dim(.......) where ..... is terms on each side, siply kill the Tor term, but does not impl the above statement. What's missing? === Subject: Re: compact 3-manifold cohomology problem >>Then use the universal coefficients theorem again to >>show that the Euler characteristic being zero with Z/2Z coefficients >>implies that the usual Euler characteristic with Z coefficients is also >>zero. > H_n(X;Z2)=H_n(X;Z)@Z2 + Tor(H_(n-1)(X;Z),Z); > looking at this and Sum_(-1)^i * dim(.......) where ..... is terms on > each side, siply kill the Tor term, but does not impl the above > statement. What's missing? This is certainly a valid question and I should have said more about it in my original post. First, let me point out that the your Tor term should be Tor(H_(n-1)(X; Z), Z/2Z). Otherwise it would just be zero since Z is (obviously) a free Z-module. However, we are actually going to use the universal coefficients theorem for cohomology. Recall that this says for any space X and group G, there is the following short exact sequence: 0 -> Ext(H_(n-1)(X), G) -> H^n(X; G) -> Hom(H_n(X); G) -> 0 In our situation of compact manifolds, all of the (co)homology groups are finitely generated. Also recall (or learn, as the case may be) the following useful facts: 1) Hom(Z, Z/nZ) = Z/nZ 2) Hom(Z/nZ, Z/mZ) = Z/dZ where d = gcd(m,n) 3) Ext(Z/nZ, Z/mZ) = Z/dZ where d = gcd(m,n) 4) Ext(Z, Z/nZ) = 0 simply because Z is a free Z-module Using 1) - 4), the short exact sequence above, the fact that our (co)homology groups are finitely generated, and the fact that H_n(X, Z/2Z) = H^n(X, Z/2Z), we can conclude the follwing: 1) Each Z summand in H_i(X) gives a Z/2Z summand in H^i(X, Z/2Z). 2) For n even, each Z/nZ summand of H_i(X) gives a Z/2Z summand in both H^i(X, Z/2Z) and H^(i+1)(X, Z/2Z), which will cancel one another in the alternating sum of the Euler characterstic (in Z/2Z coefficients). 3) For n odd, each Z/nZ contributes no Z/2Z summands to H^i(X, Z/2Z) as 2 and n are relatively prime. This now shows that sum (-1)^i rank(H_i(X)) = sum (-1)^i dim(H_i(X; Z/2Z)). Hopefully this clears some things up from before. If not, please ask. Mike === Subject: Re: compact 3-manifold cohomology problem it cleared the remaining point. thank you. I have tonnes of questions (trying to learn topology on my own), but the more a problem requires the better it makes for a thorough examination of ideas. I will look arond for involved problems and if there is something really interesting and hard, i will post again. === Subject: blue, green and red! Colour the plane with three coulors: blue, green and red! Some point must bee blue! Some point must bee green! Some point must bee red! On No line can there bee all three coulours! === Subject: Re: blue, green and red! >Colour the plane with three coulors: blue, green and red! >Some point must bee blue! >Some point must bee green! >Some point must bee red! >On No line can there bee all three coulours! x=y=0: red x*y > 0: green x*y < 0 but not x=y=0: blue There are many answers like this which have only 1 point of a certain color. Is it possible if there are at least 2 points of each color? --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: blue, green and red! In message , Keith A. Lewis > dated Fri, 28 Apr 2006 15:25:39 >>Colour the plane with three coulors: blue, green and red! >>Some point must bee blue! >>Some point must bee green! >>Some point must bee red! >>On No line can there bee all three coulours! >x=y=0: red >x*y > 0: green >x*y < 0 but not x=y=0: blue >There are many answers like this which have only 1 point of a certain color. >Is it possible if there are at least 2 points of each color? (x,0) x rational: red (x,0) x irrational: green (x,y) y <> 0 : blue -- David Hartley === Subject: forward difference operator in Zp I am working on a polynomial in Zp (p is prime) of degree d: f(x) = c_d x^d + c_(d-1) x^(d-1) + ... + c_0 I need to evaluate it in several points: g^1, g^2 ... g^L, where g is a generator of the multiplicative group Zp*. I suppose that the forward difference operator in this case would be: delta f(x) = f(gx) - f(x), but then the difference operator of order d is not equal zero, while it normally should (i.e. the degree of the polynomial remains always the same). What is the problem here and how the difference operator must be applied in this context ? Alexandre === Subject: Re: forward difference operator in Zp > I am working on a polynomial in Zp (p is prime) of degree d: > f(x) = c_d x^d + c_(d-1) x^(d-1) + ... + c_0 > I need to evaluate it in several points: g^1, g^2 ... g^L, where g is a > generator of the multiplicative group Zp*. I suppose that the forward > difference operator in this case would be: > delta f(x) = f(gx) - f(x), but then the difference operator of order d > is not equal zero, while it normally should (i.e. the degree of the > polynomial remains always the same). What is the problem here and how > the difference operator must be applied in this context ? The issue is that your forward difference operator is: delta f(x) = f(gx) - f(x) whereas the stricter analog would be: delta f(x) = f(x+g) - f(x) Since I don't know the motivation for studying the difference operator you've proposed, it's hard to guess if that is really what you should pursue. Your version removes the constant term but as you've commented, the degree remains the same. The strict analog will reduce the degree of the polynomial by (at least) one each time it's applied. === Subject: Re: Arc Distances of an Out Circle > If I have 3 points (A,B,C) that are not in a straight line: > How can I calculate the lengths of arc AB, arc BC, and arc CA of the > outcircle drawn around the triangle created by the three points > assuming the lengths of segment AB, segment BC, and segment CA are all > known? > Any assistance would be greatly appreciated. Draw the setup and it should help. The center of the circumscribed circle must be found first. Then consider each triangle formed by drawing the radii from the center of the circle to two of the known points, for example A and B. Now consider the chord AB whose length, as you said, is known. Then its half-length, the semichord, is also known. You should be able to do the rest. Grover Hughes === Subject: number theory book Hello all-- I wish to obtain a hardcopy of the book Number theory and its history, by Oystein Ore, copyright 1948 or later. I've tried Amazon and a couple of other places, and have Googled for the title, but no luck. Any ideas as to where else I might look will be very much Grover Hughes ghughes@cei.net === Subject: Re: number theory book > Hello all-- I wish to obtain a hardcopy of the book Number theory and > its history, by Oystein Ore, copyright 1948 or later. I've tried > Amazon and a couple of other places, and have Googled for the title, > but no luck. Any ideas as to where else I might look will be very much > Grover Hughes ghughes@cei.net I think B&N has that as a softcover for 10 dolars or so. It is a decent book. I would go for Shank Unsolved problems. Excelent historical apprioach. And i just ordered my copy of Weil's Historical approach. Much better than Ore's. === Subject: Re: number theory book > Hello all-- I wish to obtain a hardcopy of the book > Number theory and > its history, by Oystein Ore, copyright 1948 or > later. I've tried > Amazon and a couple of other places, and have > Googled for the title, > but no luck. Any ideas as to where else I might > look will be very much > Grover Hughes ghughes@cei.net Borders has the Dover softcover version. === Subject: Re: number theory book days. My association with the Department is that of an alumnus. >Hello all-- I wish to obtain a hardcopy of the book Number theory and >its history, by Oystein Ore, copyright 1948 or later. I've tried >Amazon and a couple of other places, and have Googled for the title, >but no luck. Any ideas as to where else I might look will be very much >appreciated. I'm not sure if by hardcopy you might mean a hardback (as opposed to a paperback). But in any case, a quick search through BookFinders www.bookfinder.com for the book gave 129 matches in two blocks (a block of 120 with New/Used, and a block of 9 used). Several of the sources are reputable and I've done regular business with them (Alibris and AbeBooks). Bookfinder is a good resource in general. It links to associated catalogs of used book stores such as Powell Books and so on (in addition to the direct sales through Amazon and the like which I don't like to use). -- === Subject: Re: Beliefs Create Reality, Even In Mathematics. On 2006-04-27 22:44:02 +0700, Tom said: Like a list of rituals I performed? Exercises I've done? Is that > what you're asking? >> And will he provide the same without any hemming/hawing or dodging? >> I doubt it. > He says he'd be happy to do so, but apparently he isn't happy enough yet. Nearly right. he said he would be happy to reciprocate. So far there is nothing to reciprocate to. Still. Nice try though. Dishonest but impressive. A === Subject: Re: Beliefs Create Reality, Even In Mathematics. On 2006-04-27 22:29:43 +0700, Tom said: >> Moved to la-la land. Oregon. >> As you wish. > Yes, as I wish. >> But I am sure not *everybody* in Oregon is potty Tommy. Or psychotic. >> Are they? > Depends. If they bother to challenge any of your unsubstantiated > claims, they may well be potty by your standards. Still avoiding saying what your experience is I see... What a surprise. A === Subject: Re: Beliefs Create Reality, Even In Mathematics. > On 2006-04-27 22:29:43 +0700, Tom said: > Moved to la-la land. >> Oregon. > As you wish. >> Yes, as I wish. > But I am sure not *everybody* in Oregon is potty Tommy. Or > psychotic. Are they? >> Depends. If they bother to challenge any of your unsubstantiated >> claims, they may well be potty by your standards. > Still avoiding saying what your experience is I see... > What a surprise. Heh, you've been avoiding posting any evidence of your experience since you have reappeared here. Why don't you lead by example instead of waiting for Tom to do and than most likely backing out when/if Tom did so. YOu have backed out on bets previously and acted dishonorably and without integrity often. -Douglas === Subject: Re: Beliefs Create Reality, Even In Mathematics. On 2006-04-27 03:43:04 +0700, Tom said: > >> You *actually* said your dog could achieve dharana and your cat could RV. I do recall mentioning that my dog can achieve dharana, one-pointed > concentration. >> There you go... I knew you would remember. Eventually... > He doesn't do it all the time, of course. >> Really? How do you know this Tom? Do you have a 'special' way of >> talking to your dog? > Are you unaware of how one may determine what a dog is concetrating his > attention on? Usually you can tell by where he's looking, for one. Is that how he told you he was achieving dharana Tom? He noticed you were watching where he was looking? > Nobody does. I still don't remember the cat part, though. Can you > cite that particular statement? Perhaps you may be in error a little > less rarely than you have estimated. Will you check? >> No. > I rather suspected you wouldn;t. You had to admit an error and > apologize last time. Plus, you don;t want to call into question your > definition of the word rare, as in I make mistakes on rare > occasions. So you dont want to think about the cat statement then? >> You sure you *didn't* say this Tommy. Perhaps you should check. > I did. I can find no such statement. I see. Not quite the same thing as saying you made no such statement though... A bit like 'I cannot recall' really, 'I can find no such statement'. LOL > I do recall something about my dog being able to achieve dharana, > one-pointed concentration, when he was shown a dog biscuit or > something. However, no competent occultist would confuse dharana with > dharma, >> >> Says who Tommy? You? Well, I am sure we all believe you Tommy... Well, since there is no doubt you are right, I will accept your > statement that a competent occultist *would* confuse dharana with > dharma. After all, you've done so and you say you're a competent > occultist. >> Well no. I didnt actually say that Tommy. Actually it was *you* that >> said it. Again. Confused? Trouble with concentration? > So if you did not confuse dharana with dharma, how come you erroneously > thought I'd claimed my dog could achieve dharma instead of dharana? > This seems quite a mystery. Your life is so very mysterious Tommy. Try not to panic. >> For the record, I have no interest in Eastern religions or Yoga. Then you don't ever bother to achieve dharana? >> By another na > Oops. Looks like you got distracted. Looks like a computer problem,. The return current no doubt. A === Subject: Re: Beliefs Create Reality, Even In Mathematics. On 2006-04-27 03:35:36 +0700, Tom said: One cannot live at all without gaining experience. However, I don't > think anything I've experienced is significant to you. >> Well no. That is what *you* said. > Of course it's what I said. Did something lead you to believe that I > was quoteing, citing, or paraphrasing someone else? >> Why not have a nice lie down and come back when you are less confused? > I'm kind of confused about why you think that sentence indicates confusion. There you go.. Confused again. Still. > I'm sure you always read carefully, but you might want to review my > statement to see just what it is I am admitting. >> You aren't sure yourself? hmmm. > That's why I checked. Have you checked yet? if not, you can simply > My Magical experience is of no significance to you whatsoever and > carries no authority by which I may demand that I be > believed by anyone without question or explanation. >> There you go... it *was* you who said it after all... All clear now? > Yes, it was me who said it. Was there some question about that? Only in your mind Tommy. You tried hard to distance yourself from your own statement but had to give up when it beecame obvious. > Do you believe I attributed that statement to you? Or are you arguing > that I am erroneously supposing that you agree with my statement? > Maybe you can clear that up, too. Do you believe that my magical > experience is not insignificant to you and is an authority by which I > may be believed by anyone without question? I believe that since you have repeatedly failed to give details, then your original statement that is insignificant and that you in fact dont have any 'except your arguments' must be accepted. >> Still waiting for yours Tom, You seem to have omitted the Druid days. >> Learn anything at all? Yes, I did learn some things, but what I learned would be insignificant > to you. You already know that. >> There you go again. Do you really not remember any of this? > It seems as though you are now attempting to argue that you believe > that I do have significant magiocal experience and can claim the > authority to be believed without question. Is that so? no. I believe you when you said you dont actually have any. What I am waiting for is detail of what you did that resulted in no actual experience. >> Well, it isnt just me who is listening Tommy. Do enlighten us all... OK. One thing I learned from the Druid days is that it's problematic > to mow a lawn around a stone altar in your back yard. A trimmer is > better. Or, if you like doing things by hand, a pair of trimming shears > works pretty well, but it's not nearly as fast. Is that the kind of > thing you had in mind that you feel everyone wants to hear? >> There you go. You join the Druids to learn great secrets. And sure enough... > No I joined the Druids to enjoy playing pagan games with some of my > friends. Nobody expected any big secrets to come from it. However, > if you think that techniques for trimming grass around a stone altar is > abig secret, you're certainly welcome to it. So the most you actually did in Magic was play pagan games with some friends. OK. I believe you. Will you 'not recall' that statement as well? > Will you be showing us how significant your own experience is? Or not? >> >> I will be very happy to reciprocate. As and when we have got to the >> bottom of the totality of your experience. Now there's a tall order. The bottom of the totality of my experience. > Like from birth or something? How will you know when it's reached > totality? >> I am sorry, I have managed to confuse you again. I really didn't know >> it was that easily done. No matter, I will explain... >> We would all like to know precisely what your *Occult* experience is >> Tommy. As opposed to what you have copied from what other people think. > Like a list of rituals I performed? Exercises I've done? Is that > what you're asking? Yes. Of course. And what you believed to be the results. I would be greatly interested in what you have done which so failed to impress you (yes, it failed to impress even you) which resulkted in you stating that dont actualy have any experience other than 'your arguments'. A === Subject: Re: Beliefs Create Reality, Even In Mathematics. > On 2006-04-27 03:35:36 +0700, Tom said: >> Do you believe that my magical experience is not insignificant to you and >> is an authority by which I may be believed by anyone without question? > I believe that since you have repeatedly failed to give details, then your > original statement that is insignificant and that you in fact dont have > any 'except your arguments' must be accepted. I think you are in too much of a hurry. Your sentence makes no grammatical sense at all. Try again, this time with more concentration. You know, that dharana stuff. > There you go again. Do you really not remember any of this? >> It seems as though you are now attempting to argue that you believe that >> I do have significant magical experience and can claim the authority to >> be believed without question. Is that so? > no. I believe you when you said you dont actually have any. I have no magical experience that you would consider significant, no. > What I am waiting for is detail of what you did that resulted in no actual > experience. The word actual is not synonymous with significant to you. Many actual experiences of mine are of no significance to you, and I'm certainly not going to try to assert that they must be. > There you go. You join the Druids to learn great secrets. And sure > enough... >> No I joined the Druids to enjoy playing pagan games with some of my >> friends. Nobody expected any big secrets to come from it. However, if >> you think that techniques for trimming grass around a stone altar is >> abig secret, you're certainly welcome to it. > So the most you actually did in Magic was play pagan games with some > friends. You are again unable to keep from fantasizing about what I tell you and feeding it back as if your fantasies are actually something I said. > OK. I believe you. No, you believe you. It's only when you believe I'm agreeing with you that you believe me, which is really just you believing you. > Will you 'not recall' that statement as well? I do not recall something I didn't say as something I said. You, on the other hand, seem to have no end of things I did not say among the things you say you recall. >> Now there's a tall order. The bottom of the totality of my experience. >> Like from birth or something? How will you know when it's reached >> totality? > I am sorry, I have managed to confuse you again. I really didn't know it > was that easily done. No matter, I will explain... > We would all like to know precisely what your *Occult* experience is > Tommy. As opposed to what you have copied from what other people think. >> Like a list of rituals I performed? Exercises I've done? Is that what >> you're asking? > Yes. Of course. And what you believed to be the results. I would be > greatly interested in what you have done which so failed to impress you > (yes, it failed to impress even you) which resulkted in you stating that > dont actualy have any experience other than 'your arguments'. You are again misremembering what I've told you. I did not say I had no experience. I said I did not expect people to believe me because I said I experienced something. You, on the other hand, have made it entirely clear that you *should* be believed because you claim to have experience, even though you have not revealed what that experience is. In fact, it's my lack of cooperation with your self-serving claim of authority that has generated most of your resentment of me. A list of rituals and exercises would mean nothing at all to any intelligent person who wished to know if my observations are true or not. Trust like that is not required or even encouraged by me. It simply doesn't matter whether or not I've done hundreds of Golden Dawn Pentagram and Hexagram rituals, Invocations to the Bornless One, the Sacred Magic of Abramelin, joined this secret society, that spiritual brotherhood, or anything else. I'm pretty sure no one on sci.skeptic or sci.math would think so. And alsmost nobody would in alt.magick, except for a couple of people who dearly need a bigger dose of skepticism in their lives === Subject: Re: Beliefs Create Reality, Even In Mathematics. On 2006-04-27 03:18:59 +0700, Tom said: Credibility is not a personal trait, ipsissimus. >> Well it certainly isnt one of yours - is that what you meant? > Nor of yours. Nor of anyone. It is the degree to which others decide > one may be trusted. It's an attribution, not a trait, and anyone's > credibility varies with each person. So the common observation that you 'lack credibility' merely confirms everybody's illiteracy then Tommy? A === Subject: Re: Beliefs Create Reality, Even In Mathematics. > On 2006-04-27 03:18:59 +0700, Tom said: >> Credibility is not a personal trait, ipsissimus. > Well it certainly isnt one of yours - is that what you meant? >> Nor of yours. Nor of anyone. It is the degree to which others decide >> one may be trusted. It's an attribution, not a trait, and anyone's >> credibility varies with each person. > So the common observation that you 'lack credibility' merely confirms > everybody's illiteracy then Tommy? How would an illiterate peson know I lacked credibility in this newsgroup unless they had read what I had written here, which they couldn't have since they can't read? You should think these accusations through a bit more before you start tossing them about. This one makes even less sense than the usual ones. === Subject: Re: Beliefs Create Reality, Even In Mathematics. On 2006-04-27 00:02:34 +0700, Lizz Holmans said: > On Wed, 26 Apr 2006 21:22:05 +0700, Apotheosis >> You are welcome. It is my plain Christian duty to help the less fortunate. > I am a Plainer Christian than most, and I would not consider what thee > has done here Christian in any fashion. Really? In what sense? > There is nothing occult about Christianity. Nothing is hidden. There > is nothing secret to be revealed to a 'higher' level of acolyte. Actually there is a very great deal in Christianity. Personally I find it isd illuminated greatly by the Qabalah, but the tradition of esoteric Christianity goes back at least as far as the 13th or 14th Century. Back as far in fact as the likely inspirations for the legends of the Rosicrucians. > Or, if thee is using 'occult' in the sloppier definition of 'mystic', > there is a long mystic tradition in Christianity, but it is in no way > occult. Well, I disagree. The GD was essentially a Christian Order and was certainly occult. > But from observation, it appears that thee is mainly more or less > demonstrating not thy logical skills or thy 'occult' knowledge but thy > evident lack of a sense of humor. Worse yet, thee has committed the > sin of boring thy audience. Bad show, old thing, bad show. Aww. I guess that is a severe scolding. I will try no to be devastated. Unto all things a due season and a good reason, even if it is not obvious. A === Subject: Re: Beliefs Create Reality, Even In Mathematics. On Fri, 28 Apr 2006 22:39:40 +0700, Archangelska >On 2006-04-27 00:02:34 +0700, Lizz Holmans said: >> On Wed, 26 Apr 2006 21:22:05 +0700, Apotheosis > You are welcome. It is my plain Christian duty to help the less fortunate. >> I am a Plainer Christian than most, and I would not consider what thee >> has done here Christian in any fashion. >Really? In what sense? I'm a member of the Religious Society of Friends. Quakers, aka 'The Plain People.' Thee surprises me. As an Apotheosis, your God-like powers should have gnosis of everything. So why did I have to tell thee? It's not like I didn't give thee any clues. Lizz 'they both contain multitudes' Holmans -- Rumpeta, rumpeta, rumpeta === Subject: Re: Beliefs Create Reality, Even In Mathematics. > On Fri, 28 Apr 2006 22:39:40 +0700, Archangelska >>On 2006-04-27 00:02:34 +0700, Lizz Holmans >>said: > On Wed, 26 Apr 2006 21:22:05 +0700, Apotheosis >> You are welcome. It is my plain Christian duty to help the less >> fortunate. > I am a Plainer Christian than most, and I would not consider what thee > has done here Christian in any fashion. >>Really? In what sense? > I'm a member of the Religious Society of Friends. Quakers, aka 'The > Plain People.' > Thee surprises me. As an Apotheosis, your God-like powers should have > gnosis of everything. So why did I have to tell thee? It's not like I > didn't give thee any clues. He stopped being an apotheosis and has gone back to his lesser rank of archangel. Perhaps the new job was too much for him, what with the unfulfillable demands for omniscience and infallibility. Some folks think his frequent changes of pseudonym are just an attempt to make it more difficult to trace just what he said and when, since he frequently makes statements that lead people to become curious about that sort of thing. He'll always be an ipsissimus to me, though. === Subject: Re: Beliefs Create Reality, Even In Mathematics. On 2006-04-26 03:29:35 +0700, Tom said: >> Well, everyone that isn't actually you, your RV'ing cat and your >> dharana-achieving dog who talks to you. > I'm still puzzled about that RV-ing cat thing. Also the talking dog > thing. We've already gotten your apology for your error about the > Dharma Dog. Look harder Tommy. Check your sent items. I know you wouldnt want to remember these lapses of sanity but they are certainly there... > Still, I've checked Google over and over and can find no statement of > mine about my dog talking or my remote-viewing cat. I expect 'they' removed them then eh Tommy? lol >> Guess you aren't in Kansas any more Tommy. > I passed through Kansas once. I didn't see any talking dogs there either. Not at all like yours then eh? lol === Subject: Re: Beliefs Create Reality, Even In Mathematics. > On 2006-04-26 03:29:35 +0700, Tom said: > Well, everyone that isn't actually you, your RV'ing cat and your > dharana-achieving dog who talks to you. >> I'm still puzzled about that RV-ing cat thing. Also the talking dog >> thing. We've already gotten your apology for your error about the Dharma >> Dog. > Look harder Tommy. Check your sent items. I know you wouldnt want to > remember these lapses of sanity but they are certainly there... Gee, I did all that already. Have you? If you have been successful where I have failed, I would expect you to be proud to demonstrate it. However, I can see why, if you also fail to find any such post of mine, you might not want to admit it. === Subject: Re: Godel's Theory. <4ae9g0Fsqr86U1@individual.net> <4aep7rFsv85pU1@individual.net> <444259F1.9040201@xortec.fi> <4af6voFstg9mU2@individual.net> <44439f79$8$fuzhry+tra$mr2ice@news.patriot.net> <444434e8$5$fuzhry+tra$mr2ice@news.patriot.net > Matters of faith can be discussed in soc.religion.christian and > similar groups; in sci.math the standard is formal constructions and > proofs. > According to string theorists, all one has to do is make pretty > mathematical models and call that science, so why can't I use the > philosophy of contexual certainty and call that mathematics? > BTW, string theory requires supersymmetry to predict fermions, and > supersymmetry has been disproven. So either it's not a good analogy, or > it's a better analogy than one could have imagined. > PS: I've given up trying to make a complete and consistent model. > Instead, I'm going the opposite direction and abandoning all > mathematics which do not describe the physical world. > Do you mean that you don't believe in uncountable sets? > It really depends on if infinity has any meaning. Nothing /measured/ > can be infinite. Well it would be more accurate to say that no measurement can be infinite: One can measure a portion of an infinite thing. > However, it really depends on the necessity of the > uncountable set to the model of physics in question. Of course, there's > always the view that it's just an approximation. :P > Do you believe > either in Euclidean or non-Euclidean geometry, but not both ... you'll > decide which to believe in when you find out which describes the > physical world? > Yes, except I'm more certain that we live in non-Euclidean geometry. > However, in the end, it all really depends on the experiments done. > (...Starblade Riven Darksquall...) === Subject: Re: Godel's Theory. <4ae9g0Fsqr86U1@individual.net> <4aep7rFsv85pU1@individual.net> <444259F1.9040201@xortec.fi> <4af6voFstg9mU2@individual.net> <44439f79$8$fuzhry+tra$mr2ice@news.patriot.net> <444434e8$5$fuzhry+tra$mr2ice@news.patriot.net> <4448377c$3$fuzhry+tra$mr2ice@news.patriot.net> <444cf34a$7$fuzhry+tra$mr2ice@news.patriot.net 04/23/2006 >I doubt these hypothesis will work unless they develop a new model >of supersymmetry. > Since your doubt is based on urban legends, I'm not much concerned by > it. >I was making a point. > Yes, with empty rhetoric. It was not empty rhetoric. I was just too busy at the time to find where I learned that standard supersymmetry was incorrect. >Apparantly you don't know what an analogy is. > Apparently you don't know that a bogus analogy is irrelevant. It's not bogus. Your logic is bogus. > Because I'm not your servant and you're the one making the claim. > Besides, I have a good idea what you're referring to, and it isn't > what you believe it is. Well I've found it for you. Proof that standard supersymmetry is violated. Happy now? >GOOD mathematics is based on the real world. > ROTF,LMAO! You don't have a clue as to what good Mathematics is. So you believe that mathematics is platonic? >You can make any old mathematics work and make it not based on the >real world, but it will not be very useful. > Tell that to G. H. Hardy. Your ignorance is showing. What makes you the authority on G. H. Hardy? > You've ceased to be entertaining, and you never were enlightening. You > show no signs of being educable, so the best thing to do is to add you > to my filters with the other ignoramuses with delusions of adequacy. > *PLONK* What cowards do when confronted with an enemy they do not want to fight. They bury their heads in the sand. Little do they know that this does not fool their enemies. -Matthew Paul Finnigan- === Subject: Re: Godel's Theory. <4ae9g0Fsqr86U1@individual.net> <4aep7rFsv85pU1@individual.net> <444259F1.9040201@xortec.fi> <4af6voFstg9mU2@individual.net> <44439f79$8$fuzhry+tra$mr2ice@news.patriot.net> <444434e8$5$fuzhry+tra$mr2ice@news.patriot.net> <4448377c$3$fuzhry+tra$mr2ice@news.patriot.net> <444cf34a$7$fuzhry+tra$mr2ice@news.patriot.net *PLONK* > What cowards do when confronted with an enemy they do not want to > fight. They bury their heads in the sand. Little do they know that this > does not fool their enemies. Hahaha, you are so good. -- Roger J. === Subject: Re: Charity addresses the fundamental problem of humanitys relationship to the universe? On 2006-04-27 23:17:45 +0700, Tom said: > On Tue, 25 Apr 2006 20:57:43 -0700, Tom >> And then there's his claim that I bragged about having a remote-viewing >> cat, too... Just out of curiosity, how would the cat tell you? >> he talks to it. > Very true. I do talk to my cat on accasion. However, the cat does no > talking at all. So how did you know *which* mousehole she was watching? It is just that you seemed quite certain on the matter... When you talk to your cat. Do you generally talk about anything specific? > My cat often doesn't come when called, either. And never talks back. Obviously respects you as much as we do then... A === Subject: Re: Charity addresses the fundamental problem of humanitys relationship to the universe? > On 2006-04-27 23:17:45 +0700, Tom said: >> Very true. I do talk to my cat on accasion. However, the cat does no >> talking at all. > So how did you know *which* mousehole she was watching? I have no idea how one would go about that. It's your fantasy, you explain it. === Subject: Re: Charity addresses the fundamental problem of humanitys relationship to the universe? On 2006-04-27 22:58:28 +0700, Tom said: > >> >> >> yes but your morale and resistance seems to be... well, destroyed. Gee, my morale seems to be just fine, but we both know that you would > have more knowledge of my inner feelings than I would, so it must be > true. As for resistance, I don't recall ever arguing that you should > be resisted. Perhaps this is just another of my many memory failures in > which I can recall no such statement that you claim I have made. >> I dont know about this Tom, > Yes, I think that is probably true. You don't know about this. yep. I hear you say it but somehow... I dont think I believe you. A === Subject: Re: Charity addresses the fundamental problem of humanitys relationship to the universe? > On 2006-04-27 22:58:28 +0700, Tom said: > yes but your morale and resistance seems to be... well, destroyed. >> Gee, my morale seems to be just fine, but we both know that you would >> have more knowledge of my inner feelings than I would, so it must be >> true. As for resistance, I don't recall ever arguing that you should >> be resisted. Perhaps this is just another of my many memory failures in >> which I can recall no such statement that you claim I have made. > I dont know about this Tom, >> Yes, I think that is probably true. You don't know about this. > yep. I hear you say it but somehow... I dont think I believe you. Fine with me. I trust you, though. When you say you don't know, I believe you to be correct. After all, you should know whether or not you don't know. However, there is the fact that you are usually very reluctant to admit when you don't know something. For instance, I have yet to hear a definite answer from you on the question of whether or not you, a person with some sort of credentials (which have yet to be revealed) that are supposed to attest to your expertise in qabalistic lore, are fluent in Hebrew. I suspect you are not. I can't think of any good reason why someone would anyone be reluctant to admit they can read and/or speak Hebrew, if they really could. I said it in Hebrew - I said it in Dutch, I said it in German and Greek, But I wholly forgot (and it vexes me much) That English is what you speak. -- Lewis Carroll === Subject: Re: Charity addresses the fundamental problem of humanitys relationship to the universe? On 2006-04-27 22:56:56 +0700, Tom said: >> strange. I dont remember telling you I was a Londoner. Has your cat >> been RV'ing me Tom? > I don't think so. She wouldn't say. You *did* ask then... Does that seem quite... normal to you Tommy? A === Subject: Re: Charity addresses the fundamental problem of humanitys relationship to the universe? > On 2006-04-27 22:56:56 +0700, Tom said: > strange. I dont remember telling you I was a Londoner. Has your cat been > RV'ing me Tom? >> I don't think so. She wouldn't say. > You *did* ask then... I don't need to ask a cat for it not to say. Try to think more clearly. === Subject: Re: Charity addresses the fundamental problem of humanitys relationship to the universe? On 2006-04-27 22:53:16 +0700, Tom said: >> On Tue, 25 Apr 2006 20:57:43 -0700, Tom And then there's his claim that I bragged about having a remote-viewing cat, > too... >> Just out of curiosity, how would the cat tell you? >> Lizz 'and, knowing cats, why would it bother?' Holmans > Just so. The only things a cat tells you is when to feed it, pet it, > or open doors it wants to go through. > I have no idea where he got the notion that my cat and dog can talk. > Perhaps he psychometrised it. Or perhaps you said it. And now cannot recall. Or cannot seem to find it. LOL A === Subject: Re: Charity addresses the fundamental problem of humanitys relationship to the universe? > On 2006-04-27 22:53:16 +0700, Tom said: > On Tue, 25 Apr 2006 20:57:43 -0700, Tom And then there's his claim that I bragged about having a remote-viewing >> cat, >> too... > Just out of curiosity, how would the cat tell you? > Lizz 'and, knowing cats, why would it bother?' Holmans >> Just so. The only things a cat tells you is when to feed it, pet it, or >> open doors it wants to go through. >> I have no idea where he got the notion that my cat and dog can talk. >> Perhaps he psychometrised it. > Or perhaps you said it. The lack of evidence for this has yet to be explained. === Subject: Re: Charity addresses the fundamental problem of humanitys relationship to the universe? On 2006-04-26 23:46:57 +0700, Lizz Holmans said: > On Wed, 26 Apr 2006 20:39:04 +0700, Apotheosis > On Tue, 25 Apr 2006 20:57:43 -0700, Tom >> >> And then there's his claim that I bragged about having a remote-viewing cat, >> too... Just out of curiosity, how would the cat tell you? >> he talks to it. > Every cat person talks to hir cats. But do they answer? > Lizz 'more luck with the vasty deep' Holmans >> A Dont know, but Tommy seems sure his cat was RV'ing the neighbours mousehole (in response to my polite query as to how many mouseholes Tommy has allowed to mme made in hi s house before calling the rodent catcher). One wonders how Tommy actually knew this without being able to talk to his cat. of course, he talks to the dog too, he already confirmed that, but the cat is a new dimension. A === Subject: Re: Charity addresses the fundamental problem of humanitys relationship to the universe? On 2006-04-26 23:44:54 +0700, Lizz Holmans said: > On Wed, 26 Apr 2006 20:37:30 +0700, Apotheosis >> would production of his brain suffice? I could send it to you. > Somehow I doubt you have legal access to his skull. probably not. Shame though. >> I have a matchbox around here somewhere. > Ah. It's our old friend Ad Hom. The last refuge of the entity with > nothing more to say. Or the last refuge of someone with something *really* important to say. A === Subject: Re: Charity addresses the fundamental problem of humanitys relationship to the universe? On Fri, 28 Apr 2006 22:32:22 +0700, Archangelska >On 2006-04-26 23:44:54 +0700, Lizz Holmans said: >> On Wed, 26 Apr 2006 20:37:30 +0700, Apotheosis would production of his brain suffice? I could send it to you. >> Somehow I doubt you have legal access to his skull. >probably not. Shame though. > I have a matchbox around here somewhere. >> Ah. It's our old friend Ad Hom. The last refuge of the entity with >> nothing more to say. >Or the last refuge of someone with something *really* important to say. Does thee really believe name-calling is an acceptable argument? That says more about thee than I've ever asked. And why did thee morph out of my killfile? Back in there, naughty entity. Lizz 'why is my killfile a lot like Walt Whitman?' Holmans -- Rumpeta, rumpeta, rumpeta === Subject: Re: Charity addresses the fundamental problem of humanitys relationship to the universe? On 2006-04-26 15:56:13 +0700, Lizz Holmans said: > On Tue, 25 Apr 2006 20:57:43 -0700, Tom And then there's his claim that I bragged about having a remote-viewing >> cat, too... > Just out of curiosity, how would the cat tell you? > who can fathom the strangeness in the crazy world of Tommy Schuler? A === Subject: Re: Charity addresses the fundamental problem of humanitys relationship to the universe? On Fri, 28 Apr 2006 22:22:32 +0700, Archangelska >On 2006-04-26 15:56:13 +0700, Lizz Holmans said: >> On Tue, 25 Apr 2006 20:57:43 -0700, Tom And then there's his claim that I bragged about having a remote-viewing > cat, too... Lizz 'just clearing up a little attribution attrition' Holmans -- Rumpeta, rumpeta, rumpeta === Subject: what is this called? I have a set of unique items (a,b,c,d), I want following results for size 2 combinations: [(a,b), (c,d)] [(a,c), (b,d)] [(a,d), (b,c)] and for size 3 combinations: [(a,b,c), (d)] [(a,b,d), (c)] [(a,c,d), (b)] [(b,c,d), (a)] Can someone give me the name of such algorithm? === Subject: Re: what is this called? You underspecified what you want. What is your input? A list of 4 items, or a list of arbitrary length? What is your desied output? Only the partitions of shape 2 2 and 3 1? In any case, you seem to want to generate partitions: look at the code of the Mathematica's package Combinatorica, for example, for ideas on how to do that. === Subject: Re: what is this called? Sorry for not being specific. I want arbitrary length. > You underspecified what you want. > What is your input? A list of 4 items, or a list of arbitrary length? > What is your desied output? Only the partitions of shape 2 2 and 3 1? > In any case, you seem to want to generate partitions: look at the code > of the Mathematica's package Combinatorica, for example, for ideas on > how to do that. === Subject: test - ignore test -- The man without a .sig === Subject: ODE in Biology Differential Equations for the Heartbeat and New Impulse and ODE in biological science Here is my e-mail rookie91@wp.pl If You've got thing I asked Send me;) I would be greatful to You for that ;) === Subject: Re: Why choose 1 as the standard? Mail-To-News-Contact: abuse@dizum.com [typical gibberish] > 1 what? > >> Just plain One: All the units cancel. >In order to answer this even close to properly one has to first ask what a >unit in this context is. >Acceleration is typically taught as the change in velocity. Velocity is >typically taught as the change in position. Surely, you mean the rate of change in ...? Normally, I wouldn't pick at such a tiny nit, but when the topic is (more or less) dimensional analysis, it's worth it. -- Michael F. Stemper #include Life's too important to take seriously. === Subject: Length of This String A cylinder 75 cm high has a circumference of 20 cm. A string makes exactly 5 complete turns round the cylinder while its two ends touch the cylinder's top and bottom. How long is the string in cm? === Subject: Re: Length of This String ETAuAhUAxSY7Ac0xBoaFBdv5vDI+1GStksYCFQDM6ZS6shPUufngslSiI6OrTf1m0g== A cylinder 75 cm high has a circumference of 20 cm. A string makes exactly 5 complete turns round the cylinder while its two ends touch the cylinder's top and bottom. How long is the string in cm? ______________________________________ Re; Unroll the cylinder and you have a plane with dimensions 20cm wide by 75cm high. The pitch of the string around the cylinder formed a uniform spiral (presumably) which now forms 5 right triangles on the plane with sides of the right angle being 20cm and 75/5=15cm. Solve the hypotenuse of one of the triangles and multiply by 5. Ans=125cm. === Subject: Re: Length of This String > A cylinder 75 cm high has a circumference of 20 cm. A string makes > exactly 5 complete turns round the cylinder while its two ends touch > the cylinder's top and bottom. How long is the string in cm? As mentioned by you if the ends touch at 5 wraps of the string then the length of the string is 5 times the circumference ...which is 100 cm === Subject: Re: Length of This String > A cylinder 75 cm high has a circumference of 20 cm. A string makes > exactly 5 complete turns round the cylinder while its two ends touch > the cylinder's top and bottom. How long is the string in cm? > As mentioned by you if the ends touch at 5 wraps of the string then the > length of the string is 5 times the circumference ...which is 100 cm You might want to reconsider that. === Subject: Re: Length of This String > A cylinder 75 cm high has a circumference of 20 cm. A string makes > exactly 5 complete turns round the cylinder while its two ends touch > the cylinder's top and bottom. How long is the string in cm? > As mentioned by you if the ends touch at 5 wraps of the string then the > length of the string is 5 times the circumference ...which is 100 cm You forgot the two ends touch the cylinder's top and bottom part. === Subject: Re: Length of This String ok then add 20 cm more .....(10cm -length of radius, at each end) which brings the length to 120cm === Subject: Re: Length of This String > ok > then add 20 cm more .....(10cm -length of radius, at each end) which > brings the length to 120cm Still wrong. === Subject: Re: Length of This String > A cylinder 75 cm high has a circumference of 20 cm. A string makes > exactly 5 complete turns round the cylinder while its two ends touch > the cylinder's top and bottom. How long is the string in cm? Roll the cylinder, without slipping, on a flat expanse of, say, sand until a mark left by string show both ends of the string. The length of such a mark equals the length of the string. === Subject: Re: Length of This String > A cylinder 75 cm high has a circumference of 20 cm. A string makes > exactly 5 complete turns round the cylinder while its two ends touch > the cylinder's top and bottom. How long is the string in cm? Is it wound uniformly? === Subject: Re: Length of This String > A cylinder 75 cm high has a circumference of 20 cm. A string makes > exactly 5 complete turns round the cylinder while its two ends touch > the cylinder's top and bottom. How long is the string in cm? Hint: slice the cylinder from top to bottom and unroll it. Rick === Subject: Re: Length of This String > A cylinder 75 cm high has a circumference of 20 cm. A string makes > exactly 5 complete turns round the cylinder while its two ends touch > the cylinder's top and bottom. How long is the string in cm? Unwrap it and you have a ramp. The vertical height of that ramp is the height of the cylinder. What is the horizontal extent? Given horizontal and vertical extents, what is the length of the ramp? - Randy === Subject: Re: Calculus XOR Probability Virgil said: > David R Tribble said: > Not really. The number of hypernaturals is as uncountable as the > reals in [0,1], even according to nonstandard accepted hyperreal > systems. > Doesn't countability in the hyperreals mean bijectable with the > hypernaturals? If so, they can hardly be uncountable. So, countability is a matter of scale? Interesting..... ... then we can map each of the hypernaturals to a corresponding > real in [0,1] ... You are attempting to define a real number line that contains > infinite reals, and then declare that it is the same as the real > number line of standard arithmetic. You're also attempting to mix > the nonstandard hyperreals in with the standard reals, and then > declare this mix is the real number line. Both of these attempts > are wrong. Wrong, morally? Wrong, how? > Logically wrong! Mathematically wrong! Each system must follow its own > rules, and what is true in one will often be false in another. So TO's > mixing of them means that he will get the same thing both true and > false simultaneously, which is the kind of contradiction that destroys > the value of any proposed system. Only if I mix contradictory ideas. > Where is this axiomatically stated that the real line has no two > points infinitely distant from each other? Let me get my crosshairs > on it... > It is a direct consequence of the Archimedean property of the reals. > This is a theorem, not an axiom, but follows from axioms. How does it follow from there being an infinite number of values between any two that there are only a finite number of naturals between any two? That assumes its conclusion, obviously. -- Smiles, Tony === Subject: Re: Calculus XOR Probability > Virgil said: David R Tribble said: Not really. The number of hypernaturals is as uncountable as the > reals in [0,1], even according to nonstandard accepted hyperreal > systems. Doesn't countability in the hyperreals mean bijectable with the > hypernaturals? If so, they can hardly be uncountable. > So, countability is a matter of scale? Interesting..... > ... then we can map each of the hypernaturals to a corresponding > real in [0,1] ... You are attempting to define a real number line that contains > infinite reals, and then declare that it is the same as the real > number line of standard arithmetic. You're also attempting to mix > the nonstandard hyperreals in with the standard reals, and then > declare this mix is the real number line. Both of these attempts > are wrong. Wrong, morally? Wrong, how? > > Logically wrong! Mathematically wrong! Each system must follow its own > rules, and what is true in one will often be false in another. So TO's > mixing of them means that he will get the same thing both true and > false simultaneously, which is the kind of contradiction that destroys > the value of any proposed system. Only if I mix contradictory ideas. Which you quite regularly do. Where is this axiomatically stated that the real line has no two > points infinitely distant from each other? Let me get my crosshairs > on it... It is a direct consequence of the Archimedean property of the reals. > This is a theorem, not an axiom, but follows from axioms. > How does it follow from there being an infinite number of values between any > two that there are only a finite number of naturals between any two? That > assumes its conclusion, obviously. That is not how one gets to that conclusion. it is a property of the naturals that there are only finitely many less that any given natural. It follows that there is a natural larger than any given rational. It follows that there is a natural larger than any Cauchy sequence of rationals. It follows that there is a natural larger that any given real. It follows that there are only finitely many naturals between any two given reals. === Subject: Re: Calculus XOR Probability Mike Kelly said: > Mike Kelly said: > Oh. Then there are only finitely many reals in [0,1)? Huh! I thought there were > more than any finite number, because for any finite number of subdivisions, > there is still a finite size to each interval, and another subdivision possible > to specify more reals. But, maybe that's just me. > There is no finite natural number to represent the number of reals in > [0,1). Then I say there are infinitely many reals in that interval. > Sounds fine. > Great! > There is no finite natural number to represent the number of natural > numbers. Then I say there are infinitely many naturals. (Why do you > disagree with that second bit, incidentally?) > The Peano axioms define an externally infinite set. > I'm sorry but I don't know what externally infinite means, maybe you > could explain. One that achieves infinity through infinite range, rather than through infinite density, because it adds new elements outside of its range, rather than within it. Does that make sense? >However, in any such > recursive definition, you are generating only a finite number of elements in > any finite number of iterations. If the naturals are restricted to finite > values, such that none is infinitely beyond any other, then I do not consider > this an infinite set, but unboundedly large while finite. > The naturals aren't generated by some iterative process that ends. They > are defined by the Peano axioms and, hey, there they are, all > infinitely many of them. What a dandy statement, chock full of new information! :D > I think you're saying no natural has an infinite number of > predecessors. This is true (if it were not then 0 would have to be the > successor to some natural -> contradiction). Each natural has a number > of predecessors equal to its value. However every natural has an > infinite number of successors. Making the set unboundedly large - you > can list forever and there are always more members you haven't listed. > That's what I call infinite. Because it's size is not-finite. > In-finite. You see? I know how the standard definitions work, and why the set of finite naturals is considered infinite, but I have other criteria. What you call infinite is literally endless, but it includes endless due to the hazy upper boundary of positions within the set finite, there are no two elements with an infinite number of elements between them. So, despite the boundlessness which makes injection into a proper subset possible, I can't consider this set to be actually infinite in size. For me countably infinite means unboundedly large but finite, and only what you call uncountable is truly infinite. If you pick two points on a line, there are truly an infinite number of points between them. That's an infinite set. > When you can prove > that no two finite naturals have an infinite number of intervening elements, > then I don't see that set as infinite. > So do your T-Naturals (or whatever it was, I forget) have some > members with infinite predecessors? That contradicts the peano axioms. No it doesn't, provided you can have an infinite sequence which ends. Now, you are going to say, infinite means NOT ending, but the reals in [0,1] start at 0 and end at 1. You move along that line from 0 to 1, and you've traversed an infinite number of points, and then stopped. besides, the Peano axioms are not the only alternative. I rather like the idea of extending the line in both directions at once, so every element has a predecessor and successor. y -> x y Further, it is inductively and trivially > provable that if one starts with {1}, having the set size and largest value > equal, and adds successive elements as the increment of the largest value, that > the size of the set and the maximal value are incremented in tandem, and remain > equal forever. There is no point where you add the next element, incrementing > both set size and maximal element, and cause those two equal values to become > unequal, especially to the extent that the set size is incremented to infinity, > while the maximal element remains finite. It's simply impossible. > Yes, it is impossible for a set of naturals of the form { 1, 2, ... , N > } to have a non-finite set size. The naturals are not a set of that > form so I don't see you've said anything that applies to the set of all > naturals. The fact is true for all n in N, that the set up to and including any one of them is finite, so there is no point where the addition of any elements produces a set of infinite size. > The identity relation between element count and value characterizes the > naturals. > I'm not sure I know what that means. I thought the Peano axioms > characterised the naturals? Identity relation between element count > and value characterises sets of naturals of the form { 1, 2, ..., N }. > The naturals are not such a set. Sure they are. You can start anywhere, and 1 is the natural place to start, especially since the element value is identical with its position in the set. It's as valid as starting at 0, and demonstrates clearly how the infinitude of the set size depends on the element values. > It does not follow that there is some number you can call the size of > the set of reals in [0,1) or the size of the set of naturals unless you > rigorously define some way of assigning sizes to infinite sets and > decide to call these sizes numbers. > That's the plan. > Cardinality is one system for doing this. The numbers used in > cardinality aren't anything much like the real numbers or the natural > numbers. You can't do any of the standard arithmetic with them. They > don't feel like numbers to me, but they are rigorously defined and > can be looked at mathematically. > Yes, but they're really not numbers, and you can't really do any math with > them. > Well, you can do math with them. Just not all of the arithmetic you'd > like to. Well, I don't like that kind of thing. :) > You've obviously rejected Cardinality as a way to describe a number > for the size of an infinite set. Your system wants to have infinite > numbers that are actually just like finite numbers except bigger, so > you can both describe infinite sets' sizes with them and do arithmetic > with them. So for you an infinite real is just a real bigger than all > other reals. One problem : there are no reals bigger than all other > reals, by the definition of real numbers. > I never made that statement, but okay, whatever. An infinite value is bigger > than all finite values. > There are no such values in the reals or naturals and you haven't > given me any other system to work in. http://mathworld.wolfram.com/HyperrealNumber.html > The only non-circular definition of infinite I've seen from you > assumes, a priori, the existence of the standard reals, specifies a > subset of these which you think is all finite reals : > 1) 0 finite(x) > 2) finite(x) -> finite(0-x) > 3) finite(x) -> finite(1/x) > 4) finite(x) -> finite(2^x) (optional) > Yes, that covers all finite reals, negative and fractional included. > I'm not sure it does but I think that's irrelevant really so let's > leave it. I am, but okay. > and then says infinite reals numbers are reals not in this set of > reals. That's fine, but the set of reals not in the set of finite > reals is empty. That is, there are no infinite numbers as defined by > you. You cannot get to infinite reals starting from the standard reals. > If you're not using the standard reals you're going to have to start > from scratch and derive whatever you think numbers are.. Even 0 > doesn't mean anything if you're not in some number system. 2^x > certainly doesn't. > 0 (or 1) is simply axiomatically declared in Peano. Do you have a problem with > that? One has to start somewhere, no? So, yes, I find I basically have to start > from scratch, and it doesn't just include number systems, but logical systems > as well, to fully integrate the foundations of math. Whew! It's better than > having nothing to do, anyway. :) > Well what are your axioms? None of the Peano naturals have infinite > predecessors and so there are no infinite naturals. Standard reals > don't have infinite values. > These ... > 1) 0 finite(x) > 2) finite(x) -> finite(0-x) > 3) finite(x) -> finite(1/x) > 4) finite(x) -> finite(2^x) (optional) > ... are not axioms. If you want to use the standard reals and say > infinite numbers are numbers bigger than any finite real then you > don't HAVE any infinite numbers. If you don't tell me what 0 1 2 > - and / mean the above rules mean nothing at all... if you want > them to be the standard numbers and operations then there aren't any > infinite numbers. What makes them not axioms, assuming the arithmetic operations and numeric primitives have been defined? > When you consider the concept that you throw all natural-numbered balls in a > vase and shake it infinitely many times and remove one, and that none has any > more chance than any other of being chosen, that sounds like a uniform > probability distribution to me. Does it sound like that to you? > If it were true that none has any more chance than any other of being chosen, > then yes, it would sound like a uniform probability distribution. However, > neither that statement nor its negation is true. The word likely simply > doesn't apply in this instance. > Oh, that's just gratuitous dismissal. Is there a chance than ball 10 will be > chosen? It might, so yes, there's a chance. > Only if there is a uniform probability distribution to choose from the > vase. There is no uniform distribution on the standard naturals (I > think you've agreed to that) and that's what is in the vase. > So, then, ball 10 has NO chance of being picked? Is that true for ALL balls? > Does NO ball have ANY chance? > There is no uniform distribution on the naturals, so the experiment > cannot be performed, even in theory. > Agreed that one cannot define any expected value. That's because one cannot > define any real size. Aleph_0 is broken. It doesn't work. One must have a size > n in order to determine individual probabilities. > I didn't say there is no expected value. I said there is *no* uniform > probability distribution on the naturals. There isn't one. It does not > exist. Uh, yeah, I know. >Well, then, I guess if one picks a ball from the > vase, their hand will be empty. That's probably because the labels aren't glued > on very well, and the balls go *poof* when exposed to light without their > labels. haha! ;) > You can think of a uniform distribution on the naturals as being > related to a vase filled with balls if it helps your intuistic way of > looking at things. But, please, don't be intentionally obtuse. It's > demeaning for you and irritating for me. The vase filled with balls, > one for each natural number, is not a physically realisable experiment. > You cannot just say well it's a vase so obviously you can shake it, > reach in and pick out a random ball. It's not a real vase. It's an > analogy for a set. > I'm not being obtuse, but reacting to nonsensical statements and a nonsensical > result. The labels are a diversion and confusion. My experiment is as > realizable as the original, is it not? Why even pretend to have an answer to > the ball and vase problem, if it's not realizable? > You're basing your argument on what you could do if it *were* > realisable. If it were realisable then you could reach in and randomly > select a ball. This doesn't imply that you can uniformly randomly > select a natural number because the experiment is not realisable. Whatever. > Unless you meant all T-natural-numbered balls? Can you get to that by > adding balls to the vase containing all the standard-natural-numbered > balls? What extra balls have to be added to the standard ones? Ones > with infinite numbers? Adding additional balls isn't going to make it > any easier to have a discrete uniform distribution. > I don't know any infinite numbers to add, either, but that's a > different problem. > It's also irrelevant. > It's irrelevant that there are no infinite natural numbers? > Interesting. >It doesn't matter if n is finite or infinite, if you are > choosing one out of n choices, then the average probability of the choices is > 1/n. > But n is *never* infinite. Says you without justification. > You haven't presented me with a system of infinite numbers such that > 1/n makes any sense whatsoever. I know what 1/n means for integers > and reals but there are no infinite numbers in the naturals or reals. > What is your alternative to the naturals and reals? Why should your > alternative make more sense than the standards? The standard natural > numbers match my intuitions about counting *very* well. The concept of > infinite numbers really doesn't, at all. > Well, then, I guess we have different intuitions. Picture the unit interval > [0,1] as a little shrunk-down real line from 0 to oo. > to oo? The real line doesn't have endpoints, surely? But [0,1] surely > does. http://mathworld.wolfram.com/AffinelyExtendedRealNumbers.html http://mathworld.wolfram.com/ProjectivelyExtendedRealNumbers.html >You don't have a very good n for the naturals. You can't say 1/aleph_0, > you can't define an average or expected value, or a probability distribution. > Huh? I don't just not have a very good n for the naturals, I don't have > *any* n which is the largest natural and thus the natural number > representing the size of the set of naturals. You can't do this > arithmetic with cardinalities and there isn't any natural number which > represents the size of the set of natural numbers. > That's right, and focusing on a value that you can't do anything with is > fruitless. That's why there's so many unanswerable questions in this theory. > Aleph_0 doesn't work. > Trying to define a uniform distribution over the naturals is indeed > fruitless, seeing as none exists.. Alpeh_0 works as a label for the > size of the naturals. That's all it claimed to be. Yes, well, if the naturals start at 1, then it's also the maximal element of the set, which makes no sense whatsoever. It's stepping on its own tail. > Do we know what that chance is? Not > unless we know how many balls are in the vase. If we know there are n balls in > the vase, then we can say, without hesitation or risk of embarrassment, that > each has an average chance of 1/n, and then all the individual probabilities > sum to 1 as they should. :D > What number represents the size of the set of naturals? Not one you can > do that kind of arithmetic with. > Right. Aleph_0 sucks, and is amorphous. That's why it sucks. > Aleph_0 sucks because it offends your intuition by not being subject to > arithmetic? OK... > Yeah, it's not subject to consistent logic of any sort. I was very impressed by > yet another novice coming up with the hard questions. > What consistent logic is it not subject to? It's just a label > describing the size of the set of naturals. It's not a number in the > sense of a natural number or a real number so it doesn't have their > sort of arithmetic. I wasn't expecting it to. Then you weren't expecting to be able to actually conclude anything about anything. Well, that's one way to avoid disappointment. > How many bits are > required to list all aleph_0 naturals? Not only can your theory not specify > this number of bits, it doesn't even have a TYPE of number that it COULD be. If > you have a finite number of bits, you get a finite number of bit strings, and > if you have a countably infinite number of bits you get an uncountably infinite > number of bit strings, and there's nothing in between to produce any countably > infinite set of bit strings. > There is no type of number to specify how many bits are needed to > represent the naturals because they go on FOREVER. That's not *that* > hard to understand? So, does that mean one needs an infinite number of bits, or a finite number? Your smallest infinity is too large, and finite's too small. There is no fix for that mess. >Specify an n you > can do arithmetic with, and the problem goes away. I've already agreed there > are problems with the standard naturals in this area. > No such n can be specified for the naturals. > Correct. Forget the naturals. Think infinite n. > I can't think infinite n if I don't know what infinite n means and > you've not explained the term. Well, you said an n larger than any > finite number. I don't know what sort of thing such an n would be. > Certainly not a natural number. You've yet to provide an alternative. Think of the number of points in a finite line segment, the reals in [0,1]. It's a larger number than any finite number, because any finite number leaves points or reals unincluded. That's why it's defined as larger than any finite. > Just so I'm clear - I don't get what an infinite n is. By your > description of infinite n I think there aren't any infinite n at > all. So we're back to finite numbers of points in the line segment? (sigh) > But, forget that concept, > say you threw all elements of a set in a bag and chose one at random. However > large that set is, the average chance for an idividual element is the > reciprocal. That breaks nothing. Aleph_0's already broken. > What if the size of the set is not a number that *has* a reciprocal? > Then I guess there is no average chance and no uniform probability > distribution. The naturals are an example of such a set. > Yeah. I know. Fuggettabouddit. Han's original statement was about infinite n. > The naturals are a diversion from the point. > There aren't any infinite n. There aren't any. No! They don't exist. No! No! Stay away from me! (yawn) > Are your T-naturals going to have some infinite number which is the > largest element of the naturals? How does that work? Naturals can be > arbitarily large. > No, there is no largest hypernatural, but there is a declared unit infinity, > which can be treated as units tend to be. > This means nothing to me, nothing whatsoever. I thought your new system > was supposed to be easy on the intuition? Maybe for a start you can > explain without vague handwaving what a declared unit infinity is? > Well, keep in mind that, while you have been building your intuition on the > standard system, it has always rubbed my intuition the wrong way. So, what's > intuitive for one may be counterintuitive for another, especially given > extensive training in a mode of thought. > The unit infinity Big'un is declared to be the number of real numbers in each > unit interval on the real line. I think we agree that's an infinite value, > since we can always place a real between any two reals and still have space to > insert more. > No I don't agree that there is some infinite value that represents > the number of reals in an interval. I think there is no finite value > that represents the number of reals in an interrval. That's not the > same thing. I don't know what an infinite value is. So, there are points in the segment, and you can subdivide the segment and count points as you go, and any finite number of subdivisions leaves finite subsegments which can be further divided, so there are more points uncounted in the segment. And you can't imagine the segment subdivided forever so that the intervals are degenerate and infinitesimal and completely fill the segment? I'm sorry. That's too bad. That must really stink. > Are your T-naturals going to include all the standard finite naturals? > Yup. > > Then there is no largest element and thus no size that you can do > arithmetic with. > Only if we declare it axiomatically, which is the plan. > A set of naturals containing all the finite naturals doesn't have a > largest member. I don't see how an axiom claiming the opposite makes > any sense. Seems contradictory. You have aleph_0, which is far more problematic than a unit infinity as the length of the real line. > If so, your T-naturals won't have a largest element and its size will > be equally amorphous. If not, which ones is it going to exclude? > The size is amorphous until declared to be specific. The number of reals in the > unit interval equals the number of unit intervals (naturals and hypernaturals) > on the infinite real line, an uncountable number, what calls hisself Big'un. > Big'un's reciprocal, Lil'un, is the unit infinitesimal. There's more than > Big'un, like the number of infinitesimal reals in [0,3) is 3* Big'un. So, I > don't need to exclude any. I just need to do the arithmetic to compare them. > OK, you've completely lost me.I can't discern this from utter nonsense. > It certainly isn't helping me to understand your thinking. A few things > I'm having trouble following... > Whatcha don' like numbers what's called Big'un an' Lil'un? Jes' wait till I > names one Hoss or Zeke. Hyuck! {:B > Firstly the phrase The size is amorphous until declared to be > specific confuses me. Are set sizes not fixed in your theory? Does the > set of naturals have many sizes? > The boundary between finite and infinite cannot be pinned down in any > substantial sense, so any set defined using finiteness as any criterion suffer > from this issue. > That didn't answer my question. Are set sizes fixed in your theory or > not? Sizes of infinite sets are comparable and orderable over a common value range. I don't know what you mean by fixed. It depends on the value range you're discussing. When you define an infinite set with a recursive definition, and then only perform a finite number of iterations, it's not an infinite set, but a finite recursive set. In general, one will compare two infinite sets over the range [0,Big'un] to get a standard measure of the set in terms of a formula on infinite units. > The infinite real line has countably many unit intervals. You're > claiming there are uncountably many naturals? > Yes, I am considering an uncountably long real line with an uncountably > infinite set of unit intervals, each corresponding to a hypernatural, finite or > infinite. > You'll have to explain to me what an infinite natural is. I don't know. > My picture of the real line is a line that goes > ..... -3, -2, -1, 0, 1, 2, 3 ..... > extending forever in either direction. You can count it by the old 0, > 1, -1, 2, -2, 3, -3 method. I don't see how it can be uncountably > long. 1,2,3....Big'un-3, Big'un-2, Big'un-1, Big'un, Big'un+1, ... Here is an infinite T-riffic number, with the limit point (the colon) at the log10(Big'un) bit: 3:333...333.333...333. That's 4*Big'un/3. :) > What are hypernaturals? Naturals larger than any finite natural? None > exist. > That's an unfounded statement of denial. You might want to look into > Nonstandard Anaylsis and the work of Abraham Robinson. > I have, briefly. It's nothing like what you're saying. > It didn't make a lot of sense to me, anyway. I could see all the > mathematical symbols lined up nicely and I'm sure it's rigorous but > that still doesn't explain to me how a number can be infinitely large > except by just defining it so. Or how an infinitesimal differs from > real numbers which can be as small as you like. The English language > descriptions use the phrase infinitely close to zero. That means > nothing to me. Do you think I can do anything about that, except express my condolences? I mean, this is like discussing the fine points of mixing curry with the McDonald's cooking staff. You sound like a devout finitist. What am I to do about that? > There are no infinitesimals in the reals, by the archimedean property. > Not in the standard reals, but in the hyperreals, there are. > So is your theory just the hyperreals of NSA? NSA doesn't contradict > standard set theory. It uses it. No, it's not, but NSA provides one rigorous basis for infinite and infinitesimal values, which you seem to disregard. >Infinitesimals can > be considered discrete, or continuity can be extended to the infinitesimal > level, leading to sub-infinitesimals. Discreteness on the infinitesimal level > does not violate the Archimedean principle on the finite level, since two > consecutive infinitesimals are not distinguishable standard reals, and > therefore do not require an intermediate value. > The real line is a continum of real numbers. There is no finite level > and a smaller infinitesimal level. The reals go allllll the way down, > as far as you like. But, in standard analysis, if there is no finite difference between two reals, they are considered equal. If you can't conceive of infinitesimal differences, then all can suggest is you consider that when you touch something, you don't really touch it. Archimedes would agree with me. :) > There is a bijection between the reals in [0,1) and [0,3) so it seems > rather unintuitive to me that there can be more reals in one interval > than the other. > Does it? It seems rather unintuitive to me that [0,3) includes all of [0,1), > plus an infinite set of additional reals in [1,3), and is no larger a set of > reals. Proper subsets are smaller than the super set. That is fundamental fact > violated by standard transfinite set theory continuously, and not very > discretely, but unabashedly and vociferously. > A fundamental fact of what set theory? Using your intuition about > finite sets to make assertions about infinite sets isn't going to get > very far. A bijection matches up elements 1-1. Which ones are supposed > to be missed out by it? It's not a matter of which are missed, but the density and/or range on the real line. The fundamental fact of the proper subset being smaller is NOT one preserved by transfinite set theory, but in my opinion, it's fundamental, and SHOULD be preserved in a theory of infinite sets. > Finally, I haven't seen you define how arithmetic operations work on > anything you've said. I still don't even know what *type* of number > BigUn is so I don't know what operations you can do on it. If it's > the number of unit intervals on the real line I'd expect it to be a > Cardinality. But I don't think that's what you want.. > Oh no, it's not a cardinality by any means. Cardinals and ordinals are not part > of my theory at all. They're varieties of phlogiston. > Big'un is an infinite unit which can be used in any arithmetic formula, the > values of which can be compared using the notion that if f(n)>g(n) for all n>m, > then f(Big'un)>g(Big'un). This allows for a full spectrum of infinite values > which can be precisely compared and ordered formulaically. The same goes for > the unit infinitesimal Lil'un. > You still didn't answer what *type* of number Bigun and Lilun are. Infinite and infinitesimal, but that apparently means nothing to you, so I don't know what to say. > Number doesn't have a mathematical definition that I know of. You can > define natural_numbers or real_numbers or rational_numbers or > complex_numbers or hyperreal_numbers but number floating around on > its own doesn't have a definition. An infinite unit which can be used > in any arithmetic formula is not a definition. I can't imagine that any definition would be considered sufficient for you, given your apparent position on the infinite anyway. So, I guess I won't worry myself too much about it. > mike. -- Smiles, Tony === Subject: Re: Calculus XOR Probability > Mike Kelly said: I'm sorry but I don't know what externally infinite means, maybe > you could explain. > One that achieves infinity through infinite range, rather than > through infinite density, because it adds new elements outside of its > range, rather than within it. Does that make sense? NO! An open real interval ( or an open rational interval, as far as that goes) is externally infinite in the TO sense but of finite range. Similarly the set {1 - 1/n: n in N} is bounded but externally infinite in TO's sense. The naturals aren't generated by some iterative process that ends. > They are defined by the Peano axioms and, hey, there they are, all > infinitely many of them. > What a dandy statement, chock full of new information! New to TO, perhaps, but things don't seem to stay with him long, so what was old yesterday is always new to him today. I think you're saying no natural has an infinite number of > predecessors. This is true (if it were not then 0 would have to be > the successor to some natural -> contradiction). Each natural has a > number of predecessors equal to its value. However every natural > has an infinite number of successors. Making the set unboundedly > large - you can list forever and there are always more members you > haven't listed. That's what I call infinite. Because it's size is > not-finite. In-finite. You see? > I know how the standard definitions work, and why the set of finite > naturals is considered infinite, but I have other criteria. What you > call infinite is literally endless, but it includes endless due to > the hazy upper boundary of finiteness. Wrong! It is an endless sequence because it has no end. That is what endless means. So do your T-Naturals (or whatever it was, I forget) have some > members with infinite predecessors? That contradicts the peano > axioms. > No it doesn't, provided you can have an infinite sequence which ends. > Now, you are going to say, infinite means NOT ending, but the reals > in [0,1] start at 0 and end at 1. When TO can arrange the reals in [0,1] as a sequence ( a mapping from the naturals to the reals, only then can he speak of it as an ending or non-ending sequence. Such deliberate conflation is deliberate fallacy. You move along that line from 0 to > 1, and you've traversed an infinite number of points, and then > stopped. besides, the Peano axioms are not the only alternative. I > rather like the idea of extending the line in both directions at > once, so every element has a predecessor and successor. Then one gets the integers as two sort of back to back sequences, which still are different from real intervals. > Yes, it is impossible for a set of naturals of the form { 1, 2, ... > , N > } to have a non-finite set size. The naturals are not a set of that > form so I don't see you've said anything that applies to the set of > all naturals. > The fact is true for all n in N, that the set up to and including any > one of them is finite, so there is no point where the addition of any > elements produces a set of infinite size. At any point 'addition' of an endless set of elements makes the set infinite. > Sure they are. You can start anywhere, and 1 is the natural place to > start 0 is even more natural, especially since it shows how false TO's argument are. > These ... 1) 0 finite(x) 2) finite(x) -> finite(0-x) 3) finite(x) > -> finite(1/x) 4) finite(x) -> finite(2^x) (optional) ... are not axioms. If you want to use the standard reals and say > infinite numbers are numbers bigger than any finite real then you > don't HAVE any infinite numbers. If you don't tell me what 0 1 > 2 - and / mean the above rules mean nothing at all... if you > want them to be the standard numbers and operations then there > aren't any infinite numbers. > What makes them not axioms, assuming the arithmetic operations and > numeric primitives have been defined? First you have to make all those definitions, which you have not done. All of the systems in which those definitions have been made are systems in which your garbage does not work. > There is no uniform distribution on the naturals, so the > experiment cannot be performed, even in theory. > Agreed that one cannot define any expected value. That's because > one cannot define any real size. Aleph_0 is broken. It doesn't > work. One must have a size n in order to determine individual > probabilities. I didn't say there is no expected value. I said there is *no* > uniform probability distribution on the naturals. There isn't one. > It does not exist. > Uh, yeah, I know. Then why do you keep arguing to the contrary? === Subject: Re: Calculus XOR Probability <444f9ccd$0$14437$6d36acad@titian.nntpserver.com> Mike Kelly said: > Oh. Then there are only finitely many reals in [0,1)? Huh! I thought there were > more than any finite number, because for any finite number of subdivisions, > there is still a finite size to each interval, and another subdivision possible > to specify more reals. But, maybe that's just me. > There is no finite natural number to represent the number of reals in > [0,1). Then I say there are infinitely many reals in that interval. > Sounds fine. > Great! > There is no finite natural number to represent the number of natural > numbers. Then I say there are infinitely many naturals. (Why do you > disagree with that second bit, incidentally?) > The Peano axioms define an externally infinite set. > I'm sorry but I don't know what externally infinite means, maybe you > could explain. > One that achieves infinity through infinite range, rather than through infinite > density, because it adds new elements outside of its range, rather than within > it. Does that make sense? What does acheive infinity mean? Sets are either infinite or they are not. Are externally infinite sets infinite? Are the rationals internally infinite, externally infinite, both or something else? >However, in any such > recursive definition, you are generating only a finite number of elements in > any finite number of iterations. If the naturals are restricted to finite > values, such that none is infinitely beyond any other, then I do not consider > this an infinite set, but unboundedly large while finite. > The naturals aren't generated by some iterative process that ends. They > are defined by the Peano axioms and, hey, there they are, all > infinitely many of them. > What a dandy statement, chock full of new information! :D > I think you're saying no natural has an infinite number of > predecessors. This is true (if it were not then 0 would have to be the > successor to some natural -> contradiction). Each natural has a number > of predecessors equal to its value. However every natural has an > infinite number of successors. Making the set unboundedly large - you > can list forever and there are always more members you haven't listed. > That's what I call infinite. Because it's size is not-finite. > In-finite. You see? > I know how the standard definitions work, and why the set of finite naturals is > considered infinite, but I have other criteria. What I said wasn't a standard definition. If there isn't a finite natural number to be the size of a set then I think it's not a finite set. That's *my* definition. >What you call infinite is > literally endless, but it includes endless due to the hazy upper boundary of > finiteness. Hazy upper bound? No, there is *no* upper bound. > positions within the set finite, there are no two elements with an infinite > number of elements between them. So, despite the boundlessness which makes > injection into a proper subset possible, I can't consider this set to be > actually infinite in size. What's the difference between not finite and actually infinite? >For me countably infinite means unboundedly large > but finite, and only what you call uncountable is truly infinite. Ok... so you want to change terminology for no reason. Fine. > If you > pick two points on a line, there are truly an infinite number of points between > them. That's an infinite set. Uh huh.. > When you can prove > that no two finite naturals have an infinite number of intervening elements, > then I don't see that set as infinite. > So do your T-Naturals (or whatever it was, I forget) have some > members with infinite predecessors? That contradicts the peano axioms. > No it doesn't, provided you can have an infinite sequence which ends. Now, you > are going to say, infinite means NOT ending, but the reals in [0,1] start at > 0 and end at 1. You move along that line from 0 to 1, and you've traversed an > infinite number of points, and then stopped. besides, the Peano axioms are not > the only alternative. I rather like the idea of extending the line in both > directions at once, so every element has a predecessor and successor. > y -> x y Further, it is inductively and trivially > provable that if one starts with {1}, having the set size and largest value > equal, and adds successive elements as the increment of the largest value, that > the size of the set and the maximal value are incremented in tandem, and remain > equal forever. There is no point where you add the next element, incrementing > both set size and maximal element, and cause those two equal values to become > unequal, especially to the extent that the set size is incremented to infinity, > while the maximal element remains finite. It's simply impossible. > Yes, it is impossible for a set of naturals of the form { 1, 2, ... , N > } to have a non-finite set size. The naturals are not a set of that > form so I don't see you've said anything that applies to the set of all > naturals. > The fact is true for all n in N, that the set up to and including any one of > them is finite, so there is no point where the addition of any elements > produces a set of infinite size. True, but so what? The naturals are not produced by adding elements. > The identity relation between element count and value characterizes the > naturals. > I'm not sure I know what that means. I thought the Peano axioms > characterised the naturals? Identity relation between element count > and value characterises sets of naturals of the form { 1, 2, ..., N }. > The naturals are not such a set. > Sure they are. You can start anywhere, and 1 is the natural place to start, > especially since the element value is identical with its position in the set. > It's as valid as starting at 0, and demonstrates clearly how the infinitude of > the set size depends on the element values. No, the naturals are *not* a set of the form { 1, 2, ..., N }. The naturals do *not* have a largest member. > It does not follow that there is some number you can call the size of > the set of reals in [0,1) or the size of the set of naturals unless you > rigorously define some way of assigning sizes to infinite sets and > decide to call these sizes numbers. > That's the plan. > > Cardinality is one system for doing this. The numbers used in > cardinality aren't anything much like the real numbers or the natural > numbers. You can't do any of the standard arithmetic with them. They > don't feel like numbers to me, but they are rigorously defined and > can be looked at mathematically. > Yes, but they're really not numbers, and you can't really do any math with > them. > Well, you can do math with them. Just not all of the arithmetic you'd > like to. > Well, I don't like that kind of thing. :) Not everything in maths is arithmetic. > You've obviously rejected Cardinality as a way to describe a number > for the size of an infinite set. Your system wants to have infinite > numbers that are actually just like finite numbers except bigger, so > you can both describe infinite sets' sizes with them and do arithmetic > with them. So for you an infinite real is just a real bigger than all > other reals. One problem : there are no reals bigger than all other > reals, by the definition of real numbers. > I never made that statement, but okay, whatever. An infinite value is bigger > than all finite values. > There are no such values in the reals or naturals and you haven't > given me any other system to work in. > http://mathworld.wolfram.com/HyperrealNumber.html But you're not working in the Hyperreals. What's *your* system? > The only non-circular definition of infinite I've seen from you > assumes, a priori, the existence of the standard reals, specifies a > subset of these which you think is all finite reals : > 1) 0 finite(x) > 2) finite(x) -> finite(0-x) > 3) finite(x) -> finite(1/x) > 4) finite(x) -> finite(2^x) (optional) > Yes, that covers all finite reals, negative and fractional included. > I'm not sure it does but I think that's irrelevant really so let's > leave it. > I am, but okay. > and then says infinite reals numbers are reals not in this set of > reals. That's fine, but the set of reals not in the set of finite > reals is empty. That is, there are no infinite numbers as defined by > you. You cannot get to infinite reals starting from the standard reals. > If you're not using the standard reals you're going to have to start > from scratch and derive whatever you think numbers are.. Even 0 > doesn't mean anything if you're not in some number system. 2^x > certainly doesn't. > 0 (or 1) is simply axiomatically declared in Peano. Do you have a problem with > that? One has to start somewhere, no? So, yes, I find I basically have to start > from scratch, and it doesn't just include number systems, but logical systems > as well, to fully integrate the foundations of math. Whew! It's better than > having nothing to do, anyway. :) > Well what are your axioms? None of the Peano naturals have infinite > predecessors and so there are no infinite naturals. Standard reals > don't have infinite values. > These ... > 1) 0 finite(x) > 2) finite(x) -> finite(0-x) > 3) finite(x) -> finite(1/x) > 4) finite(x) -> finite(2^x) (optional) > ... are not axioms. If you want to use the standard reals and say > infinite numbers are numbers bigger than any finite real then you > don't HAVE any infinite numbers. If you don't tell me what 0 1 2 > - and / mean the above rules mean nothing at all... if you want > them to be the standard numbers and operations then there aren't any > infinite numbers. > What makes them not axioms, assuming the arithmetic operations and numeric > primitives have been defined? Define them, then? They're defined in standard mathematics very rigorously but if you want to just borrow those definitions then you don't have any infinite numbers. > When you consider the concept that you throw all natural-numbered balls in a > vase and shake it infinitely many times and remove one, and that none has any > more chance than any other of being chosen, that sounds like a uniform > probability distribution to me. Does it sound like that to you? > If it were true that none has any more chance than any other of being chosen, > then yes, it would sound like a uniform probability distribution. However, > neither that statement nor its negation is true. The word likely simply > doesn't apply in this instance. > Oh, that's just gratuitous dismissal. Is there a chance than ball 10 will be > chosen? It might, so yes, there's a chance. > Only if there is a uniform probability distribution to choose from the > vase. There is no uniform distribution on the standard naturals (I > think you've agreed to that) and that's what is in the vase. > So, then, ball 10 has NO chance of being picked? Is that true for ALL balls? > Does NO ball have ANY chance? > There is no uniform distribution on the naturals, so the experiment > cannot be performed, even in theory. > Agreed that one cannot define any expected value. That's because one cannot > define any real size. Aleph_0 is broken. It doesn't work. One must have a size > n in order to determine individual probabilities. > I didn't say there is no expected value. I said there is *no* uniform > probability distribution on the naturals. There isn't one. It does not > exist. > Uh, yeah, I know. >Well, then, I guess if one picks a ball from the > vase, their hand will be empty. That's probably because the labels aren't glued > on very well, and the balls go *poof* when exposed to light without their > labels. haha! ;) > You can think of a uniform distribution on the naturals as being > related to a vase filled with balls if it helps your intuistic way of > looking at things. But, please, don't be intentionally obtuse. It's > demeaning for you and irritating for me. The vase filled with balls, > one for each natural number, is not a physically realisable experiment. > You cannot just say well it's a vase so obviously you can shake it, > reach in and pick out a random ball. It's not a real vase. It's an > analogy for a set. > I'm not being obtuse, but reacting to nonsensical statements and a nonsensical > result. The labels are a diversion and confusion. My experiment is as > realizable as the original, is it not? Why even pretend to have an answer to > the ball and vase problem, if it's not realizable? > You're basing your argument on what you could do if it *were* > realisable. If it were realisable then you could reach in and randomly > select a ball. This doesn't imply that you can uniformly randomly > select a natural number because the experiment is not realisable. > Whatever. > Unless you meant all T-natural-numbered balls? Can you get to that by > adding balls to the vase containing all the standard-natural-numbered > balls? What extra balls have to be added to the standard ones? Ones > with infinite numbers? Adding additional balls isn't going to make it > any easier to have a discrete uniform distribution. > I don't know any infinite numbers to add, either, but that's a > different problem. > It's also irrelevant. > It's irrelevant that there are no infinite natural numbers? > Interesting. >It doesn't matter if n is finite or infinite, if you are > choosing one out of n choices, then the average probability of the choices is > 1/n. > But n is *never* infinite. > Says you without justification. > You haven't presented me with a system of infinite numbers such that > 1/n makes any sense whatsoever. I know what 1/n means for integers > and reals but there are no infinite numbers in the naturals or reals. > What is your alternative to the naturals and reals? Why should your > alternative make more sense than the standards? The standard natural > numbers match my intuitions about counting *very* well. The concept of > infinite numbers really doesn't, at all. > Well, then, I guess we have different intuitions. Picture the unit interval > [0,1] as a little shrunk-down real line from 0 to oo. > to oo? The real line doesn't have endpoints, surely? But [0,1] surely > does. > http://mathworld.wolfram.com/AffinelyExtendedRealNumbers.html > http://mathworld.wolfram.com/ProjectivelyExtendedRealNumbers.html -oo and oo aren't endpoints to the real line in either of those. Affinely Extended - Note that these improper elements oo and -oo are not real numbers, and that this system of extended real numbers is not a field. How can something which isn't a real number be on the real line? Projectivley extended - Regrettably, oo is also unordered, i.e., for x in R^* it can be said neither that xoo How can something which isn't even larger or smaller than any real be on the real line? >You don't have a very good n for the naturals. You can't say 1/aleph_0, > you can't define an average or expected value, or a probability distribution. > Huh? I don't just not have a very good n for the naturals, I don't have > *any* n which is the largest natural and thus the natural number > representing the size of the set of naturals. You can't do this > arithmetic with cardinalities and there isn't any natural number which > represents the size of the set of natural numbers. > That's right, and focusing on a value that you can't do anything with is > fruitless. That's why there's so many unanswerable questions in this theory. > Aleph_0 doesn't work. > Trying to define a uniform distribution over the naturals is indeed > fruitless, seeing as none exists.. Alpeh_0 works as a label for the > size of the naturals. That's all it claimed to be. > Yes, well, if the naturals start at 1, then it's also the maximal element of > the set, which makes no sense whatsoever. It's stepping on its own tail. Say what? The Naturals don't *have* a maximal element. You're the only person I've ever heard claim that Aleph_0 must be the maximal element of the naturals. Aleph_0 isn't even a *member* of the naturals. > Do we know what that chance is? Not > unless we know how many balls are in the vase. If we know there are n balls in > the vase, then we can say, without hesitation or risk of embarrassment, that > each has an average chance of 1/n, and then all the individual probabilities > sum to 1 as they should. :D > What number represents the size of the set of naturals? Not one you can > do that kind of arithmetic with. > Right. Aleph_0 sucks, and is amorphous. That's why it sucks. > Aleph_0 sucks because it offends your intuition by not being subject to > arithmetic? OK... > Yeah, it's not subject to consistent logic of any sort. I was very impressed by > yet another novice coming up with the hard questions. > What consistent logic is it not subject to? It's just a label > describing the size of the set of naturals. It's not a number in the > sense of a natural number or a real number so it doesn't have their > sort of arithmetic. I wasn't expecting it to. > Then you weren't expecting to be able to actually conclude anything about > anything. Well, that's one way to avoid disappointment. You can't do arithemtic with everything in mathematics. > How many bits are > required to list all aleph_0 naturals? Not only can your theory not specify > this number of bits, it doesn't even have a TYPE of number that it COULD be. If > you have a finite number of bits, you get a finite number of bit strings, and > if you have a countably infinite number of bits you get an uncountably infinite > number of bit strings, and there's nothing in between to produce any countably > infinite set of bit strings. > There is no type of number to specify how many bits are needed to > represent the naturals because they go on FOREVER. That's not *that* > hard to understand? > So, does that mean one needs an infinite number of bits, or a finite number? > Your smallest infinity is too large, and finite's too small. There is no fix > for that mess. >Specify an n you > can do arithmetic with, and the problem goes away. I've already agreed there > are problems with the standard naturals in this area. > No such n can be specified for the naturals. > Correct. Forget the naturals. Think infinite n. > I can't think infinite n if I don't know what infinite n means and > you've not explained the term. Well, you said an n larger than any > finite number. I don't know what sort of thing such an n would be. > Certainly not a natural number. You've yet to provide an alternative. > Think of the number of points in a finite line segment, the reals in [0,1]. > It's a larger number than any finite number, because any finite number leaves > points or reals unincluded. That's why it's defined as larger than any finite. Or there's no number that says how many points there are? > Just so I'm clear - I don't get what an infinite n is. By your > description of infinite n I think there aren't any infinite n at > all. > So we're back to finite numbers of points in the line segment? (sigh) No, we're back to no number of points in the line segment. There's not finitely many points. So I say in-finitely many points. Infinite precisely because there is no number of points. > But, forget that concept, > say you threw all elements of a set in a bag and chose one at random. However > large that set is, the average chance for an idividual element is the > reciprocal. That breaks nothing. Aleph_0's already broken. > What if the size of the set is not a number that *has* a reciprocal? > Then I guess there is no average chance and no uniform probability > distribution. The naturals are an example of such a set. > Yeah. I know. Fuggettabouddit. Han's original statement was about infinite n. > The naturals are a diversion from the point. > There aren't any infinite n. > There aren't any. No! They don't exist. No! No! Stay away from me! (yawn) You're pretty tiresome yourself. > Are your T-naturals going to have some infinite number which is the > largest element of the naturals? How does that work? Naturals can be > arbitarily large. > No, there is no largest hypernatural, but there is a declared unit infinity, > which can be treated as units tend to be. > This means nothing to me, nothing whatsoever. I thought your new system > was supposed to be easy on the intuition? Maybe for a start you can > explain without vague handwaving what a declared unit infinity is? > Well, keep in mind that, while you have been building your intuition on the > standard system, it has always rubbed my intuition the wrong way. So, what's > intuitive for one may be counterintuitive for another, especially given > extensive training in a mode of thought. > The unit infinity Big'un is declared to be the number of real numbers in each > unit interval on the real line. I think we agree that's an infinite value, > since we can always place a real between any two reals and still have space to > insert more. > No I don't agree that there is some infinite value that represents > the number of reals in an interval. I think there is no finite value > that represents the number of reals in an interrval. That's not the > same thing. I don't know what an infinite value is. > So, there are points in the segment, and you can subdivide the segment and > count points as you go, and any finite number of subdivisions leaves finite > subsegments which can be further divided, so there are more points uncounted in > the segment. And you can't imagine the segment subdivided forever so that the > intervals are degenerate and infinitesimal and completely fill the segment? I'm > sorry. That's too bad. That must really stink. Uh, what? Yes, the reals can be infinitely subdivided. What does that have to do with a number representing how many there are? I don't know what degenerate and infinitesimal segments are. Arbitarily small? > Are your T-naturals going to include all the standard finite naturals? > Yup. > > Then there is no largest element and thus no size that you can do > arithmetic with. > Only if we declare it axiomatically, which is the plan. > A set of naturals containing all the finite naturals doesn't have a > largest member. I don't see how an axiom claiming the opposite makes > any sense. Seems contradictory. > You have aleph_0, which is far more problematic than a unit infinity as the > length of the real line. Only to you and only because you think it's some sort of number which you can do arithmetic with. > If so, your T-naturals won't have a largest element and its size will > be equally amorphous. If not, which ones is it going to exclude? > The size is amorphous until declared to be specific. The number of reals in the > unit interval equals the number of unit intervals (naturals and hypernaturals) > on the infinite real line, an uncountable number, what calls hisself Big'un. > Big'un's reciprocal, Lil'un, is the unit infinitesimal. There's more than > Big'un, like the number of infinitesimal reals in [0,3) is 3* Big'un. So, I > don't need to exclude any. I just need to do the arithmetic to compare them. > OK, you've completely lost me.I can't discern this from utter nonsense. > It certainly isn't helping me to understand your thinking. A few things > I'm having trouble following... > Whatcha don' like numbers what's called Big'un an' Lil'un? Jes' wait till I > names one Hoss or Zeke. Hyuck! {:B > > Firstly the phrase The size is amorphous until declared to be > specific confuses me. Are set sizes not fixed in your theory? Does the > set of naturals have many sizes? > The boundary between finite and infinite cannot be pinned down in any > substantial sense, so any set defined using finiteness as any criterion suffer > from this issue. > That didn't answer my question. Are set sizes fixed in your theory or > not? > Sizes of infinite sets are comparable and orderable over a common value range. > I don't know what you mean by fixed. It depends on the value range you're > discussing. When you define an infinite set with a recursive definition, and > then only perform a finite number of iterations, it's not an infinite set, but > a finite recursive set. In general, one will compare two infinite sets over the > range [0,Big'un] to get a standard measure of the set in terms of a formula on > infinite units. So your T-naturals *don't* have a size? What is a value range. What is an infinite set with a recursive definition? What is the range [0,Big'un]? What is Big'un for that matter? The length of the real line? What does length mean, then? What are infinite units? > The infinite real line has countably many unit intervals. You're > claiming there are uncountably many naturals? > Yes, I am considering an uncountably long real line with an uncountably > infinite set of unit intervals, each corresponding to a hypernatural, finite or > infinite. > You'll have to explain to me what an infinite natural is. I don't know. > My picture of the real line is a line that goes > ..... -3, -2, -1, 0, 1, 2, 3 ..... > extending forever in either direction. You can count it by the old 0, > 1, -1, 2, -2, 3, -3 method. I don't see how it can be uncountably > long. > 1,2,3....Big'un-3, Big'un-2, Big'un-1, Big'un, Big'un+1, ... > Here is an infinite T-riffic number, with the limit point (the colon) at the > log10(Big'un) bit: > 3:333...333.333...333. That's 4*Big'un/3. :) I have no idea what you are talking about. >1,2,3....Big'un-3, Big'un-2, Big'un-1, Big'un, Big'un+1, ... What on Earth is this? BigUn is the length of the real line, but you can have BigUn+1, too? What does any of this even mean? Is this supposed to be a number system you can do arithmetic with? > What are hypernaturals? Naturals larger than any finite natural? None > exist. > That's an unfounded statement of denial. You might want to look into > Nonstandard Anaylsis and the work of Abraham Robinson. > I have, briefly. It's nothing like what you're saying. > It didn't make a lot of sense to me, anyway. I could see all the > mathematical symbols lined up nicely and I'm sure it's rigorous but > that still doesn't explain to me how a number can be infinitely large > except by just defining it so. Or how an infinitesimal differs from > real numbers which can be as small as you like. The English language > descriptions use the phrase infinitely close to zero. That means > nothing to me. > Do you think I can do anything about that, except express my condolences? I > mean, this is like discussing the fine points of mixing curry with the > McDonald's cooking staff. You sound like a devout finitist. What am I to do > about that? You don't seem to know anything about NSA. If your theory is just NSA then say so, if not then you'll have to explain how *your* theory works. > There are no infinitesimals in the reals, by the archimedean property. > Not in the standard reals, but in the hyperreals, there are. > So is your theory just the hyperreals of NSA? NSA doesn't contradict > standard set theory. It uses it. > No, it's not, but NSA provides one rigorous basis for infinite and > infinitesimal values, which you seem to disregard. You seem to disregard the need for rigour, too. I have *no idea* what you mean when you say infinite number. You can't say my theory is totally unrigorous, just me waffling vaguely without defining anything, but that's OK because NSA *is* rigorous, and nothing like my theory. >Infinitesimals can > be considered discrete, or continuity can be extended to the infinitesimal > level, leading to sub-infinitesimals. Discreteness on the infinitesimal level > does not violate the Archimedean principle on the finite level, since two > consecutive infinitesimals are not distinguishable standard reals, and > therefore do not require an intermediate value. > The real line is a continum of real numbers. There is no finite level > and a smaller infinitesimal level. The reals go allllll the way down, > as far as you like. > But, in standard analysis, if there is no finite difference between two reals, > they are considered equal. If you can't conceive of infinitesimal differences, > then all can suggest is you consider that when you touch something, you don't > really touch it. Archimedes would agree with me. :) No, I can't concieve of infinitesimal distances. When you touch something forces at the subatomic level are bouncing electrons around. Where's the infinitesimal distance? You think infinitesimal means real small? > There is a bijection between the reals in [0,1) and [0,3) so it seems > rather unintuitive to me that there can be more reals in one interval > than the other. > Does it? It seems rather unintuitive to me that [0,3) includes all of [0,1), > plus an infinite set of additional reals in [1,3), and is no larger a set of > reals. Proper subsets are smaller than the super set. That is fundamental fact > violated by standard transfinite set theory continuously, and not very > discretely, but unabashedly and vociferously. > A fundamental fact of what set theory? Using your intuition about > finite sets to make assertions about infinite sets isn't going to get > very far. A bijection matches up elements 1-1. Which ones are supposed > to be missed out by it? > It's not a matter of which are missed, but the density and/or range on the real > line. The fundamental fact of the proper subset being smaller is NOT one > preserved by transfinite set theory, but in my opinion, it's fundamental, and > SHOULD be preserved in a theory of infinite sets. So you want sets which are bijectible to have different sizes? That leads to *all* sorts of contradictions. > Finally, I haven't seen you define how arithmetic operations work on > anything you've said. I still don't even know what *type* of number > BigUn is so I don't know what operations you can do on it. If it's > the number of unit intervals on the real line I'd expect it to be a > Cardinality. But I don't think that's what you want.. > Oh no, it's not a cardinality by any means. Cardinals and ordinals are not part > of my theory at all. They're varieties of phlogiston. > Big'un is an infinite unit which can be used in any arithmetic formula, the > values of which can be compared using the notion that if f(n)>g(n) for all n>m, > then f(Big'un)>g(Big'un). This allows for a full spectrum of infinite values > which can be precisely compared and ordered formulaically. The same goes for > the unit infinitesimal Lil'un. > You still didn't answer what *type* of number Bigun and Lilun are. > Infinite and infinitesimal, but that apparently means nothing to you, so I > don't know what to say. Apparently they don't mean anything to you either as you are unable to provide either a definition or an example of either. > Number doesn't have a mathematical definition that I know of. You can > define natural_numbers or real_numbers or rational_numbers or > complex_numbers or hyperreal_numbers but number floating around on > its own doesn't have a definition. An infinite unit which can be used > in any arithmetic formula is not a definition. > I can't imagine that any definition would be considered sufficient for you, > given your apparent position on the infinite anyway. So, I guess I won't worry > myself too much about it. You haven't given me *any* definitions about anything which leads me to think you don't even have any about anything. Do you think there's a mathematical definition of number? What is it? mike. === Subject: Re: Calculus XOR Probability Virgil said: > Matt Gutting said: > It *looks* (although it's not clear) as if you're saying above that for a > given bijection f between two ordered sets S and T, x = y -> f(x) = f(y), > x < y -> f(x) < f(y), and > x > y -> f(x) > f(y), No, I specifically said xy -> f(x)f(y), so that is > equivalent to (x f(x)f(y)) ^ (x>y -> f(x) f(x)>f(y)). > Which is equivalent to f being injective, provided its domain and image > set are both ordered sets and < and > are compatible with those > orderings. Uh, yes. I orignally had '<->' where I put '->' this time. Whoops. That's what happens when people start rewording things all over the place. So, it was originally a bijection between the sets. Sorry. -- Smiles, Tony === Subject: Re: Calculus XOR Probability Virgil said: > You apply an infinite number of > increments to a finite value and you get an infinite value. > No one in mathematics ever does that. > They never increment any finite value more than once. Tell that to the clock circuitry in your computer. They never ever nevever ever do that - Virgil :D -- Smiles, Tony === Subject: Re: Calculus XOR Probability > Virgil said: You apply an infinite number of > increments to a finite value and you get an infinite value. No one in mathematics ever does that. They never increment any finite value more than once. Tell that to the clock circuitry in your computer. When TO's computer clock circuitry produces an infinite value, I will. === Subject: Re: Calculus XOR Probability <74ec4$443cbc29$82a1e228$12937@news1.tudelft.nl> <443e290b$0$14403$6d36acad@titian.nntpserver.com> They never increment any finite value more than once. > Tell that to the clock circuitry in your computer. It never increments any finite value more than once. - Randy === Subject: Re: Calculus XOR Probability Virgil said: > Does the set of all finite naturals contain all the finite naturals or > not? Is it a complete, finished set? If not, what is it missing? A maximal element and an exact size. With those attributes, the problem of > probability distributions evaporates. > So that To again wants to invent a largest finite natural to fix what > is not broken. Well, isn't that silly? I say it's missing those things, and you interpret that to mean it has them. You just can't help but find ways to use that lovable catch phrase, largest finite natural. That's cute. How precocious. No Largest Finite!!! (GONNNNNNGGGGGG) Huyah huyah huyah Ommmmmmmm.....ega!! :) -- Smiles, Tony === Subject: Re: Calculus XOR Probability > Virgil said: Does the set of all finite naturals contain all the finite naturals or > not? Is it a complete, finished set? If not, what is it missing? A maximal element and an exact size. With those attributes, the problem > of > probability distributions evaporates. So that To again wants to invent a largest finite natural to fix what > is not broken. No Largest Finite!!! (GONNNNNNGGGGGG) Huyah huyah huyah Ommmmmmmm.....ega!! TO at his most articulate! === Subject: Re: Calculus XOR Probability Virgil said: > David R Tribble said: > the Inverse Function Rule. > That hoary impossibility again? That is a stupid as TO's perpetual > appeal to his largest finite natural argument. If your Inverse Function Rule means that for every unit interval > on the real line corresponding to some natural k that there is an > corresponding inverse real x in each unit interval such as > (0,1], then this is simply the mapping: > x = 1/k for all k=1,2,3,... k = 1/x for all x in (0,1]. But this mapping denumerates only some of the reals (0,1] and omits > a much larger number of them completely (e.g., x=2/3). So you > can't use this mapping to say that the number of reals in a unit > interval is the same as the number of unit naturals on the real > line. > No, David, you misunderstand the Inverse Function Rule. Here it is > again. Given a quantitative set S > What is a quantitative set other than just a set? Until this is > properly defined, the rest is nonsense. Duh-uh....like, a set of quantities? Duh. Like, not a set of symbols or strings, or nodes or branches, or stuff other than, you know, quantities. Like, numbers, Man. But, you knew that. > mapped from the naturals > What set of naturals is TO using as his domain for this function. In > defining functions which are to be invertedone must be very careful with > domains and codomains. Yours or mine. That's not important. > using f(n) for n > in !N, > What is !N? That's the hypernaturals, but it really doesn't matter. Think N. > and given g(x) s.t. f(g(x))=g(f(x))=x > This is nonsense unless functions f and g both have the same domain and > the same codomain, which is patently not the case here. For all real x. Sorry. > (g is the inverse of the > mapping function f) that the size of the set S between values A and B > is floor(g(A)-g(B)+1) > Since A and B are undefined, the above is nonsense. Undefined? They are chosen. Say I want to find out how many squares are between 100 and 400. I take floor(sqrt(400)-sqrt(100)+1)=floor(20+10+1)=31. A is 100, and B is 400. See? > (I think I remember that correctly heh). This > works for all finite sets mapped from some finite set of naturals, as > long as N and S are order isomorphic. > What N is that? and if it and S are order isomorphic, there is no need > for distinct sets, why not take S = N? We are mapping the naturals to the monotonically increasing set of reals given a function that determines the position on the real line of each element. The faster that function increases in value, the more sparse the set, and the fewer elements along the line. So, the size of the set is the inverse of the mapping function which defines it. > The rule is not itself a > mapping, but a statement regarding the relationship between the size > of the mapped set and the mapping function. > It is not only not a mapping, nor a function, it is not mathematically > recognizable as anything at all. Is that a new hairdo? I don't think I like it. It makes you look kind of fruity. Here's a hat. Better yet, here's a paper bag. ;-) Now, the equality between the number of reals in the unit interval > and the number of unit intervals on the real line does not directly > follow from this rule, but it's consistent with it. > How is something consistent with a monumental inconsistency? I give up. How? (this is going to be a good one, I can tell) > If we declare > axiomatically that the number of reals in the unit interval is > Big'un, each occupying 1/Big'un=Lil'un space within that interval, > and the number of unit intervals on the real line is Big'un, then we > can map each of the hypernaturals to a corresponding real in [0,1] > using a mapping function f(x)=x/Big'un. > And it immediately and logically follows that 2 = 1. Uh, no, that would the standard conclusion that [0,1)=[0,2), which is patently ridiculous, as I've pointed out before. You see, in my system, there are twice as many reals in [0,2) as in [0,1). Pretty weird, eh? > So, why would I claim this is the correct solution? > Overweening ego? Guess again. I hope you got all that. You miught want to mull it over a bit. :) > Not hardly! Not my loss. Have a good crumpet. -- Smiles, Tony === Subject: Re: Calculus XOR Probability > Virgil said: > Given a quantitative set S What is a quantitative set other than just a set? Until this is > properly defined, the rest is nonsense. > Duh-uh....like, a set of quantities? What are quantities? You can't have numbers before you have defined numbers, so you can't have numbers yet. Duh. Like, not a set of symbols or > strings, or nodes or branches, or stuff other than, you know, quantities. > Like, > numbers, Man. But, you knew that. Where do these alleged numbers come from? What is their justification, their basis, their properties, their axiom system? mapped from the naturals What set of naturals is TO using as his domain for this function. In > defining functions which are to be inverted one must be very careful with > domains and codomains. > Yours or mine. That's not important. It is important to mathematics, but nothing of TO's creation is so far of the least importance to mathematics. using f(n) for n > in !N, What is !N? > That's the hypernaturals, but it really doesn't matter. Think N. If that doesn't matter then the whole thing doesn't matter. TO wants to define a function for which the domain and codomain are unknown and the rule of association is unknown and then require it to have special properties which depend on the unknown domain, codomain and rule of association. and given g(x) s.t. f(g(x))=g(f(x))=x > > This is nonsense unless functions f and g both have the same domain and > the same codomain, which is patently not the case here. > For all real x. Sorry. So are f and g bijections from R to R? (g is the inverse of the > mapping function f) that the size of the set S between values A and B > is floor(g(A)-g(B)+1) Since A and B are undefined, the above is nonsense. > Undefined? They are chosen. By whom? God? >Say I want to find out how many squares are > between > 100 and 400. I take floor(sqrt(400)-sqrt(100)+1)=floor(20+10+1)=31. A is 100, > and B is 400. See? As there are only 11, not 31, I don't see at all. (I think I remember that correctly heh). This > works for all finite sets mapped from some finite set of naturals, as > long as N and S are order isomorphic. > > What N is that? and if it and S are order isomorphic, there is no need > for distinct sets, why not take S = N? > We are mapping the naturals to the monotonically increasing set of reals > given > a function that determines the position on the real line of each element. > The > faster that function increases in value, the more sparse the set, and the > fewer > elements along the line. So, the size of the set is the inverse of the > mapping > function which defines it. Only if the set is bounded. Unbounded sets of naturals all have the same cardinality. The rule is not itself a > mapping, but a statement regarding the relationship between the size > of the mapped set and the mapping function. It is not only not a mapping, nor a function, it is not mathematically > recognizable as anything at all. > Is that a new hairdo? I don't think I like it. It makes you look kind of > fruity. Here's a hat. Better yet, here's a paper bag. ;-) TO has run out of logic, so replaces it with nonsense, as usual. Now, the equality between the number of reals in the unit interval > and the number of unit intervals on the real line does not directly > follow from this rule, but it's consistent with it. How is something consistent with a monumental inconsistency? > I give up. How? (this is going to be a good one, I can tell) It's TO's problem, not mine, to make his nonsense look like sense. > If we declare > axiomatically that the number of reals in the unit interval is > Big'un, each occupying 1/Big'un=Lil'un space within that interval, > and the number of unit intervals on the real line is Big'un, then we > can map each of the hypernaturals to a corresponding real in [0,1] > using a mapping function f(x)=x/Big'un. And it immediately and logically follows that 2 = 1. > Uh, no, that would the standard conclusion that [0,1)=[0,2), which is > patently > ridiculous, as I've pointed out before. You see, in my system, there are > twice > as many reals in [0,2) as in [0,1). Pretty weird, eh? Too weird to work. All unbounded sets of naturals have equal cardinality. All real intervals of positive length have equal cardinality. Actually, all real sets containing any open interval have equal cardinalty, though that is a bit more difficult to establish. So, why would I claim this is the correct solution? Overweening ego? > Guess again. Innate idiocy? I hope you got all that. You miught want to mull it over a bit. :) Not hardly! > Not my loss. TO is already lost. === Subject: Re: Calculus XOR Probability David R Tribble said: > Mike Kelly said: >> There is no finite natural number to represent the number of natural >> numbers. Then I say there are infinitely many naturals. (Why do you >> disagree with that second bit, incidentally?) > The Peano axioms define an externally infinite set. However, in any such > recursive definition, you are generating only a finite number of elements in > any finite number of iterations. If the naturals are restricted to finite > values, such that none is infinitely beyond any other, then I do not consider > this an infinite set, but unboundedly large while finite. When you can prove > that no two finite naturals have an infinite number of intervening elements, > then I don't see that set as infinite. Further, it is inductively and trivially > provable that if one starts with {1}, having the set size and largest value > equal, and adds successive elements as the increment of the largest value, that > the size of the set and the maximal value are incremented in tandem, and remain > equal forever. There is no point where you add the next element, incrementing > both set size and maximal element, and cause those two equal values to become > unequal, especially to the extent that the set size is incremented to infinity, > while the maximal element remains finite. It's simply impossible. > The identity relation between element count and value characterizes the > naturals. > That identity relationship characterizes all the initial finite subsets > of the naturals (of the form Sn={0,1,2,...n} for any n in N), because > all of those sets have a finite size and a largest member (n). > But it does not characterize the union of all those subsets, which > is N itself, which has no largest member. You can't have an > identity relationship with something that does not exist. The identity relationship is between position and value for each and every element in the set. Do elements exist in the set? Yeah, they exist. Do they have values, and positions in the set? Yep. Yep. The problem is with your first limit ordinal omega and its incestual spawn, aleph_0. You claim to have that many elements in the set, but there's no aleph_0th one. If you have ten elements in an ordered set, is there a tenth? Does a set of a million have a millionth element? Well, yeah. Why doesn't a set with aleph_0 elements have an aleph_0th element? Because the system ultimately makes no sense. And that's what the identity relation between element position and value is all about. making sense of the set. :) -- Smiles, Tony === Subject: Re: Calculus XOR Probability > The identity relationship is between position and value for each and > every element in the set. Do elements exist in the set? Yeah, they > exist. Do they have values, and positions in the set? Yep. Yep. The > problem is with your first limit ordinal omega and its incestual > spawn, aleph_0. You claim to have that many elements in the set, but > there's no aleph_0th one. What we claim is that there is an equivalence relation on sets defined by two sets being equivalent under this relation if and only if there is a bijection between them. We claim that every finite set is equivalent under this relation to an initial set of naturals (being bounded by some natural), and that justifies our using these initial sets as representatives of the size of finite sets. We claim further that the set of all naturals cannot be put into bijection with any of these initial sets, so use it to represent the equivalence class of all sets which biject with it, and we call that equivalence class which it represents aleph_0. > If you have ten elements in an ordered set, is there a tenth? Is there a last element in that set? Yes! > Does a set of a million have a millionth element? Is there a last element in that set? Yes! > Well, yeah. Why doesn't a set with aleph_0 elements have an aleph_0th > element? Is there a last element in that set? No! So that TO wants to make the naturals have a last element again, so that the set of naturals is just another finite set. >Because the system ultimately makes no sense. It is TO's wanting what he cannot have that makes no sense. > And that's > what the identity relation between element position and value is all > about. making sense of the set. All of those sets have a first element. If TO wants something that works for all non-empty sets of naturals, he should use the first member of each set as its size. Otherwise, the equivalence relation of bijectability is the only way to go. === Subject: Re: Calculus XOR Probability > David R Tribble said: >> Mike Kelly said: >> There is no finite natural number to represent the number of natural >> numbers. Then I say there are infinitely many naturals. (Why do you >> disagree with that second bit, incidentally?) > The Peano axioms define an externally infinite set. However, in any such > recursive definition, you are generating only a finite number of elements in > any finite number of iterations. If the naturals are restricted to finite > values, such that none is infinitely beyond any other, then I do not consider > this an infinite set, but unboundedly large while finite. When you can prove > that no two finite naturals have an infinite number of intervening elements, > then I don't see that set as infinite. Further, it is inductively and trivially > provable that if one starts with {1}, having the set size and largest value > equal, and adds successive elements as the increment of the largest value, that > the size of the set and the maximal value are incremented in tandem, and remain > equal forever. There is no point where you add the next element, incrementing > both set size and maximal element, and cause those two equal values to become > unequal, especially to the extent that the set size is incremented to infinity, > while the maximal element remains finite. It's simply impossible. > The identity relation between element count and value characterizes the > naturals. >> That identity relationship characterizes all the initial finite subsets >> of the naturals (of the form Sn={0,1,2,...n} for any n in N), because >> all of those sets have a finite size and a largest member (n). >> But it does not characterize the union of all those subsets, which >> is N itself, which has no largest member. You can't have an >> identity relationship with something that does not exist. > The identity relationship is between position and value for each and every > element in the set. Do elements exist in the set? Yeah, they exist. Do they > have values, and positions in the set? Yep. Yep. The problem is with your first > limit ordinal omega and its incestual spawn, aleph_0. You claim to have that > many elements in the set, but there's no aleph_0th one. If you have ten > elements in an ordered set, is there a tenth? Does a set of a million have a > millionth element? Well, yeah. Why doesn't a set with aleph_0 elements have an > aleph_0th element? Because the system ultimately makes no sense. And that's > what the identity relation between element position and value is all about. > making sense of the set. :) It's not precisely that we have that many elements in the set. If you try to answer the question How many natural numbers are there?, the answer won't be Aleph_0. It will be There is no number that can answer that question. Aleph_0 is not the answer to the question How many natural numbers are there?, it's the answer to the question What's the cardinality of the set of natural numbers? In other words, the answer to your question Why doesn't a set with aleph_0 elements have an aleph_0th element? is simply There is no set with aleph_0 elements, because aleph_0 isn't a number. Matt === Subject: Re: Calculus XOR Probability Matt Gutting said: > David R Tribble said: >> Mike Kelly said: >> There is no finite natural number to represent the number of natural >> numbers. Then I say there are infinitely many naturals. (Why do you >> disagree with that second bit, incidentally?) > The Peano axioms define an externally infinite set. However, in any such > recursive definition, you are generating only a finite number of elements in > any finite number of iterations. If the naturals are restricted to finite > values, such that none is infinitely beyond any other, then I do not consider > this an infinite set, but unboundedly large while finite. When you can prove > that no two finite naturals have an infinite number of intervening elements, > then I don't see that set as infinite. Further, it is inductively and trivially > provable that if one starts with {1}, having the set size and largest value > equal, and adds successive elements as the increment of the largest value, that > the size of the set and the maximal value are incremented in tandem, and remain > equal forever. There is no point where you add the next element, incrementing > both set size and maximal element, and cause those two equal values to become > unequal, especially to the extent that the set size is incremented to infinity, > while the maximal element remains finite. It's simply impossible. The identity relation between element count and value characterizes the > naturals. >> That identity relationship characterizes all the initial finite subsets >> of the naturals (of the form Sn={0,1,2,...n} for any n in N), because >> all of those sets have a finite size and a largest member (n). >> But it does not characterize the union of all those subsets, which >> is N itself, which has no largest member. You can't have an >> identity relationship with something that does not exist. The identity relationship is between position and value for each and every > element in the set. Do elements exist in the set? Yeah, they exist. Do they > have values, and positions in the set? Yep. Yep. The problem is with your first > limit ordinal omega and its incestual spawn, aleph_0. You claim to have that > many elements in the set, but there's no aleph_0th one. If you have ten > elements in an ordered set, is there a tenth? Does a set of a million have a > millionth element? Well, yeah. Why doesn't a set with aleph_0 elements have an > aleph_0th element? Because the system ultimately makes no sense. And that's > what the identity relation between element position and value is all about. > making sense of the set. :) > It's not precisely that we have that many elements in the set. If you try > to answer the question How many natural numbers are there?, the answer won't > be Aleph_0. It will be There is no number that can answer that question. > Aleph_0 is not the answer to the question How many natural numbers are there?, > it's the answer to the question What's the cardinality of the set of natural > numbers? In other words, the answer to your question Why doesn't a set with > aleph_0 elements have an aleph_0th element? is simply There is no set with > aleph_0 elements, because aleph_0 isn't a number. > Matt That's funny. I thought it was a transfinite cardinal number. Maybe number in that case just means with less feeling. Hard to say. Somehow, I also thought it was supposed to be some kind of measure of the size of the set, but I guess it's just sort of a color, or aroma, of the set. Is that about right? ;) -- Smiles, Tony === Subject: Re: Calculus XOR Probability >> David R Tribble said: > Mike Kelly said: > There is no finite natural number to represent the number of natural > numbers. Then I say there are infinitely many naturals. (Why do you > disagree with that second bit, incidentally?) >> The Peano axioms define an externally infinite set. However, in any such >> recursive definition, you are generating only a finite number of elements in >> any finite number of iterations. If the naturals are restricted to finite >> values, such that none is infinitely beyond any other, then I do not consider >> this an infinite set, but unboundedly large while finite. When you can prove >> that no two finite naturals have an infinite number of intervening elements, >> then I don't see that set as infinite. Further, it is inductively and trivially >> provable that if one starts with {1}, having the set size and largest value >> equal, and adds successive elements as the increment of the largest value, that >> the size of the set and the maximal value are incremented in tandem, and remain >> equal forever. There is no point where you add the next element, incrementing >> both set size and maximal element, and cause those two equal values to become >> unequal, especially to the extent that the set size is incremented to infinity, >> while the maximal element remains finite. It's simply impossible. >> The identity relation between element count and value characterizes the >> naturals. > That identity relationship characterizes all the initial finite subsets > of the naturals (of the form Sn={0,1,2,...n} for any n in N), because > all of those sets have a finite size and a largest member (n). > But it does not characterize the union of all those subsets, which > is N itself, which has no largest member. You can't have an > identity relationship with something that does not exist. >> The identity relationship is between position and value for each and every >> element in the set. Do elements exist in the set? Yeah, they exist. Do they >> have values, and positions in the set? Yep. Yep. The problem is with your first >> limit ordinal omega and its incestual spawn, aleph_0. You claim to have that >> many elements in the set, but there's no aleph_0th one. If you have ten >> elements in an ordered set, is there a tenth? Does a set of a million have a >> millionth element? Well, yeah. Why doesn't a set with aleph_0 elements have an >> aleph_0th element? Because the system ultimately makes no sense. And that's >> what the identity relation between element position and value is all about. >> making sense of the set. :) > It's not precisely that we have that many elements in the set. If you try > to answer the question How many natural numbers are there?, the answer won't > be Aleph_0. It will be There is no number that can answer that question. > Aleph_0 is not the answer to the question How many natural numbers are there?, > it's the answer to the question What's the cardinality of the set of natural > numbers? In other words, the answer to your question Why doesn't a set with > aleph_0 elements have an aleph_0th element? is simply There is no set with > aleph_0 elements, because aleph_0 isn't a number. > Matt aleph_0 is a transfinite cardinal number. It is not a natural number, but the word number is not restricted to natural numbers. The compound noun Transfinite number appears in books, and on the web. For over a year now people have been trying to get Tony (and others) to specify exactly what they mean by How many natural numbers are there? It has been repeatedly pointed out that the cardinality of the natural numbers is aleph_0, and that that is how people will often interpret How many natural numbers are there?. But Tony apparently means something else, but given that he refuses to specify what he actually means, it is impossible to know if there is an answer or not. Stephen === Subject: Re: Calculus XOR Probability <444f9ccd$0$14437$6d36acad@titian.nntpserver.com> Mike Kelly said: >> There is no finite natural number to represent the number of natural >> numbers. Then I say there are infinitely many naturals. (Why do you >> disagree with that second bit, incidentally?) > The Peano axioms define an externally infinite set. However, in any such > recursive definition, you are generating only a finite number of elements in > any finite number of iterations. If the naturals are restricted to finite > values, such that none is infinitely beyond any other, then I do not consider > this an infinite set, but unboundedly large while finite. When you can prove > that no two finite naturals have an infinite number of intervening elements, > then I don't see that set as infinite. Further, it is inductively and trivially > provable that if one starts with {1}, having the set size and largest value > equal, and adds successive elements as the increment of the largest value, that > the size of the set and the maximal value are incremented in tandem, and remain > equal forever. There is no point where you add the next element, incrementing > both set size and maximal element, and cause those two equal values to become > unequal, especially to the extent that the set size is incremented to infinity, > while the maximal element remains finite. It's simply impossible. > The identity relation between element count and value characterizes the > naturals. > That identity relationship characterizes all the initial finite subsets > of the naturals (of the form Sn={0,1,2,...n} for any n in N), because > all of those sets have a finite size and a largest member (n). > But it does not characterize the union of all those subsets, which > is N itself, which has no largest member. You can't have an > identity relationship with something that does not exist. > The identity relationship is between position and value for each and every > element in the set. Given that by your own admission you are not very practiced at formal maths proofs, it would be a useful exercise to see if you can write this out excrutiatingly carefully, in mathematical terms. > Do elements exist in the set? Yeah, they exist. Do they > have values, and positions in the set? Yep. Yep. The problem is with your first > limit ordinal omega and its incestual spawn, aleph_0. You claim to have that > many elements in the set, but there's no aleph_0th one. Loosely, one might say there are aleph_0 elements, but it's pretty loose. No-one would claim that you can count them by counting up to aleph0 anyway. I showed you a *long* time ago (abundance of irrationals) that in the type of set you are discussing here (I've called 'unbroken'), if the cardinality of the set of naturals is P, then it includes an element with a value corresponding to P if and only if that value is the last element in the set. Here's the bit (which I don't think you ever responded to): ------------------------------------------ Let's say that an unbroken set is a subset [not necessarily proper, ydkwtm,dy?] of the natural numbers with the following property: T is an unbroken set if 1 is in T and n in T implies that any m less than n is also in T In other words it's a set starting {1, ...} and never skipping any numbers. It may or may not stop, so it may be the whole set of naturals. (A subset, not a proper subset) In the following, T is always an unbroken set. Theorem 1 No element of T is greater than the size of T. Proof: obvious Theorem 2 If an element of T is less than the size of T it is not the last element of T. Proof: obvious Theorem 3 If an element of T is equal to the size of T it _is_ the last element of T. Proof: obvious Corollary If T is finite, it includes an element equal to its size; if T is not finite, it does not. Notice how in the corollary, the distinction between the finite set and the not finite set is crucial. Let me know if you think any of the theorems is invalid. ------------------------------------------ Well, is any of them invalid? > If you have ten > elements in an ordered set, is there a tenth? Does a set of a million have a > millionth element? Well, yeah. Why doesn't a set with aleph_0 elements have an > aleph_0th element? Because the system ultimately makes no sense. Uh, no. Because the aleph_0 would have to be the last element (as all the others are). There isn't a last element. Keep smiling. Brian Chandler http://imaginatorium.org === Subject: Re: Calculus XOR Probability imaginatorium@despammed.com said: > David R Tribble said: > Mike Kelly said: >> There is no finite natural number to represent the number of natural >> numbers. Then I say there are infinitely many naturals. (Why do you >> disagree with that second bit, incidentally?) > > The Peano axioms define an externally infinite set. However, in any such > recursive definition, you are generating only a finite number of elements in > any finite number of iterations. If the naturals are restricted to finite > values, such that none is infinitely beyond any other, then I do not consider > this an infinite set, but unboundedly large while finite. When you can prove > that no two finite naturals have an infinite number of intervening elements, > then I don't see that set as infinite. Further, it is inductively and trivially > provable that if one starts with {1}, having the set size and largest value > equal, and adds successive elements as the increment of the largest value, that > the size of the set and the maximal value are incremented in tandem, and remain > equal forever. There is no point where you add the next element, incrementing > both set size and maximal element, and cause those two equal values to become > unequal, especially to the extent that the set size is incremented to infinity, > while the maximal element remains finite. It's simply impossible. > The identity relation between element count and value characterizes the > naturals. > That identity relationship characterizes all the initial finite subsets > of the naturals (of the form Sn={0,1,2,...n} for any n in N), because > all of those sets have a finite size and a largest member (n). > But it does not characterize the union of all those subsets, which > is N itself, which has no largest member. You can't have an > identity relationship with something that does not exist. > The identity relationship is between position and value for each and every > element in the set. > Given that by your own admission you are not very practiced at formal > maths proofs, it would be a useful exercise to see if you can write > this out excrutiatingly carefully, in mathematical terms. x_n=n > Do elements exist in the set? Yeah, they exist. Do they > have values, and positions in the set? Yep. Yep. The problem is with your first > limit ordinal omega and its incestual spawn, aleph_0. You claim to have that > many elements in the set, but there's no aleph_0th one. > Loosely, one might say there are aleph_0 elements, but it's pretty > loose. No-one would claim that you can count them by counting up to > aleph0 anyway. I've never heard anyoine claim that. > I showed you a *long* time ago (abundance of irrationals) that in the > type of set you are discussing here (I've called 'unbroken'), if the > cardinality of the set of naturals is P, then it includes an element > with a value corresponding to P if and only if that value is the last > element in the set. > Here's the bit (which I don't think you ever responded to): > ------------------------------------------ > Let's say that an unbroken set is a subset [not necessarily proper, > ydkwtm,dy?] of the natural numbers with the following property: > T is an unbroken set if > 1 is in T and > n in T implies that any m less than n is also in T > In other words it's a set starting {1, ...} and never skipping any > numbers. It may or may not stop, so it may be the whole set of > naturals. (A subset, not a proper subset) > In the following, T is always an unbroken set. > Theorem 1 > No element of T is greater than the size of T. > Proof: obvious > Theorem 2 > If an element of T is less than the size of T it is not the last > element of T. > Proof: obvious > Theorem 3 > If an element of T is equal to the size of T it _is_ the last element > of T. > Proof: obvious > Corollary > If T is finite, it includes an element equal to its size; if T is not > finite, it does not. > Notice how in the corollary, the distinction between the finite set and > the not finite set is crucial. > Let me know if you think any of the theorems is invalid. > ------------------------------------------ > Well, is any of them invalid? You must have sent this at a time when there were a bunch of threads going on and I couldn't keep up with them all. It doesn't look familiar. No, your theorems are all obviously true. The corollary rests, however, on the notion that an infinite set has no last element, because if it doesn't have one, then that can't be the size of the set. However, as I've said, I consider sets like the standard naturals to be unbounded but finite, and sets like the reals in [0,1] are bounded but infinite, with first and last elements. It's the middle that's endless. > If you have ten > elements in an ordered set, is there a tenth? Does a set of a million have a > millionth element? Well, yeah. Why doesn't a set with aleph_0 elements have an > aleph_0th element? Because the system ultimately makes no sense. > Uh, no. Because the aleph_0 would have to be the last element (as all > the others are). There isn't a last element. Uh, yeah, that's why it makes no sense. > Keep smiling. And the horse you rode in on! :) > Brian Chandler > http://imaginatorium.org -- Smiles, Tony === Subject: Re: Calculus XOR Probability > imaginatorium@despammed.com said: David R Tribble said: > Mike Kelly said: >> There is no finite natural number to represent the number of >> natural >> numbers. Then I say there are infinitely many naturals. (Why do you >> disagree with that second bit, incidentally?) > > The Peano axioms define an externally infinite set. However, in any > such > recursive definition, you are generating only a finite number of > elements in > any finite number of iterations. If the naturals are restricted to > finite > values, such that none is infinitely beyond any other, then I do not > consider > this an infinite set, but unboundedly large while finite. When you > can prove > that no two finite naturals have an infinite number of intervening > elements, > then I don't see that set as infinite. Further, it is inductively and > trivially > provable that if one starts with {1}, having the set size and largest > value > equal, and adds successive elements as the increment of the largest > value, that > the size of the set and the maximal value are incremented in tandem, > and remain > equal forever. There is no point where you add the next element, > incrementing > both set size and maximal element, and cause those two equal values > to become > unequal, especially to the extent that the set size is incremented to > infinity, > while the maximal element remains finite. It's simply impossible. > The identity relation between element count and value characterizes > the > naturals. > That identity relationship characterizes all the initial finite subsets > of the naturals (of the form Sn={0,1,2,...n} for any n in N), because > all of those sets have a finite size and a largest member (n). > But it does not characterize the union of all those subsets, which > is N itself, which has no largest member. You can't have an > identity relationship with something that does not exist. > The identity relationship is between position and value for each and > every > element in the set. Given that by your own admission you are not very practiced at formal > maths proofs, it would be a useful exercise to see if you can write > this out excrutiatingly carefully, in mathematical terms. > x_n=n Do elements exist in the set? Yeah, they exist. Do they > have values, and positions in the set? Yep. Yep. The problem is with your > first > limit ordinal omega and its incestual spawn, aleph_0. You claim to have > that > many elements in the set, but there's no aleph_0th one. Loosely, one might say there are aleph_0 elements, but it's pretty > loose. No-one would claim that you can count them by counting up to > aleph0 anyway. > I've never heard anyoine claim that. I showed you a *long* time ago (abundance of irrationals) that in the > type of set you are discussing here (I've called 'unbroken'), if the > cardinality of the set of naturals is P, then it includes an element > with a value corresponding to P if and only if that value is the last > element in the set. Here's the bit (which I don't think you ever responded to): ------------------------------------------ > Let's say that an unbroken set is a subset [not necessarily proper, > ydkwtm,dy?] of the natural numbers with the following property: T is an unbroken set if > 1 is in T and > n in T implies that any m less than n is also in T In other words it's a set starting {1, ...} and never skipping any > numbers. It may or may not stop, so it may be the whole set of > naturals. (A subset, not a proper subset) In the following, T is always an unbroken set. Theorem 1 > No element of T is greater than the size of T. > Proof: obvious Theorem 2 > If an element of T is less than the size of T it is not the last > element of T. > Proof: obvious Theorem 3 > If an element of T is equal to the size of T it _is_ the last element > of T. > Proof: obvious Corollary > If T is finite, it includes an element equal to its size; if T is not > finite, it does not. Notice how in the corollary, the distinction between the finite set and > the not finite set is crucial. Let me know if you think any of the theorems is invalid. > ------------------------------------------ Well, is any of them invalid? > You must have sent this at a time when there were a bunch of threads going on > and I couldn't keep up with them all. It doesn't look familiar. No, your > theorems are all obviously true. The corollary rests, however, on the notion > that an infinite set has no last element, because if it doesn't have one, > then > that can't be the size of the set. However, as I've said, I consider sets > like > the standard naturals to be unbounded but finite, and sets like the reals in > [0,1] are bounded but infinite, with first and last elements. It's the middle > that's endless. It is TO's ignorance and idiocy that are endless. === Subject: Re: Calculus XOR Probability <444f9ccd$0$14437$6d36acad@titian.nntpserver.com> (Um, aren't you new around here?) > There is no finite natural number to represent the number of natural > numbers. Then I say there are infinitely many naturals. (Why do you > disagree with that second bit, incidentally?) > The Peano axioms define an externally infinite set. > I'm sorry but I don't know what externally infinite means, maybe you > could explain. Hmm. This is a Tonyism. A total of 2419 posts have been dedicated to the argument between Sensible Person and Tony Orlow that goes: TO: There are only finitely many finite naturals, because if there were infinitely many, at least the last few would be infinitely far away. SP: But there is no last one (or few), so you argument is invalid TO) Kerfuffle, kerfuffle ... SP) ... kerfiffle, kerfoffle ['...' represents an infinity with a declared value somewhere over 2400] TO: No, because if all the values are finite, it can't be an infinite set. This is so obvious, I really can't believe I have to say it. Someone: What about the set of reals in [0, 1]? Is one of them infinite then? TO: No that's completely different. The numbers are crammed in together, and there's an internal infinity, uh, dense something. [Sorry Tony, you can edit this line if you like] So I think that an external infinity represents the case of the naturals, which are spread out, so you can only get an infinite number of them by going to an end at infinity. Or something like that. predecessors. This is true (if it were not then 0 would have to be the > successor to some natural -> contradiction). Each natural has a number > of predecessors equal to its value. However every natural has an > infinite number of successors. Making the set unboundedly large - you > can list forever and there are always more members you haven't listed. > That's what I call infinite. Because it's size is not-finite. > In-finite. You see? Hmm. No chance, while you're using the i-word. But Tony has occasional flashes of lucidity - here's what he said once: Yes, the set of bits is bounded for every specific pofnat, but there is no bound on the length of the string required for all pofnats. Tony Orlow Anyway, good luck with this. You may need it - we've seen around 10,000 posts now in TO threads, and not much progress has been made. Brian Chandler http://imaginatorium.org === Subject: Re: Calculus XOR Probability <444f9ccd$0$14437$6d36acad@titian.nntpserver.com> Mike Kelly said: > There is no finite natural number to represent the number of natural > numbers. Then I say there are infinitely many naturals. (Why do you > disagree with that second bit, incidentally?) > The Peano axioms define an externally infinite set. > I'm sorry but I don't know what externally infinite means, maybe you > could explain. > Hmm. This is a Tonyism. A total of 2419 posts have been dedicated to > the argument between Sensible Person and Tony Orlow that goes: > TO: There are only finitely many finite naturals, because if there were > infinitely many, at least the last few would be infinitely far away. > SP: But there is no last one (or few), so you argument is invalid > TO) Kerfuffle, kerfuffle ... > SP) ... kerfiffle, kerfoffle > ['...' represents an infinity with a declared value somewhere over > 2400] > TO: No, because if all the values are finite, it can't be an infinite > set. This is so obvious, I really can't believe I have to say it. > Someone: What about the set of reals in [0, 1]? Is one of them infinite > then? > TO: No that's completely different. The numbers are crammed in > together, and there's an internal infinity, uh, dense something. [Sorry > Tony, you can edit this line if you like] > So I think that an external infinity represents the case of the > naturals, which are spread out, so you can only get an infinite number > of them by going to an end at infinity. Or something like that. Fair enough.. but I still don't get than an externally infinite set isn't an infinite set, according to Tony. > predecessors. This is true (if it were not then 0 would have to be the > successor to some natural -> contradiction). Each natural has a number > of predecessors equal to its value. However every natural has an > infinite number of successors. Making the set unboundedly large - you > can list forever and there are always more members you haven't listed. > That's what I call infinite. Because it's size is not-finite. > In-finite. You see? > Hmm. No chance, while you're using the i-word. But Tony has occasional > flashes of lucidity - here's what he said once: > Yes, the set of bits is bounded for every specific pofnat, but there > is no > bound on the length of the string required for all pofnats. Tony Orlow > Anyway, good luck with this. You may need it - we've seen around 10,000 > posts now in TO threads, and not much progress has been made. Well I don't think I've seen many ideas more bizarre than Tony's. Trying to explain my own understanding of things can help me get my head straight, even if I'm not going to get through to Tony. > Brian Chandler > http://imaginatorium.org === Subject: Re: Calculus XOR Probability Virgil said: > Virgil said: Virgil said: > , > > Indeed! General order is defined by x x is remarkably similar to x x infinity). > Wrong! There are several other properties that an relation on a > set must have before it can be called an order relation on that > set. Sure, that's the transitive property, essential to any order, not > necessarily all there is to all kinds of order. The transitive property is also essential in all equivalence > relations, none of which are order relations, so transitivity isn't > enough. Order and equality aren't related? That's funny. I thought that in an > ordered set x<=y ^ y<=x -> x=y? In my land, we state this truth as > xy. It is very good. We like it. You should try it. You > will like it. :) > I do not see trichotomy anywhere in TO's claim > Indeed! General order is defined by x x When TO's error is pointed out he adds adds trichotomy as if it were > there all the time, but there are important transitive relations for > which trichotomy does not hold, so TO's mathematical ignorance once more > has lead him to make an ass of himself. > TO claimed that order is *defined by* transitivity, without any visible > requirement of trichotomy or anything else. In mathematics, such things > do not 'go without saying'. You're right, and actually I thought about this post sometime last night, and realized that trichotomy isn't actually implied in xy, unless those v's are xor's. Hmmmm... I still think the definitions for internal and external infinity are pretty good. > ..... > Doesn't bijection itself depend on order, in the sense that > saying xy <-> f(x)f(y), and that > x=y <-> f(x)=f(y)? > No! Order isomorphisms are bijections, but so are other forms > of isomorphism having nothing to do with order at all. What I said above does not imply order isomorphism. A bijection that depends on order (either preserving or reversing > it) must be an order isomorphism, so TO WAS implying order > isomorphism. But the rule above doesn't necessarily do either. It says that x and > y are different iff they map to different elements in the target set, > and that they are equal iff they map to the same element. It simply > states that the relation's a bijection. > The rule above > xy <-> f(x)f(y), and x=y <-> f(x)=f(y) > requiring that order be preserved, if a bijection, also constitutes an > order isomorphism. > So that TO is wrong! Again! To no one's surprise! No, that statement is equivalent to x<>y <-> f(x)<>f(y), which also implies x=y <-> f(x)=f(y), but not x f(x) Virgil said: Virgil said: > Virgil said: > , > > Indeed! General order is defined by x x is remarkably similar to x x infinity). > Wrong! There are several other properties that an relation on a > set must have before it can be called an order relation on that > set. > Sure, that's the transitive property, essential to any order, not > necessarily all there is to all kinds of order. The transitive property is also essential in all equivalence > relations, none of which are order relations, so transitivity isn't > enough. Order and equality aren't related? That's funny. I thought that in an > ordered set x<=y ^ y<=x -> x=y? In my land, we state this truth as > xy. It is very good. We like it. You should try it. You > will like it. :) I do not see trichotomy anywhere in TO's claim > Indeed! General order is defined by x x > When TO's error is pointed out he adds adds trichotomy as if it were > there all the time, but there are important transitive relations for > which trichotomy does not hold, so TO's mathematical ignorance once more > has lead him to make an ass of himself. TO claimed that order is *defined by* transitivity, without any visible > requirement of trichotomy or anything else. In mathematics, such things > do not 'go without saying'. > You're right, and actually I thought about this post sometime last night, and > realized that trichotomy isn't actually implied in xy, unless > those v's are xor's. Hmmmm... I still think the definitions for internal and > external infinity are pretty good. ..... > Doesn't bijection itself depend on order, in the sense that > saying xy <-> f(x)f(y), and that > x=y <-> f(x)=f(y)? > No! Order isomorphisms are bijections, but so are other forms > of isomorphism having nothing to do with order at all. > What I said above does not imply order isomorphism. A bijection that depends on order (either preserving or reversing > it) must be an order isomorphism, so TO WAS implying order > isomorphism. It simply > states that the relation's a bijection. No, it doesn't mean bijection unless one adds the additional restriction that both the domain and codomain of the function are ordered sets. For example, given two ordered sets, such as real closed intervals, which are order isomprphic, their power sets are not ordered, or even orderable, in any natural way, but are still clearly bijectable. === Subject: Re: Calculus XOR Probability <444df382$0$14522$6d36acad@titian.nntpserver.com> <444f6df4$0$14426$6d36acad@titian.nntpserver.com> Virgil said: > Virgil said: > , > > Indeed! General order is defined by x x is remarkably similar to x x infinity). This is a definition you think is rather good? Have you proof-read it? There's an unquantified variable y floating around on the right hand side - what is it supposed to be? Are you trying to say: Ax, Az, x < z -> Ey, x Order and equality aren't related? That's funny. I thought that in an > ordered set x<=y ^ y<=x -> x=y? In my land, we state this truth as > xy. It is very good. We like it. You should try it. You > will like it. :) Assuming your land means Orlovia, I wonder how many speakers Orlovian has? I notice that xy is an expression, which is true if (for example) x, y, z are integers, false if they are elements of the Klein 4-group. Ho hum. > You're right, and actually I thought about this post sometime last night, and > realized that trichotomy isn't actually implied in xy, unless > those v's are xor's. Hmmmm... I still think the definitions for internal and > external infinity are pretty good. Do remind me of the definition for external infinity... (I bet SRP has an external infinity too, hee hee!) Brian Chandler http://imaginatorium.org === Subject: Re: Calculus XOR Probability imaginatorium@despammed.com said: > Virgil said: > Virgil said: > Virgil said: > , > > Indeed! General order is defined by x x is remarkably similar to x x infinity). > This is a definition you think is rather good? Have you proof-read > it? There's an unquantified variable y floating around on the right > hand side - what is it supposed to be? Are you trying to say: > Ax, Az, x < z -> Ey, x Ey: x How is this a definition of an i-word? Consider scissors, rock, and > paper, (S, R, P) with the obvious relationship given by <. Then (as can > be written out trivially) {S, R, P} has [is?] an internal infinity. Along with trichotomy etc it works. It's not the full theory. Kerfliffle Kerfloffle... > Order and equality aren't related? That's funny. I thought that in an > ordered set x<=y ^ y<=x -> x=y? In my land, we state this truth as > xy. It is very good. We like it. You should try it. You > will like it. :) > Assuming your land means Orlovia, I wonder how many speakers Orlovian > has? Surround sound. :) > I notice that xy is an expression, which is true if (for > example) x, y, z are integers, false if they are elements of the Klein > 4-group. Ho hum. > You're right, and actually I thought about this post sometime last night, and > realized that trichotomy isn't actually implied in xy, unless > those v's are xor's. Hmmmm... I still think the definitions for internal and > external infinity are pretty good. > Do remind me of the definition for external infinity... > (I bet SRP has an external infinity too, hee hee!) y -> x Brian Chandler > http://imaginatorium.org -- Smiles, Tony === Subject: Re: Calculus XOR Probability > imaginatorium@despammed.com said: > Indeed! General order is defined by x x is remarkably similar to x x infinity). This is a definition you think is rather good? Have you proof-read > it? There's an unquantified variable y floating around on the right > hand side - what is it supposed to be? Are you trying to say: Ax, Az, x < z -> Ey, x Yes, I said that when I first put it out. Actually, what I said, > which is equivalent is Ex ^ Ex ^ x Ey: x any symbol used and assumes existence, then the quatifiers [?] are not > particular necessary, at least in this case. In any case, I made note > of the abridged notation. How is this a definition of an i-word? Consider scissors, rock, and > paper, (S, R, P) with the obvious relationship given by <. Then (as can > be written out trivially) {S, R, P} has [is?] an internal infinity. > I notice that xy is an expression, which is true if (for > example) x, y, z are integers, false if they are elements of the Klein > 4-group. Ho hum. > You're right, and actually I thought about this post sometime last night, > and > realized that trichotomy isn't actually implied in xy, > unless > those v's are xor's. Hmmmm... I still think the definitions for internal > and > external infinity are pretty good. TO's opinion of himself and his own creations is unwarranted. Note that finite open intervals, of either reals or rationals, or any set dense in the reals, are both internally and externally infinite by TO's definitions. === Subject: Re: Calculus XOR Probability >> Virgil said: >> Virgil said: > Virgil said: > , > Indeed! General order is defined by x x> is remarkably similar to x x> infinity). > This is a definition you think is rather good? Have you proof-read > it? There's an unquantified variable y floating around on the right > hand side - what is it supposed to be? Are you trying to say: > Ax, Az, x < z -> Ey, x How is this a definition of an i-word? Consider scissors, rock, and > paper, (S, R, P) with the obvious relationship given by <. Then (as can > be written out trivially) {S, R, P} has [is?] an internal infinity. That doesn't seem right. I get: S Matt Gutting said: > Well, then, I guess we have different intuitions. Picture the unit interval > [0,1] as a little shrunk-down real line from 0 to oo. >> [0,1] has two endpoints. [0,oo) doesn't. > Think extended reals, projectively if you like. >> Why do you assume that the set defined by the Peano axioms is generated? It's >> not. > Why do you say the set is not generated recursively by the axioms? It is. No, it's *described* by the axioms, not generated by them. Sets aren't generated because they don't have an inherent order. Matt === Subject: Re: Calculus XOR Probability > Matt Gutting said: > Matt Gutting said: >> proof is considered generally invalid in the limit, that is, in the infinite > case. The method of inductive proof is only considered to prove a property for > all FINITE n. However, in this sense of a limit, my suggestion is that an > equality between expressions proven inductively holds for all n, finite or > infinite. >> The problem is that you have been assuming this true in order to prove that >> there exist infinite n; that is, you have been assuming that there exist >> infinite natural numbers which will provide you with the infinite case in >> which there are infinite natural numbers. Unless you can either (a) show >> some definition of infinite n which doesn't depend on the existence of >> the infinite case, or (b) redefine what, exactly, you mean by the infinite >> case, your suggestion does not have any meaning. > One can assume things in math, and if they do not produce any contradictions, > they they are part of a consistent theory. Do I need to remind set theorists of > this, when they consistently remind those objecting to their hocus pocus that > they have axioms which work even if unjustified, and that they are therefore > justified as being consistent? Every theory has its assumptions. The infinitude > of the real line and of the reals in a finite interval is a reasonable place to > start, in my opinion. >> I'm not saying there's something wrong with assuming things, in general; I'm >> pointing out that you are assuming the existence of an entity (an infinite >> natural number n) in order to prove the existence of that entity (by providing >> a meaning for the phrase the infinite case. >> In other words, I have no problem with you assuming things; I do, however, have >> a problem when you claim to have proved a result something that is actually one >> of the things you assumed in the first place. > Okay, if that's your point, then maybe you'll recall my saying the following. > The inductive proof of the equivalence between element count and value for the > naturals starting at 1 is generally refuted with the statement that inductive > proof only holds for finite n in N, and therefore does not hold for the > infinite set. So, the set can be infinite while containing only finite values. > If that's the case, then the inductive proof that all naturals are finite only > proves that all FINITE naturals are finite, and therefore proves nothing beyond > the assumptions it started with. If it is NOT the case that inductive proof > only holds in the finite case, then the set cannot become infinite while having > only finite values, and this makes sense, because if those two numbers are > equal, and incremented in tandem simultaneously, there is no way that they can > ever become unequal, especially to the extent that one is infinitely greater > than the other. > So, what I said is that inductive proof itself does not prove whether it does > or does not apply to the infinite case. My position is that for equalities, it > does. For inequalities, one must be sure that the difference establishing the > inequality does not have a limit of 0 as n->oo. If it does, then the proof > fails in the infinite case. The problem with this is that, absent a non-circular definition of infinite, you can't assume that there *is* an infinite case to begin with. Saying that the inductive proof only holds for finite n in N is actually a mis-phrasing. It would be more apt to say that the inductive proof holds for all n in N, and the only values that can be derived from it are finite. I'm still not sure what exactly you mean by the set becoming infinite. Sets don't become anything; they just are, as they are defined. >> (And to answer your question elsewhere in the thread, a number greater >> than any finite doesn't suffice. You first have to show that there are >> numbers greater than any finite; or, if you want to define them (as extensions >> to the natural numbers), you have to describe their properties. I think you >> will run into problems dealing with what you have occasionally called The >> Twilight Zone coming between the finites and infinites [in some sense].) >> Matt > I'll have the same trouble with the Twilight Zone that the standard theory has, > if I make the same mistake of trying to pin a value on its location, like > aleph_0. Then, I'll run into contradictions. So, I'll avoid that mistake, okay? >> You're saying, if I follow you, that you won't try to pin a value on the >> location of the Twilight Zone. Does this mean that it doesn't *have* a location? >> If it does, why can't the location be specified? If it doesn't, in what sense >> does it exist? >> Here's a question: What, exactly, is the definition of this Twilight Zone? > The Twilight Zone is the transition from finite to infinite. No, there is no > location for it that can be pinned down, because the point' at which one > increments a finite and gets an infinite doesn't exist. This does not occur at > any location. It occurs over an infinite expanse and sequence of increments. > But, if both finites and infinites coexist on the number line, then this > transition zone exists, though it can never be located specifically. > The Twilight Zone is the focus of the Cantorian mantra. It's a little like the > boundary between inner and outer space. Point it out and it disappears as fast > as a ball in a vase. ;) If there is no point at which a transition occurs, then how can it occur? Does it occur a little bit at a time? If so, where does it start? How does it progress? Where does it end? Are there numbers occupying the Twilight Zone? If so, are they finite? Infinite? Some of each? If there are no numbers occupying the Twilight Zone, then how does it include an infinite ... sequence of increments? Matt >> Matt === Subject: Re: Calculus XOR Probability Matt Gutting said: > Matt Gutting said: > Matt Gutting said: >> proof is considered generally invalid in the limit, that is, in the infinite > case. The method of inductive proof is only considered to prove a property for > all FINITE n. However, in this sense of a limit, my suggestion is that an > equality between expressions proven inductively holds for all n, finite or > infinite. >> The problem is that you have been assuming this true in order to prove that >> there exist infinite n; that is, you have been assuming that there exist >> infinite natural numbers which will provide you with the infinite case in >> which there are infinite natural numbers. Unless you can either (a) show >> some definition of infinite n which doesn't depend on the existence of >> the infinite case, or (b) redefine what, exactly, you mean by the infinite >> case, your suggestion does not have any meaning. > One can assume things in math, and if they do not produce any contradictions, > they they are part of a consistent theory. Do I need to remind set theorists of > this, when they consistently remind those objecting to their hocus pocus that > they have axioms which work even if unjustified, and that they are therefore > justified as being consistent? Every theory has its assumptions. The infinitude > of the real line and of the reals in a finite interval is a reasonable place to > start, in my opinion. >> I'm not saying there's something wrong with assuming things, in general; I'm >> pointing out that you are assuming the existence of an entity (an infinite >> natural number n) in order to prove the existence of that entity (by providing >> a meaning for the phrase the infinite case. >> In other words, I have no problem with you assuming things; I do, however, have >> a problem when you claim to have proved a result something that is actually one >> of the things you assumed in the first place. Okay, if that's your point, then maybe you'll recall my saying the following. > The inductive proof of the equivalence between element count and value for the > naturals starting at 1 is generally refuted with the statement that inductive > proof only holds for finite n in N, and therefore does not hold for the > infinite set. So, the set can be infinite while containing only finite values. > If that's the case, then the inductive proof that all naturals are finite only > proves that all FINITE naturals are finite, and therefore proves nothing beyond > the assumptions it started with. If it is NOT the case that inductive proof > only holds in the finite case, then the set cannot become infinite while having > only finite values, and this makes sense, because if those two numbers are > equal, and incremented in tandem simultaneously, there is no way that they can > ever become unequal, especially to the extent that one is infinitely greater > than the other. So, what I said is that inductive proof itself does not prove whether it does > or does not apply to the infinite case. My position is that for equalities, it > does. For inequalities, one must be sure that the difference establishing the > inequality does not have a limit of 0 as n->oo. If it does, then the proof > fails in the infinite case. > The problem with this is that, absent a non-circular definition of infinite, > you can't assume that there *is* an infinite case to begin with. > Saying that the inductive proof only holds for finite n in N is actually a > mis-phrasing. It would be more apt to say that the inductive proof holds for > all n in N, and the only values that can be derived from it are finite. > I'm still not sure what exactly you mean by the set becoming infinite. Sets > don't become anything; they just are, as they are defined. >> (And to answer your question elsewhere in the thread, a number greater >> than any finite doesn't suffice. You first have to show that there are >> numbers greater than any finite; or, if you want to define them (as extensions >> to the natural numbers), you have to describe their properties. I think you >> will run into problems dealing with what you have occasionally called The >> Twilight Zone coming between the finites and infinites [in some sense].) >> Matt > I'll have the same trouble with the Twilight Zone that the standard theory has, > if I make the same mistake of trying to pin a value on its location, like > aleph_0. Then, I'll run into contradictions. So, I'll avoid that mistake, okay? >> You're saying, if I follow you, that you won't try to pin a value on the >> location of the Twilight Zone. Does this mean that it doesn't *have* a location? >> If it does, why can't the location be specified? If it doesn't, in what sense >> does it exist? >> Here's a question: What, exactly, is the definition of this Twilight Zone? The Twilight Zone is the transition from finite to infinite. No, there is no > location for it that can be pinned down, because the point' at which one > increments a finite and gets an infinite doesn't exist. This does not occur at > any location. It occurs over an infinite expanse and sequence of increments. > But, if both finites and infinites coexist on the number line, then this > transition zone exists, though it can never be located specifically. The Twilight Zone is the focus of the Cantorian mantra. It's a little like the > boundary between inner and outer space. Point it out and it disappears as fast > as a ball in a vase. ;) > If there is no point at which a transition occurs, then how can it occur? Does > it occur a little bit at a time? If so, where does it start? How does it > progress? Where does it end? Are there numbers occupying the Twilight Zone? > If so, are they finite? Infinite? Some of each? If there are no numbers > occupying the Twilight Zone, then how does it include an infinite ... sequence > of increments? > Matt It's a mystery, and not one that can be solved. There is no largest finite or smallest infinite, and no point where they meet. And yet, over an infinite number of finite steps, one goes an infinite distance. Can you say what the smallest nonzero real is? That's ultimately the same question, because that's the first real with an infinite number of predecessors. It can't be found, but you can point in the general direction. -- Smiles, Tony === Subject: Re: Calculus XOR Probability > Matt Gutting said: > Matt Gutting said: > Matt Gutting said: >> Okay, I think we agree. Now, I would pose this suggestion again. > Inductive > proof is considered generally invalid in the limit, that is, in the > infinite > case. The method of inductive proof is only considered to prove a > property for > all FINITE n. However, in this sense of a limit, my suggestion is > that an > equality between expressions proven inductively holds for all n, > finite or > infinite. >> The problem is that you have been assuming this true in order to prove >> that >> there exist infinite n; that is, you have been assuming that there >> exist >> infinite natural numbers which will provide you with the infinite >> case in >> which there are infinite natural numbers. Unless you can either (a) >> show >> some definition of infinite n which doesn't depend on the existence >> of >> the infinite case, or (b) redefine what, exactly, you mean by the >> infinite >> case, your suggestion does not have any meaning. > One can assume things in math, and if they do not produce any > contradictions, > they they are part of a consistent theory. Do I need to remind set > theorists of > this, when they consistently remind those objecting to their hocus > pocus that > they have axioms which work even if unjustified, and that they are > therefore > justified as being consistent? Every theory has its assumptions. The > infinitude > of the real line and of the reals in a finite interval is a reasonable > place to > start, in my opinion. >> I'm not saying there's something wrong with assuming things, in general; >> I'm >> pointing out that you are assuming the existence of an entity (an >> infinite >> natural number n) in order to prove the existence of that entity (by >> providing >> a meaning for the phrase the infinite case. >> In other words, I have no problem with you assuming things; I do, >> however, have >> a problem when you claim to have proved a result something that is >> actually one >> of the things you assumed in the first place. Okay, if that's your point, then maybe you'll recall my saying the > following. > The inductive proof of the equivalence between element count and value > for the > naturals starting at 1 is generally refuted with the statement that > inductive > proof only holds for finite n in N, and therefore does not hold for the > infinite set. So, the set can be infinite while containing only finite > values. > If that's the case, then the inductive proof that all naturals are finite > only > proves that all FINITE naturals are finite, and therefore proves nothing > beyond > the assumptions it started with. If it is NOT the case that inductive > proof > only holds in the finite case, then the set cannot become infinite while > having > only finite values, and this makes sense, because if those two numbers > are > equal, and incremented in tandem simultaneously, there is no way that > they can > ever become unequal, especially to the extent that one is infinitely > greater > than the other. So, what I said is that inductive proof itself does not prove whether it > does > or does not apply to the infinite case. My position is that for > equalities, it > does. For inequalities, one must be sure that the difference establishing > the > inequality does not have a limit of 0 as n->oo. If it does, then the > proof > fails in the infinite case. The problem with this is that, absent a non-circular definition of > infinite, > you can't assume that there *is* an infinite case to begin with. Saying that the inductive proof only holds for finite n in N is actually > a > mis-phrasing. It would be more apt to say that the inductive proof holds > for > all n in N, and the only values that can be derived from it are finite. I'm still not sure what exactly you mean by the set becoming infinite. > Sets > don't become anything; they just are, as they are defined. > >> (And to answer your question elsewhere in the thread, a number >> greater >> than any finite doesn't suffice. You first have to show that there >> are >> numbers greater than any finite; or, if you want to define them (as >> extensions >> to the natural numbers), you have to describe their properties. I >> think you >> will run into problems dealing with what you have occasionally called >> The >> Twilight Zone coming between the finites and infinites [in some >> sense].) >> Matt > I'll have the same trouble with the Twilight Zone that the standard > theory has, > if I make the same mistake of trying to pin a value on its location, > like > aleph_0. Then, I'll run into contradictions. So, I'll avoid that > mistake, okay? >> You're saying, if I follow you, that you won't try to pin a value on >> the >> location of the Twilight Zone. Does this mean that it doesn't *have* a >> location? >> If it does, why can't the location be specified? If it doesn't, in what >> sense >> does it exist? >> Here's a question: What, exactly, is the definition of this Twilight >> Zone? The Twilight Zone is the transition from finite to infinite. No, there is > no > location for it that can be pinned down, because the point' at which > one > increments a finite and gets an infinite doesn't exist. This does not > occur at > any location. It occurs over an infinite expanse and sequence of > increments. > But, if both finites and infinites coexist on the number line, then this > transition zone exists, though it can never be located specifically. The Twilight Zone is the focus of the Cantorian mantra. It's a little > like the > boundary between inner and outer space. Point it out and it disappears as > fast > as a ball in a vase. ;) If there is no point at which a transition occurs, then how can it occur? > Does > it occur a little bit at a time? If so, where does it start? How does it > progress? Where does it end? Are there numbers occupying the Twilight > Zone? > If so, are they finite? Infinite? Some of each? If there are no numbers > occupying the Twilight Zone, then how does it include an infinite ... > sequence > of increments? Matt It's a mystery, and not one that can be solved. There is no largest finite or > smallest infinite, and no point where they meet. And yet, over an infinite > number of finite steps, one goes an infinite distance. One never executes an infinite number of steps. The number of steps one can actually execute is always finite, but without end or bound. The only way to get infinitely many steps it to be given them by axiom. And that's what the Peano axioms do, give them to us. === Subject: Re: Calculus XOR Probability about the staircase problem vs. the Calculus XOR Probability theme > and also found that there doesn't exist a similarity between the two. Well, I agree that one glaring difference is that I give a practical definition of infinite case that is applicable to any sequence of sets of points, and which yields a well-defined mathematical object (i.e., a set of points); while you don't give any definition of infinite case at all except to simply assert that such a thing exists for the sequence {[1..n]} which is called N for the remainder of your argument, and that you propose that this N represents a situation (something that implies the ability to pick naturals at random) rather than a mathematical object (a function from the naturals to the reals, meeting certain constraints). But that is simply a difference in the way we define (or fail to define) our limits. On the other hand, I think we both use premise B... Suppose that: {x_n} is a sequence of mathematical objects; X is defined to be the the limit as n->oo of {x_n}; g is a function such that g(x_n) is a real number for all x_n; f is the function defined by f(n) = g(x_n) for all naturals n; Then: if for all n, f(n) = k constantly; then it logically follows that lim n->oo f(x_n) = f(x_oo) = g(X) = k, where : x_oo is x_n in the infinite case. ... in order to justify our erroneous results; with a contradictory result in both situations. TO's current line of thinking appears to lean towards well, the infinite case isn't captured by your limit; and that's why premise B is still true, but the answer is wrong: your limit is not the 'right' way of expressing the infinite case. That could equally apply to your argument: one could say that, because there cannot exist a function p : N-> R such that sum(p(n)) = 1 and p(n) = p(m) for all n, m in N (by the archimedean property of the reals); that therefore whatever it is you think you are claiming to be the infinite case isn't that thing at all: it is merely some limiting case. Personally, because the idea of limit is useful in its many flexible forms, my preference would be to say that infinite case must depend on a precise (and therefore, limited) definition; and that therefore premise B is false in general: there is no such thing as a truly ultimate object x_oo, which would be the infinite case of all things which would claim to approach it. Either way, I can't see any reason to accept your argument as representing a coherent mathematical statement; independently of your lack of a rigorous definition of limit. === Subject: Re: Calculus XOR Probability > Either way, I can't see any reason to accept your argument as > representing a coherent mathematical statement; independently of your > lack of a rigorous definition of limit. This is a quote from a new web page, now under construction: Do there exist mathematical theories that will _never_ reach the stage of being an idealization ? There are ! An example is in the 'sci.math' thread 'Calculus XOR Probability'. Here the Probability Theory at any finite but incredibly large interval of natural numbers {1,2,3, ... ,n} can not be generalized to all of the naturals without running into all kind of paradoxes. I've found a remedy only but after all the rest has been tried. A solution is that the limit for n -> oo cannot actually be taken. Our theory has to remain somewhere in between 'Abstractions' and 'Idealizations'. And therefore shall never become a part of ideal mainstream mathematics. So be it! Some quite advanced physical theories (like for example QED) are in very much the same position. Han de Bruijn === Subject: Re: Calculus XOR Probability <622f2$4451f890$82a1e228$1957@news2.tudelft.nl Either way, I can't see any reason to accept your argument as > representing a coherent mathematical statement; independently of your > lack of a rigorous definition of limit. > This is a quote from a new web page, now under construction: > Do there exist mathematical theories that will _never_ reach the stage > of being an idealization ? There are ! An example is in the 'sci.math' > thread 'Calculus XOR Probability'. Here the Probability Theory at any > finite but incredibly large interval of natural numbers {1,2,3, ... ,n} > can not be generalized to all of the naturals without running into all > kind of paradoxes. I've found a remedy only but after all the rest has > been tried. A solution is that the limit for n -> oo cannot actually > be taken. Our theory has to remain somewhere in between 'Abstractions' > and 'Idealizations'. And therefore shall never become a part of ideal > mainstream mathematics. So be it! Some quite advanced physical theories > (like for example QED) are in very much the same position. Yes; quantuum mechanics also suffers from the fact that there is no natural number x of protons such that 2*x = 3 protons. How will we ever be able to use such a limited theory? === Subject: Re: Calculus XOR Probability > Except the paradox of not having solved the original problem. That's true. But, as a physicist, I'm rather interested in a possible realization of the original problem in the world of physics. > The original does not require any physical possibility so that changing > it in that way eliminates the original problem. You're right. And *I* am not interested in non-applicable mathematics. Han de Bruijn === Subject: Re: Calculus XOR Probability > Except the paradox of not having solved the original problem. > That's true. But, as a physicist, I'm rather interested in a possible > realization of the original problem in the world of physics. > The original does not require any physical possibility so that changing > it in that way eliminates the original problem. > You're right. And *I* am not interested in non-applicable mathematics. > Han de Bruijn Just do not any more claim to have solved problems that you have not solved. === Subject: Re: Calculus XOR Probability >Except the paradox of not having solved the original problem. >>That's true. But, as a physicist, I'm rather interested in a possible >>realization of the original problem in the world of physics. >The original does not require any physical possibility so that changing >it in that way eliminates the original problem. >>You're right. And *I* am not interested in non-applicable mathematics. > Just do not any more claim to have solved problems that you have not > solved. Define solved. Han de Bruijn === Subject: Re: Calculus XOR Probability > >Except the paradox of not having solved the original problem. >>That's true. But, as a physicist, I'm rather interested in a possible >>realization of the original problem in the world of physics. >The original does not require any physical possibility so that changing >it in that way eliminates the original problem. >>You're right. And *I* am not interested in non-applicable mathematics. Just do not any more claim to have solved problems that you have not > solved. > Define solved. > Han de Bruijn If a problem asks a question, then the problem is solved when the question has been correctly answered. === Subject: Re: Calculus XOR Probability > link you cited, but I am a little confused for other reasons as well. Tony, uhm, that math is of no higher level than standard calculus. > Did you just prove the length of that curve is equal to 2 pi? No. I proved that the length of the renormalized curve equals 2 pi. Han de Bruijn === Subject: Re: Calculus XOR Probability Han de Bruijn said: > link you cited, but I am a little confused for other reasons as well. > Tony, uhm, that math is of no higher level than standard calculus. Yes, well, it appeared to include things I was not familiar with. I'm not a mathematician, and my calculus is a bit rusty. It doesn't get used too often. Sorry to disappoint you. Maybe if I spent more time looking it over, but I didn't have that kind of time yesterday. > Did you just prove the length of that curve is equal to 2 pi? > No. I proved that the length of the renormalized curve equals 2 pi. So, you're basically smoothing out any curvatures that would remain at the infinitesimal level, and getting a proper metric that way? That's what it sounds like. Sorry to be so dense. > Han de Bruijn -- Smiles, Tony === Subject: Re: Calculus XOR Probability Did you just prove the length of that curve is equal to 2 pi? No. I proved that the length of the renormalized curve equals 2 pi. > So, you're basically smoothing out any curvatures that would remain at the > infinitesimal level, and getting a proper metric that way? That's what it > sounds like. Sorry to be so dense. No!, HbB is changing the problem from one he either can't or does not want to solve, to one that he would rather solve. Unfortunately that leaves the original unsolved. === Subject: Re: Calculus XOR Probability > Han de Bruijn said: the > link you cited, but I am a little confused for other reasons as well. Tony, uhm, that math is of no higher level than standard calculus. > Yes, well, it appeared to include things I was not familiar with. I'm not a > mathematician, and my calculus is a bit rusty. Not to mention TO's algebra.,arithmetic geometry, et al. === Subject: Re: Calculus XOR Probability <444f9ccd$0$14437$6d36acad@titian.nntpserver.com> <44510192$0$14520$6d36acad@titian.nntpserver.com Well, then, I guess we have different intuitions. Picture the unit interval > [0,1] as a little shrunk-down real line from 0 to oo. > [0,1] has two endpoints. [0,oo) doesn't. Oh, come come, Sir. Once you've shrunk it down, you'll just look to right and left and see both ends. If either end isn't there, despite being within eyesight range, why - you'll have to declare it, won't you? Otherwise you'll get a compiler error, I expect. Brian Chandler http://imaginatorium.org === Subject: Re: Calculus XOR Probability We have seen that this is /not always true/ for limits in general; it > depends on the exact /definition/ of the limit. So premise B doesn't > follow /unless/ it follows from a clear definition of his premise A: > what does he mean by limit? > He means that as n->oo, 1/n->0, but that when n=oo, 1/n<>0. [I think 'He' is Han...] Anyway, could you clarify: you are writing <> to mean not equals? Could I suggest that you don't - I think this particular symbol comes from the world's crummiest programming language, and it is a reflection of the worldview that in part represents your problem. It seems to assume that for any x and y, either xy. This may seem part of elementary arithmetic, but isn't true in mathematics. Of course, the problem in interpreting when n=oo, 1/n!=0, is that we don't know what n=oo means. Brian Chandler http://imaginatorium.org === Subject: Re: Calculus XOR Probability Well yes, Chas, that's true. I'm gald you seem to finally getting what I'm > saying and agreeing (I think) that the definition of the limiting curve you're > using doesn't gurarantee that it is really the same object as the curve in the > limit in every respect. > And it certainly never claimed to be; because that's not what a limit > is. > Um, if it's not the same object, why would you expect it to have the same > length? When you said it really is the same object /in every respect/, I allowed that you might apply /any/ property to it, no matter how absurd (for example, the property that this set of points was produced on a Monday as opposed to this set of points was produced on a Wednesday). As far as /I/ am concerned, it is sufficient that y = x characterizes some curve as a set of points; all other properties of interest (e.g., length, connectedness, area, etc.) are well defined by that set of points, in the context of R^2 with the usual metric. So from my viewpoint, yes, my definition of limit guarantees that it really is the same object, since the objects under consideration are always and only sets of points, and equality between sets of points is based on set membership (and no other property). Thus saying my limit is the same as the diagonal follows as obviously as saying the curves y = x and 2y = 2x refer to the same object. But not every viewpoint will concur with this conclusion; just as you might claim that y = x and 2y = 2x are different things, because the first equation doesn't contain the number 2. It all depends on your /definition/ of the ultimate object. > The problem with your example, as I see it, is exactly > that, the difference between considering or not considering the derivative of > position, as well as the position, as one travels along the curve. It's a > matter of ignoring the direction of the curve in trying to measure it, not a > matter of taking an equality to be true in the infinite case. > Right; because those are two different things. > Yes, but you were blaming the infinite case for the failure of your example, > not the lack of proper metric due to non-parallel elements. I am blaming the logical inference of the form given that D is the limiting case of {C_n}, and length(C_n) = 2 for all NATURALS n, therefore length(D) = 2. That is what I am calling premise B. I have given a definition of limiting case and a proof that D is, indeed, the limiting case of {C_n} under that definition; so that premise is true. The premise that length(C_n) = 2 for all NATURALS n is obvious. So it is the therefore part of the above logical inference that contains the error. Why would someone say therefore in the above inference? Because they are implicitly applying the following more general premise: Suppose g is a function with g(C_n) being a real number for all natural n. Let f(n) = g(C_n). Then the following /always/ holds: If f(n) = k equally for all n, and D is the limiting case of {C_n}, then f(oo) = g(D) = k. That is exactly the logic you are (improperly) using when you say if something is constantly true for all finite n, then it must hold for the infinite case. If you mean something else by the limiting case of {C_n} than the limit as n->oo of {C_n} as I defined it, then you should say /exactly what you mean/ - otherwise the phrase the limiting case is meaningless; you have given it no meaning. In which case saying {whatever you wish} is true in the infinite case is equally meaningless; and it would be pointless to try to argue with you that it is either true or false. It is neither: it is meaningless. claim wait a minute, there is no such natural number x, does it make > sense for me to say yes there is; x = 3/2? Is 3/2 a natural number? > No, the naturals are not closed under division. It's a rational. Two steps forward... > Similarly, if I define the limit of a sequence of curves in a certain > way, and restrict that definition to sets of points, it doesn't make > sense to claim but if we take sets of points and directions at those > points then (blah blah blah) .... Sets of points are not the same > thing as sets of points and directions. > It makes sense if you are claiming that this set of points has a measurable > length, because one cannot measure with directionless points. One step back. Do you then doubt y = 1 - x is sufficient to describe the line with slope -1, and y intercept 1? Do you doubt that this is identical to the set of points I /defined/ as the limiting case of the staircases? > It's that claim > that made me bring up the ignored directions of the elements, since that's the > cause of the error of sqrt(2), not some magical infinity thing. You might not > have mentioned it, but it's certainly germane to the example. What's germane is not whether I mentioned that the limit curve is an accurate measurement of arc length. What's germane is your (and Han's) assuption that limiting case means anything more than the definition I gave, simply because I use the /words/ limiting case to refer to that precise definition. In particular, my definition of the limiting case of {C_n} said /absolutely nothing/ about the length of either C_n or its limit D. Why? Simply because that's /not part of the definition/ of limit. Suppose Han had said the piirjoon of the sets {1..n} as n->oo is something I will call 'N'. Since the piirjoon of {n*1/n} = 1, therefore N*(1/N) = 1. What kind of reasoning is that? Is it supposed to imply that if I say ok then, the piirjoon of (0, 00, 000, ...} is also N. Since 000 is just another way of writing 0, the piirjoon of {0, 00, 000, ...} = 0, so therefore N = 0. Therefore, with Han's result, 0*(1/0) = 1, that then is an equally valid argument? > Or if I say the real number limit as n->oo of {sin(n*x)/n} = 0, for > all x, it doesn't make sense for you to say but the limit is not a > real number, it's an indeterminate infinitesimal. /Of course/ the > limit is a real number: that's how I defined it! An indeterminate > infinitesimal is no more a real number than 3/2 is a natural number; > so such a response makes no sense. > Fine. The standard real number limit is 0, but that doesn't mean that an > infinitesimal value doesn't exist. And the fact that there does not exist a natural number x such that 2*x = 3 doesn't mean that there isn't some number system in which the meanings of 2, 3, * and = doesn't contain an x having this property. For example, modulo 7, 2*5 = 3 is a true statement; so I could claim that x = 5 could be a valid solution. Or if I define * to mean + over the naturals, then 2*1=3 is true. Or if I define * to mean subtraction in the naturals and = to mean is less than, then 2*2 = 3 is a true statement. Or if I define 2 to be the cyclic group with two elements, and 3 to be the Klein group, and * to be direct product of two groups, and = to be is isomorphic to, then again, 2*2 = 3, but for a different reason. But: SO WHAT? I'm not talking about some extended set of reals, or modulo arithmetic, or group theory, or any other thing: I'm talking about applying the /definition/ of the limit of {a_n} to some sequence of real numbers {a_n}. To divert the discussion to some other thing that doesn't address what I am talking about is /nonsensical/ in this context. > Perhaps the limit is 0 in the standard > system, Perhaps? Ah, so you still doubt that lim n -> oo {sin(n)/n} = 0 /according to the usual definition/? > but the limit is not the same as the ultimate object, as you have been > pointing out. The ultimate object is restricted by its /definition/. If by 1 foot you mean anything which is exactly the same thing as my right foot, then a football field is not 300 feet; because there is only one such thing as a foot: to whit, your right foot. If by sqrt(-1) you mean some /positive real number/ x such that x^2 = -1, then there is no such real number. If instead you mean some /complex number/ x such that x^2 = -1, then there are two complex numbers that fit the bill: i and -i. You have to /restrict the definition/ of an object in order for the definition to make sense. > So, if the standard reals fail to identify this value, then we > expand the number system, just like we produce the rationals when we discover > we have no answer for 2x=3 in the naturals. What's wrong with that? What's wrong with it is /that's not what I was talking about/. What about x = 3/2? is irrelevant to, and therefore not a valid response to, the statement does there exist a natural number such that 2*x = 3?. You are answering a different question than the one asked; so your response is silly and pointless. If what you want to ask is is there a formal mathematical sense in which the statement '2*x = 3' has meaning for some well-defined mathematical object x? that's perfectly fine. But it is missing the point to respond to the statement There is no natural number x such that 2*x = 3 with the response but there could be if x = 3/2. Similarly, to talk about infinitesimals and infinitesimal probabilities might be interesting and fun, but it has absolutley no bearing on whether or not it is possible to pick natural numbers at random in the sense that at I can pick a number from 1 to 10 at random. > If Han offered the argument you recount above, > then it suffers from that problem, as the notion of expected value does for the > same reason. But, if we leave N out of the discussion, and simply say for a set > with n elements equally likely the probability of each is 1/n, there is nothing > wrong with that statement. It doesn't even involve a limit when so stated. > In fact, it simply becomes a statement about a finite set - because the > domain is then restricted to finite sets. > Who restricted the domain to finite values? By stating that n is infinite, we > have explicitly done the opposite. Who restricted the domain to finites? You did, by using 1/n, which has an accepted, understandable value. If I say I drove my car to work today, and you say, oh really? What kind of car do you drive? and I reply, It's a bathtub full of gin, can I really blame you for looking at me askance? In what way is a bathtub full of gin a car? And what could I possibly mean by driving it? Does my bathtub have an engine in it? Or by driving it, do I really mean drinking it, e.g., that by drinking a bathtub full of gin, I obtain the fortitude to use mass transit? Or do I mean some other thing? It's impossible to tell. Similarly when you say let X be a set with n elements I assume you mean, where n is a natural number. For you to claim that n is not a natural number, but instead is a member of some other set T_N, makes me wonder what you actually mean by saying 1/n. Do you mean that for each n in T_N, there is a /unique/ element 1/n in T_N, and multiplication * defined as a binary operation on your set T_N, such that n*(1/n) = 1? Is the resulting set a field, or some other algebraic construct? Could there be other elements y in T_N such that n*y = 1? In which case, how do I tell which element you mean by 1/n? Do you mean y, or some other element? Or do by 1/n do you simply mean some other thing? It's impossible to tell. Until you define /exactly what you mean/, you might as well be saying let X be a set with n elements, where it should be obvious that n is a bathtub full of gin. On the other hand, suppose we define the function f_n, for each natural number n, as follows: if x is rational, let f_n(x) = x/n. if x is irrational, let f_n(x) = x/n^2. Given the definition of limit that I have been using, can you decide what function I mean by the limit as n -> oo {f_n}? Do you have to agree that my definition of limit is useful in every, or even any, application? Not at all; but at least you can figure out /exactly what I mean/. > The > limit comes in as n->oo, > Comes in? Limits don't come in like guests. > They come into the conversation as a topic of discussion. The piirjoon also comes in as a topic of discussion. What is a piirjoon? You'll just have to guess; perhaps gin is involved. > Limits are always > /defined/ somehow; and how that limit is defined has direct bearing on > any conclusions you draw. That's the whole point of the staircase > example: when I clearly defined what I meant by the limit, you could > see why the result either followed or it didn't. > Why should I think that as n->oo has any meaning whatsoever in this > case? You might as well be saying the piirjoon comes in as n becomes > super-mega-mega. > hrumph. I am trying to stay with the original topic, at least a little. Was the original topic mathematical limits (i.e., well-defined definitions from which we can draw conclusions) or was it piirjoons (vague abstractions based on gut feeling)? That's what I am attempting to illuminate. > and the prime question here is whether the individual > probabilities are truly 0, or some infinitesimal value equal to 1/n. > The prime question here is what the heck are you talking about? > Until you get as specific as I got when I said the limit of a sequence > of sets of points {C_n} is the set of all limit points of {C_n}, I > have no idea what you mean here. Mathematically speaking, of course. > Maybe you should go back and read Han's original post.... {:( What Han's original post lacks is explicit definitions of probability, limit, and infinite case, to start with. === Subject: Re: Calculus XOR Probability cbrown@cbrownsystems.com said: > cbrown@cbrownsystems.com said: > Well yes, Chas, that's true. I'm gald you seem to finally getting what I'm > saying and agreeing (I think) that the definition of the limiting curve you're > using doesn't gurarantee that it is really the same object as the curve in the > limit in every respect. > And it certainly never claimed to be; because that's not what a limit > is. > Um, if it's not the same object, why would you expect it to have the same > length? > When you said it really is the same object /in every respect/, I > allowed that you might apply /any/ property to it, no matter how absurd > (for example, the property that this set of points was produced on a > Monday as opposed to this set of points was produced on a > Wednesday). > As far as /I/ am concerned, it is sufficient that y = x characterizes > some curve as a set of points; all other properties of interest (e.g., > length, connectedness, area, etc.) are well defined by that set of > points, in the context of R^2 with the usual metric. > So from my viewpoint, yes, my definition of limit guarantees that it > really is the same object, since the objects under consideration are > always and only sets of points, and equality between sets of points is > based on set membership (and no other property). Thus saying my limit > is the same as the diagonal follows as obviously as saying the curves > y = x and 2y = 2x refer to the same object. > But not every viewpoint will concur with this conclusion; just as you > might claim that y = x and 2y = 2x are different things, because > the first equation doesn't contain the number 2. It all depends on > your /definition/ of the ultimate object. It's a bit different from y=x being different from 2y=2x. There is a real difference between the staircase in the limit and the diagonal, as demonstrated by your example. > The problem with your example, as I see it, is exactly > that, the difference between considering or not considering the derivative of > position, as well as the position, as one travels along the curve. It's a > matter of ignoring the direction of the curve in trying to measure it, not a > matter of taking an equality to be true in the infinite case. > Right; because those are two different things. > Yes, but you were blaming the infinite case for the failure of your example, > not the lack of proper metric due to non-parallel elements. > I am blaming the logical inference of the form given that D is the > limiting case of {C_n}, and length(C_n) = 2 for all NATURALS n, > therefore length(D) = 2. That is what I am calling premise B. > I have given a definition of limiting case and a proof that D is, > indeed, the limiting case of {C_n} under that definition; so that > premise is true. The premise that length(C_n) = 2 for all NATURALS n is > obvious. So it is the therefore part of the above logical inference > that contains the error. > Why would someone say therefore in the above inference? Because they > are implicitly applying the following more general premise: Suppose g > is a function with g(C_n) being a real number for all natural n. Let > f(n) = g(C_n). Then the following /always/ holds: If f(n) = k equally > for all n, and D is the limiting case of {C_n}, then f(oo) = g(D) = k. Perhaps that more general statement about the infinite case has problems, if one applies a deifnition of limit which doesn't preserve the measure one is looking at in the limit. But, I don't think the problem with the metric in your example has anything to do with Han's limit. > That is exactly the logic you are (improperly) using when you say if > something is constantly true for all finite n, then it must hold for > the infinite case. > If you mean something else by the limiting case of {C_n} than the > limit as n->oo of {C_n} as I defined it, then you should say /exactly > what you mean/ - otherwise the phrase the limiting case is > meaningless; you have given it no meaning. > In which case saying {whatever you wish} is true in the infinite case > is equally meaningless; and it would be pointless to try to argue with > you that it is either true or false. It is neither: it is meaningless. The meaning seems clear to me when I say that inductive proof of an equality holds for the infinite case. > Suppose I say let x be the natural number such that 2*x = 3. If you > claim wait a minute, there is no such natural number x, does it make > sense for me to say yes there is; x = 3/2? Is 3/2 a natural number? > No, the naturals are not closed under division. It's a rational. > Two steps forward... > Similarly, if I define the limit of a sequence of curves in a certain > way, and restrict that definition to sets of points, it doesn't make > sense to claim but if we take sets of points and directions at those > points then (blah blah blah) .... Sets of points are not the same > thing as sets of points and directions. > It makes sense if you are claiming that this set of points has a measurable > length, because one cannot measure with directionless points. > One step back. > Do you then doubt y = 1 - x is sufficient to describe the line with > slope -1, and y intercept 1? Nope. > Do you doubt that this is identical to the set of points I /defined/ as > the limiting case of the staircases? Yep. That limiting case has a slope of -1 nowhere. > It's that claim > that made me bring up the ignored directions of the elements, since that's the > cause of the error of sqrt(2), not some magical infinity thing. You might not > have mentioned it, but it's certainly germane to the example. > What's germane is not whether I mentioned that the limit curve is an > accurate measurement of arc length. What's germane is your (and Han's) > assuption that limiting case means anything more than the definition > I gave, simply because I use the /words/ limiting case to refer to > that precise definition. Your definition of limiting case in this instance was not appropriate for the measure you were trying to get out of it. > In particular, my definition of the limiting case of {C_n} said > /absolutely nothing/ about the length of either C_n or its limit D. > Why? Simply because that's /not part of the definition/ of limit. It certainly had nothign to do with Han's limit. > Suppose Han had said the piirjoon of the sets {1..n} as n->oo is > something I will call 'N'. Since the piirjoon of {n*1/n} = 1, therefore > N*(1/N) = 1. > What kind of reasoning is that? Inductive. > Is it supposed to imply that if I say ok then, the piirjoon of (0, > 00, 000, ...} is also N. Since 000 is just another way of writing > 0, the piirjoon of {0, 00, 000, ...} = 0, so therefore N = 0. > Therefore, with Han's result, 0*(1/0) = 1, that then is an equally > valid argument? No, that was somewhat convoluted, since Han didn't claim that was N, but the conclusion is correct, given equal 0's. :) > Or if I say the real number limit as n->oo of {sin(n*x)/n} = 0, for > all x, it doesn't make sense for you to say but the limit is not a > real number, it's an indeterminate infinitesimal. /Of course/ the > limit is a real number: that's how I defined it! An indeterminate > infinitesimal is no more a real number than 3/2 is a natural number; > so such a response makes no sense. > Fine. The standard real number limit is 0, but that doesn't mean that an > infinitesimal value doesn't exist. > And the fact that there does not exist a natural number x such that 2*x > = 3 doesn't mean that there isn't some number system in which the > meanings of 2, 3, * and = doesn't contain an x having this > property. Right. > For example, modulo 7, 2*5 = 3 is a true statement; so I could claim > that x = 5 could be a valid solution. Or if I define * to mean > + over the naturals, then 2*1=3 is true. Or if I define * to mean > subtraction in the naturals and = to mean is less than, then 2*2 > = 3 is a true statement. Or if I define 2 to be the cyclic group > with two elements, and 3 to be the Klein group, and * to be direct > product of two groups, and = to be is isomorphic to, then again, > 2*2 = 3, but for a different reason. > But: SO WHAT? Good question. > I'm not talking about some extended set of reals, or > modulo arithmetic, or group theory, or any other thing: I'm talking > about applying the /definition/ of the limit of {a_n} to some > sequence of real numbers {a_n}. To divert the discussion to some other > thing that doesn't address what I am talking about is /nonsensical/ in > this context. Almost as nonsensical as bringing in a curve measure problem with nonparallel elements when we are discussing a probability distribution over an infinite set. > Perhaps the limit is 0 in the standard > system, > Perhaps? Ah, so you still doubt that lim n -> oo {sin(n)/n} = 0 > /according to the usual definition/? No. Perhaps was used in a different way than that. Welcome to English. > but the limit is not the same as the ultimate object, as you have been > pointing out. > The ultimate object is restricted by its /definition/. > If by 1 foot you mean anything which is exactly the same thing as my > right foot, then a football field is not 300 feet; because there is > only one such thing as a foot: to whit, your right foot. > If by sqrt(-1) you mean some /positive real number/ x such that x^2 = > -1, then there is no such real number. If instead you mean some > /complex number/ x such that x^2 = -1, then there are two complex > numbers that fit the bill: i and -i. > You have to /restrict the definition/ of an object in order for the > definition to make sense. Well, then, maybe you need to restrict your notion of limit a little more. > So, if the standard reals fail to identify this value, then we > expand the number system, just like we produce the rationals when we discover > we have no answer for 2x=3 in the naturals. What's wrong with that? > What's wrong with it is /that's not what I was talking about/. What > about x = 3/2? is irrelevant to, and therefore not a valid response > to, the statement does there exist a natural number such that 2*x = > 3?. You are answering a different question than the one asked; so your > response is silly and pointless. So says the kettle. > If what you want to ask is is there a formal mathematical sense in > which the statement '2*x = 3' has meaning for some well-defined > mathematical object x? that's perfectly fine. But it is missing the > point to respond to the statement There is no natural number x such > that 2*x = 3 with the response but there could be if x = 3/2. So what? No one claims the individual probabilities Han brought up are finite. The question is whether they are truly 0, or actually infinitesimal. That's a nonstandard question. > Similarly, to talk about infinitesimals and infinitesimal > probabilities might be interesting and fun, but it has absolutley no > bearing on whether or not it is possible to pick natural numbers at > random in the sense that at I can pick a number from 1 to 10 at > random. We've already agreed the uniform probability distribution on the set of finite naturals is problematic. No point in flogging a dead horse. > If Han offered the argument you recount above, > then it suffers from that problem, as the notion of expected value does for the > same reason. But, if we leave N out of the discussion, and simply say for a set > with n elements equally likely the probability of each is 1/n, there is nothing > wrong with that statement. It doesn't even involve a limit when so stated. > In fact, it simply becomes a statement about a finite set - because the > domain is then restricted to finite sets. > Who restricted the domain to finite values? By stating that n is infinite, we > have explicitly done the opposite. > Who restricted the domain to finites? You did, by using 1/n, which > has an accepted, understandable value. Not when n is infinite, which is the topic. Sheesh! > If I say I drove my car to work today, and you say, oh really? What > kind of car do you drive? and I reply, It's a bathtub full of gin, > can I really blame you for looking at me askance? No, but I'd want to see the thing, and smell your breath. Or not. > In what way is a bathtub full of gin a car? And what could I possibly > mean by driving it? Does my bathtub have an engine in it? Or by > driving it, do I really mean drinking it, e.g., that by drinking a > bathtub full of gin, I obtain the fortitude to use mass transit? Or do > I mean some other thing? It's impossible to tell. Maybe it has an engine that runs on gin, but that sounds expensive, depending on your mileage. I'm sorry, what was your point? 8-| > Similarly when you say let X be a set with n elements I assume you > mean, where n is a natural number. For you to claim that n is not a > natural number, but instead is a member of some other set T_N, makes me > wonder what you actually mean by saying 1/n. Han was suggesting that if n is infinite, the same additive probability should hold, but standard math fails in this regard. Perhaps his point was against actual infinities, but my response is to bolster actual infinities with actual ifinitesimals. > Do you mean that for each n in T_N, there is a /unique/ element 1/n > in T_N, and multiplication * defined as a binary operation on your > set T_N, such that n*(1/n) = 1? Is the resulting set a field, or some > other algebraic construct? Could there be other elements y in T_N such > that n*y = 1? In which case, how do I tell which element you mean by > 1/n? Do you mean y, or some other element? This is, I believe, a field with all the properties of the standard reals, although with somewhat restricted arithmetic due to the inability to specify infinite strings. > Or do by 1/n do you simply mean some other thing? It's impossible > to tell. Give n people an equal piece of the pie. That's division. > Until you define /exactly what you mean/, you might as well be saying > let X be a set with n elements, where it should be obvious that n is a > bathtub full of gin. Well, the bathtub could be considered a set of gin molecules, I suppose. You shouldn't sit in there too long, by the way... > On the other hand, suppose we define the function f_n, for each natural > number n, as follows: > if x is rational, let f_n(x) = x/n. > if x is irrational, let f_n(x) = x/n^2. > Given the definition of limit that I have been using, can you decide > what function I mean by the limit as n -> oo {f_n}? Sure, y=0. In my system, rational x would produce infinitesimal y and irrational x would produce sub-infinitesimal y. > Do you have to agree that my definition of limit is useful in every, > or even any, application? Not at all; but at least you can figure out > /exactly what I mean/. Sure, the only problem is it doesn't really pertain to the problem of infinite probability distributions. > The > limit comes in as n->oo, > Comes in? Limits don't come in like guests. > They come into the conversation as a topic of discussion. > The piirjoon also comes in as a topic of discussion. What is a > piirjoon? You'll just have to guess; perhaps gin is involved. Yes, between pirjoons, bathtubs full of gin, and non-parallel metrics, you've brought a lot into the conversation. > Limits are always > /defined/ somehow; and how that limit is defined has direct bearing on > any conclusions you draw. That's the whole point of the staircase > example: when I clearly defined what I meant by the limit, you could > see why the result either followed or it didn't. > Why should I think that as n->oo has any meaning whatsoever in this > case? You might as well be saying the piirjoon comes in as n becomes > super-mega-mega. > hrumph. I am trying to stay with the original topic, at least a little. > Was the original topic mathematical limits (i.e., well-defined > definitions from which we can draw conclusions) or was it piirjoons > (vague abstractions based on gut feeling)? > That's what I am attempting to illuminate. Then you might want to address the original problem of probability spaces rather than bringing in metric digressions. > and the prime question here is whether the individual > probabilities are truly 0, or some infinitesimal value equal to 1/n. > The prime question here is what the heck are you talking about? > Until you get as specific as I got when I said the limit of a sequence > of sets of points {C_n} is the set of all limit points of {C_n}, I > have no idea what you mean here. Mathematically speaking, of course. > Maybe you should go back and read Han's original post.... {:( > What Han's original post lacks is explicit definitions of > probability, limit, and infinite case, to start with. Huh! I didn't really need those in order to understand what he was saying. Maybe I already knew it anyway. Sometimes it's hard to tell. -- Smiles, Tony === Subject: Re: Calculus XOR Probability > cbrown@cbrownsystems.com said: > Do you then doubt y = 1 - x is sufficient to describe the line > with slope -1, and y intercept 1? > Nope. Do you doubt that this is identical to the set of points I > /defined/ as the limiting case of the staircases? > Yep. That limiting case has a slope of -1 nowhere. How so? If it is a line or line segment at all (and only lines or line segments have slopes. then the slope is uniquely determined by any two distinct points of the line or line segment. For what distinct pair of points on the limit curve does one not have the slope (difference in ordinates over the corresponding difference in abscissas) equal to -1? If for every pair of distinct points on a curve in the xy-plane, the slope for the line through those two points is -1, then the curve is a line, ray, or segment of slope -1. > Your definition of limiting case in this instance was not > appropriate for the measure you were trying to get out of it. It is the only limiting case there is. > Almost as nonsensical as bringing in a curve measure problem with > nonparallel elements when we are discussing a probability > distribution over an infinite set. As the issue is what sort of properties are inherited by limits of sequences of various sorts, TO's objection is what is nonsensical, and the analogy os spot on. > So what? No one claims the individual probabilities Han brought up > are finite. The question is whether they are truly 0, or actually > infinitesimal. That's a nonstandard question. In standard probability theory, a system in which there are no such things as infinitesmals, being infintesimal is not an option. No one has claimed that there could not be some non-standard theory in which things are different, but also, no one has produced such a theory. And absent some complete set of axioms and basic definitions for such a theory, it does not exist. Who restricted the domain to finites? You did, by using 1/n, > which has an accepted, understandable value. > Not when n is infinite, which is the topic. Sheesh! Where is your system in which these allegedly infinite n's exist? === Subject: Re: Calculus XOR Probability > cbrown@cbrownsystems.com said: > What Han's original post lacks is explicit definitions of > probability, limit, and infinite case, to start with. > Huh! I didn't really need those in order to understand what he was saying. > Maybe I already knew it anyway. Sometimes it's hard to tell. To the extent that TO actually could understand what Han was saying, Han could not have been saying anything mathematical === Subject: Re: Calculus XOR Probability cbrown@cbrownsystems.com said: > difference between the staircase in the limit and the diagonal, as demonstrated > by your example. It seems to me that saying y = x and 2y = 2x are the same thing can only mean that for all pairs p = (px,py), p satisfies the first condition if and only if it satisfies the second condition. The set of all pairs satisfying these condition is the line with slope 1 going through the origin. The fact that the second equation contains a 2 and the first does not is irrelevant. It also seems to me that saying y = 1 - x; x, y >= 0 and the limit of the staircases are the same thing can only mean that for all pairs p = (px, py), p satisifies the first condition if and only if it satisfies the second condition. The set of all pairs satisfying these conditions is the diagonal D. The fact that the second condition uses the word limit while the first does not is irrelevant. are implicitly applying the following more general premise: Suppose g > is a function with g(C_n) being a real number for all natural n. Let > f(n) = g(C_n). Then the following /always/ holds: If f(n) = k equally > for all n, and D is the limiting case of {C_n}, then f(oo) = g(D) = k. > Perhaps that more general statement about the infinite case has problems, if > one applies a deifnition of limit which doesn't preserve the measure one is > looking at in the limit. The whole point of the staircase argument is to get you to consider this fact, which I will repeat for emphasis: The more general statement about the infinite case may have problems, if one applies a definition of limit which doesn't preserve the measure one is looking at in the limit. > But, I don't think the problem with the metric in your > example has anything to do with Han's limit. Keeping in mind what was just stated: Perhaps Han's statement about the infinite case has problems. How does Han justify that, when he applies his definition of limit, it preserves the measure he is looking at in the limit? is equally meaningless; and it would be pointless to try to argue with > you that it is either true or false. It is neither: it is meaningless. > The meaning seems clear to me when I say that inductive proof of an equality > holds for the infinite case. Since you haven't given a definition of in the infinite case, it is pointless to argue with you regarding whether your statement is either false or true. However, it does appear to be a statement with a form similar to the staircase argument, and to Han's argument; so one can ask the same question about the /form/ of your /argument/: Perhaps your statement about the infinite case has problems. How does you justify that, when you apply your definition of limit, it preserves the measure you are looking at in the limit? accurate measurement of arc length. What's germane is your (and Han's) > assuption that limiting case means anything more than the definition > I gave, simply because I use the /words/ limiting case to refer to > that precise definition. > Your definition of limiting case in this instance was not appropriate for the > measure you were trying to get out of it. Perhaps your definition of infinite case is not appropriate for the measure (inductive proof of an equality) you are trying to get out of it. oo is > something I will call 'N'. Since the piirjoon of {n*1/n} = 1, therefore > N*(1/N) = 1. > What kind of reasoning is that? > Inductive. > Is it supposed to imply that if I say ok then, the piirjoon of (0, > 00, 000, ...} is also N. Since 000 is just another way of writing > 0, the piirjoon of {0, 00, 000, ...} = 0, so therefore N = 0. > Therefore, with Han's result, 0*(1/0) = 1, that then is an equally > valid argument? > No, that was somewhat convoluted, since Han didn't claim that was N, but the > conclusion is correct, given equal 0's. :) Wow. So all a mathematical argument is to you is just a bunch of mathematical looking words, slung together with connectives like therefore and since, which is correct if the conclusion agrees with your intuition, and incorrect if it does not? elements when we are discussing a probability distribution over an infinite > set. The example was brought in not to bear directly on his conclusion; it was brought in to demonstrate the flaw in the /logic/ of his /argument/. or even any, application? Not at all; but at least you can figure out > /exactly what I mean/. > Sure, the only problem is it doesn't really pertain to the problem of infinite > probability distributions. It pertains to clarifying valid versus invalid methods of deduction in a mathematical argument; which in turn pertains to whether or not your or Han's arguments regarding infinite probability distributions are valid or invalid. === Subject: Re: Standard Deviation of PSIA <0001HW.C063E7CE002911B9F0284530@news.giganews.com> <0001HW.C06ACF4B002DDEFFF0284530@news.giganews.com On Tue, 18 Apr 2006 0708, Jake the Porker sobbed: > Do you know who it was who said: one Arab fingernail isn't worth a > million dead jews? > John Knight > Your ex-wife and ex-children. > Gray Shockley > -------------------------- > Shockley's Exception to Godwin's > Law: When someone is quoting > hitler and his sycophants or > advocating hitler's goals, > Godwin's Law is irrelevant. No A-Rab ever said that. It was actually a despicable criminal faggot lying jew [though I repeat myself three times] who said that. John Knight ps--but of course you know that this jew equated 1 million dead A-Rabs to his one fingernail, don't you, Schlochley? One million Arabs are not worth a Jewish fingernail Rabbi Yaacov Perrin (NY Daily News, Feb. 28, 1994, p.6) === Subject: Re: Standard Deviation of PSIA >> On Tue, 18 Apr 2006 0708, Jake the Porker sobbed: > Do you know who it was who said: one Arab fingernail isn't worth a > million dead jews? John Knight >> Your ex-wife and ex-children. >> Gray Shockley >> -------------------------- >> Shockley's Exception to Godwin's >> Law: When someone is quoting >> hitler and his sycophants or >> advocating hitler's goals, >> Godwin's Law is irrelevant. > No A-Rab ever said that. Your ex-wife and ex-children are Arabian? I'm surprised that Preacher Lindstedt and Preacher Now-in-Hell allowed you to associate with such Aryans. > It was actually a despicable criminal faggot lying jew [though I repeat > myself three times That's why Preacher Lindstedt and Preacher Now-in-Hell wouldn't let you go out with the other fellows - they each would have said five times but, apparently, they knew as well as Cary and Bob your inability to cipher over twenty, even with your pants down. That and your lack of command of the English language, of course, also proves that you are incapable of learning even to the extent that is necessary to get you up to the level of functional literacy. > who said that. > John Knight > ps--but of course you know that this jew equated 1 million dead A-Rabs > to his one fingernail, don't you, Schlochley? My, my, my. SweetNotQuiteAboy John Knight tries for an insult and only reminds everyone that both Preacher Lindstedt and Preacher Now-in-Hell tried and tried to teach Short Johnny - as his ex-wife called him - but that Short Johnny just can not learn math and English any better than any other druggie. Short Johnny should have listened to Nancy Reagan and just said, No. > One million Arabs are not worth a Jewish fingernail > Rabbi Yaacov Perrin (NY Daily News, Ah, _The New York Daily News, the motto of which is If you're too stupid to understand the New York Post. > Feb. 28, 1994, p.6) Let's see what world-shaking events John Short Johnny - as its ex-wife used to call Knight - is reading about today. --------------------------------------------------------- NY's Highest chute down King Wrong A dopey daredevil hellbent on parachuting off the Empire State Building was plucked from the brink of disaster by cops and guards yesterday in a breathtaking drama that played out on a narrow ledge more than 1,000 feet above city streets. FULL STORY Been there, leaped off that The jumping jackass who tried to parachute off the Empire State Building yesterday is not a first-time daredevil. FULL STORY Man busted in teen 'tryst' at graveyard An accused rapist who hoped to have sex with a girl he met on the Web site vampirefreaks.com was nabbed when he showed up for an illicit rendezvous at a Long Island graveyard, police said yesterday. FULL STORY She could've been the 1st Imette It could have been her. Raped, maimed and killed, her body callously dumped in a barren lot - any or all of that could have been her fate. But the 19-year-old Queens woman would not let that happen. FULL STORY Smoke, fear on East Side A menacing cloud of red smoke spread panic on a midtown Manhattan street after dye packs exploded inside a bank robber's duffel bag during the evening rush yesterday, police said. FULL STORY Toussaint gets out of jail Breaking news update: Transit Workers Union President Roger Toussaint, jailed for leading an illegal subway and bus strike last December, was released on Friday. FULL STORY Peace grannies win their war The grannies got a walk yesterday - even if some of them needed a cane to take it. A Manhattan judge acquitted 18 golden agers of breaking the law when they staged a peace protest in front of a Times Square military recruitment center. FULL STORY SickFella's sneaking cigs: feds Reputed mobster Greg DePalma already has lost one lung to cancer and says he's too sick to go to trial - but prosecutors say he can't be that ill since he's been lighting up cigarettes in his hospital room. FULL STORY The Pit finally rumbles to life Two big excavators slowly rolled down the ramp into the pit of Ground Zero yesterday morning, ready to start work on the Freedom Tower more than 4-1/2 years after 9/11. FULL STORY Building a future at Ground Zero Editorials: You've got to start somewhere, and yesterday it was with the arrival of three enormous excavating trucks at Ground Zero as two governors, a mayor, Port Authority executives and a newly civic-minded developer got to building in a partnership that was forged out of... FULL STORY Retailers pay big bucks for Manhattan hot spots Rents in Manhattan's commercial hot spots have blazed higher in the past year as retailers clamor for the attention of the millions of tourists, commuters and residents who flow through the city each day. FULL STORY Rebbe's sons plan dueling services Cops are boosting manpower in Brooklyn as they brace for a Sabbath showdown at sundown today between followers of the late Grand Rebbe Moses Teitelbaum's feuding sons. FULL STORY The Tram thing still won't work The troubled tram of Roosevelt Island has stalled twice on test runs since its high-wire breakdown last week, a tram official testified yesterday. FULL STORY Aw, foie gras just ducky by Council Foie gras is no longer on the menu in Chicago, but don't expect New York to follow the Windy City's lead and ban the pricey delicacy. For now, the City Council is too busy laughing at Chicago's aldermen - much to the annoyance of animal right activists. FULL STORY Jermaine a big pain vs. Jersey Jermaine O'Neal, who was fined $15,000 yesterday for criticizing the officiating after Game 2, unleashed his frustration on the Nets. He dominated them with 37 points, 15 rebounds and four blocks to lead Indiana to a 107-95 win in Game 3 last night at Conseco Fieldhouse. FULL STORY Jets can't pass on Matt Mike Lupica: Reggie Bush was the best college football player last season and one of the best of all time. Vince Young, the Texas quarterback, played maybe the best single game of college football that has ever been played, if you really look at the stakes and circumstances of Texas vs. USC in the Rose Bowl for the national championship. FULL STORY Seems a natural but Green will go big The Hollywood QB looked quite comfortable in a vintage New York setting and he would like to make a habit of it. With the sun-splashed Hudson River as a backdrop, USC's Matt Leinart peered into the future and tried to imagine his reaction if he's picked by the Jets. FULL STORY Chacon at home Shawn Chacon turned in his second consecutive strong performance - building on recent efforts by Mike Mussina, Randy Johnson and Chien-Ming Wang - as the Yanks rebounded from Wednesday's ugly loss with a 4-1 victory last night over the Devil Rays at the Stadium. FULL STORY Mets can be table Turner The Braves already have doused cold water on the Mets once this season, halting a feel-good 10-2 start by taking the final two games of a series in Flushing last week. Now comes a reprise or retribution. FULL STORY Home (run) cooking The sky was blue, the stadium was bluer, and the Mets were in first place. What a day to visit Shea, taste the food and watch Tom Glavine pitch against the Braves. The home team lost, but the food won me over. FULL STORY Rangers look to leave mark Survive and advance. That is what eight months of hockey - from training camp exercises at West Point to yesterday's team meeting at the MSG Training Center in Greenburgh - will come down to when the Rangers take the ice tomorrow afternoon. FULL STORY As Rosie joins, Star may take a dim 'View' Rush & Molloy: Rosie O'Donnell is about to join The View, according to our sources at ABC. Barbara Walters will deliver the news tonight as she presents a Daytime Emmy Award, they add. --------------------------------------------------------- When ShortJohnny first came to the United States through Area 51, this is the newspaper from which he/it learned English. It was suggested to it/him by a Marine Captain with a Major sense of humor but - because of ShortJohnny's lack of a sense of humor - ShortJohnny never got the joke. Instead, ShortJohnny /became/ the joke. That's one of the reasons that ShortJohnny is mistaken so often for President George W C Bush43: unfamiliarity with the English language and inability to figure out a Wal*Mart receipt with two items on it. And, of course, that raging, stage three syphillis that they share makes positive identification of Georgie and ShortJohnny not open improbable but downright impossible. Just remember: George W C Bush43 is more intelligent than Short Johnny but only by sixty or seventy IQ points. Gray Shockley ------------------ It is better to live on your feet than to die on your knees. - Robert Anson Heinlein === Subject: MetaLogic & Tarski's Undefinability Of Truth Theorem Tarski's undefinability of truth theorem proves that we can't define truth within mathematics because it is impossible to construct the set of all truths. But we can still construct a set of all the truths in mathematics using a meta-system. What is so amazing about this is you can also make your own meta-system and define truth in mathematics to be whatever you want. This proves the theory that beliefs create reality unless you believe they don't, in which case they still do. To put it another way, metasystems of mathematics create truth in mathematics, unless they aren't defined to do so, in which case they still do. This theorem extends beyond mathematics into philosophy and science. Lets look at the English language as a something like mathematics so we can see that truth isn't definable in it either. With English by itself you can obviously make sentences with correct grammar that are not true. But if we define a simple meta-system of science, which uses english we can define truth. Just make a list of axioms to define science, some of which are proofs and others disproofs. The list should tell us if any sentence in English is true and if it isn't listed, then it simply is unproven. By checking to see if a statement satisfies our list of axioms we define truth. This is almost exactly how Tarski shows how to define truth in mathematics, using a more refined metasystem. What I think you should find interesting about this however is that if we create a list of all the possible proofs and disproofs in the system then a statement like this proof is false will cause a conflict, even in our metasystem of axioms. Because in order for our system to be rich enough to talk about all the proofs in the English language it will have to talk about itself. A statement like Axiom X in our system is true would have to be listed as an axiom that was either proven or disproven. Even in mathematics we can formulate liar sentences that lead to the indefinably of truth, and that is what Tarski originally showed. We will have to create a metasystem to define our axioms as being true or false, and to do that we have to create another system that tells us when true sentence in our system are true. We can say they are true if they satisfies our axioms, and now if in our original system we have an axiom that defines a sentence such as, This proof is false, in our new metasystem we would test to see if the statement actually satisfies the axiom. Because it doesn't, and because it always contradicts the axiom we can now define it as false. This works because the sentence now reads This proof is false in system A and system B defines truth in system A in a special way, which checks for contradictions. Since the statement doesn't address if it is true or not in system B we can simply show that it is not listed as a truth in A, and in system B there is no contradictions. A regular example could be something like Birds fly south in the winter in system A. The system checks to see if this satisfies our lists of truths, and finds out that it is in the list and that it doesn't contradict anything in system A, and is therefore true. But what is so obviously interesting about how we need meta-systems to define truth in any system rich enough to talk about itself, is that we will naturally have other metasystems that define truth to be different within that system. If we create a sentence such as All sentences which define truth for A in B are true, it opens the door to creating liar sentences within the metasystem, and also contradict the truth of A within that system. This is all we need to show that because of this recursion, we will need a richer meta-system to prove the truth of B. What this means is that we can say that A is still true for B, but we can't say that B can universally define truth for itself. So naturally we can assume that another metasystem is equally valid if it defines the truth of A, and doesn't need to strictly conform to the truth in B. Because this kind of recursion extends infinitely, we can still assume that any system we adopt is true because there is always a richer metasystem we can use to define truth. This would suggest that perhaps our original meta-system accurately reflects the real truth of our original set of axioms, but because we can never reach an infinite number of metasystems to prove this it is really hard to say. In creating our metasystem we have redefined the way our original system behaves, and in doing so opened the door to redefining the definition of truth in other ways. === Subject: Re: MetaLogic & Tarski's Undefinability Of Truth Theorem > Tarski's undefinability of truth theorem proves that we can't define > truth within mathematics because it is impossible to construct the set > of all truths. Tarski's theorem says that it is not possible to define the set of true arithmetical sentences within the standard model of arithmetic. It does not say anything about defining truth within ``mathematics.'' === Subject: Re: MetaLogic & Tarski's Undefinability Of Truth Theorem > Tarski's theorem says that it is not possible to define the set of true > arithmetical sentences within the standard model of arithmetic. It says more than that. It shows that no language closed under contradictory negation for which the diagonal lemma holds can contain a truth predicate for the whole of that language; the T-scheme is inconsistent, as Jeffrey Ketland would probably put it. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: MetaLogic & Tarski's Undefinability Of Truth Theorem arithmetical sentences within the standard model of arithmetic. > It says more than that. It shows that no language closed under > contradictory negation for which the diagonal lemma holds can contain a > truth predicate for the whole of that language; the T-scheme is > inconsistent, as Jeffrey Ketland would probably put it. I would like to know what you're talking about, since I think that there is something interesting here. But I can't understand what you're saying. Let's use the standard terminology from mathematical logic. A language is a set of symbols -- on its own, it has no semantics and can contain whatever it wants. So your claim above doesn't fit this terminology. Tarski's theorem says something about the undefinability of truth predicates in some class of models. The standard formulation is the one about the standard model of arithmetic in its standard language in first order logic. I believe that there should be a more general theorem that can be stated, but I don't know a precise statement. I would appreciate it if you could give one. === Subject: Re: MetaLogic & Tarski's Undefinability Of Truth Theorem >>[Tarski's theorem] says more than that. It shows that no language closed under >>contradictory negation for which the diagonal lemma holds can contain a >>truth predicate for the whole of that language; the T-scheme is >>inconsistent, as Jeffrey Ketland would probably put it. > I would like to know what you're talking about, since I think that > there is something interesting here. But I can't understand what > you're saying. Let's use the standard terminology from mathematical > logic. My terminology was at least relatively standard, but perhaps the following will make things clear to you. We're of course talking about interpreted languages here and assume that the language has variables and open formulae. A language is closed under contradictory negation if for every sentence A there is a sentence ~A which is true just in case A is not true. To formulate the diagonal lemma we assume that some indexing of the syntactical elements (sentences, formulas, terms, etc) is given, and that the indices are directly referrable in the language, that is there is a term for each index. The diagonal lemma says that for every formula A(x) with a free variable x there is a sentence B, the fixed point of A(x), such that A(i) <--> B where i is the index of B. Tarski's result is now that if a language is closed under contradictory negation and the diagonal lemma holds for it, it can't contain a formula T(x) such that all instances of the T-scheme T(i) <--> A_i are true. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: MetaLogic & Tarski's Undefinability Of Truth Theorem following will make things clear to you. > We're of course talking about interpreted languages here and assume that > the language has variables and open formulae. A language is closed under > contradictory negation if for every sentence A there is a sentence ~A > which is true just in case A is not true. To formulate the diagonal > lemma we assume that some indexing of the syntactical elements > (sentences, formulas, terms, etc) is given, and that the indices are > directly referrable in the language, that is there is a term for each > index. The diagonal lemma says that for every formula A(x) with a free > variable x there is a sentence B, the fixed point of A(x), such that > A(i) <--> B where i is the index of B. Tarski's result is now that if a > language is closed under contradictory negation and the diagonal lemma > holds for it, it can't contain a formula T(x) such that all instances of > the T-scheme > T(i) <--> A_i > are true. Tarski's theorem... has that actually appeared in print? (I would like to know who to blame...) The usual proof of Tarski's theorem that I have seen does not go through Godel's diagonal lemma. Here is the proof I am used to. Suppose that the truth predicate of arithmetic were arithmetically definable; then it would be Sigma^0_n for some n. Thus every arithmetically definable set would be Sigma^0_n. But there are Sigma^0_{n+1} sets that are not Sigma^0_n. This is a contradiction. === Subject: Re: MetaLogic & Tarski's Undefinability Of Truth Theorem >Tarski's theorem says that it is not possible to define the set of true >arithmetical sentences within the standard model of arithmetic. >>It says more than that. It shows that no language closed under >>contradictory negation for which the diagonal lemma holds can contain a >>truth predicate for the whole of that language; the T-scheme is >>inconsistent, as Jeffrey Ketland would probably put it. > I would like to know what you're talking about, since I think that > there is something interesting here. But I can't understand what > you're saying. Let's use the standard terminology from mathematical > logic. A language is a set of symbols It is? My sources say a (finite) set of symbols is an 'alphabet', and a language is a set of strings over an alphabet. > -- on its own, it has no > semantics and can contain whatever it wants. So your claim above > doesn't fit this terminology. Try Googling some of his terms along with Tarski's theorem. He's not just makin' em up. -- --Bryan === Subject: Re: MetaLogic & Tarski's Undefinability Of Truth Theorem It is? My sources say a (finite) set of symbols is an 'alphabet', > and a language is a set of strings over an alphabet. In the sense of mathematical logic, a language is a set of constant symbols, a set of relation symbols, and a set of function symbols. These together form the language, which is also called a signature when you talk about models. This definition is useful when you want to discuss the expressiveness of certain axiomatic systems. In the sense of computer science, especially in compiler theory, an alphabet is a set of symbols and a language is the set of finite strings on those symbols. This definition is useful when you want to discuss which classes of algorithms are capable of recognizing the strings from a particular language. My goal was to get a description in the natural language I know, the one of mathematical logic. > Try Googling some of his terms along with Tarski's theorem. > He's not just makin' em up. He's not usin' em in da usual manner, neither. === Subject: Re: MetaLogic & Tarski's Undefinability Of Truth Theorem >> Tarski's theorem says that it is not possible to define the set of true >> arithmetical sentences within the standard model of arithmetic. > It says more than that. It shows that no language closed under > contradictory negation for which the diagonal lemma holds can contain a > truth predicate for the whole of that language; the T-scheme is > inconsistent, as Jeffrey Ketland would probably put it. Well, that sure clears things up for me. === Subject: Re: MetaLogic & Tarski's Undefinability Of Truth Theorem >>It says more than that. It shows that no language closed under >>contradictory negation for which the diagonal lemma holds can contain a >>truth predicate for the whole of that language; the T-scheme is >>inconsistent, as Jeffrey Ketland would probably put it. > Well, that sure clears things up for me. Great, I strive for clarity. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, dar.9fber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: MetaLogic & Tarski's Undefinability Of Truth Theorem >It says more than that. It shows that no language closed under >contradictory negation for which the diagonal lemma holds can contain a >truth predicate for the whole of that language; the T-scheme is >inconsistent, as Jeffrey Ketland would probably put it. >> Well, that sure clears things up for me. > Great, I strive for clarity. This is why I like mathematicians. None of that fancy jargon. Get straight to the point. === Subject: Re: Red, White and Blue > Is this meant to be the same problem as was discussed at > > or is the behaviour different on the first turn? > Sorry, I meant different on the first few turns, not specifically on > the first turn - where all are R anyway. > > Hi Matt, > It's the same problem, if you recall not one one math.rec could solve it. > I thought things were stated rather clearly --- but let me try again. > On any turn, if R replace by W, if W replace by B, if B discard. > Best wishes, Jim > It looks like you could set up a system of linear equations to solve > this. > Let r(a,b,c) be the probability that when you start this procedure with > a red marbles, b white marbles, and c blue marbles, the last marble > drawn is red; and define w(a,b,c) similarly. (Clearly the blue > probability will be one minus the sum of the other two.) > Then > r(1,0,0) = 1, r(0,1,0) = 0, r(0,0,1) = 0, > w(1,0,0) = 0, w(0,1,0) = 1, w(0,0,1) = 0. > Then we need a recursion for r(a,b,c) when a+b+c >= 2. With probability > a/(a+b+c), the state changes to (a-1,b+1,c), with probability > b/(a+b+c), it changes to (a,b-1,c+1), and with probability c/(a+b+c), > it changes to (a,b,c-1). Thus > r(a,b,c) = (a r(a-1,b+1,c) + b r(a,b-1,c+1) + c r(a,b,c-1))/(a+b+c) > and > w(a,b,c) = (a w(a-1,b+1,c) + b w(a,b-1,c+1) + c w(a,b,c-1))/(a+b+c) > Then you want the value of r(n,0,0), w(n,0,0), and one minus the sum. > In Maple, the necessary code is > RHS := (a,b,c,v) -> if(a+b+c=1) then if a=1 and v=r then 1 else if b=1 > and v=w then 1 else 0 fi fi else (a*v(a-1,b+1,c) + b*v(a,b-1,c+1) + > c*v(a,b,c-1))/(a+b+c) fi; > E := (a,b,c,v) -> (v(a,b,c) = RHS(a,b,c,v)); > stage := proc (N,v) local s,i,j,k; s := {}: for i from 0 to N do for j > from 0 to N-i do for k from 0 to N-i-j do if (i+j+k > 0) then s := s > union {E(i,j,k,v)} fi end do end do end do; s end proc: > vars := proc (N,v) local s,i,j,k; s := {}: for i from 0 to N do for j > from 0 to N-i do for k from 0 to N-i-j do if (i+j+k > 0) then s := s > union {v(i,j,k)} fi end do end do end do; s end proc: > marbles := N -> {solve(stage(N,r),vars(N,r)), > solve(stage(N,w),vars(N,w)) }; > And marbles(n) will find the values of r(n,0,0) and w(n,0,0) (among > others). > marbles(1) gives: r = 1, w = 0 (and b = 0) > marbles(2) gives: r = 1/4, w = 3/8 (and b = 3/8) > marbles(3) gives: r = 17/108, w = 233/648 (and b = 313/648) > marbles(4) gives: r = 6709/55296, w = 226097/663552 (and b = > 356947/663552) > marbles(5) gives: r = 439937429/4320000000, w = > 84443905877/259200000000 > marbles(6) gives: r = 3478886791/38880000000, w = > 2197473298949/6998400000000 > decimal versions: > 2 .2500000000 .3750000000 > 3 .1574074074 .3595679012 > 4 .1213288484 .3407374252 > 5 .1018373678 .3257866739 > 6 .0894775409 .3139965276 > Yes, that reassuringly agrees with the numerical answers I got when I > looked at this a while ago, which in full were: > N Red White Blue > --- ------------ ------------ ------------ > 2 0.2500000000 0.3750000000 0.3750000000 > 3 0.1574074074 0.3595679012 0.4830246914 > 4 0.1213288484 0.3407374253 0.5379337264 > 5 0.1018373678 0.3257866739 0.5723759583 > 6 0.0894775409 0.3139965276 0.5965259315 > 7 0.0808561039 0.3044736591 0.6146702370 > 8 0.0744504075 0.2965914636 0.6289581289 > 9 0.0694728560 0.2899288924 0.6405982516 > 10 0.0654736637 0.2841981319 0.6503282043 > 20 0.0465875531 0.2514676593 0.7019447876 > 30 0.0393778745 0.2357370657 0.7248850598 > 40 0.0353184655 0.2258098506 0.7388716840 > 50 0.0326260179 0.2187193375 0.7486546446 > 60 0.0306693978 0.2132805749 0.7560500273 > 70 0.0291620324 0.2089103939 0.7619275737 > 80 0.0279527398 0.2052821790 0.7667650812 > 90 0.0269532652 0.2021960393 0.7708506955 > 100 0.0261081714 0.1995214439 0.7743703847 > 200 0.0214893579 0.1837084383 0.7948022038 > 300 0.0193745213 0.1756774446 0.8049480341 Hi Chris and Matt, Best wishes, Jim === Subject: Re: Red, White and Blue Is this meant to be the same problem as was discussed at > > or is the behaviour different on the first turn? > Sorry, I meant different on the first few turns, not specifically on > the first turn - where all are R anyway. > > Hi Matt, > It's the same problem, if you recall not one one math.rec could solve > it. > I thought things were stated rather clearly --- but let me try again. > On any turn, if R replace by W, if W replace by B, if B discard. > Best wishes, Jim > It looks like you could set up a system of linear equations to solve > this. > Let r(a,b,c) be the probability that when you start this procedure with > a red marbles, b white marbles, and c blue marbles, the last marble > drawn is red; and define w(a,b,c) similarly. (Clearly the blue > probability will be one minus the sum of the other two.) > Then > r(1,0,0) = 1, r(0,1,0) = 0, r(0,0,1) = 0, > w(1,0,0) = 0, w(0,1,0) = 1, w(0,0,1) = 0. > Then we need a recursion for r(a,b,c) when a+b+c >= 2. With probability > a/(a+b+c), the state changes to (a-1,b+1,c), with probability > b/(a+b+c), it changes to (a,b-1,c+1), and with probability c/(a+b+c), > it changes to (a,b,c-1). Thus > r(a,b,c) = (a r(a-1,b+1,c) + b r(a,b-1,c+1) + c r(a,b,c-1))/(a+b+c) > and > w(a,b,c) = (a w(a-1,b+1,c) + b w(a,b-1,c+1) + c w(a,b,c-1))/(a+b+c) > Then you want the value of r(n,0,0), w(n,0,0), and one minus the sum. > In Maple, the necessary code is > RHS := (a,b,c,v) -> if(a+b+c=1) then if a=1 and v=r then 1 else if b=1 > and v=w then 1 else 0 fi fi else (a*v(a-1,b+1,c) + b*v(a,b-1,c+1) + > c*v(a,b,c-1))/(a+b+c) fi; > E := (a,b,c,v) -> (v(a,b,c) = RHS(a,b,c,v)); > stage := proc (N,v) local s,i,j,k; s := {}: for i from 0 to N do for j > from 0 to N-i do for k from 0 to N-i-j do if (i+j+k > 0) then s := s > union {E(i,j,k,v)} fi end do end do end do; s end proc: > vars := proc (N,v) local s,i,j,k; s := {}: for i from 0 to N do for j > from 0 to N-i do for k from 0 to N-i-j do if (i+j+k > 0) then s := s > union {v(i,j,k)} fi end do end do end do; s end proc: > marbles := N -> {solve(stage(N,r),vars(N,r)), > solve(stage(N,w),vars(N,w)) }; > And marbles(n) will find the values of r(n,0,0) and w(n,0,0) (among > others). > marbles(1) gives: r = 1, w = 0 (and b = 0) > marbles(2) gives: r = 1/4, w = 3/8 (and b = 3/8) > marbles(3) gives: r = 17/108, w = 233/648 (and b = 313/648) > marbles(4) gives: r = 6709/55296, w = 226097/663552 (and b = > 356947/663552) > marbles(5) gives: r = 439937429/4320000000, w = > 84443905877/259200000000 > marbles(6) gives: r = 3478886791/38880000000, w = > 2197473298949/6998400000000 > decimal versions: > 2 .2500000000 .3750000000 > 3 .1574074074 .3595679012 > 4 .1213288484 .3407374252 > 5 .1018373678 .3257866739 > 6 .0894775409 .3139965276 > Yes, that reassuringly agrees with the numerical answers I got when I > looked at this a while ago, which in full were: > N Red White Blue > --- ------------ ------------ ------------ > 2 0.2500000000 0.3750000000 0.3750000000 > 3 0.1574074074 0.3595679012 0.4830246914 > 4 0.1213288484 0.3407374253 0.5379337264 > 5 0.1018373678 0.3257866739 0.5723759583 > 6 0.0894775409 0.3139965276 0.5965259315 > 7 0.0808561039 0.3044736591 0.6146702370 > 8 0.0744504075 0.2965914636 0.6289581289 > 9 0.0694728560 0.2899288924 0.6405982516 > 10 0.0654736637 0.2841981319 0.6503282043 > 20 0.0465875531 0.2514676593 0.7019447876 > 30 0.0393778745 0.2357370657 0.7248850598 > 40 0.0353184655 0.2258098506 0.7388716840 > 50 0.0326260179 0.2187193375 0.7486546446 > 60 0.0306693978 0.2132805749 0.7560500273 > 70 0.0291620324 0.2089103939 0.7619275737 > 80 0.0279527398 0.2052821790 0.7667650812 > 90 0.0269532652 0.2021960393 0.7708506955 > 100 0.0261081714 0.1995214439 0.7743703847 > 200 0.0214893579 0.1837084383 0.7948022038 > 300 0.0193745213 0.1756774446 0.8049480341 > Hi Chris and Matt, > Best wishes, Jim Still no closed form solution though! === Subject: Re: Red, White and Blue Is this meant to be the same problem as was discussed at > > or is the behaviour different on the first turn? > Sorry, I meant different on the first few turns, not specifically on > the first turn - where all are R anyway. > It's the same problem, if you recall not one one math.rec could solve > Still no closed form solution though! Looking back at the thread, I see that I did in fact give a closed form solution in terms of a summation. How are you defining a closed form solution ? === Subject: Re: Harvey Friedman on Cantorian pseudomathematics your NAFL theory? What does it achieve that is new? Does it simplify > things, or make things more intuitively appealing? Does it expose some > flaw in conventional mathematical reasoning? (Or is it just an > exploration of a new formalism, for the sake of exploring new > formalisms, which is certainly OK) > NAFL gives, in my view, a convincing metamathematical proof > for why one ought to reject the existence of infinite sets. That's tough for me to understand. I've argued that all properties of the infinite must derive from the notion of limits, and hence infinite sets exists only in the sense that arbitrarily large sets exist. Thus the notion of existence for infinite sets is an extension of the notion of existence for finite sets (i.e. infinite sets don't exist in exactly the same sense that finite sets exist, but it is still convenient to say that they do exist, as long as it is understood what that means). I like to quote Herman Weyl: ...classical logic was abstracted from the mathematics of finite sets and their subsets...Forgetful of this limited origin, one afterwards mistook that logic for something above and prior to all mathematics, and finally applied it, without justification, to the mathematics of infinite sets. This is the Fall and original sin of set theory ... (Hermann Weyl) > Since NAFL is a new logic that restricts classical infinitary > reasoning, one has to take a fresh look at a host of classical results > and re-examine it from the NAFL viewpoint. Indeed, for example, > Cantor's diagonal argument cannot be formulated in any form in NAFL, One argument that is similar to Cantor's argument is this: given a well defined list of well defined real numbers (so that every digit of every real number can be computed), the diagonal method gives us a new number not on that list. I hope that argument works in NAFL. > Other striking results of NAFL are that Euclid's fifth postulate must > be provable from the first four and hence Euclidean geometry reigns, > while non-Euclidean geometries/relativity theories are rejected. I don't get that. The geometry where the points are the points on the surface of a sphere is a non-Euclidean geometry. What are you saying about that? > The most exciting, positive aspects of NAFL are that it conceptually > justifies some weird results of quantum mechanics, such as, quantum > superposition and quantum entanglement (see the references in the above > paper). That might be interesting. If your theory gives us an intuitively itself in a double slit experiment, then I (and many others) would sure like to hear it. > I am sure you will find NAFL very interesting, given your own stand. Maybe. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics >> Correct. But more effective is: a good -alternative-. I don't think the >> scholastics could be convinced by any argument that what they were doing >> was fantasy. The reason their activities stopped is that later >> developments made them redundant. > First thing is to not use emotionally charged language such as [...] Let me try again: What makes current set theory less of an unscientific speculation than The fact that a large group of learned men enjoy/enjoyed writing and reading the stories for 130+ years doesn't count; see the middle ages. What arguments are then left over? (I don't say such an argument doesn't exist. Just that i fail to see one, right now.) NB: my question was not rhetorical. What arguments are there? Perhaps someone knows a few good ones; i'd be all ear. Also note: with 'current set theory' i meant ZFC or NBG. These systems assume a lot more than merely the existence of infinite sets such as N or R. It claims there is P(P(P(N))), among other things. And much weirder things, too. For what it's worth: In my opinion, these systems don't contain 'indisputable knowledge' about infinite sets, but -assumptions-. And these assumptions are chosen for a reason, and they serve a purpose, namely to allow a smooth development of advanced fields of mathematics, without too much fuss. Perhaps also to allow quick an easy communication of complex ideas and constructions among colleagues. But are they really convincing as the final and definitive ultimate foundation for mathematics? -- Herman Jurjus === Subject: Re: Harvey Friedman on Cantorian pseudomathematics On Fri, 28 Apr 2006 20:44:38 +0200, Herman Jurjus said: > ... > Also note: with 'current set theory' i meant ZFC or NBG. These systems > assume a lot more than merely the existence of infinite sets such as N > or R. It claims there is P(P(P(N))), among other things. And much > weirder things, too. So you have no problem with sets like N and R. Do you have any problem with the idea of a subset of N like the set of prime numbers? No? Great, so we can reasonably say that there are subsets of N (other than N itself). But now that we're this far, what principled reason can you provide for denying that there is a set P(N) of all such subsets? I mean, they are all *there*, right? But once we have that set, exactly the same sort of reasoning leads to the conclusion that the set of all subsets of P(N) exists. And so on. Granted, you can legitimately take issue with the reasoning that justifies the power set axiom -- you'd hardly be the first -- but to dismiss power sets without argument as weird, as if it is all mumbo-jumbo and hocus-pocus, well, that's just stupid. > For what it's worth: In my opinion, these systems don't contain > 'indisputable knowledge' about infinite sets, but -assumptions-. Of course they are assumptions. Who claims otherwise? In particular, who claims that, say, infinity and powerset, are indisputable? > And these assumptions are chosen for a reason, and they serve a > purpose, namely to allow a smooth development of advanced fields of > mathematics, without too much fuss. Perhaps also to allow quick an > easy communication of complex ideas and constructions among > colleagues. > But are they really convincing as the final and definitive ultimate > foundation for mathematics? Maybe not. Who says set theory is the final, definitive, ultimate foundation for mathematics? Most mathematicians and logicians I know don't go much beyond saying simply that it does the job. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics > Do a bit of reading about the philosophy called solipsism > (which, so far as I can understand, is utterly irrefutable > even in concept) >> Actually, it's pretty easy to refute, assuming that you're not >> an infinite being. >> Did you always exist? No. Do you exist now? Yes. But facts are >> timeless and eternal. No, they're not, if solipsism is true. They would only exist as > long as the solipsist exists. So there goes your proof down the > drain. > That which is timeless is without time. Because time is irrelevant > to that which is timeless, it is meaningless to say anything about > something that's timeless in conjunction with time. > How many meters long is pi? When you understand why that's a > meaningless question, you'll understand why ``When is pi?'' is > equally meaningless. And when you realize that pi is the ratio of the circumference of a circle to it's diameter and that it can't exist if a circle can't exist, then you'll understand why pi exists even if nothing else did is meaningless. Pi is a ratio of two things. If the two things don't exist, then pi doesn't exist as that ratio. Without a circle existing, there's nothing to relate pi to (well, there's other math that also uses pi but the principle stands that there must be something in existence for pi to exist/have meaning.) It's much like your information can't exist without matter to hold it. -- Mike ------------------------------- Our enemies are innovative and resourceful, and so are we. They never stop thinking about new ways to harm our country and our people, and neither do === Subject: Re: ==off topic: choosing Caesar's wife as role model== (barely still): Harvey Friedman on Cantorian pseudomathematics [...] > I was only responding to a request by David Petry > to outline my work. I am a very infrequent poster > to sci.logic, as you can easily see for yourself. > Umm, I rather doubt that David has asked you to > explain your work in this thread nearly as many > times as you have in fact done so by conveniently > finding parallels to your work in other parts of the > discussion. There was a specific post by David Petry with a number of questions including a request to outline what is new in NAFL, why people ought to be interested in it, etc. I was responding to this specific request. Of course, ideally, everybody can look up the archives and get all the information, but few would have the interest or energy to do so. E.g. if I had replied to Petry asking him merely to look up the archives, he wouldn't have done so. [...] >> But then, some of us are busy here defending your >> right, against David Petry's urging to do only >> useful mathematics, to do exactly what you are >> doing, though perhaps not so obsessively. >> We just aren't promising to be interested in the >> results. > Again, that is your choice. But I would expect > logicians to be interested in a new logic, > philosophers to be interested in a new philosophy, > foundationalists to be interested in new > foundations, if only for purely intellectual > reasons. > In my opinion, experts ought to look very > carefully into my claims (which are simple enough > by their standards) and see first of all if they > hold up under scrutiny, and then study the > logical/philsophical/foundational implications. > Oh, my are those two sentences cut right out of > Usenet Kook Central cloth, warp and weft. > Everybody should be interested in what I've > written. Kindly take another look at what I have written above. Take another look at the list of claims that I have made in reply to David Petry. It is just human nature to expect that when you have come up with what you consider to be significant work, there ought to be a reaction to it. Are you saying that just because people refuse to acknowledge that my work exists, I am somehow supposed to infer that my work is no good? And that if I don't agree with this thesis, I am a kook? If and when my work is published, people ought to refer to it. For logics, (s)he ought to include my logic NAFL too. Ditto with philosophers, foundationalists. Physicists and mathematicians are more concerned with their own specializations, and may not immediately react to my work. I don't need to make a song and a dance about my work; neither do I have to be a professor at a leading university and take students to push my ideas further etc, in order to have people refer to my work. Once upon a time, communication of new ideas was a tedious and dreadfully slow process, but not any more. Today no one can credibly feign ignorance of new work in their field that is openly available. > Well, no. No matter what the subject matter, most > mathematicians and logicians and philosophers (among > other well educated humans, and indeed among all > humans) are so narrowly specialized in what they do > and know how to do, they'd die in screaming boredom > reading almost _anything_ attempting a paradigm > shift very far outside their own specialty. Are you justifying this state of affairs? Study of new paradigms is part of the business of foundationalists, logicians and philosophers. If they would die in screaming boredom when asked to look at new paradigms proposed in their fields, then there is something wrong with that attitude. That is my opinion, of course. And it doesn't make me a kook. I also suggest that you look at the claims of Gregory Chaitin and compare these to mine. Chaitin's digital philosophy rejects the existence of real numbers, so he has rejected all of modern physics, including relativity theory, and much of modern mathematics as well. And Chaitin admitted in a recent invited talk at a conference that his digital philosophy is fundamentally at odds with quantum mechanics. In contrast, I have shown how real analysis can be accomplished in NAFL, and also shown how NAFL justifies some of the weird aspects of quantum mechanics. So do you think my work deserves some reaction, say, 10% of the reaction that Chaitin gets from experts? (By the way, Chaitin also complains from time to time that his claims have been ignored by the logic/foundations community; but of course, he doesn't even reply to my emails, let alone react to my claims). > Experts should spend their precious time on what > I've written. > Well, perhaps, but that's _their_ choice of how to > distribute their attention, not yours, so you > shouldn't give so strong an impression that you are > trying to _impose_ such a choice by constant > drumming repetition of your calls for attention. Attention? No. A review/discussion of my work? Yes. To expect people to review my work and comment on it is reasonable, given the depth and breadth of the claims made. The availability of eprint archives and various newsgroups on the internet has made it that much easier for people to get to see what I have done and react to it. Refusal to do so is very odd behaviour. I don't mean to say that every Tom, Dick and Harry should react to my work. What I find astonishing is the universal refusal to even acknowledge that my work exists, even in informal forums like Usenet. In the face of this determined, united stand, I am supposed to wither away and fade out. Well, I don't intend to oblige. I am determined to keep going for as long as I can, and for as long as I am convinced that I am doing good work. > I am wondering why such an objective, professional > outlook seems to be fast disappearing from the > world of science and research these days. > Make that three sentences right out of Kook Central. > Anyone not interested in my work is either > unobjective, unprofessional, or both. > Well, no, not interested doesn't carry any > negative connotations whatsoever. It stands only for > itself. > Several well known kooks frequenting these > newsgroups have made that identical statement at > least meme for meme, if not word for word, many > times each. It doesn't look so pretty as I have > restated it, does it. People do reply to these alleged kooks and explain what they think is wrong with their claims. So far, no one has responded to any of my eprints on the arXiv and Philsci Archive, or to my published conference paper, or to my Usenet posts on NAFL. > Perhaps the professionals are *too* busy doing > their own thing and *too* deeply enmeshed in their > own ideologies/philosophies/pet theories to > seriously consider radically new alternatives. > [You might want to avoid semantic overfreighting > when plumping your work, another kook trademark.] I think what is tiresome is your repeated attempts to brand people as kooks, cranks, etc. Usenet is part of reality, and people will continue to post their stuff and expect experts to react. You are not going to change that with your protests. My work is not just confined to Usenet. It is already published in IJQI as a (reviewed) conference paper and also available in leading eprint archives. You can scream yourself hoarse, but when, for example, a new method for doing real analysis in a new logic (and much more) is proposed and people refuse to react, I definitely find that very odd and inexplicable. Stating this doesn't make me a kook or a crank. > And most of the rest of us as well. > Welcome to the human race. > This race is already in progress, so sit down, > strap in, hold on, and Read, Learn, Evolve before > trying to steer. > HTH > xanthian. === Subject: Re: ==off topic: choosing Caesar's wife as role model== (barely still): Harvey Friedman on Cantorian pseudomathematics > I think what is tiresome is your repeated > attempts to brand people as > kooks, cranks, etc. http://math.ucr.edu/home/baez/crackpot.html http://www.nothingisreal.com/mentifex http://www.advogato.org/person/mentifex/ http://www.scn.org/~mentifex/ If the mirror that I hold up for your use doesn't please you, perhaps one or more of the above will be more to your liking. You cannot seem to differentiate criticism of your research from criticism of your behavior, and respond to the latter as if it were the former. That is quite sad, somehow. And the usual reasons to get no response to your publications are 1) precisely that others see your behavior and decide that it is not the behavior of a person who would create something of value, or 2) that people look at your work and find it not even worth the trouble to criticise it, as being of self evident low worth not meriting comment. I can't fix what's wrong in either case, only you can do that. HTH xanthian. -- === Subject: Re: ==off topic: choosing Caesar's wife as role model== (barely still): Harvey Friedman on Cantorian pseudomathematics <4cc540e15e66186bd41dfe51874127c4.48257@mygate.mailgate.org I think what is tiresome is your repeated > attempts to brand people as > kooks, cranks, etc. > http://math.ucr.edu/home/baez/crackpot.html > http://www.nothingisreal.com/mentifex > http://www.advogato.org/person/mentifex/ > http://www.scn.org/~mentifex/ > If the mirror that I hold up for your use > doesn't please you, perhaps one or more of > the above will be more to your liking. To be frank, I am not really interesetd in crankdom. If I come across as one, I would be slightly disappointed, but not half as disapppointed as I am with the lack of a technical response to my work from the academic community. > You cannot seem to differentiate criticism > of your research from criticism of your > behavior, and respond to the latter as if > it were the former. > That is quite sad, somehow. OK, maybe I am overly frustrated with the lack of response to my work and it shows from time to time. After all I am also under various pressures to justify my work, and so on. I have to remind myself from time to time to be patient and play the waiting game, something I am not good at. Sorry if I over-reacted to your post. > And the usual reasons to get no response to > your publications are 1) precisely that others > see your behavior and decide that it is not > the behavior of a person who would create > something of value, Why should they even consider my behaviour (assuming it is as you say, which I don't agree with)? That is completely irrelevant, given that what I have done is already available in written form. What the situation demands is a straightforward technical response. > 2) that people look at > your work and find it not even worth the > trouble to criticise it, as being of self > evident low worth not meriting comment. I don't buy this. If that is the case, I would expect someone to explicitly say so. I don't have the ability to read minds, and I expect people to be straightforward in airing their opinions, whether good or otherwise. Why be devious and tiresome instead of writing a straightforward response to my claims and say, post it to the arXiv (as has been done in countless other cases)? This would save a lot of time and effort for me as well as for the many others who read the arXiv. In any case, one of my papers is now under review for nearly 10 months in a good mainstream journal and I think the referee(s) may be of a different mind-set than what you describe above. If the paper is eventually accepted, what you have said above stands refuted. Let us wait and see. I agree with you that agonizing on the Usenet is not going to be of much help to me. > I can't fix what's wrong in either case, only > you can do that. were sincere in offering it. === Subject: Re: ==off topic: choosing Caesar's wife as role model== (barely still): Harvey Friedman on Cantorian pseudomathematics > http://www.advogato.org/person/mentifex/ > If the mirror that I hold up for your use doesn't please you, perhaps one > or more of the above will be more to your liking. That is some interesting kookiness. The whole thing seems to span the divide between raving-loony kookie and he-needs-a-rest kookie. I like this quote: During the Singularity Countdown to true artificial intelligence, these diagrams have been released into the public domain and may serve as prior art to help prevent the monopolistic patenting of AI algorithms. He is thinking of the future. And he has produced hundreds of lines of dense ASCII-art diagrams, in preperation for the singularity. This is a 'new percept engram' for me, and it is effecting my 'auditory phonemes'. -- Excess generally causes reaction, and produces a change in the opposite direction, whether it be in the seasons, or in individuals, or in governments. -- Plato (427 - 347 BC) Republic, Book 3 === Subject: Re: ==off topic: choosing Caesar's wife as role model== (barely still): Harvey Friedman on Cantorian pseudomathematics >> http://www.advogato.org/person/mentifex/ > That is some interesting kookiness. The whole > thing seems to span the divide between > raving-loony kookie and he-needs-a-rest kookie. If you knew, as I do, that Arthur T. Murray has been trying to convince other people that he has solved AI and should be taken seriously by 'the experts' for approximately 4 decades, and so far all he's got that is usable by others to show for his claim to be the world's most accomplished AI researcher is > hundreds of lines of dense ASCII-art diagrams you might be more easily able to decide on which side of that divide his mental state lies. FYI xanthian. -- === Subject: Re: Harvey Friedman on Cantorian pseudomathematics Discussion, linux) <444cf6ef$0$2023$ba620dc5@text.nova.planet.nl> <444de616$0$2028$ba620dc5@text.nova.planet.nl> <685$44508b2b$82a1e228$23720@news1.tudelft.nl My English is much better than your Dutch BTW. You know, you say this an awful lot. Now, I understand it when you say it to me. After all, I've spent the last five years in your country and my Dutch skills are abysmal. Shameful even. But do you really think that Kent *should* know a damn thing about Dutch? Don't you think that there are a few languages that are probably higher on his list? Why keep pointing out that the majority of the world doesn't speak your native language as well as you speak English? What the hell is the point of that comparison? -- Jesse F. Hughes I am the barbarian at the gates. I am a revolutionary, a discoverer, a guy who didn't just try, but did, who didn't just wonder, but accomplished. -- James S. Harris gives Hollywood its tagline === Subject: ==off topic: polyglotism==: Harvey Friedman on Cantorian pseudomathematics <444cf6ef$0$2023$ba620dc5@text.nova.planet.nl> <444de616$0$2028$ba620dc5@text.nova.planet.nl> <685$44508b2b$82a1e228$23720@news1.tudelft.nl> <87ejzjt371.fsf@phiwumbda.org> My English is much better than your Dutch BTW. > You know, you say this an awful lot. Now, I understand it when you > say it to me. After all, I've spent the last five years in your > country and my Dutch skills are abysmal. Shameful even. If you can ask directions on the street, they suffice. > But do you really think that Kent *should* know a damn thing about > Dutch? Don't you think that there are a few languages that are > probably higher on his list? Well, aside from learning over twelve dozen computer languages, and employing each in ways that helped me earn a living, I can stumble around a bit in Latin, Spanish, German, and American Sign Language, having taken among the lot about 16 formal courses of study. The German allows me to read some tiny few phrases in Dutch (as very ugly German), but since I'm not even fluent in any of my second human languages (any more), I'm certainly not fluent in Han's native tongue. But even if I'd studied Dutch for twenty years, I wouldn't be so egocentric as to try to correct the usage of a native speaker of a language in which I am merely a visitor. Han, on the other hand, very much is that egocentric, as demonstrated by his willingness to do exactly that misdeed.. > Why keep pointing out that the majority of the world doesn't speak > your native language as well as you speak English? What the hell is > the point of that comparison? That's an example of handwaving to distract the audience from the original subject matter. Han doesn't seem to be a particularly honest person, and his discussion style here frequently employs unacceptable tactics such as this one. One can only hope his science work product doesn't matter particularly to anyone. A corrupt scientist is extremely dangerous to his discipline, colleagues, and neighbors.. xanthian. === Subject: Re: ==off topic: polyglotism==: Harvey Friedman on Cantorian pseudomathematics > But even if I'd studied Dutch for twenty years, I wouldn't be so > egocentric as to try to correct the usage of a native speaker of a > language in which I am merely a visitor. That's irrelevant because the meaning of abstract is not typical for the English language. Again: abstract is the same word, with the same (Latin) roots, in Dutch and many, many other languages. > Han, on the other hand, very much is that egocentric, as > demonstrated by his willingness to do exactly that misdeed.. Entirely off-topic. > That's an example of handwaving to distract the audience from the > original subject matter. Han doesn't seem to be a particularly honest > person, and his discussion style here frequently employs unacceptable > tactics such as this one. One can only hope his science work product > doesn't matter particularly to anyone. A corrupt scientist is extremely > dangerous to his discipline, colleagues, and neighbors.. Entirely off-topic. And insulting as well. But what else than a violent response can you expect from USA military, huh? Han de Bruijn === Subject: Re: ==off topic: polyglotism==: Harvey Friedman on Cantorian pseudomathematics >> But even if I'd studied Dutch for twenty years, I wouldn't be so >> egocentric as to try to correct the usage of a native speaker of a >> language in which I am merely a visitor. > That's irrelevant because the meaning of abstract is not typical > for the English language. Again: abstract is the same word, with > the same (Latin) roots, in Dutch and many, many other languages. There is a serious problem with this approach to meanings of words. Meanings change over time. For example, the first three meanings of abstract as an adjective in the Oxford English Dictionary, second edition, are all marked either obs (obsolete) or arch (archaic). That means the word used to have those meanings, but no longer has the obsolete meanings, and is no longer used in normal writing with the archaic meaning. Moreover, at least in English, very few words have exactly one meaning. English words are not defined by their Latin roots, or by the meanings of their cognates in other languages. They are defined, in general, by how speakers of English use them. You can make a rough guess at the meaning of a word from roots or cognates, but it is only a guess. I can guess the meaning of some Spanish words from their Latin roots and/or French cognates, but that is not reliable. If you want to know EXACTLY how a word was used in some specific writing, in any language, the best way to find out is to ask the author for clarification. That is also a very practical way in Usenet discussions. Patricia === Subject: Re: ==off topic: polyglotism==: Harvey Friedman on Cantorian pseudomathematics > But even if I'd studied Dutch for twenty years, I wouldn't be so > egocentric as to try to correct the usage of a native speaker of a > language in which I am merely a visitor. >> That's irrelevant because the meaning of abstract is not typical >> for the English language. Again: abstract is the same word, with >> the same (Latin) roots, in Dutch and many, many other languages. > There is a serious problem with this approach to meanings of words. > Meanings change over time. For example, the first three meanings of > abstract as an adjective in the Oxford English Dictionary, second > edition, are all marked either obs (obsolete) or arch (archaic). > That means the word used to have those meanings, but no longer has the > obsolete meanings, and is no longer used in normal writing with the > archaic meaning. > Moreover, at least in English, very few words have exactly one meaning. > English words are not defined by their Latin roots, or by the meanings > of their cognates in other languages. They are defined, in general, by > how speakers of English use them. You can make a rough guess at the > meaning of a word from roots or cognates, but it is only a guess. I can > guess the meaning of some Spanish words from their Latin roots and/or > French cognates, but that is not reliable. > If you want to know EXACTLY how a word was used in some specific > writing, in any language, the best way to find out is to ask the author > for clarification. That is also a very practical way in Usenet discussions. True. But I want to go back to the roots. Han de Bruijn === Subject: Re: ==off topic: polyglotism==: Harvey Friedman on Cantorian pseudomathematics ... > True. But I want to go back to the roots. So do I, enough so to have a copy of the OED. Just realize that that has nothing to do with the meaning of the word in modern English writing. Patricia === Subject: Re: ==off topic: polyglotism==: Harvey Friedman on Cantorian pseudomathematics > True. But I want to go back to the roots. http://hdebruijn.soo.dto.tudelft.nl/QED/klassiek.htm Han de Bruijn === Subject: Re: Harvey Friedman on Cantorian pseudomathematics >>My English is much better than your Dutch BTW. > You know, you say this an awful lot. Now, I understand it when you > say it to me. After all, I've spent the last five years in your > country and my Dutch skills are abysmal. Shameful even. > But do you really think that Kent *should* know a damn thing about > Dutch? Don't you think that there are a few languages that are > probably higher on his list? > Why keep pointing out that the majority of the world doesn't speak > your native language as well as you speak English? What the hell is > the point of that comparison? Don't you understand that it's *very impolite* if you say that someone who is doing his best to speak *your* language isn't quite succesful in doing so? Worse, to use that as a cheap argument to win a debate ... Han de Bruijn === Subject: Re: Harvey Friedman on Cantorian pseudomathematics > >>My English is much better than your Dutch BTW. You know, you say this an awful lot. Now, I understand it when you > say it to me. After all, I've spent the last five years in your > country and my Dutch skills are abysmal. Shameful even. But do you really think that Kent *should* know a damn thing about > Dutch? Don't you think that there are a few languages that are > probably higher on his list? Why keep pointing out that the majority of the world doesn't speak > your native language as well as you speak English? What the hell is > the point of that comparison? > Don't you understand that it's *very impolite* if you say that someone > who is doing his best to speak *your* language isn't quite succesful in > doing so? Worse, to use that as a cheap argument to win a debate ... > Han de Bruijn So why is HdB being so impolite to JFH? As a cheap argument to win a debate? === Subject: Re: Harvey Friedman on Cantorian pseudomathematics Discussion, linux) which is not a computational model. Say, you seem to have missed a few of my questions. Let me helpfully reprint them here. For simplicity's sake, let me also change from the Busy Beaver function to the halting problem. Is the phrase The halting problem cannot be computed meaningful? If not, then how does it justify what followed in the top sentence? If it is meaningful, then how does one verify its truth? (Alternatively: What computable predictions does it make?) -- Jesse F. Hughes Radicals are interesting because they were considered 'radical' by modern mathematics depends on. --Another JSH history lesson === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <87k69bt5cw.fsf@phiwumbda.org Say, you seem to have missed a few of my questions. Let me helpfully > reprint them here. For simplicity's sake, let me also change from the > Busy Beaver function to the halting problem. > Is the phrase The halting problem cannot be computed meaningful? We've already discussed this question in this thread. Notice that all by itself, the statement The halting problem cannot be computed does not deny the possibility that there is a very short, simple, and efficient algorithm that takes any computer program of less than, say, a zillion squared lines, and tells us whether or not that program halts. Doesn't that suggest to you that that statement has nothing concrete to say about the phenomena we can observe? === Subject: Re: Harvey Friedman on Cantorian pseudomathematics Originator: tchow@lebesgue.mit.edu.mit.edu (Timothy Chow) >It's very hard for me to believe that believing in the existence of a >world of the infinite far beyond the world of computation will lead to >computationally meaningful assertions. Believing in the consistency of >the formalism is another matter. The two are not mutually exclusive. To take a trivial example, believing in a contradiction will, according to classical logic, lead to computationally meaningful assertions (as well as, of course, assertions that are meaningless). So far, you've only been asserting that it's a *necessary* condition, for something to be meaningful, that it have computationally meaningful consequences. You have not been arguing that it's sufficient, nor have I been trying to imply that you have. All I've been saying is that the large cardinal axioms have observable consequences, so you can't rule them out based on your necessary condition. The consistency of a large cardinal axiom also has computationally meaningful assertions, of course. Now, as the example of a contradiction implies anything makes clear, you probably want to assert something stronger, along the lines of, something has to *naturally* or *relevantly* imply some computationally meaningful consequences to be meaningful. This could, of course, lead into a debate as to what constitutes a natural or relevant implication. It is certainly open to you to argue that the implication from the large cardinal axiom to the Pi^0_1 sentence is contrived. However, the implication is a finitely checkable proof in a standard axiomatic system, so this would be a fairly desperate move on your part. >> So are you actively pursuing this goal, beyond arguing on USENET? >No, not for almost 15 years. May I ask why not? >> No. Cantorian set theory has observable consequences, as Friedman's >> examples illustrate. >The alleged consistency of set theory has observable consequences. >Your authoritative assertions don't inspire in me confidence that you >are telling the truth. (I don't doubt that Friedman et al. believe that >what you are saying is true) I am certainly not trying to convince you by sounding authoritative. What would it take to convince you that Cantorian set theory has observable consequences? Do you agree that logicians have successfully captured *the way set theorists talk* about Cantorian set theory in a set of axioms that can be manipulated on a computer? This doesn't commit you to saying that those assertions are meaningful, only that when someone claims to have a proof that X implies Y in Cantorian set theory, you can mechanically check that they have correctly followed the rules of logic in making the inference. If you agree, then you should agree that *in principle* someone could derive a Pi^0_1 statement, or even a Pi^0_0 statement (obtained by setting n to a low number in the Pi^0_1 statement) from ZFC + a large cardinal axiom, and that you could computationally verify that the derivation is logically correct. Yes? If we get this far, then the only question is whether *this particular* inference is logically valid. Here there is certainly room for debate. As I said, Friedman hasn't even published his proof. Even if he had, one could cast doubt on whether the proof could be formalized to the point where you could literally run a computer program to check it. There are people actively working on formalizing mathematics to the point where it is literally checkable by computer, but the state of the art is closer to Jeremy Avigad's formal proof of the prime number theorem, which is considerably simpler than Friedman's theorem. So, there is certainly plenty of leeway for you to deny that Friedman's result is really an instance of X implies Y. However, this again strikes me as a fairly desperate move. >There's something that bothers me about all this. The idea that you >need to invoke the very latest unpublished results of Friedman in order >to argue that my position is invalid is troubling. I'm not in a >position to evaluate those results, but I have to suspect that you are >telling a subtle lie (not in the sense that you know you are telling >the lie). I'm not sure why you say that I need to invoke these results. I could, for example, have invoked earlier results of Friedman along the same lines. He has a paper in the Annals of Math a few years ago, on the necessary use of large cardinals. I invoked the latest one simply because it's the most natural-sounding one available. In any case, what makes you uneasy seems to be large cardinal axiom implies Pi^0_1 sentence. Perhaps you're uneasy because you suspect me of trying to insinutate the opposite implication: Pi^0_1 sentence implies large cardinal axiom. That implication is indeed a lie, and your intuition that something so finite can't imply the existence of a large cardinal is basically correct. That's not what's relevant to the current argument, however. For the purposes of this argument, I'm happy to stick with the *consistency* of the large cardinal axiom, which you seem to have no problems with. Right? The consistency of a large cardinal axiom is computationally meaningful. No problems there, I assume? Then I ask again: The study of Cantorian set theory has led to the discovery of useful, computationally meaningful facts, namely the consistency of large cardinal axioms. Why then do you oppose Cantorian set theory? Would it satisfy you if set theorists obeyed a gag order, docilely agreeing that they will never publicly assert the actual existence of large cardinals, but asserting only that such-and-such statements *follow from the consistency* of large cardinals? That is, they can go ahead doing research as they normally do, as long as they obey a politically correct code of speech? -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <4450d393$0$24622$b45e6eb0@senator-bedfellow.mit.edu> <44515da0$0$560$b45e6eb0@senator-bedfellow.mit.edu> So are you actively pursuing this goal, beyond arguing on USENET? >No, not for almost 15 years. > May I ask why not? As a graduate student, I was unsuccessful in getting anyone at all interested in what I was doing, but I did manage to get a lot of people upset by what I was doing. > What > would it take to convince you that Cantorian set theory has observable > consequences? For starters, you would have to prove to me that you understand what my argument is, before I will believe you have refuted it. > Do you agree that logicians have successfully captured *the > way set theorists talk* about Cantorian set theory in a set of axioms that > can be manipulated on a computer? Yes, or at least, I have no reason to doubt it. > Then I ask again: The study of Cantorian set theory has led to the > discovery of useful, computationally meaningful facts, namely the > consistency of large cardinal axioms. Why then do you oppose Cantorian > set theory? Here's what I'm arguing: Cantorian set theory includes a mythology about a world beyond the world of computation. It is implausible to me that we *need* that mythology to prove things about the world of computation, and there are very good reasons for getting rid of it if possible. > Would it satisfy you if set theorists obeyed a gag order, > docilely agreeing that they will never publicly assert the actual > existence of large cardinals, but asserting only that such-and-such > statements *follow from the consistency* of large cardinals? That is, > they can go ahead doing research as they normally do, as long as they > obey a politically correct code of speech? Of course you're being silly, but it would be a good idea if the set theorists admitted that set theory is not the *foundation* of mathematics. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics Originator: tchow@lebesgue.mit.edu.mit.edu (Timothy Chow) >> What would it take to convince you that Cantorian set theory has >> observable consequences? >For starters, you would have to prove to me that you understand what my >argument is, before I will believe you have refuted it. I *am* trying to understand what your argument is. I am trying to understand what you mean when you deny that Cantorian set theory has observable consequences. I presented an example of what, according to the way most people use those terms, is an example of an observable consequence of Cantorian set theory. I spelled out for you why Cantorian set theory implies such-and-such an observable consequence is itself observable under your computationalist philosophy. Therefore I am puzzled now as to why you seem to be dodging the question of whether the example is correct. I hope it's not the case that your desire to ensure that your argument is not refuted is simply outweighing your desire to help others understand your argument? In any case, acknowledging that the example is correct does not require you to acknowledge that I have refuted your argument. The example shows only that the mere requirement that something have observable consequences does not by itself disqualify Cantorian set theory. It is still open to you to argue against Cantorian set theory on other grounds. >> Then I ask again: The study of Cantorian set theory has led to the >> discovery of useful, computationally meaningful facts, namely the >> consistency of large cardinal axioms. Why then do you oppose Cantorian >> set theory? >Here's what I'm arguing: Cantorian set theory includes a mythology >about a world beyond the world of computation. It is implausible to me >that we *need* that mythology to prove things about the world of >computation, and there are very good reasons for getting rid of it if >possible. O.K., so now you've shifted the emphasis of your argument. Previously, your emphasis was on truth necessarily has observable consequences. I tried to respond to that by showing that this principle does not, by itself, exclude Cantorian set theory, because Cantorian set theory has observable consequences. Apparently you disagree, but are unable or unwilling to point out where exactly the example I provided goes wrong. So perhaps that line of discussion is fruitless. Now, your emphasis has shifted. Your argument now seems to be, even if we grant for the sake of argument that Cantorian set theory has observable consequences, those observable consequences can be derived from weaker assumptions. This is roughly Feferman's complaint about Friedman's examples. Given the choice between a theory that has both observable and unobservable consequences, and a theory that has only observable consequences, we should prefer the latter. Or something like that. Yes? Assuming this is your argument, then what I would like you to clarify is whether you agree that the *consistency* of large cardinals lies within the world of computation, has observable consequences, and involves no mythology about a world beyond the world of computation. It would seem to me that your answer is yes, though I'm beginning to wonder, since I've asked this question (more or less) several times without getting any explicit affirmation from you. The closest I've seen is your statement that Believing in the consistency of the formalism is another matter, but even this is not an explicit affirmation that you in fact do believe in the meaningfulness of the consistency of the formalism. If you agree, then the next question is whether the consistency of large cardinals is not only meaningful, but potentially useful for AI. I'll assume that the answer is yes, since your concern that said consistency might have only irrelevant or computationally infeasible consequences is addressed by Friedman's examples. If we get this far, then my real question is, don't you think that trying to rid the world of Cantorian set theory is likely, as a practical matter, to eliminate these useful consistency statements too? Granted, it is logically conceivable that one could arrive at these consistency statements without studying and believing in Cantorian set theory itself. However, that is not, historically, how those consistency statements were arrived at. They were arrived at by people pursuing Cantorian set theory. And maybe there's something about the human brain that makes it very difficult to arrive at these consistency statements other than by pursuing Cantorian set theory. Doesn't it seem likely that you'll throw out the baby with the bathwater? In the end, it seems to me that all you want to impose is a gag rule that stops set theorists from saying that set theory is the foundation of mathematics, and similar statements. You presumably want to harvest the useful, computational fruits of their labor, provided that they toe the line ideologically. Yes? -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <44515da0$0$560$b45e6eb0@senator-bedfellow.mit.edu> <445226f9$0$569$b45e6eb0@senator-bedfellow.mit.edu> What would it take to convince you that Cantorian set theory has >> observable consequences? >For starters, you would have to prove to me that you understand what my >argument is, before I will believe you have refuted it. > I *am* trying to understand what your argument is. I am trying to > understand what you mean when you deny that Cantorian set theory has > observable consequences. *If* I said that, I was being sloppy. My claim is that Cantorian set theory includes a mythology about a world beyond the world of computation, and that world is not observable. > Now, your emphasis has shifted. Your argument now seems to be, even if we > grant for the sake of argument that Cantorian set theory has observable > consequences, those observable consequences can be derived from weaker > assumptions. This is roughly Feferman's complaint about Friedman's > examples. Given the choice between a theory that has both observable > and unobservable consequences, and a theory that has only observable > consequences, we should prefer the latter. Or something like that. Yes? I totally agree that we should strongly prefer the theory that has only observable consequences. Basically, I agree with that whole paragraph. > Assuming this is your argument, then what I would like you to clarify is > whether you agree that the *consistency* of large cardinals lies within > the world of computation, has observable consequences, and involves no > mythology about a world beyond the world of computation. I assume that when you say consistency of large cardinals, you mean that the axiom asserting the existence of large cardinals is consistent with ZFC. Then yes, such a statement has observable implications. > If you agree, then the next question is whether the consistency of large > cardinals is not only meaningful, but potentially useful for AI. I don't really know what potentially useful for AI means. > If we get this far, then my real question is, don't you think that trying > to rid the world of Cantorian set theory is likely, as a practical matter, > to eliminate these useful consistency statements too? It's very difficult to argue with as a practical matter. Not only that, we really haven't agreed on what useful means in the context where you use it. Your question is somewhat clever in that it really can't be argued against. > Granted, it is > logically conceivable that one could arrive at these consistency statements > without studying and believing in Cantorian set theory itself. However, > that is not, historically, how those consistency statements were arrived > at. They were arrived at by people pursuing Cantorian set theory. And > maybe there's something about the human brain that makes it very difficult > to arrive at these consistency statements other than by pursuing Cantorian > set theory. Doesn't it seem likely that you'll throw out the baby with the > bathwater? That's more of an emotional plea than a rational argument. I have argued previously that the pursuit of Cantor's mythology has held back progress in AI as much as thirty years or more, and progress in AI could have truly enormous benefits for humanity. You're arguing that without the pursuit of Cantor's mythology, we might not have certain consistency statements which in some vague sense may be useful. I have no way of proving beyond a shadow of a doubt that the world would be almost infinitely better off without Cantor's mythology, but that is something I believe is true, and I have given it a lot of thought. Nothing you have said would cause me to think you have a better idea. > In the end, it seems to me that all you want to impose is a gag rule that > stops set theorists from saying that set theory is the foundation of > mathematics, and similar statements. You presumably want to harvest the > useful, computational fruits of their labor, provided that they toe the > line ideologically. Yes? You sure do have a way of spinning things to your advantage. I have to grudgingly admire that talent. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics Originator: tchow@lebesgue.mit.edu.mit.edu (Timothy Chow) >> I *am* trying to understand what your argument is. I am trying to >> understand what you mean when you deny that Cantorian set theory has >> observable consequences. >*If* I said that, I was being sloppy. My claim is that Cantorian set >theory includes a mythology about a world beyond the world of >computation, and that world is not observable. O.K., fine. >I assume that when you say consistency of large cardinals, you mean >that the axiom asserting the existence of large cardinals is consistent >with ZFC. Then yes, such a statement has observable implications. O.K., good. >> If you agree, then the next question is whether the consistency of large >> cardinals is not only meaningful, but potentially useful for AI. >I don't really know what potentially useful for AI means. That the investigation of such consistency statements might lead to progress in AI, the kind of progress that you claim has been impeded by Cantor's mythology, and that could have benefits for humanity. >That's more of an emotional plea than a rational argument. I have >argued previously that the pursuit of Cantor's mythology has held back >progress in AI as much as thirty years or more, and progress in AI >could have truly enormous benefits for humanity. You're arguing that >without the pursuit of Cantor's mythology, we might not have certain >consistency statements which in some vague sense may be useful. I >have no way of proving beyond a shadow of a doubt that the world would >be almost infinitely better off without Cantor's mythology, but that is >something I believe is true, and I have given it a lot of thought. >Nothing you have said would cause me to think you have a better idea. O.K. At least it seems you agree with, or have no refutation of, the claim that certain consistency statements arising from the pursuit of Cantor's mythology are (1) computationally meaningful, and (2) *might* lead to progress in AI. Of course, there might be harmful results of pursuing Cantor's mythology that outweigh any such nebulous benefits. I'm not particularly interested in entering that debate. What got me interested in this thread in the first place was the mention of Friedman's results, and I don't think I can argue anything more on the basis of those results than I already have. By the way, I messed up the statement about directed graphs slightly. The second subgraph, isomorphic to H, is not to be found in the complement of A, but in GA, which by definition is the set of successor vertices of the vertices in A. Anyway, the precise statement can be found in the >> In the end, it seems to me that all you want to impose is a gag rule that >> stops set theorists from saying that set theory is the foundation of >> mathematics, and similar statements. You presumably want to harvest the >> useful, computational fruits of their labor, provided that they toe the >> line ideologically. Yes? >You sure do have a way of spinning things to your advantage. I have to >grudgingly admire that talent. That sounds like a yes, which surprises me. I would have guessed that you would want to impose more than a gag rule, but to ban the pursuit of Cantorian mythology entirely. And I would have expected you to argue that any baby (consistency statement) that might be thrown out with the bathwater would probably be no great loss, and would be more than compensated for by the utopian joys of a Cantorian-set-theory-free world. But anyway, as I said, I'm not so interested in pursuing the debate along these lines. -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <445226f9$0$569$b45e6eb0@senator-bedfellow.mit.edu> <44527625$0$560$b45e6eb0@senator-bedfellow.mit.edu> If you agree, then the next question is whether the consistency of large >> cardinals is not only meaningful, but potentially useful for AI. >I don't really know what potentially useful for AI means. > That the investigation of such consistency statements might lead to > progress in AI, the kind of progress that you claim has been impeded by > Cantor's mythology, and that could have benefits for humanity. But isn't it obvious that I cannot deny that as a possibility? >> In the end, it seems to me that all you want to impose is a gag rule that >> stops set theorists from saying that set theory is the foundation of >> mathematics, and similar statements. You presumably want to harvest the >> useful, computational fruits of their labor, provided that they toe the >> line ideologically. Yes? >You sure do have a way of spinning things to your advantage. I have to >grudgingly admire that talent. > That sounds like a yes, which surprises me. I would have guessed that > you would want to impose more than a gag rule, but to ban the pursuit of > Cantorian mythology entirely. The Cantorian mythology has been woven into the mathematics (e.g. real analysis, functional analysis, algebra etc.) taught in our universities, or at least in the graduate schools of our universities. I believe that it is doing real harm there. It blinds students to the observable implications of the mathematics they are doing, and it diverts resources to the pursuit of fantasies without observable implications. Not to mention the real social injustice of imposing a fantasy on students who don't want to partake of it. I would never tell those who are convinced that they want to pursue set theory that they can't. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics On 28 Apr 2006 13:31:06 -0700, david petry said: > ... > In the end, it seems to me that all you want to impose is a gag > rule that stops set theorists from saying that set theory is the > foundation of mathematics, and similar statements. You presumably > want to harvest the useful, computational fruits of their labor, > provided that they toe the line ideologically. Yes? >>You sure do have a way of spinning things to your advantage. I have to >>grudgingly admire that talent. >> That sounds like a yes, which surprises me. I would have guessed that >> you would want to impose more than a gag rule, but to ban the pursuit of >> Cantorian mythology entirely. > The Cantorian mythology has been woven into the mathematics (e.g. real > analysis, functional analysis, algebra etc.) taught in our > universities, or at least in the graduate schools of our universities. > I believe that it is doing real harm there. It blinds students to the > observable implications of the mathematics they are doing, and it > diverts resources to the pursuit of fantasies without observable > implications. Not to mention the real social injustice of imposing a > fantasy on students who don't want to partake of it. Dramatis personae: Arthur (a.k.a, Tim Chow) Black Knight (a.k.a, David Petry) ARTHUR: You fight with the strength of many men, Sir Knight. [pause] I am Arthur, King of the Britons. [pause] I seek the finest and the bravest knights in the land to join me in my court at Camelot. [pause] You have proved yourself worthy. Will you join me? [pause] You make me sad. So be it. Come, Patsy. BLACK KNIGHT: None shall pass. ARTHUR: What? BLACK KNIGHT: None shall pass. ARTHUR: I have no quarrel with you, good Sir Knight, but I must cross this bridge. BLACK KNIGHT: Then you shall die. ARTHUR: I command you, as King of the Britons, to stand aside! BLACK KNIGHT: I move for no man. ARTHUR: So be it! ARTHUR and BLACK KNIGHT: Aaah!, hiyaah!, etc. [ARTHUR chops the BLACK KNIGHT's left arm off] ARTHUR: Now stand aside, worthy adversary. BLACK KNIGHT: 'Tis but a scratch. ARTHUR: A scratch? Your arm's off! BLACK KNIGHT: No, it isn't. ARTHUR: Well, what's that, then? BLACK KNIGHT: I've had worse. ARTHUR: You liar! BLACK KNIGHT: Come on, you pansy! [clang] Huyah! [clang] Hiyaah! [clang] Aaaaaaaah! [ARTHUR chops the BLACK KNIGHT's right arm off] ARTHUR: Victory is mine! [kneeling] We thank Thee Lord, that in Thy mer-- BLACK KNIGHT: Hah! [kick] Come on, then. ARTHUR: What? BLACK KNIGHT: Have at you! [kick] ARTHUR: Eh. You are indeed brave, Sir Knight, but the fight is mine. BLACK KNIGHT: Oh, had enough, eh? ARTHUR: Look, you stupid bastard. You've got no arms left. BLACK KNIGHT: Yes, I have. ARTHUR: Look! BLACK KNIGHT: Just a flesh wound. [kick] ARTHUR: Look, stop that. BLACK KNIGHT: Chicken! [kick] Chickennn! ARTHUR: Look, I'll have your leg. [kick] Right! [whop] [ARTHUR chops the BLACK KNIGHT's right leg off] BLACK KNIGHT: Right. I'll do you for that! ARTHUR: You'll what? BLACK KNIGHT: Come here! ARTHUR: What are you going to do, bleed on me? BLACK KNIGHT: I'm invincible! ARTHUR: You're a looney. BLACK KNIGHT: The Black Knight always triumphs! Have at you! Come on, then. [whop] [ARTHUR chops the BLACK KNIGHT's last leg off] BLACK KNIGHT: Oh? All right, we'll call it a draw. ARTHUR: Come, Patsy. BLACK KNIGHT: Oh. Oh, I see. Running away, eh? You yellow bastard! Come back here and take what's coming to you. I'll bite your legs off! === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <44515da0$0$560$b45e6eb0@senator-bedfellow.mit.edu> <445226f9$0$569$b45e6eb0@senator-bedfellow.mit.edu In the end, it seems to me that all you want to impose is a gag rule that > stops set theorists from saying that set theory is the foundation of > mathematics, and similar statements. You presumably want to harvest the > useful, computational fruits of their labor, provided that they toe the > line ideologically. Yes? Given David's history from usenet archives, it is clear to see he has been saying the same thing over and over for 15 years or so. You are unlikely to get different answers than all those before who have tried. I think he might have an interesting research topic if he only stripped away the subjective, emotionally charged language and actually tried to make precise what he means. That is, it would be interesting to study out what is possible to know if one somehow limits oneself to only computationally feasible things. But one would first have to make precise what is computationally feasible (and what does that mean). Then one could say that in this theory the meaningful statements are those that can be computed in such a way. It is entierly possible to believe that perhaps those statements that in the end have some practical value outside of mathematics will belong to this category. Of course the problem with David's approach is that he is not interested in actually doing this. He is only interested in somehow refuting set theory, whatever that means. And he is only interested in making categorical philosophical statements rather then doing any mathematics. I mean if he were really interested in something enough to rant about it for 15 years, he would get somewhere in this time. Jiri === Subject: Re: Harvey Friedman on Cantorian pseudomathematics Originator: tchow@lebesgue.mit.edu.mit.edu (Timothy Chow) >That is, it would be interesting to study out what is possible to know >if one somehow limits oneself to only computationally feasible things. Well, I gave the pointer to Sazonov's work. Here it is again, for convenience: http://www.csc.liv.ac.uk/~sazonov/papers/lcc.ps David Petry sounded at least mildly interested, so maybe something will come of that. -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <445226f9$0$569$b45e6eb0@senator-bedfellow.mit.edu> <445276f9$0$560$b45e6eb0@senator-bedfellow.mit.eduThat is, it would be interesting to study out what is possible to know >if one somehow limits oneself to only computationally feasible things. > Well, I gave the pointer to Sazonov's work. Here it is again, for > convenience: > http://www.csc.liv.ac.uk/~sazonov/papers/lcc.ps > David Petry sounded at least mildly interested, so maybe something will > come of that. I've been looking at the paper, but I don't like what I see. The author seems to think that 2^1000 is big enough to call infinity. It's true that a scientist, for example, would never need to know the exact value of a number in that range, but he would need to use numbers that big and much bigger. For example, a geneticist wants to know how many possible strands of DNA there are, and it is a much bigger number. My goal has always been to produce something of practical value to applied mathematicians in general. It's possible that Sazonov's feasible numbers may be of limited use to computer scientists, but it's hardly the kind of thing that every applied mathematician needs to learn. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics ... >>What >>would it take to convince you that Cantorian set theory has observable >>consequences? > For starters, you would have to prove to me that you understand what my > argument is, before I will believe you have refuted it. Perhaps a doable short term project might be trying to help people understand your argument, by working on definitions other people can apply, and get the same results as you do. For example, you don't seem to agree with my view of the scope of the world of natural number computation as including anything that has been, or can be, digitized. Perhaps you could work on a definition that other people can apply, and get the same results as you do. I still don't know your attitude to functions whose computability has not been proved, or for which there is no proved upper performance bound. Are they OK to study? If so, what should they be called? If not, how will we ever add to the set of computable, performance bounded, functions? As Barb Knox pointed out, if you start with only primitive recursive performance bounds, all your functions will be primitive recursive. Is there any non-PR function you are prepared to accept as being computable? Why not work on getting a solid set of definitions, for the sake of better communication? Patricia === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <4450d393$0$24622$b45e6eb0@senator-bedfellow.mit.edu> <44515da0$0$560$b45e6eb0@senator-bedfellow.mit.edu> not been proved, or for which there is no proved upper performance > bound. Are they OK to study? So you're asking me, like, is it OK to explore the unknown? Well, gee, let me think about that a while ... uh... sure, why not? Yes! Permission granted. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics > I still don't know your attitude to functions whose computability has > not been proved, or for which there is no proved upper performance > bound. Are they OK to study? > So you're asking me, like, is it OK to explore the unknown? Well, gee, > let me think about that a while ... uh... sure, why not? Yes! > Permission granted. Then all those restrictions DP wants to impose are phony. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics >>I still don't know your attitude to functions whose computability has >>not been proved, or for which there is no proved upper performance >>bound. Are they OK to study? > So you're asking me, like, is it OK to explore the unknown? Well, gee, > let me think about that a while ... uh... sure, why not? Yes! > Permission granted. Now what do you think they should be called, since you seem to be restricting function to computable functions with known performance bounds? Patricia === Subject: Re: Harvey Friedman on Cantorian pseudomathematics >> So are you actively pursuing this goal, beyond arguing on USENET? >No, not for almost 15 years. > May I ask why not? > As a graduate student, I was unsuccessful in getting anyone at all > interested in what I was doing, but I did manage to get a lot of people > upset by what I was doing. > What > would it take to convince you that Cantorian set theory has observable > consequences? > For starters, you would have to prove to me that you understand what my > argument is, before I will believe you have refuted it. > Do you agree that logicians have successfully captured *the > way set theorists talk* about Cantorian set theory in a set of axioms that > can be manipulated on a computer? > Yes, or at least, I have no reason to doubt it. > Then I ask again: The study of Cantorian set theory has led to the > discovery of useful, computationally meaningful facts, namely the > consistency of large cardinal axioms. Why then do you oppose Cantorian > set theory? > Here's what I'm arguing: Cantorian set theory includes a mythology > about a world beyond the world of computation. It is implausible to me > that we *need* that mythology to prove things about the world of > computation, and there are very good reasons for getting rid of it if > possible. > Would it satisfy you if set theorists obeyed a gag order, > docilely agreeing that they will never publicly assert the actual > existence of large cardinals, but asserting only that such-and-such > statements *follow from the consistency* of large cardinals? That is, > they can go ahead doing research as they normally do, as long as they > obey a politically correct code of speech? > Of course you're being silly, but it would be a good idea if the set > theorists admitted that set theory is not the *foundation* of > mathematics. Set theorists do not say that in general. What they do tend to say is that set theory is the best foundation SO FAR. As no one, including Petry, has yet come up with a better one, I imagine they will continue saying it. === Subject: Re: Harvey Friedman on Cantorian pseudomathematics <4450d393$0$24622$b45e6eb0@senator-bedfellow.mit.edu> <44515da0$0$560$b45e6eb0@senator-bedfellow.mit.edu> that set theory is the best foundation SO FAR. Just out of curiosity, which set theorist claims that set theory is the best foundation so far but does not make the additional claim that set theory is the best foundation period? === Subject: Re: How to prove that a special matrix is invertible. >Hi all, >I am trying to prove that a special matrix is invertible. I've spent >quite some time without success. I know the matrix is invertible >because I've simulated it in Matlab and observed that the determinant >is non-zero unless a paramter (epsilon) is not integer other than zero. >This is a KxK (Toeplitz) matrix with entries as >A(i,j)=sin(pi*epsilon)/sin(pi*(i-j+epsilon)/N) >( N is a large number, say 100: K=1, 2, 3, ... < N ) >I've derived the determinant, but it becomes very complicated as K >becomes larger. >What should I try? Any help is greatly appreciated. # This Maple script lets you test your conjecture # for small values of K (up to about 8 in a one-minute run). # Try to generalize the patterns you see. with(linalg); K := 9; # fixed value a := array(1..K, 1..K); cos(piN) := (wN + 1/wN)/2; # wN = exp(pi*I/N) sin(piN) := (wN - 1/wN)/(2*I); cos(piepsN) := (we + 1/we)/2; # we = exp(pi*I*eps/N) sin(piepsN) := (we - 1/we)/(2*I); # Express matrix elements in terms of we and wN. for i from 1 to K do for j from 1 to K do # The factor sin(pi*epsilon) in the numerator is constant # and does not affect the rank. a[i,j] := normal(1/expand(sin(piN*(i-j) + piepsN))); od; od; # Take the determinant. Factor it. d1 := factor(normal(det(a)/(2*I*we^K)^K)); # Numerator of d1 appears to be a polynomial in wN, # which factors as a product of cyclotomic polynomials. # Denominator of d1 has factors we +- (power of wN). -- VP Cheney Burr-ed his gun as a bird flew past The nation responds burr as we await bird flu shots and fight a real cold war. pmontgom@cwi.nl Microsoft Research and CWI Home: Bellevue, WA === Subject: Solving trignometry by 'hand' Hi all, I'm having some problems at solving this by hand without a computer: 1 = 1/(sqrt(4sin(t)^2 + 1))(sqrt(3sin(t)) + cos(t)) Any suggestions? I'm not even sure if this is possible. I just got here by playing with some equations from Physics and motion in two dimensions. Paulo Matos === Subject: Re: Solving trignometry by 'hand' >Hi all, >I'm having some problems at solving this by hand without a computer: >1 = 1/(sqrt(4sin(t)^2 + 1))(sqrt(3sin(t)) + cos(t)) >Any suggestions? I'm not even sure if this is possible. I just got here >by playing with some equations from Physics and motion in two >dimensions. I interpret this as 1 = (1/factor1) * factor2. It would have been clearer to write either 1 = factor1 * factor2 or factor1 = factor2. Eliminate fractions and square both sides 4sin(t)^2 + 1 = (sqrt(3sin(t)) + cos(t))^2 Subtract 3 sin(t) + cos(t)^2 from both sides to get (quadratic in sin(t)) = 2*cos(t)*sqrt(3sin(t)) Square both sides to get a quartic in s = sin(t). That is s (25s^3 - 18 s^2 + 9s - 12) = 0. The real roots are s = 0 and s ~ 0.906. Use the equation following the first squaring to determine the sign of cos(t). Then confirm the original equation (to guard against extraneous roots). -- VP Cheney Burr-ed his gun as a bird flew past The nation responds burr as we await bird flu shots and fight a real cold war. pmontgom@cwi.nl Microsoft Research and CWI Home: Bellevue, WA === Subject: Re: Solving trignometry by 'hand' > Hi all, > I'm having some problems at solving this by hand without a computer: > 1 = 1/(sqrt(4sin(t)^2 + 1))(sqrt(3sin(t)) + cos(t)) > Any suggestions? I'm not even sure if this is possible. I just got here > by playing with some equations from Physics and motion in two > dimensions. > Paulo Matos I don't see a way to do this by hand. If I draw a graph, it looks like t is approximately 2.6. Dave === Subject: Re: is mathematics just a tautology? <87lktv4ono.fsf@phiwumbda.org> <87d5f74awu.fsf@phiwumbda.org> <87acaa4gmd.fsf@phiwumbda.org> <87aca93y7f.fsf@phiwumbda.org> tautology means no news. === Subject: Re: is mathematics just a tautology? <87lktv4ono.fsf@phiwumbda.org> <87d5f74awu.fsf@phiwumbda.org> <87acaa4gmd.fsf@phiwumbda.org> <87aca93y7f.fsf@phiwumbda.org> <87y7xrtlko.fsf@phiwumbda.org> my posts are right on the point. if you can't handle the truth, leave the kitchen. === Subject: A company with a great idea only to be blocked by the US gov. Go to their web page and click on hydrogen fuel systems. Read about their fuel system for automobiles and most important their latest news release! Sounds like big oil again controlling any new breakthroughs of alternative energy! Dan === Subject: (strict sense) stationarity of AR processes? HI all, I have a question about the stationary of AR processes in the strict sense? Take AR(1), X_n=a*X_(n-1) + e_n, |a|<1. where e_n's are white noise process(uncorrelated with common mean 0 and common sigma^2), but not neccessarily IID process, i.e., each e_n may not need to be independently and identically distributed. (This is a definition from Brockwell's Time Series book). My question is: is X_n stationary in the strict sense? My guess is that it is not strictly stationary. So I just have to provide a counter-example: Suppose the process starts from X0 which is a random variable. Then X1=a * X0 + e1, will not have same distribution as X0, (this looks to me like a triviality, but I am not sure if there will be some hidden traps...) But what if the process starts from minus infinity? Staring from minus infinity, will X0 and a * X0 + e1 have the same distribution? === Subject: Is MMSE estimate biased or unbiased? HI all, I got very confused about the biasedness or unbiasedness of the very basic MMSE estimate and Best Linear Estimate. My (obviously wrong) conjecture is that they are all inherently biased. There is no unbiased MMSE and Best Linear Estimate except some very trivial case. My thought process is as follows: Observe X, want to estimate Y: 1. MMSE estimate Y_hat = E(Y | X) According to the definition of biasedness = E(Y_hat) - Y's true value. But E(Y_hat)=E(E(Y|X))=E(Y) which will never equal Y itself unless Y is a constant. Thus in most cases, Y_hat=E(Y|X) is a biased estimate. 2. Best Linear Estimate: Y_hat=meanY+ Cov(Y, X)(Cov(X, X)^(-1))*(X - meanX) E(Y_hat)=meanY, which will never equal Y itself unless Y is a constant. ------------------------------------ I am guessing that my definition of biasedness and unbiasedness for the estimate is wrong. The definition of biasedness = E(Y_hat) - Y's true value may only work for population parameter estimate, which is not the Random Variable Y here. Please give me some enlightenment! === Subject: Under what condition is the Best Linear Estimate the same as the Best conditional Estimate? HI all, Observe X, and want to estimate Y. Aside from the jointly Gaussian case, are there any other conditions that can allow me to conclude the best linear estimate is the same as the global best which is the conditional estimate E(Y|X)? === Subject: Re: Length of This String Isn't this just a matter of finding the longest side of a right-angled triangle 75 cm high and 100 cm along the horizontal base? Am I missing something? === Subject: Betti's numbers Hi. My name is Angelo. I have a question about Betti's numbers. I know that for dimension 3 beta_0 = # connected component; beta_1 = # tunnels or holes; beta_2 = # cavities. Is there some results about interpretation of Betti's number for dimension bigger than 3. === Subject: Re: picking balls >> If I have a bin containing some N balls. I randomly pick a >> ball with uniform distribution (and put it aside, no repetition). >> Now someone adds a certain number of balls to the bin (0 >> or more). I again randomly pick a ball with uniform >> distribution and put aside, and someone again puts 0 or >> more balls to the bin. This process continuously repeats >> itself. Assume the number of balls in the bin are >> guaranteed to always be finite. >> Question: Will every ball eventually be picked? It's a possible outcome, although one with probability essentially 0. The issue is in your requirement that the ball being picked is chosen with uniform distribution. Without loss of generality, let's imagine this bin as a queue, each ball numbered in order as it is thrown in. If by freak occurrence, the randomly picked ball is always the one with the lowest number, then yes, all balls will be picked. If, on the other hand, by equally freakish occurrence the one with the highest number is always picked, then no, not every ball gets picked. Since the ball being picked is chosen by uniform distribution, we look at the space of all possible pick choices. Within that space, the set of those leaving no ball behind is non-empty (and thus a possible outcome), but has measure 0 (and thus not one you want to bet on). Problems with possible outcomes of probability 0 are easy to construct. The simplest one I know is this: God randomly picks a real number from [0,1] with uniform distribution. What is the probability that the chosen number is rational? Jonathan Hoyle === Subject: Re: picking balls > number of balls added before the kth draw (so S1 = N). The probability > that a fixed ball amongst the first N is never drawn is My problem is slightly different. I am not asking about the initial N balls. But any fixed ball at any given instant moment. So let say I pick only of the balls at time k=100, and my question should also apply to that ball. But I guess that problem is identical to the one you just pointed to, as you take your N initial balls to be the balls S_k-k at a particular k, and apply the same reasoning. Just to summarize and ask a new question: 1) Pr[picking a fixed ball b]: 1 - Pr[never picking b] 2) Pr[never picking b] = prod(k=1, infty, (S_k-k)/(S_k-k+1)) So if we can show that 2) goes to zero, we are done, right? So my question is when does a infinite product of series where each factor is <1.0 (in our case each factor is always less than 1.0) go to 0 and when does it not? === Subject: Re: picking balls >If I have a bin containing some N balls. I randomly pick a ball with >uniform distribution (and put it aside, no repetition). Now someone adds >a certain number of balls to the bin (0 or more). I again randomly pick >a ball with uniform distribution and put aside, and someone again puts 0 >or more balls to the bin. This process continuously repeats itself. >Assume the number of balls in the bin are guaranteed to always be finite. ... and never 0, I hope. >Question: Will every ball eventually be picked? Let N_k be the number of balls after the k'th iteration. For any given ball which is in the bin after the k'th iteration, the probability that it is picked in the next iteration is 1/N_k. The probability that it survives forever is product_{j=k}^infty (1 - 1/N_j), which is nonzero iff sum_k 1/N_k < infinity. Thus: a) if sum_k 1/N_k = infinity, then almost surely every ball is eventually picked. b) if sum_k 1/N_k < infinity, there is epsilon > 0 such that for each ball, the probability that this ball is never picked is at least epsilon. Suppose exactly one of the initial balls is painted white, and when a white ball is picked, one of the other balls in the bin (if necessary one that is added at that iteration) is painted white (thus after each iteration there is exactly one white ball in the bin). Then in case (b), Borel-Cantelli says that almost surely only finitely many white balls will be drawn. In particular, in case (b) almost surely some balls will never be picked. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: picking balls > probability that it survives forever is product_{j=k}^infty (1 - 1/N_j), Ok. > which is nonzero iff sum_k 1/N_k < infinity. Thus: How do you get the iff between those two propositions (first quoted line and second quoted line). On some intuitive level I can understand, if the sum grows fast it cancels the 1 in the product making the probability of never picking that ball 0. But I cannot understand it more precisely, or formally. === Subject: Re: picking balls <44520DD1.8020403@hotmail.com probability that it survives forever is product_{j=k}^infty (1 - 1/N_j), > Ok. > which is nonzero iff sum_k 1/N_k < infinity. Thus: I should have also required all N_k > 1. > How do you get the iff between those two propositions (first quoted > line and second quoted line). On some intuitive level I can understand, > if the sum grows fast it cancels the 1 in the product making the > probability of never picking that ball 0. But I cannot understand it > more precisely, or formally. A standard result on convergence of infinite products. If 0 < a_j < 1, then product_{j=1}^infty (1 - a_j) converges (to a nonzero value) iff sum_j a_j converges. To prove it, take logarithms and bound: -2 a_j < log(1 - a_j) < -a_j for 0 < a_j < 0.7 say. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: simplifying sqrt(14 -6sqrt(5))+(26-15sqrt(3))^(1/3) +(173+76sqrt(3)+60sqrt(5)+44sqrt(15))^(1/4) >Which method will you propose to work out >and if possible simplify this expression with radicals ? > sqrt(14 -6*sqrt(5) ) + (26-15*sqrt(3) )^ (1/3) + >(173+76*sqrt(3) + 60*sqrt(5)+44*sqrt(15) )^ (1/4) Maple's simplify does that one quite well. But try sqrt(7 + sqrt(46)) + sqrt(10 - sqrt(46) + 2*sqrt(21 - 3*sqrt(46))) - sqrt(14 + 2*sqrt(3)) Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: simplifying sqrt(14 -6sqrt(5))+(26-15sqrt(3))^(1/3) +(173+76sqrt(3)+60sqrt(5)+44sqrt(15))^(1/4) >>Which method will you propose to work out >>and if possible simplify this expression with radicals ? >>sqrt(14 -6*sqrt(5) ) + (26-15*sqrt(3) )^ (1/3) + >>(173+76*sqrt(3) + 60*sqrt(5)+44*sqrt(15) )^ (1/4) > Maple's simplify does that one quite well. But try > sqrt(7 + sqrt(46)) + sqrt(10 - sqrt(46) + 2*sqrt(21 - 3*sqrt(46))) > - sqrt(14 + 2*sqrt(3)) > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada sqrt(3), of course === Subject: Re: simplifying sqrt(14 -6sqrt(5))+(26-15sqrt(3))^(1/3) +(173+76sqrt(3)+60sqrt(5)+44sqrt(15))^(1/4) >Which method will you propose to work out >and if possible simplify this expression with radicals ? >sqrt(14 -6*sqrt(5) ) + (26-15*sqrt(3) )^ (1/3) + >(173+76*sqrt(3) + 60*sqrt(5)+44*sqrt(15) )^ (1/4) >> Maple's simplify does that one quite well. But try >> sqrt(7 + sqrt(46)) + sqrt(10 - sqrt(46) + 2*sqrt(21 - 3*sqrt(46))) >> - sqrt(14 + 2*sqrt(3)) >> Robert Israel israel@math.ubc.ca >> Department of Mathematics http://www.math.ubc.ca/~israel >> University of British Columbia Vancouver, BC, Canada >sqrt(3), of course We can confirm this if we observe sqrt(10 - sqrt(46) + 2*sqrt(21 - 3*sqrt(46)) = sqrt(3) + sqrt(7 - sqrt(46)) sqrt(7 + sqrt(46)) + sqrt(7 - sqrt(46)) = sqrt(14 + 2*sqrt(3)) Each of these can be checked by squaring both sides. -- VP Cheney Burr-ed his gun as a bird flew past The nation responds burr as we await bird flu shots and fight a real cold war. pmontgom@cwi.nl Microsoft Research and CWI Home: Bellevue, WA === Subject: Re: simplifying sqrt(14 -6sqrt(5))+(26-15sqrt(3))^(1/3) +(173+76sqrt(3)+60sqrt(5)+44sqrt(15))^(1/4) > Which method will you propose to work out > and if possible simplify this expression with radicals ? > sqrt(14 -6*sqrt(5) ) + (26-15*sqrt(3) )^ (1/3) + > (173+76*sqrt(3) + 60*sqrt(5)+44*sqrt(15) )^ (1/4) 14 - 6.sqr 5 = 9 - 6.sqr 5 + 5 = (3 - sqr 5)^2 26 - 15.sqr 3 = 8 - 3*4.sqr 3 + 3*2*3 - 3.sqr 3 = (2 - sqr 3)^3 5 - sqr 3 - sqr 5 + (173+76*sqrt(3) + 60*sqrt(5)+44*sqrt(15) )^ (1/4) (a.sqr 3 + b.sqr 5)^4 = 9a^4 + 4*3.a^3 b.sqr 15 + 6*3*5.a^2 b^2 + 4*5.ab^2 sqr 15 + 5b^4 = r + s.sqr 15 Shucks. Look at (c + a.sqr 3 + b.sqr 5)^4 ??? How about (1 + sqr 3 + sqr 5)^4 = ?? === Subject: Re: simplifying sqrt(14 -6sqrt(5))+(26-15sqrt(3))^(1/3) +(173+76sqrt(3)+60sqrt(5)+44sqrt(15))^(1/4) Merci beaucoup , For all your replies . I am expecting simple algebraic 'recipes' and wish avoiding heavier ones like polynomials roots, norms ,theory of fields and biquadratic extensions , is it possible to rely only upon two tools : 1Á) simple separatable forms : (a +b*sqrt(c) )^ n = e +f*sqrt(c) (a +b*sqrt(c) + d*sqrt(e) )^n =f +g*sqrt(c) + h*sqrt(c*e) (a+b*c^(1/3))^n = d + e*c^(1/3)+f*c^(2/3) ... 2Á) bijective real functions fi(x) and fi^[ - 1](x) **** **************************************** *** Here is my problem's building :: (3 -sqrt(5)) + (2 -sqrt(3)) + (1+sqrt(3)+sqrt(5)) = 6 f1^[-1](f1 (3 -sqrt(5))) + f2^[-1](f2(2 -sqrt(3)) ) +f3^[-1](f3(1+sqrt(3)+sqrt(5))) = 6 with f1(u) =u^2 , f2(v) =v^3 , f3(w) = w^4 , we obtain : sqrt(14 -6*sqrt(5) ) + (26-15*sqrt(3) )^ (1/3) + (173+76*sqrt(3) + 60*sqrt(5)+44*sqrt(15) )^ (1/4) Alain === Subject: Re: simplifying sqrt(14 -6sqrt(5))+(26-15sqrt(3))^(1/3) +(173+76sqrt(3)+60sqrt(5)+44sqrt(15))^(1/4) > Which method will you propose to work out > and if possible simplify this expression with radicals ? > sqrt(14 -6*sqrt(5) ) + (26-15*sqrt(3) )^ (1/3) + > (173+76*sqrt(3) + 60*sqrt(5)+44*sqrt(15) )^ (1/4) Mathematica simplifies it to 6. === Subject: Re: simplifying sqrt(14 -6sqrt(5))+(26-15sqrt(3))^(1/3) +(173+76sqrt(3)+60sqrt(5)+44sqrt(15))^(1/4) > Which method will you propose to work out > and if possible simplify this expression with radicals ? > sqrt(14 -6*sqrt(5) ) + (26-15*sqrt(3) )^ (1/3) + > (173+76*sqrt(3) + 60*sqrt(5)+44*sqrt(15) )^ (1/4) > Alain Well. I dunno about general methods, but most of the time simply looking at it for a long enough time tells me, whether there's something there (at least when the solution is simple). Let's work term by term: sqrt(14-6sqrt5) is either of the form a+b*sqrt5 (a,b rational) or cannot be simplified at all. Simply computing the square of the theory of algebraic integers we can infer that a and b must either integers or half-integers. If you want to get a little bit fancier you can try to utilize the norms. The norm of 14-6sqrt5 is (14-6sqrt5)(14+6sqrt5)=14^2-5*6^2=16, which is equal to 4^2. So the norm of the square root (if it exists in the field Q(sqrt5) must be 4 or -4. So we must have a^2-5b^2=4 or -4, this might give you an idea of what to guess. In this case I would simply observe that 14 - 6*sqrt5= 9 - 6*sqrt5 + 5 = 3^2 - 6*sqrt5 +(sqrt5)^2, and wait for the penny of the binomial formula to drop. Similarly with the second term. The norm of 26-15*sqrt3 is 676-3*225=1, so this is a unit in the ring Z[sqrt3]. Again either the cubic root is also a unit in the ring Z[sqrt3], or it is in a bigger field (in which case no simplification is possible). The number a+b*sqrt3 has norm a^2-3b^2, which we now assume is equal to 1 . The solution a=2,b=1 immediately suggests itself, but (2+sqrt3)^3=8+12sqrt3+18+3sqrt3=26+15*sqrt3, so this isn't quite the correct guess, but you can certainly see the required change for yourself. The last term is the trickiest of this lot as now we looking for something in a biquadratic extension Q(sqrt3,sqrt5) of the field of rationals. You may be again simple be able to make a guess that the fourth root is of the form a+b*sqrt3+c*sqrt5+d*sqrt15, and attempt to solve for the coefficients more or less by trial and error (since this is given as an exercise the solution probably isn't very complicated). You may be able to get some ideas by computing various and sundry norms: To compute the (relative) norm in the field Q(sqrt5) you multiply (173+76*sqrt(3) + 60*sqrt(5)+44*sqrt(15) with the conjugate (replace sqrt3 with -sqrt3 throughout) (173-76*sqrt(3) + 60*sqrt(5)-44*sqrt(15) ). To compute its relative norm in the field Q(sqrt3) you multiply it with the conjugate (173+76*sqrt(3) - 60*sqrt(5)-44*sqrt(15) ) and finally you may compute the norm of either of the relative norms in Q. As all the norm maps are multiplicative, they take fourth powers to fourth power, so you may be able to make deductions about the various norms of the putative fourth root. This wasn't a well thought out guide - simply listing some ad hoc methods I apply to this type of simplifications myslef. I encourage you to study field theory and algebraic number theory, if you want to learn more. Jyrki === Subject: Re: simplifying sqrt(14 -6sqrt(5))+(26-15sqrt(3))^(1/3) +(173+76sqrt(3)+60sqrt(5)+44sqrt(15))^(1/4) > Which method will you propose to work out > and if possible simplify this expression with radicals ? > sqrt(14 -6*sqrt(5) ) + (26-15*sqrt(3) )^ (1/3) + > (173+76*sqrt(3) + 60*sqrt(5)+44*sqrt(15) )^ (1/4) You can write this as the sum of three algebraic numbers, each of which it is easy to find a minimal polynomial for. Using results or whatever, you can then find a polynomial for the sum. === Subject: maps of projective space RP2n for any self map f: RP2n -> RP2n, there is always a fixed point. Is this a result of Borsuk-Ulam theorem? === Subject: taylor decomposition Let P a polynomial in K[x], K field. We know that for any a in K we can have the taylor decomposition: P(x) = P(a) + P_1(x)(x-a)+P_2(x)(x-a)^2+...... where P_1,P_2,P_3... are polynomials of degree less than deg P ( for example in characteristic 0, P_i = P^(i)/(i!)). My question is : can we generalize the decomposition when we do not have x-a but h(a), h of degree bigger than 1? Suppose that h has degree bigger than 1. Can we have: P = P(a)+S_1(x)h(a)+S_2(x)h(a)^2+.....? where S_1,S_2,S_3 have degree less than deg P? (it's not homework) === Subject: Re: maintaining balance on bike in forward motion is easier than at ... ETAtAhUAlMWDmuUCvlOpmCuV/EyYiLq3YU8CFFFwcQlMPHWqiuQEEldxtvhVCMRo The main stabilizing component of a bike is the inclined fork of the front wheel. Indeed, try to ride a bike moving backward, you will see rapidly that angular momentum doesn't help. __________________________________ Re; It's called caster; the same concept is applied to automotive steering as well, for self-centering, stability, and over-all control. Dan Akers === Subject: very basic question that confused me Hi all, Define b = a / |a|, a and b are vectors in an inner product space, |a| is the norm of the vector < , > is the inner product operator sqrt = square root In the following derivation, * ** |b| = sqrt() = sqrt()=sqrt(/|a||a|)=1 Anyone please tell me how does it go from * to **?? === Subject: Re: very basic question that confused me > Hi all, > Define b = a / |a|, a and b are vectors in an inner product space, > |a| is the norm of the vector > < , > is the inner product operator > sqrt = square root > In the following derivation, > * ** > |b| = sqrt() = sqrt()=sqrt(/|a||a|)=1 > Anyone please tell me how does it go from * to **?? For any vector, v, in an inner product space, |v|^2 = === Subject: Re: very basic question that confused me > Hi all, > Define b = a / |a|, a and b are vectors in an inner product space, > |a| is the norm of the vector > < , > is the inner product operator > sqrt = square root > In the following derivation, > * ** > |b| = sqrt() = sqrt()=sqrt(/|a||a|)=1 > Anyone please tell me how does it go from * to **?? As others have noted, this relies on the properties of the inner product, esp. compatibility with scalar products in the arguments (take out both factors of 1/|a|). Let me point out that this derivation is only valid for nonzero vectors a, and that a is not the zero vector if and only if its norm squared = |a|^2 is nonzero. === Subject: Re: very basic question that confused me > Hi all, > Define b = a / |a|, a and b are vectors in an inner product space, > |a| is the norm of the vector > < , > is the inner product operator > sqrt = square root > ... > |b| = sqrt() = sqrt()=sqrt(/|a||a|)=1 > Anyone please tell me how does it go from * to **?? sqrt()=sqrt(/|a||a|) because sqrt() = sqrt( * (1/|a|) ) = sqrt( * (1/|a|) * (1/|a|)) (which is sqrt(1)) where we have used this axiom of an inner product: = = k LH === Subject: Re: very basic question that confused me > Hi all, > Define b = a / |a|, a and b are vectors in an inner product space, > |a| is the norm of the vector > < , > is the inner product operator > sqrt = square root > In the following derivation, > * ** > |b| = sqrt() = sqrt()=sqrt(/|a||a|)=1 > Anyone please tell me how does it go from * to **?? sqrt(/|a||a|) = sqrt()/sqrt(|a||a|) = |a|/|a| = 1 === Subject: Re: very basic question that confused me Sorry, I meant how does it go from sqrt() to sqrt(/|a||a|) Sorry if I didn't make that clear. === Subject: Geometric Sequences and Series I'm having a bit of trouble with solving this sequence: (I need to supply the tenth term) (I'm putting extra spaces in between because I find it difficult to read otherwise) c ^ - 4, -c ^ - 2, 1... the difference appears to be - c ^ -2 - correct? If so, when I try and solve for the tenth term, I have the following: term 10 = term1 + (n-1) d = c ^ - 4 + (9) (- c ^ - 2) = c ^ - 4 + (- 9 c ^ - 2) = c ^ - 4 - 9 c ^ - 2 Do I have this correct so far? If so, I'm clueless what to do next. I do know that c ^ - 4 = 1/c^4 but don't know if this helps at all here. Any advice would be much appreciated, Shaynelle === Subject: Re: Geometric Sequences and Series > I'm having a bit of trouble with solving this sequence: (I need to > supply the tenth term) > (I'm putting extra spaces in between because I find it difficult to > read otherwise) > c ^ - 4, -c ^ - 2, 1... > the difference appears to be - c ^ -2 - correct? If so, when I try and > solve for the tenth term, I have the following: > term 10 = term1 + (n-1) d > = c ^ - 4 + (9) (- c ^ - 2) > = c ^ - 4 + (- 9 c ^ - 2) > = c ^ - 4 - 9 c ^ - 2 > Do I have this correct so far? If so, I'm clueless what to do next. I > do know that c ^ - 4 = 1/c^4 but don't know if this helps at all here. I think you're confused. It shows in your notation, and in the approach you've tried to use to get the 10th term. I also think you've just finished studying arithmetic sequences, because you're using words and techniques that seem to come from that type of problem. > Any advice would be much appreciated, > Shaynelle First, there is notation. If you have a complicated expression, learn to use parentheses to make the meaning clear. Second, there is the word difference. In an informal sense, it means how two items differ from each other. In mathematics, it almost always means the result of subtracting one thing from another. When you stated the difference appears to be ..., you must have been using the informal meaning, since (I hope) you must realize that when you subtract c^(-4) from -c^(-2), you just don't get -c^(-2). Replace the variable c by some numbers, and you'll see what I mean. In ordinary language, you can afford to be imprecise. That's what makes ordinary language flexible, and capable of expression. In mathematical language, you need to be precise. Is it precise to say that the difference appears to be -c^(-2)? Hardly. Your topic suggests you should know what's going on: you are writing about geometric sequences and geometric series. What makes a sequence geometric is what? A *constant ratio* of one term to its predecessor. A constant *ratio*, NOT a constant *difference* Let's look at the sequence again (I'll use parentheses to make it more legible): c^(-4), -c^(-2), 1, ... and look at the ratio of each term to its predecessor: 2nd/1st: (-c^(-2))/c^(-4) = -c^(-2) * c^4 = -c^2 3rd/2nd: 1 / (-c^(-2)) = - 1/c^(-2) = -c^2 Notice that the two ratios are the same. If it's a geometric sequence, that must be true. If those were different, I'd do two things: first, make sure I copied the problem correctly, and second, make sure I did my algebra correctly. And yes, I do check my work. Now, you need to find the 10th term? Think about it: Let's imagine we have the geometric sequence T1 T2 T3 ... (the number is just part of the name, not something multiplied by T) Q. How do you go from the 1st term to the second? A. The ratio of terms 2nd/1st is some constant, I'll call it R for ratio: T2/T1 = R We find T2 by multiplying both sides by T1: T1*(T2/T1) = T1*R notice that T1 cancels on the left side: T2 = T1*R So, we get the second term by *multiplying* the first term by the constant R. Q. How do you go from the second term to the third? A. The ratio T3/T2 is that same constant R: T3/T2 = R T2*(T3/T2)= T2*R (notice how T2 cancels) T3 = T2*R = (T1*R)*R (substituting from the above) = T1*R^2 (simplifying) We get the third term by multiplying the second term by the constant R. Let's get smart: we could go on and on, but let's find a pattern. At each step, you get the next term by multiplying by the constant R. So, if your first term is T1, the second one T1*R, the third T1*R*R = T1*R^2, the fourth T1^R^2*R = T1*R^3, and so forth: T1 = T1*R^0 (R^0 is another name for 1) T2 = T1*R^1 (R^1 is another name for R) T3 = T1*R^2 T4 = T1*R^3 What's the pattern here? It's not difficult. Now, you already know the first term of your sequence: c^(-4) You also know the constant ratio: -c^2 Can you find the 10th term of the sequence now? That's what my contribution is. Dale === Subject: Re: Geometric Sequences and Series > I'm having a bit of trouble with solving this sequence: (I need to > supply the tenth term) > (I'm putting extra spaces in between because I find it difficult to > read otherwise) > c ^ - 4, -c ^ - 2, 1... > the difference appears to be - c ^ -2 - correct? If so, when I try and > solve for the tenth term, I have the following: > term 10 = term1 + (n-1) d > = c ^ - 4 + (9) (- c ^ - 2) > = c ^ - 4 + (- 9 c ^ - 2) > = c ^ - 4 - 9 c ^ - 2 > Do I have this correct so far? NO! You seem to be confusing geometric sequences with arithmetic sequences, and sequences with series. In the geometric sequence c^(-4), c^(-2), 1, ..., each term is a multiple of the previous term, so you have to figure out what the common ratio of successive terms is. Compare the sequence a, a*r, a*r*r = a*r^2, a*r*r*r=a*r^3, ... versus the sequence a, a+d, a+d+d = a+2*d, a+d+d+d = a + 3*d, ... The first of these is a geometric sequence (each new term is formed by multiplying the previous term by the same fixed number). The second is an arithmetic sequence ( each new term is formed by adding to the previous term the same fixed number) For a geometric sequence, the common ratio (r in the above) can be found by dividing any term by the immediately previous term. For an arithmetic sequence the common increment or difference can be found by subtracting form any term the immediately previous term. For a series, one adds together a number of successive terms of a sequence. A finite geometric series takes the terms a, a*r, a*r^2, ..., a*r^(n-1) of a finite geometric sequence and adds them together as a + a*r + a*r^2 + ... + a*r^(n-1), where n is the number of terms. For values on r strictly between -1 and 1, one can also have an infinite geometric series with a finite value. > If so, I'm clueless what to do next. I > do know that c ^ - 4 = 1/c^4 but don't know if this helps at all here. > Any advice would be much appreciated, > Shaynelle === Subject: Re: Geometric Sequences and Series >I'm having a bit of trouble with solving this sequence: (I need to >supply the tenth term) >(I'm putting extra spaces in between because I find it difficult to >read otherwise) >c ^ - 4, -c ^ - 2, 1... I think it would be better to use parentheses: c^(-4), c^(-2), 1, ... >the difference appears to be - c ^ -2 - correct? No -- try it for a particular value of c, for example c=2. >If so, when I try and >solve for the tenth term, I have the following: >term 10 = term1 + (n-1) d >= c ^ - 4 + (9) (- c ^ - 2) >= c ^ - 4 + (- 9 c ^ - 2) >= c ^ - 4 - 9 c ^ - 2 >Do I have this correct so far? No. >If so, I'm clueless what to do next. > I do know that c ^ - 4 = 1/c^4 but don't know if this helps at all here. It might help, but there's an easier way ... Hints: Rewrite 1 as a power of c. The notice the pattern of the exponents. quasi === Subject: Re: Geometric Sequences and Series > I'm having a bit of trouble with solving this sequence: (I need to > supply the tenth term) > (I'm putting extra spaces in between because I find it difficult to > read otherwise) > c ^ - 4, -c ^ - 2, 1... > the difference appears to be - c ^ -2 - correct? If so, when I try and > solve for the tenth term, I have the following: > term 10 = term1 + (n-1) d > = c ^ - 4 + (9) (- c ^ - 2) > = c ^ - 4 + (- 9 c ^ - 2) > = c ^ - 4 - 9 c ^ - 2 > Do I have this correct so far? No. If c=2, your given terms are 2^(-4) = 1/16, 2^(-2) = 1/4, 1, ... The difference between 1/16 and 1/4 is not 1/4, In fact, the difference between the first two numbers is 1/4 - 1/16 = 3/16, while the difference between the second and third numbers is 1 - 1/4 = 3/4, etc. The differences are NOT constant. > If so, I'm clueless what to do next. I Learn the difference between an _arithmetic_ series and a _geometric_ series. What kind do you have here? R.G. Vickson > do know that c ^ - 4 = 1/c^4 but don't know if this helps at all here. > Any advice would be much appreciated, > Shaynelle === Subject: Re: Geometric Sequences and Series > I'm having a bit of trouble with solving this sequence: (I need to > supply the tenth term) What exactly do you mean by solving a sequence. I know about solving problems, but what is there to solve? > (I'm putting extra spaces in between because I find it difficult to > read otherwise) > c ^ - 4, -c ^ - 2, 1... even with extra spaces in between it is still difficult to comprehend. If I try to solve the sequence 1,2,3 Is 1,2,3,4,..n a solution ? Or is 1,2,3,5,7,11,13, one? > the difference appears to be - c ^ -2 - correct? If so, when I try and > solve for the tenth term, I have the following: > term 10 = term1 + (n-1) d > = c ^ - 4 + (9) (- c ^ - 2) > = c ^ - 4 + (- 9 c ^ - 2) > = c ^ - 4 - 9 c ^ - 2 > Do I have this correct so far? If so, I'm clueless what to do next. I > do know that c ^ - 4 = 1/c^4 but don't know if this helps at all here. try c=0 and c=1 ... or c = 10^207 > Any advice would be much appreciated, > Shaynelle === Subject: Re: Geometric Sequences and Series > I'm having a bit of trouble with solving this sequence: (I need to > supply the tenth term) > (I'm putting extra spaces in between because I find it difficult to > read otherwise) > c ^ - 4, -c ^ - 2, 1... > the difference appears to be - c ^ -2 - correct? If so, when I try and > solve for the tenth term, I have the following: > term 10 = term1 + (n-1) d > = c ^ - 4 + (9) (- c ^ - 2) > = c ^ - 4 + (- 9 c ^ - 2) > = c ^ - 4 - 9 c ^ - 2 > Do I have this correct so far? If so, I'm clueless what to do next. I > do know that c ^ - 4 = 1/c^4 but don't know if this helps at all here. > Any advice would be much appreciated, > Shaynelle In an arithmetic sequence, the DIFFERENCE of consecutive terms is the same. For example, 1, 3, 5, 7, .... (difference is 2). In a geometric sequence, the RATIO of consecutive terms is the same. For example, 1, 2, 4, 8, 16, 32, ... (ratio is 2). You are looking at the DIFFERENCE. Look at the RATIO (one term divided by another). What is (c^(-2))/(c^(-4))? What is 1/(c^(-2))? What should the next term be to get the same ratio? === Subject: Re: Geometric Sequences and Series > I'm having a bit of trouble with solving this sequence: (I need to > supply the tenth term) > (I'm putting extra spaces in between because I find it difficult to > read otherwise) > c ^ - 4, -c ^ - 2, 1... > the difference appears to be - c ^ -2 - correct? No, that is the ratio > If so, when I try and > solve for the tenth term, I have the following: > term 10 = term1 + (n-1) d This is the formula for an arithmetic sequence, not a geometric sequence. > = c ^ - 4 + (9) (- c ^ - 2) > = c ^ - 4 + (- 9 c ^ - 2) > = c ^ - 4 - 9 c ^ - 2 > Do I have this correct so far? If so, I'm clueless what to do next. I > do know that c ^ - 4 = 1/c^4 but don't know if this helps at all here. > Any advice would be much appreciated, > Shaynelle === Subject: Re: On the Non Computability of Numbers On Fri, 28 Apr 2006 19:57:46 -0500, Nicky in >> On Thu, 27 Apr 2006 20:14:24 -0500, Nicky in >> On Thu, 27 Apr 2006 13:41:51 -0500, Nicky in > On Thu, 27 Apr 2006 10:05:23 -0500, Nicky in > On the Non Computability of Numbers >> ~v~~ >> In order to judge the non computability of numbers we first ask what >> is meant by the computability of numbers and proceed from there to >> determine non computability. >> The computability of numbers is decided on a framework of Cartesian >> geometry, Cantorian arithmetic, and Turing von Neumann mechanics. >> In other words we take straight lines intersecting at right angles >> and >> uniform subdivisions of this space as computable and determine that >> all things operated on with TvN mechanics within it are computable >> numbers and that anything not within the space can not be operated >> on with TvN mechanics and thus are not computable numbers. >> So now we must ask whether everything represents a computable >> number? no. >> Everything doesn't represent a computable number or we must not ask >> the question? If the former perhaps you could explain how you know it? Read your question, ...everything represents a computable number? >> And it is obvious you didn't read it at all. >What does it mean to you ? Just what it says. > Can you state it more clearly ? No. >What is everything? >> What does it sound like? >say it louder, I can't hear you. >> Well you see I used the word everything to indicate everything >> except one particular grain of sand. >Then your statements are utter non-sense. Of course they are. They're just not quite so utter as your nonsense. > Do you intending to include sand grain > # > K on beach B >at location C ? >> Beats me. What does it sound like? >so you do not know what you are talking about >(sound part -See Above.) >> Of course I don't. Why else would I be talking about it? I just happen >> to know more about what I'm talking about than you do. >No, you don't. Prove it. I just did. >You only exist because I have allow it. Huh? Perhaps you would care to rephrase that? Assuming you're a native english speaker. Or maybe a little ESL is in order. >> For example are space, motion through space and even intelligence >> just >> so many variations on computable numbers? no. >> Is this something besides an opinion? Your statement is advanced speculation and fails on the first reading. >> That's nice. Do you often include reasons for your opinions? >Since you acknowledged you failed, you now get an F. (you would have >passed >with a D) >> Whereas you pass gas with an A. You still haven't posted reasons for >> your opinions. You just post more opinions. >Why did you decide to get an F ? You could pass gass with an A too. But no, >you have decided that failure is your main path, like numbering >everything. Better get started on that one. Yeah. It definitely looks like ESL is in order. >Try again. >> Why? >Why not? >> Why? >Because you will learn, and that would be good for you. Well that's another matter of opinion. >> To answer this question we take a sequence of two or more computable >> numbers, for example 01101 and 10011 and ask what lies in between? 01011, 01100, 01101, 01110, 01111, 10000, 10001, 10010 that's trivial >> What's trivial? Answering your minor question about numbering. >> What minor question about what? >Where? >> Where what? >Where is your numbering ability at? have you started pasting numbers on >things ? Not yet. >>The enumeration of some computable numbers? If so I >> agree. If you are referring to something else as trivial I suggest you >> explain what and why. see above >> Why? >Wrong, it is How? How Much? >> Not very. >Actually completely. Wrong So you say. So you don't prove. Another opinion. >> And >> obviously we must in fact have some things which are not computable >> or >> we could not have distinct groupings in the sequence. In other words >> there must be something between 01101 and 10011 not computable or >> we could not have distinct groups of binary digits. ? >> I thought you just said the problem was trivial. It is, the answer is above, you have my permission to write it down on a >piece of paper and keep it with you. >> On the other hand if we maintain that what lies between computable >> groupings and even digits is computable we are then faced with an >> impossible situation. >?? >> ???? ?? means, what are you talking about ?? >> And ???? means what are you talking about ???? >What are you talking about !!!! >> What are you talking about !!!! ???? >no, what are *YOU*, talking about what ?? Yeah sure. >>Let's say that within the group of binary digits >> 01101 we ask once again what lies between individual digits. >> Obviously >> that cannot also be computable or between any combination of >> computable digits we will have further computable digits ad >> infinitum. just a scaling problem, trivial. >> I see. So where does the scaling problem begin and end exactly? add a decimal point, no problem. >> Why? We're not talking decimals. >Yes we are, and a decimal point is the solution to your problem. >> Funny I rather imagined I was talking binary. It's a little hard to >> tell what you're talking other than more undemonstrated opinions. >binary also uses the decimal point. I rather imagined binary uses a binary point. >you wrongly assumed integer binary. I did? >I correct your mistakes for free this time, but don't let it happen again. I didn't let your mistakes happen the first time. >> Scaling only applies to scaled phenomena which computable numbers are >> not since what's between computable numbers aren't computable numbers. Add a decimal point. >> I never suggested that they are not computable. I suggested you are >> not computable. You are however inscrutable. >You are the one that stated ...computable numbers aren't computable >numbers. >> Of course I did. I just can't tell where. Perhaps you can point out >> the relevant quote. If not you can just go on making up fairy tales. >Sure, check for yourself; you. >Let me know which one it is, is not. >> As soon as you point out the reference in context I'll be happy to >> point out which one it is and which one it is not. >See Above Ditto. > Then they are computable. >> I suspect they're computable in any event. >01101 divided by 10011 is what ? >> You? >Google for it, it is out there. Bet you did not know that. >> And I'll bet all you know is what Google tells you. >what is 10110.01101 ? Can you tell me in base 10, octal or hex ? >(hint: floating point) Hint: Go yourself. >> Therefore we are completely justified in assuming that not >> everything >> represents a computable number because whatever lies between such >> numbers cannot be computable. you have two systems mixed, reality and discrete numbers. >> Really? Then perhaps you'd care to explain the difference in >> mechanical terms. Also analog and digital. >> Unintelligent and intelligent? With you as the former and me as the >> latter. >you mean lard-der. >> I mean fathead. >you are just learning, and it is OK, you will go far. Au contraire, asshole. > But learn about >floating point representations in various number systems first, it will >spair you from numbering everything and then have to go back and number >the things you missed in between. Gee. I may be mistaken here but I seem to recollect programming decimal/binary floating point conversions in ASSEMBLER before you were born. >>And furthermore we are completely >> justified in assuming computable numbers merely represent only a >> mechanical subset of whatever lies between computable numbers. whatever >> I couldn't agree more. >> ~v~~ A good learning lesson for you. Try harder next time. >> Whereas you'll still just be trying all the time. ~v~~ === Subject: Re: Evidence For The Randomness Of Prime Numbers I also write a short paper for the randomness of prime number. But the conclusion is that the prime numbers is not strict random. Here is its abstract: The prime numbers look like a randomly chosen sequence of natural numbers, but there is still no strict theory to determine 'Randomness'. In these years, cryptography has developed a battery of statistical tests for randomness. In this paper, we just apply these methods to study the distribution of primes. Here the binary sequence constructed by second difference of primes is used as samples. We find this sequence can't reach all the 'random standard' of FIPS 140-1/2, but still show obvious random feature. The interesting self-similarity is also observed in this sequence. These results add the evidence that prime numbers is a chaos system. May be interesting. Full paper: http://arxiv.org/abs/math.NT/0603450 > Martin Winer and myself have written an internet > publication about the > randomness of prime numbers based on the Last Digit > Distribution. The > abstract is given below. > -----------------