mm-3079 === Subject: Re: Even and Odd Problem It seems also that with a, b, c, pairwise coprime, any FLT counterexample would have 2 odds, but I know that's not true, since 2a, 2b, 2c would also be a counterexample. We did a bit on proof in Discrete Math. Direct, Indirect, and By Contradiction. But I had to drop 'cause of the summer heat and the commute. It was dreadful. Where's that damn FLT For Amateurs book? Doug >In any counterexample to FLT, a^n + b^n = c^n. What can be said about >a, b, and c and their even/odd status? >I know any power of even is even, and any power of odd is odd, and even >+ odd is odd, etc... > Pretending for a moment that FLT has not been proven and therefore could > possibly have counterexamples... > There can't be an odd number of odd numbers in the equation, so there must > be 0 or 2 odd in (a,b,c). > If there are 0 odd and 3 even, the case can be reduced by dividing each > number by 2. > a^n + b^n = c^n > (1/2)^n * a^n + (1/2)^n * b^n = (1/2)^n * c^n > (a/2)^n + (b/2)^n = (c/2)^n > Repeat until you get 2 odds. So if there's a FLT counterexample, there's > one with 2 odds. The same goes for Pythagorean triples. > --Keith Lewis klewis {at} mitre.org > The above may not (yet) represent the opinions of my employer. === Subject: Re: Even and Odd Problem > Where's that damn FLT For Amateurs book? There are several of them - one by Ribenboim, one by van der Pooreten, one by Aczel, one by Singh.... -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Even and Odd Problem > It seems also that with a, b, c, pairwise coprime, any FLT > counterexample would have 2 odds, but I know that's not true, since 2a, > 2b, 2c would also be a counterexample. It is true for the primitive solutions in which a and b are coprime, which 2*a and 2*b are not. Coprimeness of a pair of integers requires that they have a greatest common divisor of 1. === Subject: Re: Even and Odd Problem This is getting interesting. We are now getting near a discussion of proof styles. I would like to discuss that some as I did well in Logic so many years ago. (25 years ago) > It seems also that with a, b, c, pairwise coprime, any FLT > counterexample would have 2 odds, but I know that's not true, since 2a, > 2b, 2c would also be a counterexample. > It is true for the primitive solutions in which a and b are coprime, > which 2*a and 2*b are not. Coprimeness of a pair of integers requires > that they have a greatest common divisor of 1. Is the method of proof by contradiction the most popular for the many failed, incomplete, or flawed proofs of FLT? Did the successful proof of Wiles et al. use this method, or was it a direct or indirect proof instead? FLT by Contradiction: Theorem n>2 / 0 a^n + b^n # c^n First, am I stating that right? Since we can work out the few cases of n,a,b,c by hand, would 2 a^n + b^n # c^n be essentially equivalent, given those cases? Proof structure: Suppose a^n + b^n = c^n. Hands wave, chalk dust flies, and we find a contradiction such as 2 = 3. Therefore a^n + b^n # c^n since there are only two cases for the (in) equality? FLT by Direct Proof: If one could prove a^n + b^n irrational (it's not!) and c^n rational (it's an integer, we know that), one would have shown by direct proof that a^n + b^n # c^n because rational # irrational. I am sure there are other possiblities for direct proof, althought I don't know what they are. FLT by Indirect Proof: Likewise, and I am less sure of this, an indirect proof would directly prove the converse, which is that a^n + b^n = c^n --> ~(n>2 / 0 ~(n>2) / ~(0 ~(n>2) or a^n + b^n = c^n --> ~(0It seems also that with a, b, c, pairwise coprime, any FLT >counterexample would have 2 odds, but I know that's not true, since 2a, >2b, 2c would also be a counterexample. 2a, 2b and 2c are not pairwise coprime. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Even and Odd Problem counterexample if a^n + b^n = c^n. >It seems also that with a, b, c, pairwise coprime, any FLT >counterexample would have 2 odds, but I know that's not true, since 2a, >2b, 2c would also be a counterexample. > 2a, 2b and 2c are not pairwise coprime. Well, 2a, 2b, and 2c are not pairwise coprime, but going back a bit: If there were a FLT countexample. then there would be one with 2 odds. a^n + b^n = c^n in that case 2^n( a^n + b^n) = 2^n*c^n and (2a)^n + (2b)^n = (2c)^n would also be a counterexample, no? === Subject: Re: Question about modular function related to the projective line > I'm learning about modular functions and related topics, and I have > computed the following as an example in order to understand better. > Consider the projective line and define: > f_p(T):= exp( Sum_{k=1..infty} (N(p^k)*T^k/k) ) > where p is any prime and N(x)=x+1 (this is just counting the number of > points in the projective line on finite fields). > Then I form the product over all the primes: > f(s)=prod_p f_p(p^(-s)) = Sum_n a(n).n^(-s) > and I get a(n)=1,3,4,7,6,12,8,14,13,18,12,28,... I also know that this > is equal to z(s).z(s-1), z being the classical zeta function. Now, you > can use these coefficients to define a modular function: > f(q)=Sum a(n).q^n where q=exp(2Pi.i.z) > I haven't been able to find out anything about this modular form in the > literature, although I'm pretty sure this is well known... anyone can > say anything about this function? Does it have any meaning at all? Are you sure that 14 isn't a 15? If it is a 15, maybe your a(n) is just sigma(n), the sum of the divisors of n. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Question about modular function related to the projective line is just sigma(n), the sum of the divisors of n. > -- > Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) Actually, it is very easy to prove that z(s).z(s-1) gives sigma(n) as you suggest, so I probably made a computational mistake (truncating some series too soon or something). So, what is known about the modular function f(q) = Sum_n sigma(n).q^n ? I'm calling it a modular function but it's just a series for now, strictly speaking, since I know nothing about it's invariance under the action of a group, other than z->z+1, of course. David === Subject: Re: Another Galois theory problem Say f(x) is irreducible in F[x] and E is a normal extension of F. If >f(x) = f_1(x)f_2(x)...f_k(x), each f_i(x) in E[x] and irreducible over >E, show each degree f_i(x) = degree f_k(x). > Have you looked at all at automorphisms of E over F? Try to see if, > for every i, you can find an automorphism of E over F that maps f_1(x) > to f_i(x). To do this, I would want to use two facts, which I put in quotes because I cannot find them anywhere as theorems and yet one of them seem to be part of that giant body of Galois theory common sense which one seems to have to simply be born with; and the other I know isn't even true, but it seems the only thing which would accomodate your suggestion.. 1. An automorphism sends roots of same irreducible factors to roots of same irreducible factors 2. A permutation of roots of a polynomial determined an automorphism Assuming these were true, one would simply split f(x) over an algebraic closure, take the automorphism induced (using 2) by permuting some root of f_1(x) with some root of f_i(x), and then (by 1) this would have to send f_1(x) to f_i(x) like you suggest. Of course rule 2 above is patently false since then every polynomial with distinct roots would have galois group being the symmetric group. And nowhere have I cited normality... Is rule 1 even correct? It is the kind of thing which TAs throw around at recitations but there is no trace of it anywhere in literature. My hope is that rule 1 is correct and that some weaker version of rule 2 somehow holds with some normality conditions. Like I said in my last post, I think there is some heuristic which I am just not getting when it comes to normal extensions, because right now when someone tells me The extension is normal, this sheds no illumination whatsoever, beyond the raw definitions. === Subject: Re: Another Galois theory problem days. My association with the Department is that of an alumnus. >Say f(x) is irreducible in F[x] and E is a normal extension of F. If >f(x) = f_1(x)f_2(x)...f_k(x), each f_i(x) in E[x] and irreducible over >E, show each degree f_i(x) = degree f_k(x). > Have you looked at all at automorphisms of E over F? Try to see if, >> for every i, you can find an automorphism of E over F that maps f_1(x) >> to f_i(x). >To do this, I would want to use two facts, which I put in quotes >because I cannot find them anywhere as theorems and yet one of them >seem to be part of that giant body of Galois theory common sense >which one seems to have to simply be born with; and the other I know >isn't even true, but it seems the only thing which would accomodate >your suggestion.. >1. An automorphism sends roots of same irreducible factors to roots of >same irreducible factors True and trivial. Let f(x) be an irreducible with coefficients in F, let r be a root of f(x) in E, and let s be an automorphism of E over F; then s fixes F pointwise. Write f(x) = x^n + a_{n-1}x^{n-1} + ... + a_1 x + a_0 with a_i in F. Then 0 = f(r) = r^n + a_{n-1}r^{n-1} + ... + a_1 r + a_0 Then 0 = s(0) = s(r^n + a_{n-1}r^{n-1} + ... + a_1 r + a_0) = s(r)^n + a_{n-1}s(r)^{n_1} + ... + a_1s(r) + a_0 = f(s(r)) so if r is a root, so is s(r). >2. A permutation of roots of a polynomial determined an automorphism False. An automorphism determines a permutation of the roots of a given irreducible polynomial; but even in the irreducible case (a key hypothesis you dropped here), not every permutation need be induced by an automorphism. What ->is<- true, and you can most certainly find it in books, is that if f(x) is an irreducible polynomial with coefficients in F, and E is a splitting field of f, then for any two roots r1 and r2 of f(x) in E (not necessarily distinct), there exists an automorphism of E over F that sends r1 to r2 (the action on the roots is transitive). >Is rule 1 even correct? Yes. >Like I said in my last post, I think there is some heuristic which I >am just not getting when it comes to normal extensions, because right >now when someone tells me The extension is normal, this sheds no >illumination whatsoever, beyond the raw definitions. I think there is quite a bit about extensions in general that you are not getting, not just normal ones. I do not mean this as a put down, but as an observation. In an earlier post you commented about enlarging an extension to make it algebraic. This betrays a deep misunderstanding about algebraicity and extensions in general. You cannot make an extension algebraic by extending it, if it was not algebraic to begin with. Usually, you need to ->restrict<- it, if it was not algebraic to begin with, in order to consider an algebraic extension (thus, rather than consider the complex numbers over Q, you need to consider the smaller field of algebraic numbers); even then it may not always work. I think you need to review all of that stuff again. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Another Galois theory problem Say f(x) is irreducible in F[x] and E is a normal extension of F. If >f(x) = f_1(x)f_2(x)...f_k(x), each f_i(x) in E[x] and irreducible over >E, show each degree f_i(x) = degree f_k(x). > Have you looked at all at automorphisms of E over F? Try to see if, > for every i, you can find an automorphism of E over F that maps f_1(x) > to f_i(x). > To do this, I would want to use two facts, which I put in quotes > because I cannot find them anywhere as theorems and yet one of them > seem to be part of that giant body of Galois theory common sense > which one seems to have to simply be born with; and the other I know > isn't even true, but it seems the only thing which would accomodate > your suggestion.. > 1. An automorphism sends roots of same irreducible factors to roots of > same irreducible factors > 2. A permutation of roots of a polynomial determined an automorphism > Assuming these were true, one would simply split f(x) over an algebraic > closure, take the automorphism induced (using 2) by permuting some root > of f_1(x) with some root of f_i(x), and then (by 1) this would have to > send f_1(x) to f_i(x) like you suggest. Of course rule 2 above is > patently false since then every polynomial with distinct roots would > have galois group being the symmetric group. And nowhere have I cited > normality... > Is rule 1 even correct? It is the kind of thing which TAs throw around > at recitations but there is no trace of it anywhere in literature. > My hope is that rule 1 is correct and that some weaker version of rule > 2 somehow holds with some normality conditions. > Like I said in my last post, I think there is some heuristic which I > am just not getting when it comes to normal extensions, because right > now when someone tells me The extension is normal, this sheds no > illumination whatsoever, beyond the raw definitions. After posting this, I realized I ought to use the theorem, Galois group of an irreducible polynomial acts transitively on the roots (which I found browsing sci.math archive, it is not in my textbook anywhere). So take a splitting field of f, it necessarily has an automorphism which fixes F and sends a root of f_1(x) to a root of f_i(x), and hence (if rule 1 above is true) sends f_1(x) to f_i(x). But if this is the case, I am further confused: what about the normality hypothesis? === Subject: Re: Another Galois theory problem days. My association with the Department is that of an alumnus. >Say f(x) is irreducible in F[x] and E is a normal extension of F. If >f(x) = f_1(x)f_2(x)...f_k(x), each f_i(x) in E[x] and irreducible over >E, show each degree f_i(x) = degree f_k(x). Me: > Have you looked at all at automorphisms of E over F? Try to see if, > for every i, you can find an automorphism of E over F that maps f_1(x) > to f_i(x). >After posting this, I realized I ought to use the theorem, Galois >group of an irreducible polynomial acts transitively on the roots >(which I found browsing sci.math archive, it is not in my textbook >anywhere). So take a splitting field of f, it necessarily has an >automorphism which fixes F and sends a root of f_1(x) to a root of >f_i(x), and hence (if rule 1 above is true) sends f_1(x) to f_i(x). >But if this is the case, I am further confused: what about the >normality hypothesis? Sigh... Take a splitting field K of f containing E. Take the automorphism of K over F that sends a root of f1 to a root of fi. This automorphism must restrict to an automorphism of E, ->BY NORMALITY<-. Work from there. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Name of chapter dependency diagrams sought > At the beginning of some textbooks, especially mathematics texts, there > is a diagrammatic representation of a directed acyclic graph showing, > for each chapter, which earlier chapters it assumes you understand. > What are these diagrams called? > In particular, there is a German word for these things. Anyone know > what it is? Leitfaden. See, e.g., Serre's Local Fields. -- m === Subject: Re: Product of matrices of zero trace >Prove that any matrix is product of matrices of trace 0. Do you have >any ideas ? I assume these are n x n matrices where n >= 2 (of course it's not true for n=1), and you're talking about the product of two matrices of trace 0. Let C be any n x n matrix. Case 1: C is diagonal. Let P be the matrix for a permutation with no fixed points, and write C = P (P^(-1) C), noting that P and P^(-1) C have all 0's on the diagonal. Case 2: C is not diagonal. If C_{ij} <> 0 where i <> j, then I claim C = A B where A is obtained from C and B from I by changing the i'th columns. Using a renumbering of the rows and columns if needed, we can assume WLOG i=n and write C in block-matrix form as [ W x ] C = [ y' z ] with y' <> 0, where W is the top left (n-1) by (n-1) submatrix. Take t = -trace(W), and write [ W u ] [ I v ] [ W Wv + (1-n)u ] A B = [ y' t ] [ 0 1-n ] = [ y' y'v + (1-n)t ] Let v be any vector with y'v = z - (1-n)t, and u = (x - Wv)/(1-n), and this will work. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Product of matrices of zero trace Prove that any matrix is product of matrices of trace 0. Do you have >any ideas ? > I assume these are n x n matrices where n >= 2 (of course it's not > true for n=1), and you're talking about the product of two matrices of > trace 0. > Let C be any n x n matrix. > Case 1: C is diagonal. Let P be the matrix for a permutation with > no fixed points, and write C = P (P^(-1) C), noting that P and > P^(-1) C have all 0's on the diagonal. > Case 2: C is not diagonal. If C_{ij} <> 0 where i <> j, then > I claim C = A B where A is obtained from C and B from I > by changing the i'th columns. Using a renumbering of > the rows and columns if needed, we can assume WLOG i=n > and write C in block-matrix form as > [ W x ] > C = [ y' z ] > with y' <> 0, where W is the top left (n-1) by (n-1) submatrix. > Take t = -trace(W), and write > [ W u ] [ I v ] [ W Wv + (1-n)u ] > A B = [ y' t ] [ 0 1-n ] = [ y' y'v + (1-n)t ] > Let v be any vector with y'v = z - (1-n)t, and u = (x - Wv)/(1-n), > and this will work. How about this nice problem : Every non-invertible matrix can be represented as a product of nilpotent matrices. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada === Subject: Re: Product of matrices of zero trace >Prove that any matrix is product of matrices of trace 0. Do you have >any ideas ? I have an idea that this isn't true for the one-by-one identity matrix, at least. Lee Rudolph === Subject: JSH: Way too interesting So I have this neat result using congruences which is so easy and trivial that I can just put it out there and watch what happens! And on this group, surprising even me, there is still the usual reaction. I can check with other groups and see what happens, shifting how I present the mathematics. Far more interesting than I thought possible. It's like a study of the world with the most powerful intellectual microscope ever built--a simple solution to the factoring problem versus a social view that I'm just some crackpot. === Subject: Re: Way too interesting > So I have this neat result using congruences which is so easy and > trivial that I can just put it out there and watch what happens! > And on this group, surprising even me, there is still the usual > reaction. > I can check with other groups and see what happens, shifting how I > present the mathematics. > Far more interesting than I thought possible. > It's like a study of the world with the most powerful intellectual > microscope ever built--a simple solution to the factoring problem > versus a social view that I'm just some crackpot. > You're labeled a crackpot largely in part to how you act. Are you that stupid that you can't realize that? === Subject: Re: Way too interesting So I have this neat result using congruences which is so easy and > trivial that I can just put it out there and watch what happens! > And on this group, surprising even me, there is still the usual > reaction. > I can check with other groups and see what happens, shifting how I > present the mathematics. > Far more interesting than I thought possible. > It's like a study of the world with the most powerful intellectual > microscope ever built--a simple solution to the factoring problem > versus a social view that I'm just some crackpot. > > You're labeled a crackpot largely in part to how you act. Are you that > stupid that you can't realize that? > Dave But actions are irrelevant to mathematical proof. And besides, I have often noted that a lot of my postings are for entertainment value, or are out of boredom, so it does not quite make sense that the hostility that can be read repeatedly in replies to me, like in your posting here, is just about my behavior. There is something more, some kind of reaction formation? The crucial difference now, which allows the possibility of answers, is the simple factoring proof. How is your brain reacting to the new element? === Subject: Re: Way too interesting So I have this neat result using congruences which is so easy and >trivial that I can just put it out there and watch what happens! > And on this group, surprising even me, there is still the usual >reaction. > I can check with other groups and see what happens, shifting how I >present the mathematics. > Far more interesting than I thought possible. > It's like a study of the world with the most powerful intellectual >microscope ever built--a simple solution to the factoring problem >versus a social view that I'm just some crackpot. > > > You're labeled a crackpot largely in part to how you act. Are you that > stupid that you can't realize that? > Dave > But actions are irrelevant to mathematical proof. And they would be, if you had ... Ah, skip it. > And besides, I have often noted that a lot of my postings are for > entertainment value, or are out of boredom, so it does not quite make > sense that the hostility that can be read repeatedly in replies to me, > like in your posting here, is just about my behavior. Then why are you posting here, and not at some place like alt.psychology? > There is something more, some kind of reaction formation? > The crucial difference now, which allows the possibility of answers, is > the simple factoring proof. You have an equation which, when solved, solves the factoring problem. Well, so do I. It is (a + 2) (b + 2) = T, where a and b are allowed to be nonnegative integers. You have given no _algorithm_, no precise procedure, for factoring an arbitrary integer. Any competant computer programmer knows this. OTOH, maybe that's why you got fired from Alltel ... Say ... Many years back, my brother got a bill from Alltel which was ten times too large. Fortunately the company saw that it was a mistake and recalculated it for him. You didn't have a hand in that, by any chance? > How is your brain reacting to the new element? You mean the plutonium atom? Yes, it has turned me into a super-genius. --- Christopher Heckman === Subject: Re: Way too interesting > So I have this neat result using congruences which is so easy and > trivial that I can just put it out there and watch what happens! > And on this group, surprising even me, there is still the usual > reaction. > I can check with other groups and see what happens, shifting how I > present the mathematics. > Far more interesting than I thought possible. > It's like a study of the world with the most powerful intellectual > microscope ever built--a simple solution to the factoring problem > versus a social view that I'm just some crackpot. You're labeled a crackpot largely in part to how you act. Are you that >> stupid that you can't realize that? >> Dave > But actions are irrelevant to mathematical proof. > And besides, I have often noted that a lot of my postings are for > entertainment value, or are out of boredom, so it does not quite make > sense that the hostility that can be read repeatedly in replies to me, > like in your posting here, is just about my behavior. > There is something more, some kind of reaction formation? > The crucial difference now, which allows the possibility of answers, is > the simple factoring proof. > How is your brain reacting to the new element? > When I read a proof, I try to analyze its logic; I often find myself getting lost in your proofs. Does that make me any less a mathematician? If you don't like the reaction you get, either change your style or don't post. It's simple. === Subject: Re: JSH: Way too interesting > So I have this neat result using congruences which is so easy and > trivial that I can just put it out there and watch what happens! > And on this group, surprising even me, there is still the usual > reaction. > I can check with other groups and see what happens, shifting how I > present the mathematics. Everybody duck! He's going to start spamming other newsgroups again. > Far more interesting than I thought possible. > It's like a study of the world with the most powerful intellectual > microscope ever built--a simple solution to the factoring problem > versus a social view that I'm just some crackpot. I don't think anyone needs a microscope for that. > === Subject: Re: JSH: Way too interesting > So I have this neat result using congruences which is so easy and > trivial that I can just put it out there and watch what happens! > And on this group, surprising even me, there is still the usual > reaction. > I can check with other groups and see what happens, shifting how I > present the mathematics. > Everybody duck! He's going to start spamming other newsgroups again. > Far more interesting than I thought possible. > It's like a study of the world with the most powerful intellectual > microscope ever built--a simple solution to the factoring problem > versus a social view that I'm just some crackpot. > I don't think anyone needs a microscope for that. I am curious about the type of thinking that would allow the denial of my research for over 4 years, as can it just be as simple as a person like yourself knowing the truth but just lying about it? I would like to think, no. So, for some reason, you believe that I am wrong, but why? And now with a result that cannot be long denied because of its practical relevance, whatever rationalizations you have used to justify your behaviour SHOULD I would think, collapse. Yet here you are, still posting in defiance of that analysis. What am I missing? What is going on inside your head? === Subject: Re: JSH: Way too interesting > So, for some reason, you believe that I am wrong, but why? I believe you are wrong because you are extremely stupid. === Subject: Re: JSH: Way too interesting >> So, for some reason, you believe that I am wrong, but why? > I believe you are wrong because you are extremely stupid. No, thats simply not true! We all know that he is a genius and far beyond his time!!!!!! Why not admit it!!! Hell, We all should be willing to say he's right just to shut him up if it would work!! === Subject: Re: JSH: Way too interesting > So, for some reason, you believe that I am wrong, but why? >> I believe you are wrong because you are extremely stupid. > No, thats simply not true! We all know that he is a genius and far beyond > his time!!!!!! Why not admit it!!! Hell, We all should be willing to say > he's right just to shut him up if it would work!! Yes, I too am totally convinced that JSH is a complete Genius, just like Gauss or Newton but in the factoring field. If he simply factored one number, then he would prove to the world he has been wronged all these years, and he will show them, he will show them, and they will quiver at his mighty wrath. === Subject: Re: JSH: Way too interesting >> So, for some reason, you believe that I am wrong, but why? > I believe you are wrong because you are extremely stupid. >> No, thats simply not true! We all know that he is a genius and far beyond >> his time!!!!!! Why not admit it!!! Hell, We all should be willing to say >> he's right just to shut him up if it would work!! > Yes, I too am totally convinced that JSH is a complete Genius, just like > Gauss or Newton but in the factoring field. > If he simply factored one number, then he would prove to the world he has > been wronged all these years, and he will show them, he will show them, > and they will quiver at his mighty wrath. No! He has already done it! He has proven that every number can be factored in 2*4 or 3*19.43828! Thats what his surrogate factoring is all about! Its quite amazing if you would take the time to understand it. I'm just now realizing its true potential! I think it is the best discovery of all humanity! === Subject: Re: Irrationality and the Fundamental Theorem of Arithmetic > But I then came across a simpler, seemingly more basic proof >> ascribed to Niven (which doesn't invoke the FTAr): >> >> If r is rational then >> there is a least integer b such that br is an integer >> but (br-b) is less than b (since r is btwn 1 and 2) >> and then (br-b)r also is an integer, >> which contradicts the assumption that b was the least >> and so r is irrational. According to Jorn Steuding, Diophantine Analysis, page 15, >this idea for proving the irrationality of sqrt 2 is due >to Estermann. A reference is Math Gazette 59 (1975), page 110. > > That proof is much, much older than 1975. > I'm shocked any number theorist could believe it so new. > Are you sure the reference is not to a different proof? > The London Mathematical Society published a long obituary > for Estermann. It's on the web, and one page of it, > http://www.numbertheory.org/obituaries/LMS/estermann/page11.html > is relevant here. It says, In retirement, Estermann discovered > [1975] an elegant proof of Pythagoras's theorem which is actually > simpler than the original and is sufficiently short to be > included verbatim. [note - from what follows, it's evident > that they don't mean the theorem about the square of the hypotenuse, > they mean the irrationality of the square root of two] > They then give Estermann's proof, which is essentially > the one ascribed to Niven up near the top of this post. > I wouldn't be a bit surprised to learn that Niven (or someone) > got there before Estermann, but I don't think anyone in this > thread has cited a reference predating Estermann. I've had a bit of a look for the proof. In a thread on sci.math.research in September, 1998, Jim Propp attributed the proof to Niven, without any references. Michael Hardy claimed that the proof was actually older than the odd/even and unique factorization proofs, but I think he was talking about a geometric proof. The Well-Ordering proof may be nothing more than the algebraic formulation of that geometric proof, but I don't accept that the geometric proof *is* the well-ordering proof. You can find the Propp and Hardy posts under the Subject header, The Book. I didn't find the proof in Niven's book, Numbers: Rational and Irrational, in the New Mathematical Library series, nor in his book, Irrational Numbers, in the Carus Mathematical Monographs series. I searched Math Reviews for anything that had both Niven and irrational in it, and found nothing that looked like a reference to this proof. David Bloom published (a version of) the proof in Mathematics Magazine (A one-sentence proof that sqrt2 is irrational, Math of a geometric argument in a math history book by Eves, and it was presented by Niven at a lecture in 1985. In short, I still haven't found any reference to Niven or any- one else preceding Estermann's paper in 1975. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Irrationality and the Fundamental Theorem of Arithmetic > <14183393.1152611628375.JavaMail.jakarta@nitrogen.mathforum.org>, > > I like the Estermann proof because it also does not seem to rely on the > FTAr, and therefore does seem simpler than the usual proofs. > > However, I do not think it is the same as the Niven proof. At least I > cannot > massage the algebra to make them look the same. The Niven proof uses the > property that the square root of two is between one and two; the Estermann > proof uses geometry to motivate an identity. > > Both proofs leave me in a quandry. I used to tell my students that the > irrationality was a consequence of the FTAr, now I will just have to say it > is a `fact' with no deeper conceptual diagnosis. > You can still tell your students that the irrationality of the square > root of two is a consequence of the Unique Factorization Theorem, > because it is - the proof using UFT doesn't become invalid, just because > there are proofs that don't use UFT. Moreover, the proof using UFT > generalizes to a proof that if n and k are integers and the k-th root > of n isn't an integer then it is irrational - it's a bit harder to get > this out of the other proofs, nice as they are. You can even go farther; > if a polynomial with integer coefficients is not satisfied by any > rational with numerator a factor of the constant term and denominator > a factor of the leading coefficient, then all of its roots are > irrational. And you prove this using UFT. Although I've worked out how to do a Well-Ordering Principle proof of the polynomial result. Theorem: Let f be a monic polynomial with integer coefficients. Let r be a solution of f(r) = 0. Then if r is not an integer, it is irrational. Proof. Assume the hypotheses, assume r is rational, and let n be the smallest positive integer such that n r^j is an integer for all j less than the degree of f. Then n {r} (where {z} means the fractional part of z) is a smaller positive integer with the same property, contradiction, QED. Justification of this proof is a good exercise for the reader, and the reader who wants some good exercise should skip the rest of this post & figure out the justification on her own. First of all, if r is rational, say, r = a / b with a and b integers, b > 0, then b^(d - 1), where d is the degree of f, has the property that multiplication by r^j gives an integer for all j less than the degree of f. Thus, the set of all such positive integers is not empty. By Well-Ordering, there is a least such positive integer, which we call n. As r is not an integer, we have 0 < {r} < 1, so 0 < n {r} < n, so n {r} is a smaller positive integer. We write {r} = r - k for some integer k. [Aside: we're using the mathematician's definition of fractional part, where, e.g., {- 0.3} = 0.7] Now n {r} r^j = n r^(j + 1) - k n r^j is clearly an integer so long as j + 1 is less than the degree of f. But the equation f(r) = 0 tells us how to write r^d as a sum of integer multiples of smaller powers of r, so even in the case j = d - 1 we get that n {r} r^j is an integer. This completes the justification of the proof. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Irrationality and the Fundamental Theorem of Arithmetic <20434773.1152690104218.JavaMail.jakarta@nitrogen.mathforum.org>, > Yes, I now agree that the proof I thought was Niven's > to that last page of the obituary. > Does this show that the FTAr is not a necessary ingredient of > proofs of irrationality? If so, what conceptual diagnosis > of irrationality can we offer our students? JK that you can use the Unique Factorization Theorem to prove that the base-10 logarithm of 2 is irrational? There may be a way to prove that without the UFT, but I've never seen it done. I'm not entirely sure what you mean by a conceptual diagnosis of irrationality, On the first page of Chapter 1 of Burger & Tubbs, Making Transcendence Transparent, it says, The Fundamental Principle of Number Theory: There are no integers between zero and one. Time after time, proofs of irrationality and/or transcendence come down to this simple principle. Is that a good enough conceptual diagnosis of irrationality for you? What do you want? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: order of a number mod p ? [Bart] > Is there are formula for finding the order of a number a mod p > Eg what is order of > a ( mod p ) > where p is prime ? [Tim Peters] >> AFAIK, there isn't an _algorithm_ more efficient than this one in general >> (and assuming p doesn't divide a): >> Find the prime factorization of p-1, as the product of q_i^e_i. >> result := p-1 >> for each q^e in p-1's prime factorization: >> result := result / q^e >> residue := mod(a^result, p) >> while residue > 1: >> result := result * q >> residue := mod(residue^q, p) >> return result >> IOW, we know in advance that a^(p-1) is congruent to 1, and that the >> order of a must divide p-1, so this just looks for the smallest >> divisor of p-1 that still works, one prime factor of p-1 at a time. [Doug Goncz] > I wonder how efficient that is for large p, and so large p-1. Isn't the > factoring problem, er, a problem? Sure is! I didn't say it was efficient in an absolute sense, just that I don't know of an algorithm _more_ efficient than that one in general. OTOH, I'm not aware of a proof that finding the order is as difficult as factoring, and offhand don't see a way to prove that. Part of the apparent problem is that sometimes it's obviously easy; e.g., the order of 1 is 1 and the order of p-1 is 2. > Can anyone provide other algorithms for finding the order of a mod p? Assuming a and p are coprime, this one makes the above look very efficient ;-): result := 1 residue := mod(a, p) while residue > 1: result := result + 1 residue := mod(residue * a, p) return result BTW, I'd love to see an algorithm more efficient than the one at the top, or a reference to a proof that the problem is as hard as factoring. > Sure, the above is efficient if we *know* the prime factorization of > p-1, but if we don't, then what? Then we're screwed if we need certainty. If not, then, for example, if (p-1)/2 is very hard to factor (has no small factors, for a start), and a^((p-1)/2) is not congruent to 1 modulo p, then it's a good _guess_ that a's order is in fact p-1. If we can put a lower bound on the smallest prime factor of (p-1)/2, the probability of guessing correctly can be quantified. > Also, for this algorithm p doesn't divide a. How do you write that? I think it's best to check whether gcd(a, p) > 1, and bail (raise an exception; return a nonsense value like 0 or -1; ...) if so. That generalizes nicely then to composite p too. For general integer p > 1, a necessary & sufficient condition for a to _have_ an order modulo p is that a and p be coprime, and checking gcd(a, p) is an efficient way to check that. In the algorithm at the top, add that, and change p-1 to phi(p) everywhere, and you'll have an algorithm for the general case. Note that in the general case, the most efficient way known to compute phi(p) requires factoring p -- so, in the general case, the algorithm requires factoring both p and phi(p). === Subject: Re: Factoring and residues <44b447c6$0$17991$892e7fe2@authen.yellow.readfreenews.net > Can you figure out something wrong with the following? > Start with the simple: > x^2 - y^2 = S - 2*x*k > You can solve for x and y in terms of the other variables easily as >that is just > x^2 + 2*x*k + k^2 = y^2 + S + k^2 > But all that that says is > 2*x*k = S > Actually, all that it says is x^2 + 2 x k = y^2 + S. That is the starting equation with a minor re-arranging, but I also add later that x^2 - y^2 = 0 mod T which forces a second congruence relationship. > Mod dosen't even enter into it. > JSH discovered modular arithmetic and quadratic residues last week, and > he is thinking that that technique will save Surrogate Factoring. He > stubbornly refuses to accept that determining whether r is a quadratic > residue of N is just as hard as factoring N. > --- Christopher Heckman But there is no determination done, as you pick x_res, that is, you pick the residue of x modulo T. So you pick x mod T, which of course picks x^2 mod T, but not directly. One interesting point is that y is never directly related to anything as instead y^2 is. Why is that significant? The method is solid. Because only the residue of x is picked there is no way for y to be forced to equal that residue, so the mathematics can choose as it wills, which SHOULD mean, it chooses as if it just does not care. After all, it is mathematics. Why should it care? === Subject: Re: Factoring and residues <44b447c6$0$17991$892e7fe2@authen.yellow.readfreenews.net >Can you figure out something wrong with the following? >> Start with the simple: >> x^2 - y^2 = S - 2*x*k >> You can solve for x and y in terms of the other variables easily as >that is just >> x^2 + 2*x*k + k^2 = y^2 + S + k^2 > But all that that says is > 2*x*k = S > Actually, all that it says is x^2 + 2 x k = y^2 + S. > That is the starting equation with a minor re-arranging, but I also add > later that > x^2 - y^2 = 0 mod T without saying what T is. My best guess, based on your posts, is T = S - 2 x k, although the modular equation above is also true for T = 1. > which forces a second congruence relationship. >Mod dosen't even enter into it. > JSH discovered modular arithmetic and quadratic residues last week, and > he is thinking that that technique will save Surrogate Factoring. He > stubbornly refuses to accept that determining whether r is a quadratic > residue of N is just as hard as factoring N. > But there is no determination done, as you pick x_res, that is, you > pick the residue of x modulo T. > So you pick x mod T, which of course picks x^2 mod T, but not directly. > One interesting point is that y is never directly related to anything > as instead y^2 is. So how good is your method if it says that, say, y^2 = 183? > Why is that significant? > The method is solid. Because only the residue of x is picked there is > no way for y to be forced to equal that residue, so the mathematics can > choose as it wills, which SHOULD mean, it chooses as if it just does > not care. > After all, it is mathematics. Why should it care? Mathematics doesn't care whether there is an efficient factoring algorithm. Computer Science does. --- Christopher Heckman === Subject: Re: JSH: Factoring and residues [added JSH: to subject; cut sci.crypt & alt.math] [jstevh@msn.com] >> ... >> That is the starting equation with a minor re-arranging, but I also add >> later that >> x^2 - y^2 = 0 mod T [Proginoskes] > without saying what T is. T is the integer you want to factor. It's another independent variable here. > My best guess, based on your posts, is > T = S - 2 x k, T isn't defined in terms of any of the other 4 (S, k, x, y) variables here. It's introduced for the first time as a constraint: x^2 - y^2 = 0 mod T on the possible values for x and y; it has nothing to with S or k, except to the extent that they're indirectly constrained via this constraint on x & y. > although the modular equation above is also true for T = 1. His wild hope is that: gcd(x +/- y, T) in the end must reveal non-trivial factors of T. >> ... >> One interesting point is that y is never directly related to anything >> as instead y^2 is. > So how good is your method if it says that, say, y^2 = 183? He hasn't gotten that far yet, because he hasn't tried it, and can't think any straighter than he ever can when overwhelmed by the nervous thrill of impending victory. Leaving aside that the instructions for finding y: >> x+k = sqrt(y^2 + S + k^2) >> >> and finding y is just a matter of factoring (S+k^2)/4. don't make a lick of sense, he hasn't even gotten as far as noticing that there's no reason to imagine S+k^2 is divisible by 4. Now that's funny :-) In a different set of newsgroups, he started to factor T=35, and picked S=x_res=1 (giving k=18), but dropped it immediately after saying: [JSH] Then y is found by factoring (1+18^2)/4 and then you have x as well. He didn't notice that 325 isn't divisible by 4? Or he did notice it, and that's _why_ he dropped his attempt to factor 35 at that point? Or this is another method where he's happy to settle for arbitrary real factors? Or ...? The funny thing is that there's no flattering answer :-( >> ... >> After all, it is mathematics. Why should it care? > Mathematics doesn't care whether there is an efficient factoring > algorithm. Computer Science does. LOL! How true. Another funny thing is that whenever James has made a math doesn't care argument in the past, somehow or other math always managed to care after all, and favored the outcome he didn't want. By now, he should suspect that math has a deep grudge against him personally ;-) === Subject: JSH: So how can they lie about math? I am desperately in need of understanding how some people can lie so easily about mathematics, so they reply to me as if it only matters that they disagree!!! Of course with my other research that could be effective. But how could anyone reasonably expect simply disagreeing with me and calling me names to stop people from noticing a possible solution to the factoring problem? What kind of thought process is going on here? I do know that some mathematicians show signs of mental breakdowns when confronted with my proofs in such a way that they cannot simply deny. Is that it? Simply some kind of bizarrre mental break? How long can it last? And what happens when mathematical reality and social reality--as the factoring problem is of major social importance to the point that lies of this type cannot long succeed--forces through? === Subject: Re: JSH: So how can they lie about math? > But how could anyone reasonably expect simply disagreeing with me and > calling me names to stop people from noticing a possible solution to > the factoring problem? Can you see the truth, at last? There are over 4000 subscribers to sci.math. No more than 20 of them regularly respond to your posts, always negatively. The rest of us are definitely not under their control, so if there was anything in your posts that we could respond to positively, then we would. Draw your own conclusions. -- === Subject: Re: JSH: So how can they lie about math? > There are over 4000 subscribers to sci.math. Where do you get this figure? -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: JSH: So how can they lie about math? There are over 4000 subscribers to sci.math. > Where do you get this figure? That's what it says in the list of Google groups for sci.math: High activity, 4200 subscribers, Usenet. -- === Subject: Re: JSH: So how can they lie about math? > There are over 4000 subscribers to sci.math. >> Where do you get this figure? > That's what it says in the list of Google groups for > sci.math: High activity, 4200 subscribers, Usenet. are subscribed to this group. They can, I suppose, put a minimum figure on it based on how many registered users of Google Groups have subscribed to to through Google Groups. Otherwise, they have no idea who else reads the group, other than trying to guess from the number of people posting. Which ignores lurkers and spammers and sock puppets... -- David Taylor === Subject: Re: JSH: So how can they lie about math? > There are over 4000 subscribers to sci.math. > Where do you get this figure? >> That's what it says in the list of Google groups for >> sci.math: High activity, 4200 subscribers, Usenet. > are subscribed to this group. > They can, I suppose, put a minimum figure on it based on how many > registered > users of Google Groups have subscribed to to through Google Groups. > Otherwise, they have no idea who else reads the group, other than > trying to guess from the number of people posting. Which ignores lurkers > and spammers and sock puppets... > -- > David Taylor They would know posts per day, and only Google Group subscribers, but not all the other posters via other Usenet servers, which there are quite a lot. 8,000. And of that 0/8000 = 0.00 % respond positively to JSH A 100% loser. think he would get the message But he thinks he is a mathematition, and they always lie. I am not a mathematition. True or False ? === Subject: Re: JSH: So how can they lie about math? > There are over 4000 subscribers to sci.math. >> Where do you get this figure? > That's what it says in the list of Google groups for > sci.math: High activity, 4200 subscribers, Usenet. Then I wonder what kind of subscribers that refers to and how the figure was obtained. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: JSH: So how can they lie about math? > But how could anyone reasonably expect simply disagreeing with me and > calling me names to stop people from noticing a possible solution to > the factoring problem? > You mean like this? >If you ing morons think that I will let you get away with not >giving me credit for my ing math discoveries then you have another >ing thing coming. >What the ??!!! >Where's pure math now, huh? Where's loving math for the ing >beauty of it now you ing s??!!! >LOOK AT IT!!! >Here is the partial difference equation and instructions for >integrating. >dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, >sqrt(y-1))], >S(x,1) = 0. >And p(x, y) = floor(x) - S(x, y) - 1, and you get S as the sum of dS >from dS(x,2) to dS(x,y). >http://mathforprofit.blogspot.com/ >You ing s. I will get credit for my discovery and get ing >paid, and you best believe that I will not ing let you stupid >s get away with your ing stupid bull--pure math my >ass--without me coming at you ers with some ing PURE ing >PURE AS math that you stupid s have been ting on for over >a ing YEAR!!! >You goddamn S!!! >What the is wrong with you s??!!! Don't you even believe in >your own stupid ? Where's pure math now? >Where is it? > === Subject: Re: JSH: So how can they lie about math? >I am desperately in need of understanding how some people can lie so >easily about mathematics, so they reply to me as if it only matters >that they disagree!!! >Of course with my other research that could be effective. >But how could anyone reasonably expect simply disagreeing with me and >calling me names to stop people from noticing a possible solution to >the factoring problem? >What kind of thought process is going on here? >I do know that some mathematicians show signs of mental breakdowns when >confronted with my proofs in such a way that they cannot simply deny. >Is that it? Simply some kind of bizarrre mental break? Jesus. Yes, that's it. You've driven the entire mathematical community stark raving mad. Every single one of us is crazy. Maybe there's another explanation. Of I course I _would_ say that, wouldn't I... >How long can it last? And what happens when mathematical reality and >social reality--as the factoring problem is of major social importance >to the point that lies of this type cannot long succeed--forces >through? > ************************ David C. Ullrich === Subject: Re: JSH: So how can they lie about math? Any day now James is going to factor a number. It's so close I can almost taste it! === Subject: Re: So how can they lie about math? >I am desperately in need of understanding how some people can lie so > easily about mathematics, You are the liar. You INSULT everyone in this newsgroup. now Understand it, crackpot. === Subject: Re: So how can they lie about math? >I am desperately in need of understanding how some people can lie so > easily about mathematics, so they reply to me as if it only matters > that they disagree!!! But,........... you are ALWAYS WRONG !!!. > Of course with my other research that could be effective. > But how could anyone reasonably expect simply disagreeing with me and > calling me names to stop people from noticing a possible solution to > the factoring problem? > **** IT DOES NOT WORK !! **** <<<< > What kind of thought process is going on here? > I do know that some mathematicians show signs of mental breakdowns when > confronted with my proofs in such a way that they cannot simply deny. YOU HAVE NEVER SHOWN A PROOF OF ANYTHING ! YOU DO NOT KNOW WHAT A PROOF IS ! > Is that it? Simply some kind of bizarrre mental break? > How long can it last? And what happens when mathematical reality and > social reality--as the factoring problem is of major social importance > to the point that lies of this type cannot long succeed--forces > through? FACTOR ONE NUMBER !! OR GO CRAWL BACK UNDER THAT ROCK !! > === Subject: Re: So how can they lie about math? <44b5cc5e$0$17990$892e7fe2@authen.yellow.readfreenews.net >I am desperately in need of understanding how some people can lie so > easily about mathematics, so they reply to me as if it only matters > that they disagree!!! > But,........... you are ALWAYS WRONG !!!. > Of course with my other research that could be effective. > But how could anyone reasonably expect simply disagreeing with me and > calling me names to stop people from noticing a possible solution to > the factoring problem? > **** IT DOES NOT WORK !! **** <<<< > What kind of thought process is going on here? > I do know that some mathematicians show signs of mental breakdowns when > confronted with my proofs in such a way that they cannot simply deny. > YOU HAVE NEVER SHOWN A PROOF OF ANYTHING ! > YOU DO NOT KNOW WHAT A PROOF IS ! He doesn't seem to know what an algorithm is, either, which is more disturbing. Maybe that's why he got fired from Alltel ... --- Christopher Heckman > Is that it? Simply some kind of bizarrre mental break? > How long can it last? And what happens when mathematical reality and > social reality--as the factoring problem is of major social importance > to the point that lies of this type cannot long succeed--forces > through? > FACTOR ONE NUMBER !! > OR GO CRAWL BACK UNDER THAT ROCK !! > === Subject: Re: So how can they lie about math? <44b5cc5e$0$17990$892e7fe2@authen.yellow.readfreenews.net > YOU HAVE NEVER SHOWN A PROOF OF ANYTHING ! > YOU DO NOT KNOW WHAT A PROOF IS ! > He doesn't seem to know what an algorithm is, either, which is more > disturbing. Maybe that's why he got fired from Alltel ... Don't you start :p -- Larry Lard Replies to group please === Subject: Re: So how can they lie about math? >I am desperately in need of understanding how some people can lie so > easily about mathematics, so they reply to me as if it only matters > that they disagree!!! > Of course with my other research that could be effective. > But how could anyone reasonably expect simply disagreeing with me and > calling me names to stop people from noticing a possible solution to > the factoring problem? > What kind of thought process is going on here? > I do know that some mathematicians show signs of mental breakdowns when > confronted with my proofs in such a way that they cannot simply deny. > Is that it? Simply some kind of bizarrre mental break? > How long can it last? And what happens when mathematical reality and > social reality--as the factoring problem is of major social importance > to the point that lies of this type cannot long succeed--forces > through? Man, thats ridiculously awesome!! I didn't know the kazoo was such a beautiful instrument! http://oozak.caseyporn.com/ === Subject: Re: So how can they lie about math? >I am desperately in need of understanding how some people can lie so > easily about mathematics, so they reply to me as if it only matters > that they disagree!!! > Of course with my other research that could be effective. > But how could anyone reasonably expect simply disagreeing with me and > calling me names to stop people from noticing a possible solution to > the factoring problem? > What kind of thought process is going on here? > I do know that some mathematicians show signs of mental breakdowns when > confronted with my proofs in such a way that they cannot simply deny. > Is that it? Simply some kind of bizarrre mental break? > How long can it last? And what happens when mathematical reality and > social reality--as the factoring problem is of major social importance > to the point that lies of this type cannot long succeed--forces > through? > It's easy. YOU'RE SIMPLY WRONG! I can say 2+2=5 all day and it still doesn't make it right, even if I can back it up. If I claim 2+2=5 and mathematicians disagree with me, does that mean they're out to get me like you insinuate? === Subject: Re: JSH: So how can they lie about math? > I am desperately in need of understanding how some people can lie so > easily about mathematics, so they reply to me as if it only matters > that they disagree!!! Everybody hates you. Duh. > Of course with my other research that could be effective. > But how could anyone reasonably expect simply disagreeing with me and > calling me names to stop people from noticing a possible solution to > the factoring problem? Everybody hates you. Duh. > What kind of thought process is going on here? Everybody hates you. Duh. > I do know that some mathematicians show signs of mental breakdowns when > confronted with my proofs in such a way that they cannot simply deny. > Is that it? No. Everybody hates you. Duh. > Simply some kind of bizarrre mental break? No. Everybody hates you. Duh. > How long can it last? Untill you go off the deep end. > And what happens when mathematical reality and > social reality--as the factoring problem is of major social importance > to the point that lies of this type cannot long succeed--forces > through? Wait and see. > === Subject: Re: JSH: So how can they lie about math? > I am desperately in need of understanding how some people can lie so > easily about mathematics, so they reply to me as if it only matters > that they disagree!!! > Everybody hates you. Duh. > Of course with my other research that could be effective. > But how could anyone reasonably expect simply disagreeing with me and > calling me names to stop people from noticing a possible solution to > the factoring problem? > Everybody hates you. Duh. > What kind of thought process is going on here? > Everybody hates you. Duh. Actually the behavior is like that of an angry child in this case. Some kind of reduction to the infantile by the pressure of the results? But why exactly? And why would this poster think his behavior ok on this forum? Is it not at least somewhat embarrassing? The discussion is in no way like that of adult conversation. I think that is at least quasi deliberate, but maybe it is more real than the poster is willing to admit? Possibly my research makes others feel less about themselves? === Subject: Re: JSH: So how can they lie about math? > Possibly my research makes others feel less about themselves? What research? Are you talking about that *duck crap* you sling in here, wanting someone else to prove it might be right so you can take the credit for it, crackpot ? === Subject: Re: non-reflexive symmetric and transitive relations > .... > A relation C -> AxA on some set A is both symmetric and transitive. > i.e., > for all a,b in A, aCb ---> bCa > for all a,b,d in A, aCb, bCd -----> aCd > If d=a, then, > for all a,b in A, aCb ---> bCa > for all a,b in A, aCb, bCa -----> aCa....(1) > IOW, a symmetric and transitive relation is ALWAYS reflexive, which we > know to be untrue. > I am trying to grasp where I am going wrong.... Plenty of people have shed light on your original question, but you may like to see the flaw in your own proof above. The reflexive law which you're trying to prove is (for each a in A) aCa. To make the above argument work, you need a preliminary step (for each a in A)(there exists b in A) aCb. But if some element a isn't related to anything whatever, then there's no such b, and your argument can't get off the ground. What you have actually proved is aCb ---> aCa. HTH Ken Pledger. === Subject: Re: non-reflexive symmetric and transitive relations > Are there any examples of relations which are both symmetric and > transitive but not reflexive? > Any inputs welcome!!! If a symmetric transitive relation R over a set S, is not reflexive, then some a in S with (a,a) not in R. Let S = { a,b,c }. Then { (b,b), (c,c), (b,c), (c,b) } is not empty, symmetric, transitive and not reflexive relation over S. In general when nonnul A subset S, any equivalence relation over SA, is a symmetric, transitive, not reflexive relation over S. A (binary) relation over S is a subset of SxS === Subject: Re: Liberal Atheists damned to Hellfire -----BEGIN PGP SIGNED MESSAGE----- > Jesus loves you...everyone else thinks you're an asshole. Just because Jesus loves him doesn't mean Jesus doesn't think he's an asshole. === Subject: Question from an old algebraic topology qualifying exam I am struggling on an old qual, it seems difficult. I am not sure how to solve these. Any help would be appreciated. Problem; Let Y be a subset of 3 dimentinal Euclid space such that Y = {(x,y,z) in R^3 : 9 =< x^2 + y^2 + z^2 =< 16, and x^2 + z^2 >= 1, y^2 + z^2 >= 1 } Let the boundary of Y be X and consider the continuous function f: X -> X ; (x, y, z) -> (-y, x, z) 1)Find the integer coefficient homology group H_* (X; Z). 2)Suppose f induces self-homomorphism on the first dimentional integer coefficient homology group f_* : H_1 (X; Z) -> H_1 (X;Z). Find the characterisic polynomial of its matrix. === Subject: Re: Question from an old algebraic topology qualifying exam >Let Y be a subset of 3 dimentinal Euclid space such that >Y = {(x,y,z) in R^3 : 9 =< x^2 + y^2 + z^2 =< 16, >and x^2 + z^2 >= 1, y^2 + z^2 >= 1 } >Let the boundary of Y be X Okay. So can you describe X in some other way, as an example of a space you are (or should be) familiar with? One (natural) way to get this description is to look at the description above and take it bit by bit. Thus: {(x,y,z) in R^3 : 9 =< x^2 + y^2 + z^2 =< 16} describes the closed region between two spheres (of radii 3 and 4), so *its* boundary is precisely those two spheres. The inequalities x^2 + z^2 < 1, y^2 + z^2 < 1 (complementary to the other two inequalities in the description of Y) each describe an open region that is a tubular (product) neighborhood of a straight line in R^3, namely, the y axis and the x axis respectively. Y is the region between those two spheres with those open regions both *removed*, and (because the radius of the tubes is 1 which is less than 3) its boundary X is the union of the following parts: (1) the two spheres, each with four (round) disks removed; (2) four cylindrical surfaces, each bounded by one (round) circle on the outer sphere and one on the inner sphere. So X is a surface (2-dimensional manifold). *What* surface? Count and find out. >and consider the continuous >function >f: X -> X ; (x, y, z) -> (-y, x, z) This function, being the restriction to X of an orthogonal linear transformation A of R^3, and the spheres, cylinders, etc., mentioned above all behaving nicely with respect to A, can now be described--EASILY--in terms of how it permutes the two 4-punctured 2-spheres in (1) and the four cylinders in (2) . >1)Find the integer coefficient homology group H_* (X; Z). As soon as you have answered *What* surface?, you will know this *abstractly*. If you think a bit more, you should know it *concretely*, in the sense that you will actually be able to find, first, a geometrically described set of 1-cycles on X whose homology classes generated the homology group, and, second, the (fairly obvious!) relations among them--so you'll have the group given by generators and relations. >2)Suppose f induces self-homomorphism on the first >dimentional integer coefficient homology group >f_* : H_1 (X; Z) -> H_1 (X;Z). Why suppose? It does, and that's an end on't. >Find the characterisic polynomial of its matrix. Continue as indicated. Lee Rudolph === Subject: Re: Question from an old algebraic topology qualifying exam >Let Y be a subset of 3 dimentinal Euclid space such > that >Y = {(x,y,z) in R^3 : 9 =< x^2 + y^2 + z^2 =< 16, >and x^2 + z^2 >= 1, y^2 + z^2 >= 1 } >Let the boundary of Y be X > Okay. So can you describe X in some other way, as an > example of a space you are (or should be) familiar > with? > One (natural) way to get this description is to look > at the description above and take it bit by bit. > Thus: > {(x,y,z) in R^3 : 9 =< x^2 + y^2 + z^2 =< 16} > describes > the closed region between two spheres (of radii 3 and > 4), > so *its* boundary is precisely those two spheres. > The inequalities x^2 + z^2 < 1, y^2 + z^2 < 1 > (complementary to the other two inequalities in the > description of Y) each describe an open region that > is a tubular (product) neighborhood of a straight > line in R^3, namely, the y axis and the x axis > respectively. Y is the region between those two > spheres with those open regions both *removed*, > and (because the radius of the tubes is 1 which > is less than 3) its boundary X is the union of > the following parts: (1) the two spheres, each > with four (round) disks removed; (2) four cylindrical > surfaces, each bounded by one (round) circle on the > outer sphere and one on the inner sphere. So X is > a surface (2-dimensional manifold). *What* surface? > Count and find out. I thimk it is the torus. >and consider the continuous >function >f: X -> X ; (x, y, z) -> (-y, x, z) > This function, being the restriction to X of an > orthogonal > linear transformation A of R^3, and the spheres, > cylinders, > etc., mentioned above all behaving nicely with > respect to > A, can now be described--EASILY--in terms of how it > permutes the two 4-punctured 2-spheres in (1) and the > four cylinders in (2) >1)Find the integer coefficient homology group H_* > (X; Z). > As soon as you have answered *What* surface?, you > will know > this *abstractly*. If you think a bit more, you > should know it > *concretely*, in the sense that you will actually be > able to > find, first, a geometrically described set of > 1-cycles on X whose > homology classes generated the homology group, and, > second, > the (fairly obvious!) relations among them--so you'll > have the > group given by generators and relations. If X is torus then, H_0 (X; Z) = Z H_1 (X; Z) = Z (+) Z H_q (X; Z) = 0 ( q > 1 ) >2)Suppose f induces self-homomorphism on the first >dimentional integer coefficient homology group >f_* : H_1 (X; Z) -> H_1 (X;Z). > Why suppose? It does, and that's an end on't. >Find the characterisic polynomial of its matrix. > Continue as indicated. I trnaslated wrong. >Find characrteristic equation of its matrix is right. Is this just a characteristic equation of next matrix? [0, -1, 0] [1, 0, 0] [0, 0, 1] i.e. its characteristic equation is (lamda)^2*(1-lamda) + 1 = 0 > Lee Rudolph === Subject: Re: Question from an old algebraic topology qualifying exam ... >> Y is the region between those two >> spheres with those open regions both *removed*, >> and (because the radius of the tubes is 1 which >> is less than 3) its boundary X is the union of >> the following parts: (1) the two spheres, each >> with four (round) disks removed; (2) four cylindrical >> surfaces, each bounded by one (round) circle on the >> outer sphere and one on the inner sphere. So X is >> a surface (2-dimensional manifold). *What* surface? >> Count and find out. >I thimk it is the torus. I think you should draw a picture. Lee Rudolph === Subject: Period of random number generators I am confused about the period of the random number generator (a^n + b^n) mod c. Assuming a, b, and c pairwise coprime, as is usually done to maximize period, that is... t is used to represent the period of a random number generator in many writings. t((a^n) mod c) | c-1 where | means divides. That's has something to do with 0 being forbidden; if (a,c)=1 then a^n =/= 0 mod c. It just can't happen for any power of a. Er, any postive power? Arithmetically (a^n + b^n) mod c = ((a^n) mod c + (b^n) mod c) mod c 0 is not forbidden because (a^n) mod c + (b^n) mod c can equal c. So does t( ( a^n + b^n ) mod c ) divide c, or c-1, or what? Doug === Subject: Re: Period of random number generators Anybody, under which condition would it be correct to write t((a^n) + b^n) mod c) = lcm ( t((a^n) mod c, t((b^n) mod c) ? > I am confused about the period of the random number generator (a^n + > b^n) mod c. > Assuming a, b, and c pairwise coprime, as is usually done to maximize > period, that is... > t is used to represent the period of a random number generator in many > writings. Doug === Subject: Re: Period of random number generators > Anybody, under which condition would it be correct to write > t((a^n) + b^n) mod c) = lcm ( t((a^n) mod c, t((b^n) mod c) ? > I am confused about the period of the random number generator (a^n + > b^n) mod c. > Assuming a, b, and c pairwise coprime, as is usually done to maximize > period, that is... > t is used to represent the period of a random number generator in many > writings. To adopt your notation, if gcd( t((a^n) mod c, t((b^n) mod c ) = 1 then t((a^n) + b^n) mod c) = lcm ( t((a^n) mod c, t((b^n) mod c). However it is not true in general as the following example shows, even if a,b,c are coprime. Let a = 4, b = 7, c = 9. Then a,b are both of multiplicative order 3 in Z/9Z*, but the sequence a^n + b^n mod c is simply constant: (4 + 7) mod 9 = 2 (16 + 49) mod 9 = 2 (64 + 343) mod 9 = 2, etc. === Subject: Re: Period of random number generators Er, I guess then t((a^n) + (b^n) mod c) | lcm ( t((a^n) mod c, t((b^n) mod c) right, Chip? Doug > Anybody, under which condition would it be correct to write > t((a^n) + b^n) mod c) = lcm ( t((a^n) mod c, t((b^n) mod c) ? > I am confused about the period of the random number generator (a^n + >b^n) mod c. >Assuming a, b, and c pairwise coprime, as is usually done to maximize >period, that is... > t is used to represent the period of a random number generator in many >writings. > To adopt your notation, if gcd( t((a^n) mod c, t((b^n) mod c ) = 1 > then t((a^n) + b^n) mod c) = lcm ( t((a^n) mod c, t((b^n) mod c). > However it is not true in general as the following example shows, > even if a,b,c are coprime. > Let a = 4, b = 7, c = 9. Then a,b are both of multiplicative order > 3 in Z/9Z*, but the sequence a^n + b^n mod c is simply constant: > (4 + 7) mod 9 = 2 > (16 + 49) mod 9 = 2 > (64 + 343) mod 9 = 2, > etc. === Subject: Re: Period of random number generators > Er, I guess then > t((a^n) + (b^n) mod c) | lcm ( t((a^n) mod c, t((b^n) mod c) > right, Chip? > Doug >Anybody, under which condition would it be correct to write > t((a^n) + b^n) mod c) = lcm ( t((a^n) mod c, t((b^n) mod c) ? > >I am confused about the period of the random number generator (a^n + >b^n) mod c. >Assuming a, b, and c pairwise coprime, as is usually done to maximize >period, that is... >> t is used to represent the period of a random number generator in many >writings. > To adopt your notation, if gcd( t((a^n) mod c, t((b^n) mod c ) = 1 > then t((a^n) + b^n) mod c) = lcm ( t((a^n) mod c, t((b^n) mod c). > However it is not true in general as the following example shows, > even if a,b,c are coprime. > Let a = 4, b = 7, c = 9. Then a,b are both of multiplicative order > 3 in Z/9Z*, but the sequence a^n + b^n mod c is simply constant: > (4 + 7) mod 9 = 2 > (16 + 49) mod 9 = 2 > (64 + 343) mod 9 = 2, > etc. > Yes, that much we can prove! === Subject: Re: Period of random number generators > Er, I guess then > t((a^n) + (b^n) mod c) | lcm ( t((a^n) mod c, t((b^n) mod c) > right, Chip? > Yes, that much we can prove! Awsome! I take nap now... zzzzzz.... Doug === Subject: Re: Period of random number generators Gerry and Chip have shown t( ( (a^n) + (b^n) ) mod c) | phi (c) or c-1. Are any values forbidden? It seems to me that exactly one value must be forbidden for this RNG to have such possible periods. Doug > I am confused about the period of the random number generator (a^n + > b^n) mod c. > Assuming a, b, and c pairwise coprime, as is usually done to maximize > period, that is... > t is used to represent the period of a random number generator in many > writings. > t((a^n) mod c) | c-1 where | means divides. > That's has something to do with 0 being forbidden; if (a,c)=1 then > a^n =/= 0 mod c. It just can't happen for any power of a. Er, any > postive power? Yes, we are talking about postive powers here. > Arithmetically (a^n + b^n) mod c = ((a^n) mod c + (b^n) mod c) mod c > 0 is not forbidden because (a^n) mod c + (b^n) mod c can equal c. > So does t( ( a^n + b^n ) mod c ) divide c, or c-1, or what? > Doug === Subject: Re: Period of random number generators > Gerry and Chip have shown t( ( (a^n) + (b^n) ) mod c) | phi (c) or c-1. > Are any values forbidden? It seems to me that exactly one value must be > forbidden for this RNG to have such possible periods. > Doug > I am confused about the period of the random number generator (a^n + > b^n) mod c. > Assuming a, b, and c pairwise coprime, as is usually done to maximize > period, that is... > t is used to represent the period of a random number generator in many > writings. > t((a^n) mod c) | c-1 where | means divides. > That's has something to do with 0 being forbidden; if (a,c)=1 then > a^n =/= 0 mod c. It just can't happen for any power of a. Er, any > postive power? > Yes, we are talking about postive powers here. > Arithmetically (a^n + b^n) mod c = ((a^n) mod c + (b^n) mod c) mod c > 0 is not forbidden because (a^n) mod c + (b^n) mod c can equal c. > So does t( ( a^n + b^n ) mod c ) divide c, or c-1, or what? > Doug The value c-1 is correct only if c is prime, when phi(c) = c-1. === Subject: Re: Period of random number generators > I am confused about the period of the random number generator (a^n + > b^n) mod c. > Assuming a, b, and c pairwise coprime, as is usually done to maximize > period, that is... > t is used to represent the period of a random number generator in many > writings. > t((a^n) mod c) | c-1 where | means divides. > That's has something to do with 0 being forbidden; if (a,c)=1 then > a^n =/= 0 mod c. It just can't happen for any power of a. Er, any > postive power? > Arithmetically (a^n + b^n) mod c = ((a^n) mod c + (b^n) mod c) mod c > 0 is not forbidden because (a^n) mod c + (b^n) mod c can equal c. > So does t( ( a^n + b^n ) mod c ) divide c, or c-1, or what? Assuming that the period you are asking about refers to what value of n gives rising to a repeating sequence, think in terms of Fermat's little theorem. Since a,b are relatively prime to c, we have (a,b,c > 1): a^phi(c) = 1 mod c b^phi(c) = 1 mod c where phi(c) is the Euler phi function of c, ie. the number of integers between 1 and c-1 that are coprime to c. So the period is a divisor of phi(c), because: a^0 + b^0 = a^phi(c) + b^phi(c) mod c and similarly with any exponent that is a multiple of ph(c). === Subject: Re: Period of random number generators > I am confused about the period of the random number generator (a^n + > b^n) mod c. > Assuming a, b, and c pairwise coprime, as is usually done to maximize > period, that is... > t is used to represent the period of a random number generator in many > writings. > t((a^n) mod c) | c-1 where | means divides. > That's has something to do with 0 being forbidden; if (a,c)=1 then > a^n =/= 0 mod c. It just can't happen for any power of a. Er, any > postive power? > Arithmetically (a^n + b^n) mod c = ((a^n) mod c + (b^n) mod c) mod c > 0 is not forbidden because (a^n) mod c + (b^n) mod c can equal c. > So does t( ( a^n + b^n ) mod c ) divide c, or c-1, or what? Working mod c, a^(n + c - 1) + b^(n + c - 1) = a^n + b^n, so certainly the period divides c - 1. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Odd Squares, 8 (2^3), and the Sum of the First n Positive Integers I did see it noted that the odd squares are 1 (mod 8) (http://mathworld.wolfram.com/SquareNumber.html). I imagine it is noted elsewhere, but they also have a relationship to the sum of the first n positive integers. If it is noted that the sum of the first n positive integers = n(n+1)/2, then ((2^3)(((n(n+1))/2))+1=(2n+1)^2 (8(n^2+n)/2)+1=4(n^2)+2n+2n+1 (4(n^2+n))+1=4(n^2)+4n+1 4(n^2)+4n+1=4(n^2)+4n+1 for example: the sum of the first positive integer, 1, is 1 then, (8*1)+1=9 or ((2*1)+1)^2=3^2 the sum of the first two positive integers, 1+2, is 3 then, (8*3)+1=25 or ((2*2)+1)^2=5^2 the sum of the first three positive integers, 1+2+3, is 6 then, (8*6)+1=49 or ((2*3)+1)^2=7^2 the sum of the first four positive integers, 1+2+3+4, is 10 then, (8*10)+1=81 or ((2*4)+1)^2=9^2 the sum of the first five positive integers, 1+2+3+4+5, is 15 then, (8*15)+1=121 or ((2*5)+1)^2=11^2 etc. === Subject: Re: Odd Squares, 8 (2^3), and the Sum of the First n Positive Integers Am 13.07.2006 03:58 schrieb rer: > I did see it noted that the odd squares are 1 (mod 8) > (http://mathworld.wolfram.com/SquareNumber.html). I imagine it is noted > elsewhere, but they also have a relationship to the sum of the first n > positive integers. > If it is noted that the sum of the first n positive integers = > n(n+1)/2, then > ((2^3)(((n(n+1))/2))+1=(2n+1)^2 > (8(n^2+n)/2)+1=4(n^2)+2n+2n+1 > (4(n^2+n))+1=4(n^2)+4n+1 > 4(n^2)+4n+1=4(n^2)+4n+1 .......... = (2n +1)^2 by binomial expansion. Gottfried Helms === Subject: Re: Clifford Numbers in Mathcad > I do know | 3x + 4y + 12z | = 13, but I haven't learned how to do even > simple Clifford operations like 1x * 1y = 1xy in my head with clarity. > I think I got that one right but I don't see it. I need a Clifford > Arithmetic Primer! Not a text, I have access to those, I am talking > about doing simple operations on paper. You might try starting with this: http://en.wikipedia.org/wiki/Exterior_algebra Note that what you are calling Clifford numbers is usually called the exterior algebra or Grassmann algebra, whereas mathematicians mean something a little more complicated by Clifford algebra. Why the geometric algebra people seem bent on renaming things I don't know. === Subject: Re: Clifford Numbers in Mathcad Gene's reference to it is *much* more useful than any I have found hyperlinked and will *really* help me study this algebra. Kudos to Gene! > I do know | 3x + 4y + 12z | = 13, but I haven't learned how to do even > simple Clifford operations like 1x * 1y = 1xy in my head with clarity. > I think I got that one right but I don't see it. I need a Clifford > Arithmetic Primer! Not a text, I have access to those, I am talking > about doing simple operations on paper. > You might try starting with this: > http://en.wikipedia.org/wiki/Exterior_algebra > Note that what you are calling Clifford numbers is usually called the > exterior algebra or Grassmann algebra, whereas mathematicians mean > something a little more complicated by > Clifford algebra. Why the geometric algebra people seem bent on > renaming things I don't know. I am the Cult Of, I am the Cult Of, Per ! Son ! Al ! I ! Ty ! (a song I like; who performs it?) Maybe it's a Hestenesian cult buried in acadmeia, with a secret handshake, membership rites, and hazing, too. ;) I have read that whoever comes up with an idea in math or physics is likely not to be remembered in the history of science as the originator; that usually someone else's extension of the original makes the splash, or that the historians just don't get the derviations because after all they are historians, not mathemeticians or physicists... Doug === Subject: Hard analysis problem (epsilon deltas) I need help proving the following: Suppose f: [0,infty) -> R is continuous, int_0^{infty} f(t) dt exists, and for every x>0, int_0^{infty} f(t) e^{-xt} dt exists. Show int_0^{infty} f(t) e^{-xt} dt --> int_0^{infty} f(t) dt as x --> 0 from the right. I'm told it is possible to do this with just epsilon-delta, but it seems impossible to me. Even with integration by parts and mean value theorems for integrals, I cannot solve the problem. The problem is that it is really a double limit in disguise, since we're really talking about int_0^T f(t) e^{-xt} dt as x-->0 and T-->infty, and the values we want to choose for x and T end up depending on eachother in nontrivial ways! Any help would be appreciated. =) S.P. === Subject: Re: Hard analysis problem (epsilon deltas) On 12 Jul 2006 19:07:59 -0700, Snis Pilbor Suppose [i] >f: [0,infty) -> R is continuous, >int_0^{infty} f(t) dt exists, and [ii] >for every x>0, >int_0^{infty} f(t) e^{-xt} dt exists. Actually [ii] follows from [i], as the proof below shows. >Show >int_0^{infty} f(t) e^{-xt} dt --> int_0^{infty} f(t) dt as x --> 0 from >the right. >I'm told it is possible to do this with just epsilon-delta, but it >seems impossible to me. Even with integration by parts and mean value >theorems for integrals, I cannot solve the problem. >The problem is that it is really a double limit in disguise, since >we're really talking about int_0^T f(t) e^{-xt} dt as x-->0 and >T-->infty, and the values we want to choose for x and T end up >depending on eachother in nontrivial ways! Say I = int_0^infinity f. Let F(0) = 0 and F' = f, so that F(t) -> I as t -> infinity. A careful integration by parts (using that fact that F is bounded!) shows that for every x > 0 int_0^infinity f(t) exp(-xt) dt = 1/x int_0^infinity F(t) exp(-xt) dt (and now the fact that F is bounded shows that the second integral above exists, hence the first one does as well). Note that 1/x int_0^infinity exp(-xt) dt = 1. Hence (*) int_0^infinity f(t) exp(-xt) dt - I = 1/x int_0^infinity (F(t) - I) exp(-xt) dt. Now let eps > 0. Choose A so that |F(t) - I| < eps for all t > A. Split the right side of (*) into int_0^A + int_A^infinity... >Any help would be appreciated. =) >S.P. ************************ David C. Ullrich === Subject: Re: =?ISO-8859-1?Q?Grundz=FCge?= der Mathematik: V1 Translation Error? > de.sci.mathematik? > Could somebody please help me with this? I apologize for communicating in > English. My ability to write in German is very limited. What is your problem? If you post your question on the newsgroup I supplied, even when it is in English, I would think there would be response, and I have posted in English in a 'de' newsgroup and never have seen a bad response. But news-server): Hatto von Aquitanien asks the following question: > If someone happens to have Volume 1 of the German language original > of Grundz.9fge der Mathematik, by Behnke, et al, may I please ask you > to provide the correct form of the expression in Part B, Chapter 1, > Section 4.1 a > couple paragraphs after equation (56)? What is written in the > English translation is the following syntactic gibberish. > a_0.a_1a_2a_3... + g^-1 Sum[a_i g^-i + g^-k, i=0 ,k] > = a_0,a_1...a_k-1(a_k + 1)000... -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Prime numbers >> What's everyone's opinion of the site >> www.primenumbersformula.com? Is the >> author for real, or just a crank? > At least a non-negligible proportion of crank. > The only sensible material on that page is vast > swathes of copyrighted material lifted from > Professor Caldwell's Prime Pages, such as > http://primepages.org/notes/faq/why.html > Heh, he didn't even bother to correct the grammar > error when he lifted the material! If you think that's badly lifted from the Prime Pages then compare these: http://www.csam.iit.edu/~cs549/cs549/project/Contrastprimalitytests.htm http://primes.utm.edu/prove/merged.html -- Jens Kruse Andersen === Subject: Re: Prime numbers > Two quotes: > 1) I have discovered the formula of the prime numbers after 20 years of > research and injury. > 2) I am Seyyed Mohammad Reza Hashemi Moosavi and university professor that I > chosen as a superior investigator in superiors and initiators festival in > 1383. > What's 1) supposed to mean? That he got injured after finding the formula or > that he was continuously injured during those 20 years as a result of his > research? I will admit, that's pretty funny. I wonder if it's like the slipper at school - this will hurt me more than you. It's probably true, I didn't suffer any permanent damage from viewing his webpage. Quite what he suffered, one is reticent to find out. > According to 2) this guy is at least 623 years old. Perhaps his formula was > the elixir for eternal life. Nope, he's well over 1300 years old. They use AUC in Iran, didn't you know? Phil -- The man who is always worrying about whether or not his soul would be damned generally has a soul that isn't worth a damn. -- Oliver Wendell Holmes, Sr. (1809-1894), American physician and writer === Subject: Re: Prime numbers > Heh, he didn't even bother to correct the grammar > error when he lifted the material! The editors have been informed... Phil -- The man who is always worrying about whether or not his soul would be damned generally has a soul that isn't worth a damn. -- Oliver Wendell Holmes, Sr. (1809-1894), American physician and writer === Subject: Re: Prime numbers >If the author is correct, his work would be publicized widely in the mathematics community. So far it hasn't happened, so I am skeptical. > Anyone who's learned computability knows how to create an > almost unending stream of prime number generation functions. > There is no novel 'discovery' in that direction any more. > Hi Phil, > do you actually know how to interpret this function? > I was trying to read it and not sure what the floor like functions > mean. I must also be misinterpreting something because it looks like > most stuff cancels out. > Any ideas as this would be easy to test? > However, right now it looks like crank/troll work! I didn't look closely - I'd be surprised if there was anything pithy in the first equation I saw. However, I do know that floor is a 'cheat' used when counting things, such as factors. If you take floor(x/a)-floor((x-1)/a) for example you get the predicate 'x is divisibly by a' evaluated as 1 (true) or 0 (false). Thus 'sieves' can be built by summing such expressions, and the result 0 would imply passing the sieve - i.e. primeness. Counting the number of values that pass that test gives you the Pi function. It's all fairly trivial. Fun when you first see it, but not deep. Phil -- The man who is always worrying about whether or not his soul would be damned generally has a soul that isn't worth a damn. -- Oliver Wendell Holmes, Sr. (1809-1894), American physician and writer === Subject: Re: Prime numbers [amzoti, on www.primenumbersformula.com] > Hi Phil, > do you actually know how to interpret this function? > I was trying to read it and not sure what the floor like functions > mean. I'm sure they mean the floor. > I must also be misinterpreting something because it looks like > most stuff cancels out. I expect that's intentional ;-) > Any ideas as this would be easy to test? Well, let's look at it top down. It has the structure: H(m) = 2*((2m+1)/2)^e(m) for a messy function e(m), which we'll get to later. _Assume_ for the moment that: e(m) = 1 if 2m+1 is prime = 0 if 2m+1 is not prime If that's true, then: H(m) = 2*((2m+1)/2)^1 = 2m+1 if 2m+1 is prime = 2*((2m+1)/2)^0 = 2 if 2m+1 is not prime So, if that's true, H(m) _obviously_ generates all and only the primes as m goes from 1 to infinity, and generates each prime exactly once except for 2. 2 is generated for each m s.t. 2m+1 is not prime. If you look at the sequence he gives below it: it generates all the prime numbers (3,5,7,2,11,13,2,17,19, ...) that matches the above. At m=1, 2m+1=3 and is prime. Likewise for 5 and 7 at m=2 and m=3. At m=4, 2m+1=9 is not prime, and that's the first 2 in his sequence. m=5 gives 11 and m=6 gives 13, then at m=7 2m+1=15 is not prime and another 2 shows up in his sequence. Etc. So it just boils down to finding a function e(m) s.t. e(m) = 1 if 2m+1 is prime = 0 if 2m+1 is not prime His e(m) works, and is really a pretty standard kind of trick if you're attracted to silly ;-) functions like this: e(m) = floor(a * floor(b/a) / b) where: a = 2m+1 b = (2m)!+1 Remember that Wilson's theorem says p is prime if and only if p divides (p-1)!+1. Substitute 2m+1 for p to get: 2m+1 is prime if and only if 2m+1 divides (2m!)+1 or, using the variable names just above, 2m+1 is prime if and only if a divides b Suppose a does divide b (2m+1 is prime). Then b = q*a for some integer q, and floor(a * floor(b/a) / b) = floor(a * floor(q*a/a) / (q*a)) = floor(a * q / (q*a)) = floor(1) = 1 So e(m) is 1 if 2m+1 is prime. OTOH, if a does not divide b (2m+1 is not prime), then b = q*a + r for some integers q and r with 0 < r < a, and floor(b/a) = floor((q*a + r)/a) = floor(q + r/a) = since 0 < r/a < 1 q so floor(a * floor(b/a) / b) = floor(a * q / (q*a + r)) = since r > 0, q*a+r > q*a 0 So e(m) is 0 if 2m+1 is not prime, and we're done. > However, right now it looks like crank/troll work! It does seem overly impressed with itself. As an exercise, now figure out why this function (found on Wikipedia) works: f(n) = 2 + mod(2*n!, n+1) That also generates all and only the primes as n goes from 1 to infinity, and also generates each prime exactly once except for 2 (which is generated infinitely often). IMO, that's more elegant. === Subject: Re: What happened to the Wolfram site? > Also, I was looking at the prices of the CAS program - whew! > How in the heck do they expect the average joe to pay for this? I just > like tinkering with stuff in a quick fashion and have been using > Mathematica for years this way - but just cannot afford to keep uo with > updates. > So - I looked at Maple 10 and it now also costs and arm and a leg. > Student versions or free versions of CAS SW - here I come! What the world needs badly is an Open Source software CAS system equivalent in power to Maple or Mathematica. Actually, there is at least one -- Maxima -- which is an open source derivative of Macsyma. But it doesn't seem to have caught on. Just dreaming, I suppose, but it would be neat if the mathematics community would see it as a worthwhile project to transfer all mathematical knowledge to the computer, and create an Open Source combination computer algebra system and automated theorem prover together with powerful numeric algorithms and some artificial intelligence. === Subject: Re: What happened to the Wolfram site? > Just dreaming, I suppose, but it would be neat if the mathematics > community would see it as a worthwhile project to transfer all > mathematical knowledge to the computer, and create an Open Source > combination computer algebra system and automated theorem prover > together with powerful numeric algorithms and some artificial > intelligence. Assuming that artificial intelligence is to intelligence as artificial leg is to leg---I think we already have it. As for the prices. Well. WRI has always been motivated by that great corporate motivator, greed. I'm disappointed that the Maple implementors saw this and decided to follow the same path. (With nothing like the increase in power which their PRICE increases would indicate.) -- Ron Bruck === Subject: Re: What happened to the Wolfram site? > what has happened to the Wolfram site? Great Greasy Greed. > www.wolfram.com For it to open, you mean http://www.wolfram.com > Also, I was looking at the prices of the CAS program - whew! > How in the heck do they expect the average joe to pay for this? Your first born will suffice. > I just like tinkering with stuff in a quick fashion and have been using > Mathematica for years this way - but just cannot afford to keep uo with > updates. Why bother? Updates are worse than blind dates. > So - I looked at Maple 10 and it now also costs and arm and a leg. > Student versions or free versions of CAS SW - here I come! === Subject: Re: Properties of weakly continuous maps in a Hilbert space > On Mon, 10 Jul 2006 20:59:26 -0400, Jim Burns >That doesn't seem at first to answer the question, which >was about /weakly/ continuous functions. > How can you possibly even _conjecture_ that this does > not answer the question, since, as you say below, you > don't know what weakly continuous means? I hope you will believe me when I tell you I was not trying to upset you, or irritate you or get you. I'm sorry if I have. As to how I could _conjecture_ (or, as I would prefer, guess), I had already used what resources were readily available and had failed at finding out what weakly continuous meant (see below). At that point, my choices were to leave the post for another day (which my history shows often means never) or post with incomplete information and make it clear how incomplete my information was. Below you write if you'd asked someone would have said, no problem. But, part of the reason I posted was to get a clarification of what weakly continuous meant -- I just didn't use question marks. I have noticed over the years that it's much easier to get a correction on USEnet than an answer to a question. Perhaps that's not true of you -- then, good for you, but I'd still maintain you are not typical. What would you have done if you were me? Notice that I am not asking what you would have done in my place, but to imagine what it's like to be me and be faced with a similar choice. You are a professional mathematician, I am no kind of academic at all. You apparently take your appearance on USEnet very seriously, while I, having very little in the way of reputation to start with, have less to lose. You probably also have, beyond your mental mathematical resources, texts, references and colleagues practically at your fingertips. Well, I don't. Would you just dropped the topic and moved on? Maybe I would have, too, if I hadn't already invested half an hour or so in trying to beat a definition of weakly continuous function out of the World Wide Web. I hope you're not upset that I sorta-kinda-maybe-if-God- willing-I'm-right accused you of making a mistake. You do make mistakes, you know, though it's very rare, I'm sure. This was not one of them, but anyone who knew the correct definition would know that, including the original poster. Yes, I thought of that before I posted. If I had had a chance of misleading him, I would have let it drop. But if, by chance, I were right, I could have done a bit of good. I hope that answers your question. Yours, Jim Burns > However, the two >proofs are so similar in structure that one might consider >it a good clue. >If I'm right in thinking that /weakly/ continuous functions >f and g will map weakly convergent series to weakly convergent >series and limit point to limit point, then g will take >a weakly convergent series as input, then pass a weakly >convergent series to f, which returns yet another weakly >convergent series. Likewise for limit points. Therefore, >yes, the composition of weakly continuous functions is >weakly continuous. >I didn't know what weakly continuous meant, so I attempted >weakly continuous function less so. (The series x_n is >weakly convergent if the inner product < x_n, y > converges >for all y in the Hilbert space H.). >However, I strongly suspect the definition weakly continuous >function is analogous to continuous function, as I >expressed it above. If I'm wrong, I would appreciate being >told. (Too late if I've made a fool of myself, of course.) > If you don't know what weakly continuous means probably > you should determine that first, before attempting to > figure out questions regarding weakly continuous > functions? > Yes, you've made a fool of yourself. Nothing foolish > about not knowing what weakly continuous means - if > you'd asked someone would have said, no problem. But > asking a question about weakly continuous functions > _before_ finding a definition of the phrase, and > without asking for the definition, that's definitely > making a fool of yourself. As is conjecturing that > an answer is incorrect, before learning the > definition of the terms involved. > A Hilbert space has a weak topology. A weakly continuous > function is precisely a function which is continuous with > respect to this topology. And hence what I said _does_ > answer the question. >Jim Burns > ************************ > David C. Ullrich === Subject: Re: Properties of weakly continuous maps in a Hilbert space > On Mon, 10 Jul 2006 20:59:26 -0400, Jim Burns >That doesn't seem at first to answer the question, which >was about /weakly/ continuous functions. > How can you possibly even _conjecture_ that this does > not answer the question, since, as you say below, you > don't know what weakly continuous means? I hope you will believe me when I tell you I was not trying to upset you, or irritate you or get you. I'm sorry if I have. As to how I could _conjecture_ (or, as I would prefer, guess), I had already used what resources were readily available and had failed at finding out what weakly continuous meant (see below). At that point, my choices were to leave the post for another day (which my history shows often means never) or post with incomplete information and make it clear how incomplete my information was. Below you write if you'd asked someone would have said, no problem. But, part of the reason I posted was to get a clarification of what weakly continuous meant -- I just didn't use question marks. I have noticed over the years that it's much easier to get a correction on USEnet than an answer to a question. Perhaps that's not true of you -- then, good for you, but I'd still maintain you are not typical. What would you have done if you were me? Notice that I am not asking what you would have done in my place, but to imagine what it's like to be me and be faced with a similar choice. You are a professional mathematician, I am no kind of academic at all. You apparently take your appearance on USEnet very seriously, while I, having very little in the way of reputation to start with, have less to lose. You probably also have, beyond your mental mathematical resources, texts, references and colleagues practically at your fingertips. Well, I don't. Would you just dropped the topic and moved on? Maybe I would have, too, if I hadn't already invested half an hour or so in trying to beat a definition of weakly continuous function out of the World Wide Web. I hope you're not upset that I sorta-kinda-maybe-if-God- willing-I'm-right accused you of making a mistake. You do make mistakes, you know, though it's very rare, I'm sure. This was not one of them, but anyone who knew the correct definition would know that, including the original poster. Yes, I thought of that before I posted. If I had had a chance of misleading him, I would have let it drop. But if, by chance, I were right, I could have done a bit of good. I hope that answers your question. Yours, Jim Burns > However, the two >proofs are so similar in structure that one might consider >it a good clue. >If I'm right in thinking that /weakly/ continuous functions >f and g will map weakly convergent series to weakly convergent >series and limit point to limit point, then g will take >a weakly convergent series as input, then pass a weakly >convergent series to f, which returns yet another weakly >convergent series. Likewise for limit points. Therefore, >yes, the composition of weakly continuous functions is >weakly continuous. >I didn't know what weakly continuous meant, so I attempted >weakly continuous function less so. (The series x_n is >weakly convergent if the inner product < x_n, y > converges >for all y in the Hilbert space H.). >However, I strongly suspect the definition weakly continuous >function is analogous to continuous function, as I >expressed it above. If I'm wrong, I would appreciate being >told. (Too late if I've made a fool of myself, of course.) > If you don't know what weakly continuous means probably > you should determine that first, before attempting to > figure out questions regarding weakly continuous > functions? > Yes, you've made a fool of yourself. Nothing foolish > about not knowing what weakly continuous means - if > you'd asked someone would have said, no problem. But > asking a question about weakly continuous functions > _before_ finding a definition of the phrase, and > without asking for the definition, that's definitely > making a fool of yourself. As is conjecturing that > an answer is incorrect, before learning the > definition of the terms involved. > A Hilbert space has a weak topology. A weakly continuous > function is precisely a function which is continuous with > respect to this topology. And hence what I said _does_ > answer the question. >Jim Burns > ************************ > David C. Ullrich === Subject: Re: Properties of weakly continuous maps in a Hilbert space > On Mon, 10 Jul 2006 20:59:26 -0400, Jim Burns >That doesn't seem at first to answer the question, which >was about /weakly/ continuous functions. > How can you possibly even _conjecture_ that this does > not answer the question, since, as you say below, you > don't know what weakly continuous means? I hope you will believe me when I tell you I was not trying to upset you, or irritate you or get you. I'm sorry if I have. As to how I could _conjecture_ (or, as I would prefer, guess), I had already used what resources were readily available and had failed at finding out what weakly continuous meant (see below). At that point, my choices were to leave the post for another day (which my history shows often means never) or post with incomplete information and make it clear how incomplete my information was. Below you write if you'd asked someone would have said, no problem. But, part of the reason I posted was to get a clarification of what weakly continuous meant -- I just didn't use question marks. I have noticed over the years that it's much easier to get a correction on USEnet than an answer to a question. Perhaps that's not true of you -- then, good for you, but I'd still maintain you are not typical. What would you have done if you were me? Notice that I am not asking what you would have done in my place, but to imagine what it's like to be me and be faced with a similar choice. You are a professional mathematician, I am no kind of academic at all. You apparently take your appearance on USEnet very seriously, while I, having very little in the way of reputation to start with, have less to lose. You probably also have, beyond your mental mathematical resources, texts, references and colleagues practically at your fingertips. Well, I don't. Would you just dropped the topic and moved on? Maybe I would have, too, if I hadn't already invested half an hour or so in trying to beat a definition of weakly continuous function out of the World Wide Web. I hope you're not upset that I sorta-kinda-maybe-if-God- willing-I'm-right accused you of making a mistake. You do make mistakes, you know, though it's very rare, I'm sure. This was not one of them, but anyone who knew the correct definition would know that, including the original poster. Yes, I thought of that before I posted. If I had had a chance of misleading him, I would have let it drop. But if, by chance, I were right, I could have done a bit of good. I hope that answers your question. Yours, Jim Burns > However, the two >proofs are so similar in structure that one might consider >it a good clue. >If I'm right in thinking that /weakly/ continuous functions >f and g will map weakly convergent series to weakly convergent >series and limit point to limit point, then g will take >a weakly convergent series as input, then pass a weakly >convergent series to f, which returns yet another weakly >convergent series. Likewise for limit points. Therefore, >yes, the composition of weakly continuous functions is >weakly continuous. >I didn't know what weakly continuous meant, so I attempted >weakly continuous function less so. (The series x_n is >weakly convergent if the inner product < x_n, y > converges >for all y in the Hilbert space H.). >However, I strongly suspect the definition weakly continuous >function is analogous to continuous function, as I >expressed it above. If I'm wrong, I would appreciate being >told. (Too late if I've made a fool of myself, of course.) > If you don't know what weakly continuous means probably > you should determine that first, before attempting to > figure out questions regarding weakly continuous > functions? > Yes, you've made a fool of yourself. Nothing foolish > about not knowing what weakly continuous means - if > you'd asked someone would have said, no problem. But > asking a question about weakly continuous functions > _before_ finding a definition of the phrase, and > without asking for the definition, that's definitely > making a fool of yourself. As is conjecturing that > an answer is incorrect, before learning the > definition of the terms involved. > A Hilbert space has a weak topology. A weakly continuous > function is precisely a function which is continuous with > respect to this topology. And hence what I said _does_ > answer the question. >Jim Burns > ************************ > David C. Ullrich === Subject: Re: Properties of weakly continuous maps in a Hilbert space >I didn't know what weakly continuous meant, so I attempted >weakly continuous function less so. (The series x_n is >weakly convergent if the inner product < x_n, y > converges >for all y in the Hilbert space H.). >However, I strongly suspect the definition weakly continuous >function is analogous to continuous function, as I >expressed it above. If I'm wrong, I would appreciate being >told. (Too late if I've made a fool of myself, of course.) I can understand your confusion. What you didn't know (and what may be hard to find out just by googling) is that the phrase weakly continuous does not mean a special kind of continuity is involved; it's perfectly ordinary continuity, but with respect to a particular topology (the weak topology). You see, continuity *always* refers to some topology or other, because the concept is meaningless without it. It's just that in very many situations, there is only one *standard* topology, and therefore it's assumed the standard topology is the one intended, in the absence of any statement to the contrary. Not so in Hilbert space, where there are several different topologies in common use. Some of these are called the strong, weak, and norm topologies. So, the phrase weakly continuous means continuous in the weak topology. I forget now what was the exact context of your question, but I think it may have been whether a composition of weakly continuous functions is weakly continuous. The point is, a composition of continuous functions is always continuous, no matter what the underlying topology may be. Of course, a function that is continuous in one topology may be discontinuous in some other topology, but that doesn't matter here. As long as you use the same topology throughout, any composition of functions that are continous in the given topology will always be continuous in that same topology. -- Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. === Subject: Re: Properties of weakly continuous maps in a Hilbert space On Wed, 12 Jul 2006 22:42:53 -0400, Jim Burns On Mon, 10 Jul 2006 20:59:26 -0400, Jim Burns >That doesn't seem at first to answer the question, which >was about /weakly/ continuous functions. >> How can you possibly even _conjecture_ that this does >> not answer the question, since, as you say below, you >> don't know what weakly continuous means? >I hope you will believe me when I tell you I was not trying to >upset you, or irritate you or get you. I'm sorry if I have. >As to how I could _conjecture_ (or, as I would prefer, >guess), I had already used what resources were readily >available and had failed at finding out what weakly >continuous meant (see below). At that point, my choices were >to leave the post for another day (which my history shows >often means never) or post with incomplete information >and make it clear how incomplete my information was. Those were not your only choices. You could have simply asked what the phrase meant! >Below you write if you'd asked someone would have said, >no problem. But, part of the reason I posted was to get a >clarification of what weakly continuous meant -- I just >didn't use question marks. I have noticed over the years that >it's much easier to get a correction on USEnet than an answer >to a question. Perhaps that's not true of you -- then, good >for you, but I'd still maintain you are not typical. That's not true of sci.math. When you ask a question involving the phrase weafly continuous people are going to assume you know what the phrase means. If you ask what it means then people will tell you. >What would you have done if you were me? Notice that I am not >asking what you would have done in my place, but to imagine >what it's like to be me and be faced with a similar choice. When you put the question that way it's a little silly. If I were you then by definition everything I do would be the same as what you did. What you _should_ have done is simply ask what the term meant before asking a question _about_ it. (Certainly before explaining why someone else's statements about it were wrong...) >You are a professional mathematician, I am no kind of academic >at all. You apparently take your appearance on USEnet very >seriously, while I, having very little in the way of >reputation to start with, have less to lose. You probably >also have, beyond your mental mathematical resources, texts, >references and colleagues practically at your fingertips. >Well, I don't. What's your point? Yes, I'm a professional mathematician. Happens all the time around here that I see people talking about things I know nothing about - if I can't find a definition quickly on the web I ask, and someone usually explains. sci.math is a _huge_ resource, and your access to it is the same as mine. >Would you just dropped the topic and moved on? Maybe I would >have, too, if I hadn't already invested half an hour or so >in trying to beat a definition of weakly continuous >function out of the World Wide Web. Again, I don't see your point. Lemme see: What does it mean to say that a map from a Hilbert space to itself is `weakly continuous'? That didn't take a half hour for me to type. >I hope you're not upset that I sorta-kinda-maybe-if-God- >willing-I'm-right accused you of making a mistake. You >do make mistakes, you know, though it's very rare, I'm sure. >This was not one of them, but anyone who knew the correct >definition would know that, including the original poster. >Yes, I thought of that before I posted. If I had had a chance >of misleading him, I would have let it drop. But if, by >chance, I were right, I could have done a bit of good. Just trying to help - that's noble. But you got it wrong. Here's the deal: sci.math can be a very useful thing. A lot of people learn about a lot of things here; I know that _I've_ learned a lot of interesting stuff here, for example. sci.math is a useful thing because there are a lot of people who know what they're talking about, giving away free information. But it's also true that there are a lot of people who have no idea what they're talking about, posting a lot of nonsense. sci.math would be a lot more useful if that were not so. When you volunteer a correction even though you _know_ that you don't know what you're talking about that counts as being part of the _problem_, not part of the solution. Again: yes, anyone can make a mistake. I certainly do, more often than some posters here. When someone says something false that's too bad, because it can cause confusion. Usually errors get corrected, and we just hope that anyone who sees the error will also see the correction. So anyone can make a mistake, there's no way to ensure that everything posted here is correct. Fine. _But_ when you say something on a topic even though you _know_ that you don't know what you're talking about, that's _not_ in the anyone-can-make-a-mistake category! Simply not making assertions about things you know you know nothing about is a very simple rule that will prevent you from making _that_ sort of mistake. >I hope that answers your question. >Yours, >Jim Burns > However, the two >proofs are so similar in structure that one might consider >it a good clue. >If I'm right in thinking that /weakly/ continuous functions >f and g will map weakly convergent series to weakly convergent >series and limit point to limit point, then g will take >a weakly convergent series as input, then pass a weakly >convergent series to f, which returns yet another weakly >convergent series. Likewise for limit points. Therefore, >yes, the composition of weakly continuous functions is >weakly continuous. >I didn't know what weakly continuous meant, so I attempted >weakly continuous function less so. (The series x_n is >weakly convergent if the inner product < x_n, y > converges >for all y in the Hilbert space H.). >However, I strongly suspect the definition weakly continuous >function is analogous to continuous function, as I >expressed it above. If I'm wrong, I would appreciate being >told. (Too late if I've made a fool of myself, of course.) >> If you don't know what weakly continuous means probably >> you should determine that first, before attempting to >> figure out questions regarding weakly continuous >> functions? >> Yes, you've made a fool of yourself. Nothing foolish >> about not knowing what weakly continuous means - if >> you'd asked someone would have said, no problem. But >> asking a question about weakly continuous functions >> _before_ finding a definition of the phrase, and >> without asking for the definition, that's definitely >> making a fool of yourself. As is conjecturing that >> an answer is incorrect, before learning the >> definition of the terms involved. >> A Hilbert space has a weak topology. A weakly continuous >> function is precisely a function which is continuous with >> respect to this topology. And hence what I said _does_ >> answer the question. >Jim Burns >> ************************ >> David C. Ullrich ************************ David C. Ullrich === Subject: Re: Properties of weakly continuous maps in a Hilbert space > I didn't know what weakly continuous meant, so I attempted > Weak continuity means, between the weak topologies. The weak > topology for a Hilbert space has as a subbase for the > neighborhoods of 0, the sets of the form {x : (w,x) < 1}. > This means you take finite intersections of sets of this form > (to get the base of the neighborhood system at 0), then > translates of these, then arbitrary unions of THOSE, to get > the topology. > David's comment was meant to convey the fact that > compositions of continuous mappings (between compatible > topologies) are ALWAYS continuous. > There are many variations on weak continuity: demicontinuity > [from strong to weak topology], hemicontinuity [from strong to > weak on line segments--or weak to weak, if you prefer (same > thing)], etc. etc. But one thing you need to understand: > these variations were introduced BECAUSE THERE ARE VERY FEW > WEAKLY CONTINUOUS FUNCTIONS!! of convergence and continuity to my list of things I want to loom into mrore closely, when I get around to it. Jim Burns === Subject: Re: How can you determine if two matrices are significantly different statistically? > You have two different points confounded. > If your matrices you want to test for difference is NOT either a > covariance matrix OR a correlation matrix, then Morrison's book > and AFAIK none of the standard multivariate statistics books will > be able to help you. > If you DO have two covariance matrices, then there are different > tests for them to be the same. You got that point but missed the > one in my preceding paragraph. > It appears that your data in the matrices are, as you said, some > kind of SIMILARITY or DISSIMILARITY that are not covariances > or correlations. I got that part from someone else's reply to you, but EEG data is much too general an label to imply anything specific. In particular, if you had described your data as you do now in the two paragraphs below, I would have known immediately that they were NOT any of the matrices I thought you might have had. > It's > voltages collected on 128 electrodes over time. So a row is the voltage > at time points 0 - 2047 / second. A column is the voltage at a > timepoint for every electrode. Sounds like a multiple time-series to me, where the notion of a matrix is not even necessary! > I have two matrices: one for data collected for one trial and one for > data collected for another trial. You have two sets of multiple time series collected at different trials. Those are rather different horses from what I thought you were talking about in your OP and later posts, until now. > I also create matrices that are various permutations of the raw data > such as doing an fft for each channel and placing the 128 ffts in a > channel X frequency matrix. Or doing a correlation (cross correlogram) > of eacht row of the two matrices (correlate data from electrode 1, > trial 1, with electrode 2, trial 2, ....)and putting the resulting > vectors in a matrix. Then I might want to compare two such matrices. I > hoped that there might be a method to simply compare two matrices of > numbers regardless of what the numbers represent, that I could use with > my raw data as well as with various derived quantities. I'll bow out of this thread, because this is NOT one of the areas I can claim expertise, and I firmly believe Silence is Golden on those occasions for me. > Are they even symmetric? That is, is A(i,j) = A (j,i) for all (i,j)? > Multidimensional Scaling (MDS) is a way of representing the > objects as points in a Euclidean Space, but since there are > no distributional assumptions in MDS solutions, there is no way > for one to statistically TEST if two such configurations are > significantly different. > Understood. Is this true for discriminant methods? Discriminant methods are horses of a different color from any of the methods mentioned so far, > Besides, that's getting far afield from formal, standard statistical > methods. >Am >I correct in thinking you are talking about a sort of second order >test, in which correlation(s) is assessed, and then the differences >between two sets of correlation is compared? > That would be if your A(i,j) is a correlation matrix, and B(i,j) is a > different correlation matrix. There is nothing second order about > it. If they are NOT correlations under some assume model from > which they came, there's not much you can do about them. >I hope for a single >measure of difference...in the same way a t-test is a way to determine >the significance of the difference between two numbers (given certain >assumptions). > There the assumptions are typically data from normal populations > between the MEANS. If they have different variances, then you have > the Behrens-Fisher (unsolved) problem which can have approximate > solutions only. Finally when you weaken the assumptions to no > distributional assumption on the data, there are nonparametric methods > for testing differences of locations. > In short, everything depends on what is ASSUMED to be known, and > what is to be compared (tested), >I want to test the difference between two matrices of >numbers. Perhaps that doesn't make sense. > That makes sense. It's just that statisticians have not found a > way to tackle that sensible question yet. :-) > Heh, for once I'm on the cutting edge. Did it hurt? :-) Actually no. You merely mis-described your PROBLEM as one of testing the equality of two matrices! Multiple time series are hardly new -- that's practically a whole field by itself. > In fact, there are tons of sensible questions that statisticians > don't know how to deal with, and some of them are unlikely to be > ever solvable statistically. > JIm > Nada, > -- Reef Fish Bob. > Jim You're welcome. Now I'll let others take over your NEW problem, or a revised description of your original problem. -- Reef Fish Bob. === Subject: Dirac delta with complex argument What is the meaning of Dirac delta when the argument is complex. delta(z) where z is a complex number is dealt with in the following two papers: (1) Generalisation of the Dirac-delta impulse extending Laplace and z-transform domains Corinthios, M.J.; Vision, Image and Signal Processing, IEE Proceedings- (2) Complex-variable distribution theory for Laplace and z-transforms Corinthios, M.J.; Vision, Image and Signal Processing, IEE Proceedings- Volume 152, Issue 1, 28 Feb. 2005 Page(s):97 - 106 I am still trying to understand and digest the ideas mentioned in the above papers. It is not clear whether delta(z) is the same as a two-dimensional impulse or different from it. In http://www.theorie.physik.uni-muenchen.de/~serge/no_distrib_limit.pdf it is argued Why the Dirac delta function cannot be applied to complex arguments. What is the final word? vv === Subject: Re: Dirac delta with complex argument > What is the meaning of Dirac delta when the argument is complex. > delta(z) where z is a complex number is dealt with in the following two > papers: > (1) Generalisation of the Dirac-delta impulse extending Laplace and > z-transform domains > Corinthios, M.J.; > Vision, Image and Signal Processing, IEE Proceedings- > (2) Complex-variable distribution theory for Laplace and z-transforms > Corinthios, M.J.; > Vision, Image and Signal Processing, IEE Proceedings- > Volume 152, Issue 1, 28 Feb. 2005 Page(s):97 - 106 > I am still trying to understand and digest the ideas mentioned in the > above papers. It is not clear whether delta(z) is the same as a > two-dimensional impulse or different from it. In > http://www.theorie.physik.uni-muenchen.de/~serge/no_distrib_limit.pdf > it is argued Why the Dirac delta function cannot be applied to complex > arguments. > What is the final word? > vv Every distribution S in D'(R) has an analytic representation, i.e. a function s analytic in CR satisfying lim_(y -> 0+) s(x + iy) - s(x - iy) = S(x) in the distributional sense. Two analytic representations of the same distribution differ only by an entire function which is 0 on the real line. Maybe these ideas can be useful to define delta(z), but I don't know. === Subject: Re: Dirac delta with complex argument > What is the meaning of Dirac delta when the argument is complex. > delta(z) where z is a complex number is dealt with in the following two > papers: > (1) Generalisation of the Dirac-delta impulse extending Laplace and > z-transform domains > Corinthios, M.J.; > Vision, Image and Signal Processing, IEE Proceedings- > (2) Complex-variable distribution theory for Laplace and z-transforms > Corinthios, M.J.; > Vision, Image and Signal Processing, IEE Proceedings- > Volume 152, Issue 1, 28 Feb. 2005 Page(s):97 - 106 > I am still trying to understand and digest the ideas mentioned in the > above papers. It is not clear whether delta(z) is the same as a > two-dimensional impulse or different from it. In > http://www.theorie.physik.uni-muenchen.de/~serge/no_distrib_limit.pdf > it is argued Why the Dirac delta function cannot be applied to complex > arguments. > What is the final word? > vv The best I can come with is sign(Im(z)) / 2 pi i z + delta(Im(z)) delta(Re(z)) or something like that. Stephen === Subject: A question on Spec Z[x] In the course of describing a typical prime ideal P in Z[x], one considers the ideal : P intersect Z which (in case P is nonprincipal) is a maximal ideal in Z. Say P intersect Z = (p), where p is a prime integer. The field Z/(p) naturally embeds in Z[x]/P. I read online that this embedding is a finite field extension, which I take to mean that Z[x]/P is a field which has finite dimension over the field Z/(p) as a vector space. ..But I don't see how Z[x]/P is even a field! If you believe that what I read was wrong then maybe you can help me discover what P looks like. dan === Subject: Re: A question on Spec Z[x] > In the course of describing a typical prime ideal P in Z[x], one > considers the ideal : P intersect Z > which (in case P is nonprincipal) is a maximal ideal in Z. Say P > intersect Z = (p), where p is a prime integer. > The field Z/(p) naturally embeds in Z[x]/P. > I read online that this embedding is a finite field extension, which I > take to mean that > Z[x]/P is a field which has finite dimension over the field Z/(p) as a > vector space. > ..But I don't see how Z[x]/P is even a field! Assume P is a non-principal prime in Z[x] such that P intersect Z = (p) with p a prime in Z. Then there is a surjective map Z[X]/(p) -> Z[X]/P coming from the inclusion (p) --> P. The kernel of this map is P/(p), which is not zero because we are assuming that P is not principal. Clearly, P/(p) is a prime ideal of Z[X]/(p). Since Z[X]/(p) = (Z/(p))[X] is a PID, P/(p) is principal, so there exists f in Z[X] such that its image g in (Z/(p))[X] generates P/(p); this implies in particular that P = (f, p). Note that g is irreducible in (Z/(p))[X] because P/(p) is prime. Now, Z[X]/P = (Z/(p))[X]/(g) with g irreducible in (Z/(p))[X], so it is a field of finite degree over Z/(p). HTH. -- m === Subject: Re: Is sum of two normally distributed variables normally distributed? >>That's how I explained to the Surrealists on the matter of LOGIC >>which they seem to be lacking, in their obsessed search for >>pathological example, in response to a SIMPLE question. >Eh? The simple question was, is it possible for X and Y to >be normally distributed while X+Y is not? >>Except that WASN'T the OP's question! > In my newsreader, I find the following wind-up question at the > end of OP's post: >Is it possible at all to create >X ~ N(mu_X,sigma_X) and Y ~ N(mu_Y,sigma_Y) >so that X+Y !~ N(mu,sigma), >where !~ denotes not distributed as? > Perhaps that was cut off in your newsreader. > - Randy It is amazing to me that Mr. Reef Fish Bob waisted pages on concocting his answers to the question he neglected to read and comprehend. I suppose his first academic appointment began too long ago (in 1977 by his own admission) and he does not read questions from the beginning anymore :) No wonder he gets easily confused with what's Real and Surreal in this world. BTW, for all respondents I strongly suggest reading this book (http://www.pupress.princeton.edu/titles/7929.html) before engaging in lengthy and unproductive exchange of abusive e-mails. - Janusz. === Subject: Re: Is sum of two normally distributed variables normally distributed? <44B1F69B.7040800@netscape.net> <44B1FE30.6050101@netscape.net> his answers to the question he neglected to read and comprehend. On your allegation of neglected to read and comprehend. 1. You neglected to read that I was the FIRST to respond to the OP, and I asked WHY he asked, and he gave this response and thanked me. Those were the first THREE posts of this thread until the pathologists entered: OP>> However, i have reasons to believe that it is not totally OP>> true. RF> What are your reasons? OP> Strong belief in it combined with the fact that i've OP> always used it and didn't die, hehe. I'll be done OP> with my master thesis this winter so i'm sort of OP> embarassed by such a mistake... :) > I suppose his first academic appointment began too long ago (in 1977 by > his own admission) and he does not read questions from the beginning > anymore :) 2. What was cited above were in the FIRST THREE posts of this thread. Who does not read questions from the beginning? 3. his first academic appointment began (in 1977 by his own admission? By 1977 I was already a tenured FULL Professor -- hardly his first academic appointment, Who failed to comprehend what he read? Janusz Kawczak, who has a LIFE TIME posting history of 18 posts Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 2005 (1 in Oct) 2006 1 3 5 2 2 4 2 16 of which in newsgroups I DO NOT READ nor participate (except in response to cross-posted posts, in sci.stat.math), came out bull ting about my neglected to read and comprehend. I demonstrated above with THREE concrete evidence that HE was the one who neglected to read and comprehend. > BTW, for all respondents I strongly suggest reading this book > (http://www.pupress.princeton.edu/titles/7929.html) . > - Janusz. That must be Janusz's credential for his bull mastery by citing Harry G. Frankfurt's book On Bull. -- Reef Fish Bob. === Subject: Re: Is sum of two normally distributed variables normally distributed? Regarding the citation, it is precisely to the contrary of your supposition. Therein, you may find comments on how to identify it and not to produce it in abundance, as you so demonstrated. - Janusz. >>It is amazing to me that Mr. Reef Fish Bob waisted pages on concocting >>his answers to the question he neglected to read and comprehend. > On your allegation of neglected to read and comprehend. > 1. You neglected to read that I was the FIRST to respond to the OP, > and I asked WHY he asked, and he gave this response and > thanked me. Those were the first THREE posts of this thread > until the pathologists entered: > OP>> However, i have reasons to believe that it is not totally > OP>> true. > RF> What are your reasons? > OP> Strong belief in it combined with the fact that i've > OP> always used it and didn't die, hehe. I'll be done > OP> with my master thesis this winter so i'm sort of > OP> embarassed by such a mistake... :) >>I suppose his first academic appointment began too long ago (in 1977 by >>his own admission) and he does not read questions from the beginning >>anymore :) > 2. What was cited above were in the FIRST THREE posts of > this thread. Who does not read questions from the beginning? > 3. his first academic appointment began (in 1977 by his own > admission? By 1977 I was already a tenured FULL > Professor -- hardly his first academic appointment, > Who failed to comprehend what he read? > Janusz Kawczak, who has a LIFE TIME posting history of 18 posts > Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec > 2005 (1 in Oct) > 2006 1 3 5 2 2 4 2 > 16 of which in newsgroups I DO NOT READ nor participate > (except in response to cross-posted posts, in sci.stat.math), came > out bull ting about my neglected to read and comprehend. > I demonstrated above with THREE concrete evidence that HE > was the one who neglected to read and comprehend. >>BTW, for all respondents I strongly suggest reading this book >>(http://www.pupress.princeton.edu/titles/7929.html) . >>- Janusz. > That must be Janusz's credential for his bull mastery by > citing Harry G. Frankfurt's book On Bull. > -- Reef Fish Bob. === Subject: Re: Is sum of two normally distributed variables normally distributed? Reef Fish, I am curious why are you hiding under this silly pseudonym? What are you afraid of? To me, any person hiding under pseudonym and tries to give meritorious answer bears no credentials of any kind. So, what is your real name? - Janusz. >>It is amazing to me that Mr. Reef Fish Bob waisted pages on concocting >>his answers to the question he neglected to read and comprehend. > On your allegation of neglected to read and comprehend. > 1. You neglected to read that I was the FIRST to respond to the OP, > and I asked WHY he asked, and he gave this response and > thanked me. Those were the first THREE posts of this thread > until the pathologists entered: > OP>> However, i have reasons to believe that it is not totally > OP>> true. > RF> What are your reasons? > OP> Strong belief in it combined with the fact that i've > OP> always used it and didn't die, hehe. I'll be done > OP> with my master thesis this winter so i'm sort of > OP> embarassed by such a mistake... :) >>I suppose his first academic appointment began too long ago (in 1977 by >>his own admission) and he does not read questions from the beginning >>anymore :) > 2. What was cited above were in the FIRST THREE posts of > this thread. Who does not read questions from the beginning? > 3. his first academic appointment began (in 1977 by his own > admission? By 1977 I was already a tenured FULL > Professor -- hardly his first academic appointment, > Who failed to comprehend what he read? > Janusz Kawczak, who has a LIFE TIME posting history of 18 posts > Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec > 2005 (1 in Oct) > 2006 1 3 5 2 2 4 2 > 16 of which in newsgroups I DO NOT READ nor participate > (except in response to cross-posted posts, in sci.stat.math), came > out bull ting about my neglected to read and comprehend. > I demonstrated above with THREE concrete evidence that HE > was the one who neglected to read and comprehend. >>BTW, for all respondents I strongly suggest reading this book >>(http://www.pupress.princeton.edu/titles/7929.html) . >>- Janusz. > That must be Janusz's credential for his bull mastery by > citing Harry G. Frankfurt's book On Bull. > -- Reef Fish Bob. === Subject: Re: Is sum of two normally distributed variables normally distributed? <44B1F69B.7040800@netscape.net> <44B1FE30.6050101@netscape.net> What are you afraid of? To me, any person hiding under pseudonym > and tries to give meritorious answer bears no credentials of any kind. Do you think I really care what YOU think? Just like the rest of the morons (at least a hundred of them over the years) who had asked > So, what is your real name? > - Janusz. Why don't you learn how to use Google search, and find out enough about Reef Fish for a 20 page biography. Do YOUR own homework. I took me seconds to find out your life time posting record of 18 posts over a period of 8 months. If you have something of statistical substance to say, say it. Otherwise, just keep you big mouse SHUT to keep you foot from getting in there all the time. -- Reef Fish Bob, >>It is amazing to me that Mr. Reef Fish Bob waisted pages on concocting >>his answers to the question he neglected to read and comprehend. > On your allegation of neglected to read and comprehend. > 1. You neglected to read that I was the FIRST to respond to the OP, > and I asked WHY he asked, and he gave this response and > thanked me. Those were the first THREE posts of this thread > until the pathologists entered: > OP>> However, i have reasons to believe that it is not totally > OP>> true. > RF> What are your reasons? > OP> Strong belief in it combined with the fact that i've > OP> always used it and didn't die, hehe. I'll be done > OP> with my master thesis this winter so i'm sort of > OP> embarassed by such a mistake... :) >>I suppose his first academic appointment began too long ago (in 1977 by >>his own admission) and he does not read questions from the beginning >>anymore :) > 2. What was cited above were in the FIRST THREE posts of > this thread. Who does not read questions from the beginning? > 3. his first academic appointment began (in 1977 by his own > admission? By 1977 I was already a tenured FULL > Professor -- hardly his first academic appointment, > Who failed to comprehend what he read? > Janusz Kawczak, who has a LIFE TIME posting history of 18 posts > Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec > 2005 (1 in Oct) > 2006 1 3 5 2 2 4 2 > 16 of which in newsgroups I DO NOT READ nor participate > (except in response to cross-posted posts, in sci.stat.math), came > out bull ting about my neglected to read and comprehend. > I demonstrated above with THREE concrete evidence that HE > was the one who neglected to read and comprehend. >>BTW, for all respondents I strongly suggest reading this book >>(http://www.pupress.princeton.edu/titles/7929.html) . >>- Janusz. > That must be Janusz's credential for his bull mastery by > citing Harry G. Frankfurt's book On Bull. > > -- Reef Fish Bob. === Subject: Re: Is sum of two normally distributed variables normally distributed? And, why would I waste my time on looking for some unimportant Reef Fish Bob and use Google's valuable bandwidth for it? That's a complete misuse of resources. Make sure to swim in deep waters! Otherwise, you can really get stuck on some unpleasant corals :) - Janusz. >>Reef Fish, I am curious why are you hiding under this silly pseudonym? > That question is now a unmistakable sign of CLUELESSNESS. > The Reef Fish name was explained in 1992. It's still Google > retrievable. > As of a few year ago your inuendo and question was common. > NOW there's hardly anyone who has read sci.stat.math for more than > a few month does NOT know who Reef Fish is. >>What are you afraid of? To me, any person hiding under pseudonym >>and tries to give meritorious answer bears no credentials of any kind. > Do you think I really care what YOU think? Just like the rest of the > morons (at least a hundred of them over the years) who had asked >>So, what is your real name? >>- Janusz. > Why don't you learn how to use Google search, and find out enough > about Reef Fish for a 20 page biography. Do YOUR own homework. > I took me seconds to find out your life time posting record of 18 posts > over a period of 8 months. > If you have something of statistical substance to say, say it. > Otherwise, > just keep you big mouse SHUT to keep you foot from getting in there > all the time. > -- Reef Fish Bob, >>It is amazing to me that Mr. Reef Fish Bob waisted pages on concocting >>his answers to the question he neglected to read and comprehend. >On your allegation of neglected to read and comprehend. >1. You neglected to read that I was the FIRST to respond to the OP, > and I asked WHY he asked, and he gave this response and > thanked me. Those were the first THREE posts of this thread > until the pathologists entered: >OP>> However, i have reasons to believe that it is not totally >OP>> true. >RF> What are your reasons? >OP> Strong belief in it combined with the fact that i've >OP> always used it and didn't die, hehe. I'll be done >OP> with my master thesis this winter so i'm sort of >OP> embarassed by such a mistake... :) >>I suppose his first academic appointment began too long ago (in 1977 by >>his own admission) and he does not read questions from the beginning >>anymore :) >2. What was cited above were in the FIRST THREE posts of > this thread. Who does not read questions from the beginning? >3. his first academic appointment began (in 1977 by his own > admission? By 1977 I was already a tenured FULL > Professor -- hardly his first academic appointment, > Who failed to comprehend what he read? >Janusz Kawczak, who has a LIFE TIME posting history of 18 posts > Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec >2005 (1 in Oct) >2006 1 3 5 2 2 4 2 >16 of which in newsgroups I DO NOT READ nor participate >(except in response to cross-posted posts, in sci.stat.math), came >out bull ting about my neglected to read and comprehend. >I demonstrated above with THREE concrete evidence that HE >was the one who neglected to read and comprehend. >>BTW, for all respondents I strongly suggest reading this book >>(http://www.pupress.princeton.edu/titles/7929.html) . >>- Janusz. >That must be Janusz's credential for his bull mastery by >citing Harry G. Frankfurt's book On Bull. >-- Reef Fish Bob. === Subject: Re: Probability question - please help! ... > Here's the second question: Relating to probability, I am trying to > determine the probability of two events: E1 is the probability that I > will score less than or equal to a given score (in other words, using > my historical data to predict how probable it is that I will shoot a > given score). E2 is the probability that I will score exactly the > given score. ... > So I'm thinking, well, if I say I want to score exactly a 58 for E2, > then the likelihood would be one over the total number of possible > discrete scores from 9-72, right? Wouldn't my probability then be > 1/64? Or, is 9 too low to use as the lower bound, since I would never > score it? How about 58, would that be my lower bound, since I scored > it once already? ... You seem to have assumed that all the scores from 9 to 72 are equally likely to occur, but such an assumption is unrealistic. Scores close to your average score are more likely to occur than scores distant from your average, and different scores will arise with different probabilities that probably cannot be reasonably computed from first principles. Anyway, golf scores should be the result of skilled play rather than random processes. I think the best you can do is to model your scores on individual holes, and then find the probability that holes 1...9 (or 10...18) add up to 58. For example, if in ten rounds you've scored 3 3's, 5 4's, and 2 5's on hole 1, you might model hole 1 probabilities as P(3)=0.3, P(4)=0.4, P(5)=0.2. With such a model for each hole, use random numbers in a Monte Carlo process to play 10000 rounds, then report the number of rounds for each total that occurred in the simulation. -jiw === Subject: Re: Probability question - please help! <44B5C9EA.1F1A5FB9@pat7.com> Hi James, Just one last question about your suggested methodology - you are mentioning calculating my probabilities for each hole based on historical data, which is fair enough. How would that differ or compare to calculating the probabilities of each hole being equally likely? In other words, like how I first described, instead of considering the total score, I consider the discrete scores possible for each hole over a [min,max] and assign equal probabilities to them. Tell me what is wrong with my thinking in that approach. Mike > ... > Here's the second question: Relating to probability, I am trying to > determine the probability of two events: E1 is the probability that I > will score less than or equal to a given score (in other words, using > my historical data to predict how probable it is that I will shoot a > given score). E2 is the probability that I will score exactly the > given score. > ... > So I'm thinking, well, if I say I want to score exactly a 58 for E2, > then the likelihood would be one over the total number of possible > discrete scores from 9-72, right? Wouldn't my probability then be > 1/64? Or, is 9 too low to use as the lower bound, since I would never > score it? How about 58, would that be my lower bound, since I scored > it once already? > ... > You seem to have assumed that all the scores from 9 to 72 are > equally likely to occur, but such an assumption is unrealistic. > Scores close to your average score are more likely to occur > than scores distant from your average, and different scores > will arise with different probabilities that probably cannot > be reasonably computed from first principles. Anyway, golf > scores should be the result of skilled play rather than > random processes. > I think the best you can do is to model your scores on individual > holes, and then find the probability that holes 1...9 (or 10...18) > add up to 58. For example, if in ten rounds you've scored 3 3's, > 5 4's, and 2 5's on hole 1, you might model hole 1 probabilities > as P(3)=0.3, P(4)=0.4, P(5)=0.2. With such a model for each hole, > use random numbers in a Monte Carlo process to play 10000 rounds, > then report the number of rounds for each total that occurred in > the simulation. > -jiw === Subject: Re: Probability question - please help! ... > Just one last question about your suggested methodology - you are > mentioning calculating my probabilities for each hole based on > historical data, which is fair enough. How would that differ or > compare to calculating the probabilities of each hole being equally > likely? In other words, like how I first described, instead of > considering the total score, I consider the discrete scores possible > for each hole over a [min,max] and assign equal probabilities to them. > Tell me what is wrong with my thinking in that approach. ... With that approach, as Randy Poe noted, your chance of scoring 9 holes in one is exactly the same as your chance of scoring 9 pars. In my previous post, I said golf is a game of skill, which implies that outcomes are not totally random. I think this logically forces the conclusion that a simple equally-likely-outcomes approach is wrong. More pragmatically, it is obvious that on most holes for most golfers, par is more likely than a birdie, and far more likely than an eagle. So it doesn't make sense to use equal probabilities for all possible scores. I think there is no completely-theoretical way of figuring out the probability of different scores. Even with deep study of the factors you mentioned in another post -- real-life data and phenomena [...] equipment characteristics, ball characteristics, environmental characteristics (wind, landscape, etc.) -- figuring out the chances would be problematic. Note, major league baseball teams are using computer programs, during games, to figure out in real time how to play (eg, whether to bunt, steal, etc.); I think you can find a lot of info about this on the web; it might suggest ways to predict golf scores. Statistically, you could compute correlation of first hole scores (or first plus second, etc.) with totals, or could look for indicator holes, where the score for a hole correlates well with total score. -jiw === Subject: Re: Probability question - please help! <44B5C9EA.1F1A5FB9@pat7.com> <44B67079.FBBB5CFF@pat7.com> Hi James, Ah, okay, now we are getting somewhere. You are speaking my language (which is a big deal to me actually, when talking technically like this). I understand that just to fill in a couple of cells in a spreadsheet, there is really no practical way to do the complex computer modeling I am asking about. But equally invalid is the assumption that hole scores are equally likely. So, I will take the Monte Carlo/individual approach and see what that yields. The reason I started thinking about this was because, I'm an Excel fanatic but don't often (until this year at work) get to really dig into the function library, and I figured I probably better look there to see what the experts put in there for me, rather than trying to do some Einstein-ing and thinking about the probabilities in that way. I thought perhaps that this is a very well-known problem and there should be a pretty straightforward way to compute what I want. Now, you mention the term correlation, and I understand that that is not just an analyst's looking at data and coming to some conclusion(s) but rather, a real mathematical-statistical process/function. I don't really know what it means, but I would like for you to explain correlation to me if you could. Also, I think I misstated in a previous post - I had the s.d. and variance (which one is the other squared) backwards. Finally, I noticed this skewness factor, and the best I can make sense of it graphically is that, as I played with a sample spreadsheet given to me by someone else and looked at varying skewness factors online, it became apparent to my mind what skewness really measures -- the direction of shift of the distribution hump if you will. Maybe that's not right but for now that's how I'm thinking of skewness. I imagine small skewness (like my golf scores indicate) means the curve is fairly well equal about some mean and not humped up to one side or another. Mike > ... > Just one last question about your suggested methodology - you are > mentioning calculating my probabilities for each hole based on > historical data, which is fair enough. How would that differ or > compare to calculating the probabilities of each hole being equally > likely? In other words, like how I first described, instead of > considering the total score, I consider the discrete scores possible > for each hole over a [min,max] and assign equal probabilities to them. > Tell me what is wrong with my thinking in that approach. > ... > With that approach, as Randy Poe noted, your chance of scoring 9 > holes in one is exactly the same as your chance of scoring 9 pars. > In my previous post, I said golf is a game of skill, which implies > that outcomes are not totally random. I think this logically forces > the conclusion that a simple equally-likely-outcomes approach is wrong. > More pragmatically, it is obvious that on most holes for most golfers, > par is more likely than a birdie, and far more likely than an eagle. > So it doesn't make sense to use equal probabilities for all possible > scores. > I think there is no completely-theoretical way of figuring out the > probability of different scores. Even with deep study of the > factors you mentioned in another post -- real-life data and phenomena > [...] equipment characteristics, ball characteristics, environmental > characteristics (wind, landscape, etc.) -- figuring out the chances > would be problematic. Note, major league baseball teams are using > computer programs, during games, to figure out in real time how to > play (eg, whether to bunt, steal, etc.); I think you can find a lot > of info about this on the web; it might suggest ways to predict golf > scores. > Statistically, you could compute correlation of first hole scores (or > first plus second, etc.) with totals, or could look for indicator > holes, where the score for a hole correlates well with total score. > -jiw === Subject: Re: Probability question - please help! <44B5C9EA.1F1A5FB9@pat7.com Hi James, > Just one last question about your suggested methodology - you are > mentioning calculating my probabilities for each hole based on > historical data, which is fair enough. How would that differ or > compare to calculating the probabilities of each hole being equally > likely? In other words, like how I first described, instead of > considering the total score, I consider the discrete scores possible > for each hole over a [min,max] and assign equal probabilities to them. > Tell me what is wrong with my thinking in that approach. It means your chance of scoring 9 holes in one is exactly the same as your chance of scoring 9 pars. - Randy === Subject: Re: Probability question - please help! <44B5C9EA.1F1A5FB9@pat7.com> Hi Randy, I'm not sure quite what you mean, but I know that the probabilities for the events you describe are pretty much zero! I do have an additional question though -- how would anyone model the true probabilities for the scores per hole, factoring in real-life data and phenomena? For example, equipment characteristics, ball characteristics, environmental characteristics (wind, landscape, etc.)? If one were to attempt to do that, does it cease to be a probabilistic problem and then become a deterministic problem? Could you please elaborate on this? I am interested. Mike > Hi James, > Just one last question about your suggested methodology - you are > mentioning calculating my probabilities for each hole based on > historical data, which is fair enough. How would that differ or > compare to calculating the probabilities of each hole being equally > likely? In other words, like how I first described, instead of > considering the total score, I consider the discrete scores possible > for each hole over a [min,max] and assign equal probabilities to them. > Tell me what is wrong with my thinking in that approach. > It means your chance of scoring 9 holes in one is exactly > the same as your chance of scoring 9 pars. > - Randy === Subject: Re: Probability question - please help! <44B5C9EA.1F1A5FB9@pat7.com> Hi James, Yes, I know it is a highly unrealistic assumption, which is why I'm glad you answered me. :) I like your suggestion. I was just thinking perhaps I was trying to make something up which had little or no basis in reality, and was going to look and see what the official probability modeling for this situation is (i.e. are there any equations of this seemingly well-known problem which I should be using in my spreadsheet to calculate the probabilities of interest?) The Monte Carlo suggestion is a cool one. I think I am going to try this weekend (or tonight if I'm so motivated) and see what I can come up with. It's a shame you can't post attachments in Google groups, otherwise I'd post/send my results. FYI, it's golf day! Mike > ... > Here's the second question: Relating to probability, I am trying to > determine the probability of two events: E1 is the probability that I > will score less than or equal to a given score (in other words, using > my historical data to predict how probable it is that I will shoot a > given score). E2 is the probability that I will score exactly the > given score. > ... > So I'm thinking, well, if I say I want to score exactly a 58 for E2, > then the likelihood would be one over the total number of possible > discrete scores from 9-72, right? Wouldn't my probability then be > 1/64? Or, is 9 too low to use as the lower bound, since I would never > score it? How about 58, would that be my lower bound, since I scored > it once already? > ... > You seem to have assumed that all the scores from 9 to 72 are > equally likely to occur, but such an assumption is unrealistic. > Scores close to your average score are more likely to occur > than scores distant from your average, and different scores > will arise with different probabilities that probably cannot > be reasonably computed from first principles. Anyway, golf > scores should be the result of skilled play rather than > random processes. > I think the best you can do is to model your scores on individual > holes, and then find the probability that holes 1...9 (or 10...18) > add up to 58. For example, if in ten rounds you've scored 3 3's, > 5 4's, and 2 5's on hole 1, you might model hole 1 probabilities > as P(3)=0.3, P(4)=0.4, P(5)=0.2. With such a model for each hole, > use random numbers in a Monte Carlo process to play 10000 rounds, > then report the number of rounds for each total that occurred in > the simulation. > -jiw === Subject: Re: Probability question - please help! Okay, that was completely useless. > Okay, so I was thinking > I enrolled at MIT in 1970, but dropped out in 1971 because > I decided to be a generalist and look at the big picture, > instead of specializing in a narrow field. MIT doesn't offer > degrees in seeing the big picture. > As I see it, after 35 years of looking, sustainability of the > global economy is the prime question of the day. Lowering > fuel consumption is vital to making the economy sustainable. > A good way to use less fuel would be to have suburbanites > move back in to the cities, so they wouldn't have to drive > so much. > The inner city poor would have to be relocated to make room. > First, we would have to legalize drugs to take away the black > market jobs. Then, we would need to offer them jobs and > housing on farms and factories outside the cities. That way, > more food and goods could be produced locally, for additional > fuel savings. > But if the population keeps increasing, all bets are off. No, > we have to freeze, and eventually decrease the population, too. > A good starting point would be to stop suppressing influenza. > To those who would object, I point out that we're living much > better today than even royalty did centuries ago. There should > be no complaints about having limits to longevity. There are > other creatures and future generations to think about here too. > Sustainability is all about justice, not just us. > -- === Subject: Re: Computing pdf/cdf for X^2 when X is normally distributed >> Suppose we have a normally distributed variable, i.e. >> X ~ N (0,s2) and we wish to learn more about its >> boring, square buddy X^2. >> X^2 / s2 has chi-squared distribution with 1 degree of freedom. > Or, to put it another way, X^2 has gamma distribution with shape > parameter 1/2 and scale parameter > 1 / (2 s2). Why not to use the standard tool of characteristic function approach to resolve this and many other of this type of questions? Of course, the CDF approach is the simplest method. However, due to the fact that often one does not get a closed form formula for the CDF, it may become tricky to recognize what is the background density that gave a rise to this particular CDF. In the above case it is trivial to recognize, but the above case is rather an exception than the rule. - Janusz. === Subject: Re: Improvements to my building based on group consensus. <53u2b2tu6lr3kmd3td35mjoe1rr4dhpn2o@4ax.com> Very good -- so we are ok with mode then. My next question is since I have to start with 20 unique independent ideas -- how many rounds of surveys are needed to arrive at the single solution. Also, how do I deal with cases when there isn't a single mode. Should I re-run the survey till a mode is reached? And, how many should I pick at each stage. - Anil === Subject: Re: The gcd of a set A, and of P, the set of primes [Doug Goncz] > With apologies to Tim Peters for the mess I made of A GCD Question: I wasn't offended :-) > If (a,b) = gcd(a,b) = the greatest integer dividing both integers a and > b, then > ( {a,b} ) = the gcd of set with integer elements a,b, and > (set A) = ({A}) = (A) = the greatest integer dividing every member of > integer set A, Ah, so _that's_ what you're after? If so, note that, when viewed as a binary operator, gcd is associative and commutative, and has an identity element (gcd(i, 0) = i for all i). So it's no harder to define gcd of a finite set than it is to define the sum or product of a finite set (addition & multiplication are also associative, commutative, and have identity elements). If the set S consists of s_1, s_2, ..., s_n, then define: ({}) = 0 ({s_1, s_2, ...} = gcd(s_1, ({s_2, ...})) same way you'd define: sum({}) = 0 sum({s_1, s_2, ...}) = s_1 + sum({s_2, ...}) or: product({}) = 1 product({s_1, s_2, ...}) = s_1 * product({s_2, ...}) You can easily prove then that (S) so defined is the largest integer dividing every element in S. For example, gcd({6, 12, 45}) = 3 and it makes no difference which order you compute that in (which is a matter of easy proof, same way you'd prove that it doesn't matter in which order you compute the sum or product of a finite set using + or *): gcd(6, gcd(12, 45)) = gcd(6, 3) = 3 gcd(gcd(6, 12), 45) = gcd(6, 45) = 3 gcd(12, gcd(6, 45)) = gcd(12, 3) = 3 etc. > then > (a,b) = ({a,b}) as it should, doesn't it? Yup. > || means such that > | means divides > / means AND > / means OR > |A| means the measure or size of set A; |A| = n > (A) = k < m || > ( k | A.1) & (k | A.2) ...) & > ~(m | A.1) / (m | A.2)... ) ( I think I have that right) If that last one is trying to say that (A) is the largest k such that k divides every element in A, yes. > We'll call the set of primes P. Do you really need to define this for infinite sets? That takes a lot more work. > (P) = 1, for sure! (by definition: no prime divides different prime; no > n#1 divides a prime) > That is, we can define (A) so that (P) = 1. The definition above would give this, if suitably decorated with enough verbiage & proofs to make the operation well-defined on countably infinite sets. If you need to do that, a key property is that if min(a, b) >= 1, 1 <= gcd(a, b) <= min(a, b) That is, so long as 0 isn't in the mix, gcd applied to a finite prefix of a (possibly infinite) sequence of elements never increases, and is bounded below by 1. This is better-behaved than addition or multiplication, and makes reasoning about infinite gcds (if truly necessary) easier. Note that if the gcd of a finite prefix is 1, then the gcd of every extension of that prefix is also 1 (much like if the product of a finite prefix is 0, the product of every extension of that prefix is also 0). > A in P is any subset of the primes (as usual, without duplication) > (A) should be 1, shouldn't it? If and only if |A| >= 2. > Side note: ({}) = 0 ? ( the greatest number dividing no number is zero? > Or should this be 1? ) Things always work best if a reduction applied to an empty set returns the identity element for the underlying operation, so 0 is best here. 1 has the notable property that gcd(1, i) = 1 for all i, much as i*0 = 0 for all i. > {but ({p}) = p as pointed out by Tim; the gcd of single integer is that > integer > ({p,p}) = p but ({p,p}) isn't a standard nonredundant set so we can set > that concern aside > AxA = the Cartesian product of set A with n^2 elements > |A| = n --> AxA = > { (A.1, A.1) , (A.1, A.2) , (A.2, A.1) ... (A.1, A.n), ... (A.n, A.1) > ... (A.n, A.n) } > We can say > (A) = product with i = 1 to n^2 of gcd(AxA.i) / ( product with i = 1 > to n of A.n ) To my eyes, this looks as strained, and as unlikely to work nicely, as trying to define sum(S) or product(S) in terms of + or * applied to the |S|^2 elements of SxS. In all cases, it only requires |S|-1 applications of the underlying binary operation to compute the result. Note that you were quite comfortable writing product with i = 1 to n of A.n, and you should be just as comfortable with gcd with i = 1 to n of A.n. > and that could be useful, but then > ({p}) = 1 instead of p That's one of its problems, yes. For another, consider that under this definition: ({a, b}) = gcd(a, a) * gcd(a, b) * gcd(b, a) * gcd(b, b) / (gcd(a, a) * gcd(b, b)) = gcd(a, b) * gcd(b, a) = gcd(a, b) * gcd(a, b) = gcd(a, b)^2 The best approach isn't to wrap a square root around the formula too ;-) > Given an algorithm for gcd(a,b), such as Euclid's, making gcd(a,b) > computable, then Then you can apply it to reduce a set the same way you can apply binary + or * to compute the sum or product of a set. Maybe all you're really missing here is the exercise of proving that applying plain-old two-argument gcd |S|-1 times is all it takes to find the greatest common divisior of all the elements in S. > ??How to define ({A}) = (A) = gcd of set A in only one consistent, > computable, useful way, and how also to resolve these other > difficulties? The above works for me :-) === Subject: Re: Let An Intellectual Visionary Save England : Turing Numbers German Machine. <69lva29vibo3h1g9oq5bck713ekqth27do@4ax.com >excellent william. >but there is more hidden !! > [alphametic snipped] >more ?? >today's date and more.. > http://en.wikipedia.org/wiki/Discordianism#The_Law_of_Fives > The Law of Fives is summarized on page 00016 of the Principia > Discordia: But, i am waiting to decode the subject line. very reward. Re: Let An Intellectual Visionary Save England : Turing Numbers German Machine. verbal mathematics alphametic cryptarithm poem etc.. Turing Numbers.. german machine ??!! don.mcdonald 13.7.06 > The Law of Fives states simply that: ALL THINGS HAPPEN IN FIVES, > OR ARE DIVISIBLE BY OR ARE MULTIPLES OF FIVE, OR ARE SOMEHOW > DIRECTLY OR INDIRECTLY APPROPRIATE TO 5. > The Law of Fives is never wrong. > The Law of Fives includes the word Five five times. > Like most of Discordianism, the Law of Fives appears on the surface to > be either some sort of weird joke, or bizarre supernaturalism; but > under this, it may help clarify the Discordian view of how the human > mind works; Lord Omar is quoted later on the same page as having > written, I find the Law of Fives to be more and more manifest the > harder I look. > Appendix Beth of Robert Shea's and Robert Anton Wilson's The > Illuminatus! Trilogy considers some of the numerology of > Discordianism, and the question of what would happen to the Law of > Fives if everyone had six fingers on each hand. > Another way of looking at the Law of Fives is as a symbol for the > observation of reality changing that which is being observed in the > observer's mind. Just as how when one looks for fives in reality, one > finds them, so will one find conspiracies, ways to determine when the > apocalypse will come, and so on and so forth when one decides to look > for them. It cannot be wrong, because it proves itself reflexively > when looked at through this lens. === Subject: Re: Let An Intellectual Visionary Save England : Turing Numbers German Machine. >But, i am waiting to decode the subject line. >very reward. >Re: Let An Intellectual Visionary Save England : Turing Numbers German >Machine. >verbal mathematics >alphametic >cryptarithm >poem etc.. >Turing Numbers.. >german machine ??!! It is an enigma, that's for sure. === Subject: Re: perhps it is end of the struggle > I think I have found a way to finish the struggles about prime numbers > and also too many problems about them that we have no prove about them > for yet. > But my thesis is not just a mathematical proof, I have used some of our > biology and physics. > I am working on it since June,2000. > I am looking for someone that is intrested in this case(to use our > physical > and biological knowledge to prove some of mathematical problems). > I want you to help me about. Using some of our biology and physics should set off warning bells in your head. A mathematical proof should work, no matter what your body chemistry is. --- Christopher Heckman === Subject: Re: Surprised even me > You people are totally incompetent. > All these years I thought that at some level maybe some of you knew > mathematics, but now I see you don't. > I am actually silly enough to still be further disappointed. > You people have no real interest in mathematics. > You are play-acting a role of what you think a mathematician would be > like, but if Gauss were alive today, he would not talk to any of you, I > am sure. of course not, he can't speak English, he's German. === Subject: Re: Four Color Theorem >>Hypothesis: A simple loopless maximal planar graph with a 5-degree >>vertex cannot be a >>minimal counter example to the four color theorem!. > It follows from the four colour theorem. However, proving it directly >> is another matter. A possible way is induction on the minimum vertex >> degree of a maximal planar graph. > Any triangulation with minimum degree 1,2,3 or 4 is cannot be a minimal >> counter example by induction on the number of vertices. Assume that >> none of the triangulations with minimum vertex degree n-1 is a minimal >> counter example and prove the result for n. > This would prove the general version of your statement and the four >> colour theorem. But this will take some doing as inductive arguments to >> prove four colour theorem and its reformulations have always failed. >> My proof is based on the assumption that the following graph is not 4-chroma. >> [The drawing below is correct in the print mode} >> If you're going to post plain text which is supposed to line up, use a >proportional font, like Courier. >> |..............| Internal chord >> |--------------| External {B-C-B} Kempe chain >> D <== A-----B-----C-----B------D ==> A >> |--------------|--------------| External {A-C) & {C-D} >> Kempe chains >> The graph is planar. The {B-C-B} chain may cross the {A-C} & {C-D} chains >> at common 'C' vertices. >> It looks like you made Kempe's mistake again. > The graph is an attempt to find a feasible method of forcing a 4-coloring on a > chordless 5-cycle graph, without adding any chords. Without the internal chord > It is supposed to be a graph that cannot be 3-colored! With the internal chord, > theoretically, it cannot be properly 4-colored.. Theoretically; but (as I've told you before), if you have a planar graph with a face of size >= 4, then there is more than one coloring. All you have done is made one coloring fail to work; you haven't done anything to the rest, which may still work. (In fact, in the graph G with a pentagon, there are at least 2^(5-3) = 8 colorings. I'm not sure whether this counts permutation of the colors or not.) --- Christopher Heckman > I expect that someone will reveal the flaw. Then I will observe that > you can't force a 4-coloring without at least 3 external chains and > that this is the only way to get three chains. > With only two chains, the colors of only three vertices may be forced > . [Each > vertex has only two neighbors. Therefore, there are at least two > color options > for each vertex. Forcing means removing all but one option!] > This leaves 2 vertices whose colors cannot be forced by chains; but by some > unknown configuration? > Before I go any further, does anyone care to comment on the 3-chain configuration? > Also, any suggestions as to how a specific vertex can be forced to be a specific > color? >Exactly what was Kempe's mistake? > See http://www.stonehill.edu/compsci/LC/Four-Color/Four-color.htm . Two > graphs for which his proof fails are given by: > http://mathworld.wolfram.com/KittellGraph.html (23 vertices) > http://mathworld.wolfram.com/ErreraGraph.html (17 vertices) > I think that Kempe's mistake must have been that he could not prove > that his method could properly 3-color the cycle graph? I don't care > if > the cycle graph can be 3-colored or not! > ---Bill J. > --- Christopher Heckman === Subject: Re: Four Color Theorem > I know that paper which is again based on dischraging methods. I am > going to think of your question. Meanwhile I am trying to give answers > to these: 1) For any given triangulated planar graph in what way we can > delete minimum number of edges so that the resulting planar G' graph is > 3-colorable, It depends. Any even triangulation (a triangulation where every vertex has even degree) is 3-colorable. In fact, there is an old characterization of when a planar graph G is 3-colorable: iff there is an even triangulation T which contains G. > 2) Can we use spiral-chains in finding or chracterizing > the planar graph G' and 3) Is the class of 3-colorable planar graphs > {G'} cover all planar graphs without 3- and 4-cycles? I think you mean 4- and 5-cycles here; any triangle-free planar graph is 3-colorable (Grotzsche's Theorem, 1959; Carsten Thomassen has given a 3-page proof of this fact). --- Christopher Heckman >I am not close Steinberg's Conjecture file yet. I hope we will again >working on it again. I hope our lemma (corrected by yourself) on the >triangulated ring would some way be used together with the spiral-chain >coloring in settleing that conjecture as well. By the way have you >heard anything new on Steinberg's conjecture newly? > I've been collecting information about Steinberg's Conjecture, along > with properties of such graphs. (Some day I might turn it into a > webpage.) The latest result that I know of is: > O. V. Borodin, A. N. Glebov, A. Raspaud, and M. R. Salavatipour, > Planar Graphs without Cycles of Length from 4 to 7 are 3-colorable, > J. of Combinatorial Theory (Series B), 93:303-311, 2005. > Link: http://www.cs.ualberta.ca/~mreza/research/stein3.ps > I've been wondering about a question that Craig Tovey asked me during > my doctoral defense: If you have a planar graph G without 4- and > 5-cycles, can you use the difficult graphs technique to find out how > big of an independent set can you find? If the answer is at least n/3, > then you have a weaker version of Steinberg's Conjecture. OTOH, if the > answer is less than n/3, you've disproven Steinberg's Conjecture. > --- Christopher Heckman >> I must be careful when I give an answer to your question. Yes my proof >> has not yet published but I think it has been passed through extensive >> public review in the last two years. I have given several seminars in >> the mathematical departments. The bm line is that, no one yet and >> I trust, no one will be given serious comments against its correctness. >> I have my doubts, especially in Cahit's paper which settles >Steinberg's Conjecture. In particular, I derived the conditions under >which a triangulated ring is 3-colorable, and they do not match >Cahit's. That, and my increased workload during the academic year, have >kept me from looking at more of the problem. >> It is my opinion that the proofs should be worked out in more detail. >> --- Christopher Heckman >> It is true that in the final publisheable form I will have to add a few >> things and delete some unneccessary figures and explain my spiral-chain >> coloring algorithm in more detail, that is three-coloring of the spiral >> segments and additional stuff on three-plus-one spiral chain coloring >> and also adding may be a new section for the three-edge coloring of >> (bridgeless) cubic planar graphs. I will going to present my proof in >> the ICM2006, Madrid in August ( >> http://icm2006.org/v_f/web_UC.php?CodiSeccio=14&CodiTipusEvent=4sc&Ordenacio = ae&Format=UC4sc) >> and based on the feedback I receive I am going to submit to a proper >> journal. The only thing that I would say about Dharwadker's proof of >> the 4CT is that there is no description of an coloring algorithm in his >> proof. >> Cahit >>What became of your proof of the four colour theorem? Usually people >> tend to make the same old mistakes in proving the elusive theorem. Have >> you found any errors yet or do you still think that your proof is >> correct? Another person by the name of Ashay Dharwadker has claimed to >> have proven the four colour theorem but his proof never made it to a >> journal. What is the current status of your proof? >May be the answer to Bill's question is in my response in another group >>at > http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_hard;ac t ion=display;num=1150590683 > Cahit > Hypothesis: A simple loopless maximal planar graph with a 5-degree > vertex cannot be a > minimal counter example to the four color theorem!. >> It follows from the four colour theorem. However, proving it directly > is another matter. A possible way is induction on the minimum vertex > degree of a maximal planar graph. >> Any triangulation with minimum degree 1,2,3 or 4 is cannot be a minimal > counter example by induction on the number of vertices. Assume that > none of the triangulations with minimum vertex degree n-1 is a minimal > counter example and prove the result for n. >> This would prove the general version of your statement and the four > colour theorem. But this will take some doing as inductive arguments to > prove four colour theorem and its reformulations have always failed. >> But a variation of induction is the backbone to reduction, a technique >>that both Appel and Haken, and Robertson Sanders Seymour Thomas used to >>find a proof. >> --- Christopher Heckman === Subject: Re: Four Color Theorem I am aware of these results, i.e., even triangulations and triangle-free planar graphs are 3-colorable. But there are some other planar graphs which are not even triangulation or triangle-free but still 3-colorable. Take for example a chromatic critical (planar) graph G (here critical means by deletion of any vertex result a decrease in chromatic index of the original graph G). Now consider spiral-chain coloring of G which is 4 colorable. I suspect that except the last vertex in the spiral-chain all other vertices of G are 3-colorable or perhaps I will need an counterexample that chromatic critical planar graphs with more than one four colored vertices. Then we may develop a method to show that all planar graphs without 4 and 5-cycles are 3-colorable. I put a figure in my map coloring site about this idea in the photo gallery at http://www.flickr.com/photos/49058045@N00/ under the title Steinberg. > I know that paper which is again based on dischraging methods. I am > going to think of your question. Meanwhile I am trying to give answers > to these: 1) For any given triangulated planar graph in what way we can > delete minimum number of edges so that the resulting planar G' graph is > 3-colorable, > It depends. Any even triangulation (a triangulation where every > vertex has even degree) is 3-colorable. In fact, there is an old > characterization of when a planar graph G is 3-colorable: iff there is > an even triangulation T which contains G. > 2) Can we use spiral-chains in finding or chracterizing > the planar graph G' and 3) Is the class of 3-colorable planar graphs > {G'} cover all planar graphs without 3- and 4-cycles? > I think you mean 4- and 5-cycles here; any triangle-free planar graph > is 3-colorable (Grotzsche's Theorem, 1959; Carsten Thomassen has given > a 3-page proof of this fact). > --- Christopher Heckman >I am not close Steinberg's Conjecture file yet. I hope we will again >working on it again. I hope our lemma (corrected by yourself) on the >triangulated ring would some way be used together with the spiral-chain >coloring in settleing that conjecture as well. By the way have you >heard anything new on Steinberg's conjecture newly? > I've been collecting information about Steinberg's Conjecture, along >with properties of such graphs. (Some day I might turn it into a >webpage.) The latest result that I know of is: > O. V. Borodin, A. N. Glebov, A. Raspaud, and M. R. Salavatipour, >Planar Graphs without Cycles of Length from 4 to 7 are 3-colorable, >J. of Combinatorial Theory (Series B), 93:303-311, 2005. > Link: http://www.cs.ualberta.ca/~mreza/research/stein3.ps > I've been wondering about a question that Craig Tovey asked me during >my doctoral defense: If you have a planar graph G without 4- and >5-cycles, can you use the difficult graphs technique to find out how >big of an independent set can you find? If the answer is at least n/3, >then you have a weaker version of Steinberg's Conjecture. OTOH, if the >answer is less than n/3, you've disproven Steinberg's Conjecture. > --- Christopher Heckman >> I must be careful when I give an answer to your question. Yes my proof >> has not yet published but I think it has been passed through extensive >> public review in the last two years. I have given several seminars in >> the mathematical departments. The bm line is that, no one yet and >> I trust, no one will be given serious comments against its correctness. >> I have my doubts, especially in Cahit's paper which settles >> Steinberg's Conjecture. In particular, I derived the conditions under >> which a triangulated ring is 3-colorable, and they do not match >> Cahit's. That, and my increased workload during the academic year, have >> kept me from looking at more of the problem. >> It is my opinion that the proofs should be worked out in more detail. >> --- Christopher Heckman >>It is true that in the final publisheable form I will have to add a few >> things and delete some unneccessary figures and explain my spiral-chain >> coloring algorithm in more detail, that is three-coloring of the spiral >> segments and additional stuff on three-plus-one spiral chain coloring >> and also adding may be a new section for the three-edge coloring of >> (bridgeless) cubic planar graphs. I will going to present my proof in >> the ICM2006, Madrid in August ( >> http://icm2006.org/v_f/web_UC.php?CodiSeccio=14&CodiTipusEvent=4sc&Ordenacio = ae&Format=UC4sc) >> and based on the feedback I receive I am going to submit to a proper >> journal. The only thing that I would say about Dharwadker's proof of >> the 4CT is that there is no description of an coloring algorithm in his >> proof. > Cahit > What became of your proof of the four colour theorem? Usually people >>tend to make the same old mistakes in proving the elusive theorem. Have >>you found any errors yet or do you still think that your proof is >>correct? Another person by the name of Ashay Dharwadker has claimed to >>have proven the four colour theorem but his proof never made it to a >>journal. What is the current status of your proof? >>May be the answer to Bill's question is in my response in another group >>at >> http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_hard;ac t ion=display;num=1150590683 >> Cahit >Hypothesis: A simple loopless maximal planar graph with a 5-degree >vertex cannot be a >minimal counter example to the four color theorem!. > It follows from the four colour theorem. However, proving it directly > is another matter. A possible way is induction on the minimum vertex > degree of a maximal planar graph. > Any triangulation with minimum degree 1,2,3 or 4 is cannot be a minimal > counter example by induction on the number of vertices. Assume that > none of the triangulations with minimum vertex degree n-1 is a minimal > counter example and prove the result for n. > This would prove the general version of your statement and the four > colour theorem. But this will take some doing as inductive arguments to > prove four colour theorem and its reformulations have always failed. >> But a variation of induction is the backbone to reduction, a technique > that both Appel and Haken, and Robertson Sanders Seymour Thomas used to > find a proof. > > --- Christopher Heckman === Subject: how to rotate a multivariate Gaussian I would like to rotate (e.g. with respect to the z-axis) a multivariate Gaussian, which is described by the mean mu and the covariance matrix Sigma. Rotating the mean mu is straight forward. But how to rotate the matrix Sigma? I am particular interessted in the 3D case. I tried to multiply Sigma with a standard rotation matrix R_z=(cos(theta) -sin(theta) 0 | sin(theta) cos(theta) 0| 0 0 1). I tried the same with the correlation matrix Omage = Sigma^-1. Both seemed not to work. I am grateful for every hint. === Subject: Re: how to rotate a multivariate Gaussian > I would like to rotate (e.g. with respect to the z-axis) a multivariate > Gaussian, which is described by the mean mu and the covariance matrix > Sigma. Rotating the mean mu is straight forward. > But how to rotate the matrix Sigma? I am particular interessted in the > 3D case. > I tried to multiply Sigma with a standard rotation matrix > R_z=(cos(theta) -sin(theta) 0 | sin(theta) cos(theta) 0| 0 0 1). I > tried the same with the correlation matrix Omage = Sigma^-1. Both > seemed not to work. Did you multiply Sigma with the rotation matrix R on both sides i.e.: R^(-1) Sigma R ? > I am grateful for every hint. You're welcome. Han de Bruijn === Subject: Re: how to rotate a multivariate Gaussian > I would like to rotate (e.g. with respect to the z-axis) a multivariate > Gaussian, which is described by the mean mu and the covariance matrix > Sigma. Rotating the mean mu is straight forward. > But how to rotate the matrix Sigma? I am particular interessted in the > 3D case. > I tried to multiply Sigma with a standard rotation matrix > R_z=(cos(theta) -sin(theta) 0 | sin(theta) cos(theta) 0| 0 0 1). I > tried the same with the correlation matrix Omage = Sigma^-1. Both > seemed not to work. > I am grateful for every hint. Hint: If x is a random vector with mean vector m and covariance matrix C, and if y = A*x + b, where A is a given matrix and b is a given vector, what are the mean vector and covariance matrix of y? === Subject: my current problmee i have 5 series of natural numbers all starting from 5 and end at 100. Each sum of each series doesnt differs from eahc other, SUMseries1=SUMseries2=.... lenght of series is 50, that means there are 50 numbers in each serries, each number can be repeated. Help me to prove that there is no series that is the same as the other Any ideasa thankfull === Subject: Re: Constructing a (GF(2^m)) polynomial from its roots using FFT techniques? Jaco Versfeld === Subject: Re: Inverting a submatrix, when the matrix and inverse of the matrix is known? Jaco Versfeld === Subject: A POSSIBLE FACTORING ALGORITHM Let the equations: y^2=120*n1+7^2 (I) y^2=120*n+1, where n=n2+1 (II) y=p1*p2*p3* .pn (III), where n1, n2 are positive integers, y is real number, p1, p2, p3, pn are prime numbers Statement: If the outcome y is real (not integer) then we omit the solution else y is prime or y is the product of previous primes found for the values of n1, n2, as equation (III) states. (IV) Clarification note that equation (III) is valid only for integer solutions of (I), or y has factor 7. (Has to be proved or disproved) If the solutions are from (II) then this can be a composite prime power product. Note the primes 2,3,5 are not predicted by this method. The question: Could anyone find values(integers) of n1 or n2 that violete the above statement (IV). === Subject: Re: Affine spaces as posets > Let F be a finite field and A a (finite) affine space over F, where its > vectors a in A have the length k. By length k do you mean dimension k? > Define a total order < on the elements of F. Notice that this order in incompatible with the field operations. > This induces a partial order << on the vectors a of the affine space A > as follows: a << a' hold if and only if a =/= a' and a[i] <= a'[i] holds > for all coordinates i = 1, ..., k. Hence (A, <<) is a poset. Simple product order. > I know something of affine spaces since I had a good linear algebra > teacher and I also know something of posets and lattices. > QUESTION: Is there something known of affine spaces as posets, which is > not known about affine spaces alone or posets alone? I'm of the opinion your amalgam is brittle and lacks tensile strength. The interval topology induced by the product order is discrete. === Subject: Re: Seeking Beautiful Single-Variable Calculus Problems > I'm a math team coach and homeschooling parent looking for some extra depth for > the math learners in my care. Most of the young people on my math team get their > main math instruction from the U. of MN Talented Youth Mathematics Program, > using the Stewart Early Transcendentals textbook with an excellent supplementary > materials. Even though they've got some good instruction already, I'm looking > for suggestions of calculus problems that really make a learner THINK about > calculus concepts and go beyond the usual school homework exercise. That's the > best way to help those kids thrive in their main program and go on to higher > mathematics study. What I especially need, as someone largely clueless about > calculus myself, is to find exercises that have numerical or symbolic answers, > or better still full solutions, so that I can check them even if I can't (as is > likely) work through them myself from first principles. My students work beyond > my own level of ability, and they are eager to learn more. > Here's a list of calculus books I own, from which it would be especially easy to > find problems that you suggest are good, challenging problems. The list is > ordered by publication date. > Burington, Richard Stevens & Torrance, Charles Chapman, Higher Mathematics: With > Applications to Science and Engineering (New York: McGraw-Hill, 1939). > Randolph, John F., Analytic Geometry and Calculus (New York: Macmillan, 1949). > Hart, William, Calculus (Boston: D.C. Heath, 1955). > Thomas, George B., Calculus and Analytic Geometry (Reading, MA: Addison-Wesley, > 4th ed. 1968). > Kline, Morris, Calculus: An Intuitive and Physical Approach (Mineola, NY: Dover > c. 1977). > Thomas, George B. & Finney, Ross L., Elements of Calculus and Analytic Geometry > (Menlo Park, CA: Addison-Wesley, 1981) (and separate solution manual for this > title). > Marsden, Jerrold & Weinstein, Alan, Calculus I Second Edition (New York: > Springer-Verlag, 1985). > Marsden, Jerrold & Weinstein, Alan, Calculus II Second Edition (New York: > Springer-Verlag, 1985). > Purcell, Edwin J & Varberg, Dale, Calculus with Analytic Geometry (Englewood > Cliffs, NJ: Prentice Hall, 5th ed. 1987) (and separate solution manual for this > title). > Spivak, Michael, Calculus (Houston, TX: Publish or Perish, 3rd ed. 1994). > Finney, Ross L., Thomas, George B., Demana, Franklin & Waits, Bert K., Calculus: > Graphical, Numerical, Algebraic: Single Variable Version (Reading, MA: > Addison-Wesley, 1995). > Ostebee, Arnold & Zorn, Paul, Calculus from Graphical, Numerical, and Symbolic > Points of View Volume 1 (Fort Worth, TX: Saunders College Publishing, > preliminary ed. 1995) (and separate selected answers volume for this title). > Priestley, William McGowen, Calculus: A Liberal Art (New York: Springer-Verlag, > 2nd ed. 1998). > Anton, Howard, Calculus: A New Horizon (New York: John Wiley, 6th ed. 1999). > Stewart, James, Calculus: Early Transcendentals (Belmont, CA: Brooks/Cole, 5th > Hughes-Hallett, Deborah, et al., Calculus (New York: Wiley, 4th ed. 2005). > I have had some other really good books out of libraries, such as Apostol, > Strang, and Courant. And if you know of great problems with solutions from > online sources, I'm sure that would be of interest to many of us. Please let me > know about your favorite problems for the student who wants to go beyond the > usual syllabus. > -- > Karl M. Bunday P.O. Box 1456, Minnetonka MN 55345 > Learn in Freedom (TM) http://learninfreedom.org/ > remove .de to email I do not see Kleppner Kolenkow in there. If you want some good problems, involving calculus, try this. And you can try Hassani for more physical applications. === Subject: Re: Seeking Beautiful Single-Variable Calculus Problems > .... I'm looking > for suggestions of calculus problems that really make a learner THINK about > calculus concepts and go beyond the usual school homework exercise.... This classic little fallacious argument may be too easy for them. Since x^2 is positive, you would expect the integral from -1 to +1 of 1/(x^2) dx to be positive. However, the usual calculation using an antiderivative gives [-1/x] from -1 to +1 = - 1 - (- (- 1)) = - 2 which is negative. Why? Ken Pledger. === Subject: Re: Seeking Beautiful Single-Variable Calculus Problems .... I'm looking > for suggestions of calculus problems that really make a learner THINK about > calculus concepts and go beyond the usual school homework exercise.... > This classic little fallacious argument may be too easy for them. > Since x^2 is positive, you would expect > the integral from -1 to +1 of 1/(x^2) dx > to be positive. However, the usual calculation using an antiderivative > gives > [-1/x] from -1 to +1 > = - 1 - (- (- 1)) > = - 2 which is negative. > Why? It's an IMP_____ INTEGRAL. --- Christopher Heckman === Subject: Military logistic problem First of all I'm not sure if this belongs here, if not, could you suggest a more appropriate group. I'm having some problems researching the possibility of explicit solutions to military logistics problems. Although, I think the actual problem is more general. I have a time-step simulation tool that simulates strategic-lift (aircraft, ships, etc...) moving stuff from a start port to an end port. Obviously aircraft make many flights back and forth. While a time-step simulation suits this problem, it is quite expensive in time. The question I need answering is whether there exists an explicit solution to such a problem. I'm pretty sure it's not linear, and perhaps chaotic behaviour prevents such a solution. By the way, when I say solution I mean things like finding: a) How long it takes to move everything with various amounts of start-lift, port infrastructure (#berths, #AC handlers, etc...) and so on; b) The optimum number of the above to produce the quickest time; etc... I would be very grateful if you could point me in a direction to find if such a thing is possible and if so, whether the complexity of such a solution would make it impractical. confusing!) Chris Riddle === Subject: Re: Military logistic problem >First of all I'm not sure if this belongs here, if not, could you >suggest a more appropriate group. >I'm having some problems researching the possibility of explicit >solutions to military logistics problems. This falls in the field of Operations Research aka OR, so I have added sci.op-research to the newsgroups list. >Although, I think the actual problem is more general. I have a >time-step simulation tool that simulates strategic-lift (aircraft, >ships, etc...) moving stuff from a start port to an end port. >Obviously aircraft make many flights back and forth. >While a time-step simulation suits this problem, it is quite expensive >in time. >The question I need answering is whether there exists an explicit >solution to such a problem. I'm pretty sure it's not linear, and >perhaps chaotic behaviour prevents such a solution. >By the way, when I say solution I mean things like finding: >a) How long it takes to move everything with various amounts of >start-lift, port infrastructure (#berths, #AC handlers, etc...) and so >on; >b) The optimum number of the above to produce the quickest time; etc... >I would be very grateful if you could point me in a direction to find >if such a thing is possible and if so, whether the complexity of such a >solution would make it impractical. >confusing!) >Chris Riddle For a less compute-intensive solution, figure out where the smallest bottleneck in the system is, and plan to fill it just under 100%. I apologize if I'm oversimplifying. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Military logistic problem First of all I'm not sure if this belongs here, if not, could you >suggest a more appropriate group. >I'm having some problems researching the possibility of explicit >solutions to military logistics problems. > This falls in the field of Operations Research aka OR, so I have added > sci.op-research to the newsgroups list. To which I'll add that if you can search the Ops Research literature, particularly the journal Operations Research, you'll likely find a wealth of approaches to logistical problems. - Randy === Subject: Re: Military logistic problem >> Although, I think the actual problem is more general. I have a >> time-step simulation tool that simulates strategic-lift (aircraft, >> ships, etc...) moving stuff from a start port to an end port. >> Obviously aircraft make many flights back and forth. >> While a time-step simulation suits this problem, it is quite expensive >> in time. >> The question I need answering is whether there exists an explicit >> solution to such a problem. I'm pretty sure it's not linear, and >> perhaps chaotic behaviour prevents such a solution. >> By the way, when I say solution I mean things like finding: >> a) How long it takes to move everything with various amounts of >> start-lift, port infrastructure (#berths, #AC handlers, etc...) and so >> on; >> b) The optimum number of the above to produce the quickest time; etc... >> I would be very grateful if you could point me in a direction to find >> if such a thing is possible and if so, whether the complexity of such a >> solution would make it impractical. >> confusing!) >> Chris Riddle You have not indicated whether your simulation is stochastic or deterministic. If the problem is deterministic (or you're willing to pretend it is), you can likely achieve what you want with a linear program, mixed integer linear program, or possibly network program. Transport problems tend to be (plausibly) linear in terms of things like time, fuel consumption, etc., and in the absence of wrinkles that gets you network model (easily solvable even in quite large instances) or at worst a linear program (also easily solvable). If you run into things like fixed-charge issues (we can use port X, but only if we pay a lease fee that is not proportional to our usage, or truckload v. LTL shipping), or either-or sorts of constraints (either we use sealift or we use airlift, but not both), then integer variables enter the picture and solution times go up. Sorry, can't be more specific without more detail on the problem. /Paul === Subject: Re: Quadratic eigenvalue problem C6L1V@shaw.ca escreveu: > Do you mean [A*r^2 + B*r + C] v = 0? Here r is the eigenvalue and v > is the eigenvector. In general, the equation [A*r^2 + B*r + C] v = 0 > could have a solution, but the matrix equation A*r^2 + B*r + C = 0 > might not have any solution at all. The problem is [A*r^2 + B*r + C] v = 0. > I don't understand this statement. Isn't the nullspace of C the > complete solution to the eigenvalue problem for eigenvalue = 0? No. The zero eigenvalue has an algebraic multiplicity of 12, the characteristic equation is of the form: lambda^12*F(lambda)=0 On the other hand the geometric multiplicity of lambda=0, i.e the dimension of the nullspace of C, is 4. The eigenvalue is therefore defective, which means a set of generalized eigenvectors forming a Jordan chain should be computed A set of vectors d0,d1,d2 and d3 form a Jordan chain associated with lambda=0 if the following relations hold C*d0=0 B*d0+C*d1=0 A*d2+B*d1+C*d0=0 A*d3+B*d2+C*d1=0 where d0 <> 0 is an eigenvector and d1,d2 and d3 are the corresponding generalized eigenvectors. > Have you tried to give numerical values to E, G, etc., to see whether > Maple can then obtain the nullspace? It works with numerical values, but I was interested in a symbolic solution. Ricardo === Subject: Re: Quadratic eigenvalue problem <120720061227396323%bruck@math.usc.edu> Sorry, instead of A*lambda^2+B*lambda+C=0 I should have written (A*lambda^2+B*lambda+C)v=0. The problem is to find a scalar lambda (eigenvalue) and an associated vector v (eigenvector) that satisfy the following equation: Q(lambda)v=0 with Q(lambda)=(A*lambda^2+B*lambda+C) The scalar lambda is determined through the solution of the characteristic equation of the polynomial matrix Q(lambda): det(A*lambda^2+B*lambda+C)=0 Ricardo === Subject: Absorbing sets Hi all: A subset A of a real vector space V is said to be absorbing if each _v_ in V belongs to t*A if _t_ is large enough. Given a Banach space V, it is easy to prove that the following property holds: (P) Every closed absorbing subset of V has an interior point. This is an easy consequence of the Baire theorem. A consequence of this fact is that the following property also holds: (P') Every closed, absorbing, and convex subset of V is a neighborhood of 0. Now, here are my questions: 1) Can someone please provide an example of an absorbing subset of a Banach space with empty interior? 2) Given a normed vector space V, is it true that (P) holds iff V is a Banach space? 2) Given a normed vector space V, is it true that (P') holds iff V is a Banach space? Jose Carlos Santos === Subject: Re: Absorbing sets > 1) Can someone please provide an example of an absorbing subset of a > Banach space with empty interior? Hint: it is possible in R^2 -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Absorbing sets >> 1) Can someone please provide an example of an absorbing subset of a >> Banach space with empty interior? > Hint: it is possible in R^2 I know that this will make me bang my head against a wall but... can you please post such an example? Jose Carlos Santos === Subject: Re: Absorbing sets > Hi all: > A subset A of a real vector space V is said to be absorbing if each v > in V belongs to t*A if t is large enough. > Given a Banach space V, it is easy to prove that the following property > holds: > (P) Every closed absorbing subset of V has an interior point. > This is an easy consequence of the Baire theorem. A consequence of this > fact is that the following property also holds: > (P') Every closed, absorbing, and convex subset of V is a neighborhood > of 0. > Now, here are my questions: > 1) Can someone please provide an example of an absorbing subset of a > Banach space with empty interior? > 2) Given a normed vector space V, is it true that (P) holds iff V is a > Banach space? > 2) Given a normed vector space V, is it true that (P') holds iff V is a > Banach space? > Jose Carlos Santos Does this put food on the table? === Subject: Re: Set Theory: Should you believe? > that has caused a bit of discussion in some logic circles. > My claims in short: 1) most of `elementary mathematics' is not > sufficiently well understood by the mathematical establishment, leading to > weaknesses in K12 and college curriculum, 2) the current theory of `real > numbers' is a joke, and sidesteps the crucial issue of understanding the > computational specification of the continuum, and 3) `infinite sets' are a > metaphysical concept, and unnecessary for correct mathematics. > Analysts and set theorists are welcome to send me reasoned responses. > Assoc Prof N J Wildberger > School of Maths > UNSW I need to mull over what your paper says before I attempt to provide much in the way of analysis. I tend to agree with the spirit of the presentation, though I am not quite as dismissive of the established dogma. Somewhere in the dust piles of my neglected papers is a one or two page derivation of Kepler's laws beginning with no other assumption than basic arithmetic and the belief that I could reasonably well communicate such concepts as the Pythagorean theorem through simple drawings. The presentation provides an illustration of what taking a derivative means; IIRC using epsilon delta arguments. It gives a definition of an ellipse, of Newton's laws of motion, and of the inverse square law of gravitation. When I consider that Russell and Whitehead took some 360 pages to get around to 1+1=2, I have to wonder if they really chose the best set of axioms. I acknowledge that their objective was perfect rigor, where mine was intuitive clarity, but I am inclined to believe the two objectives are far closer to one another than is suggested by the current formalisms of so-called foundational mathematics. For now I will make only one comment about your paper. Regarding the proof that multiplication is associative; that fact is proved in the volume I am currently struggling through. This set of volumes attempts to outline the state of the art of foundational mathematics in its day. Perhaps it is nothing more than the formalization of that to which you object, but it was intended to address many of the concerns you appear to be expressing. Fundamentals of Mathematics, Volume I Foundations of Mathematics: The Real Number System and Algebra Edited by H. Behnke, F. Bachmann, K. Fladt, W. Suss and H. Kunle Translated by S. H. Gould Fundamentals of Mathematics represents a new kind of mathematical publication. While excellent technical treatises have been written about specialized fields, they provide little help for the nonspecialist; and other books, some of them semipopular in nature, give an overview of mathematics while omitting some necessary details. Fundamentals of Mathematics strikes a unique balance, presenting an irreproachable treatment of specialized fields and at the same time providing a very clear view of their interrelations, a feature of great value to students, instructors, and those who use mathematics in applied and scientific endeavors. Moreover, as noted in a review of the German edition in Mathematical Reviews, the work is ?designed to acquaint [the student] with supplied with references to the literature, both current and ?classical.?? The outstanding pedagogical quality of this work was made possible only by the unique method by which it was written. There are, in general, two authors for each chapter: one a university researcher, the other a teacher of long experience in the German educational system. (In a few cases, more than two authors have collaborated.) And the whole book has been coordinated in repeated conferences, involving altogether about 150 authors and coordinators. Volume I opens with a section on mathematical foundations. It covers such topics as axiomatization, the concept of an algorithm, proofs, the theory of sets, the theory of relations, Boolean algebra, and antinomies. The closing section, on the real number system and algebra, takes up natural numbers, groups, linear algebra, polynomials, rings and ideals, the theory of numbers, algebraic extensions of a fields, complex numbers and quaternions, lattices, the theory of structure, and Zorn?s lemma. Volume II begins with eight chapters on the foundations of geometry, followed by eight others on its analytic treatment. The latter include discussions of affine and Euclidean geometry, algebraic geometry, the Erlanger Program and higher geometry, group theory approaches, differential geometry, convex figures, and aspects of topology. Volume III, on analysis, covers convergence, functions, integral and measure, fundamental concepts of probability theory, alternating differential forms, complex numbers and variables, points at infinity, ordinary and partial differential equations, difference equations and definite integrals, functional analysis, real functions, and analytic number theory. An important concluding chapter examines ?The Changing Structure of Modern Mathematics.? The part of the first volume specifically dedicated to foundational mathematics strikes me as fairly agnostic regarding which of several possible approaches are ideal in establishing the proper foundations of mathematics. It is so condensed as to be cryptic. Each subsection could fill a chapter, each section could fill a book, and each chapter could fill a book shelf if they were elaborated upon to the point of covering their topics comprehensively. -- Nil conscire sibi === Subject: Re: Set Theory: Should you believe? > that has caused a bit of discussion in some logic circles. > My claims in short: 1) most of `elementary mathematics' is not sufficiently > well understood by the mathematical establishment, leading to weaknesses in > K12 and college curriculum, 2) the current theory of `real numbers' is a > joke, and sidesteps the crucial issue of understanding the computational > specification of the continuum, and 3) `infinite sets' are a metaphysical > concept, and unnecessary for correct mathematics. > Analysts and set theorists are welcome to send me reasoned responses. > Assoc Prof N J Wildberger > School of Maths > UNSW After casually reading his notes, I think that he is saying this. The axioms of modern set theory are too burdensome for actual mathematicians, who in practice take a somewhat Platonic view of their subject. But he is unsatisfied with the pragmatic Platonist approach that we take, particular in the manner in which mathematics is taught at pre-college levels. He thinks we need to roll up our sleeves and rethink foundations so that we get something that really is usuable, so that we can finally truly rid ourselves of Platonism, superstition, instinct, gut reaction, religion, etc, etc. Personally I really like the Platonic approach, using set theory as a highly convenient crutch. I'm not going to preclude the possibility that one day a set of foundations for mathematics will be found, that will greatly simplify and advance the extent that we will be able to think about mathematics, but I think it will take a great genius, and also some crisis of cicumstance (perhaps the discovery of some horribly unresolvable contradiction). But on the whole, even if his tone was not exactly politic, I liked a lot of what he said. But I don't think it is going to change the world. Stephen === Subject: Re: Set Theory: Should you believe? that has caused a bit of discussion in some logic circles. > My claims in short: 1) most of `elementary mathematics' is not sufficiently > well understood by the mathematical establishment, leading to weaknesses in > K12 and college curriculum, 2) the current theory of `real numbers' is a > joke, and sidesteps the crucial issue of understanding the computational > specification of the continuum, and 3) `infinite sets' are a metaphysical > concept, and unnecessary for correct mathematics. > Analysts and set theorists are welcome to send me reasoned responses. > Assoc Prof N J Wildberger > School of Maths > UNSW > After casually reading his notes, I think that he is saying this. The > axioms of modern set theory are too burdensome for actual > mathematicians, who in practice take a somewhat Platonic view of their > subject. But he is unsatisfied with the pragmatic Platonist approach > that we take, particular in the manner in which mathematics is taught at > pre-college levels. He thinks we need to roll up our sleeves and > rethink foundations so that we get something that really is usuable, so > that we can finally truly rid ourselves of Platonism, superstition, > instinct, gut reaction, religion, etc, etc. > Personally I really like the Platonic approach, using set theory as a > highly convenient crutch. I'm not going to preclude the possibility > that one day a set of foundations for mathematics will be found, that > will greatly simplify and advance the extent that we will be able to > think about mathematics, but I think it will take a great genius, and > also some crisis of cicumstance (perhaps the discovery of some horribly > unresolvable contradiction). > But on the whole, even if his tone was not exactly politic, I liked a > lot of what he said. But I don't think it is going to change the world. > Stephen question: What of what he said did you like and for what reasons? I would like to understand what the appeal of such a rant as Wildberger's is. In my view, it is evidently wrong in many parts, wrong under light scrutiny in many others and falacious for the most part of the rest. The only statement I can think of which is undebatable is the fact that education in mathematics is ever more deficient. But this is also strongly suggested by a lot of other factors, as the ever growing number of crackpots, the ever more present wrong notion that mathematics is dependent on computability and the expanding belief that mathematics is somehow subjected to the constrains of physical reality. === Subject: Re: Set Theory: Should you believe? >that has caused a bit of discussion in some logic circles. >My claims in short: 1) most of `elementary mathematics' is not sufficiently >well understood by the mathematical establishment, leading to weaknesses in >K12 and college curriculum, 2) the current theory of `real numbers' is a >joke, and sidesteps the crucial issue of understanding the computational >specification of the continuum, and 3) `infinite sets' are a metaphysical >concept, and unnecessary for correct mathematics. >Analysts and set theorists are welcome to send me reasoned responses. >Assoc Prof N J Wildberger >School of Maths >UNSW >>After casually reading his notes, I think that he is saying this. The >>axioms of modern set theory are too burdensome for actual >>mathematicians, who in practice take a somewhat Platonic view of their >>subject. But he is unsatisfied with the pragmatic Platonist approach >>that we take, particular in the manner in which mathematics is taught at >>pre-college levels. He thinks we need to roll up our sleeves and >>rethink foundations so that we get something that really is usuable, so >>that we can finally truly rid ourselves of Platonism, superstition, >>instinct, gut reaction, religion, etc, etc. >>Personally I really like the Platonic approach, using set theory as a >>highly convenient crutch. I'm not going to preclude the possibility >>that one day a set of foundations for mathematics will be found, that >>will greatly simplify and advance the extent that we will be able to >>think about mathematics, but I think it will take a great genius, and >>also some crisis of cicumstance (perhaps the discovery of some horribly >>unresolvable contradiction). >>But on the whole, even if his tone was not exactly politic, I liked a >>lot of what he said. But I don't think it is going to change the world. >>Stephen > question: What of what he said did you like and for what reasons? > I would like to understand what the appeal of such a rant as > Wildberger's is. In my view, it is evidently wrong in many parts, wrong > under light scrutiny in many others and falacious for the most part of > the rest. The only statement I can think of which is undebatable is the > fact that education in mathematics is ever more deficient. But this is > also strongly suggested by a lot of other factors, as the ever growing > number of crackpots, the ever more present wrong notion that > mathematics is dependent on computability and the expanding belief that > mathematics is somehow subjected to the constrains of physical reality. Well let me start by saying that I am only stating an opinion so there is no need to get upset or mad about it. I do admit that I didn't deeply scrutinize his writings, but I didn't feel that he said anything mathematically false. So it really all boils down to feelings about the current state of mathematics. Next, while he was definitely ranting, and perhaps venting a certain anger he also feels, I think it is important to listen to the message rather than the messenger or the style in which the message is communicated. There is a clear distinction between someone who expresses discomfort with the present way we do mathematics which is at worst an unpopular view, and saying that Cantor's diagonal argument is logically flawed which at best is a crackpot position to hold. This web site is clearly in the former category. The view that mathematics should try to constrain itself to physical reality is, in my opinion, not a crackpot position. I think this is what Morris Kline was getting at in his books Mathematics: Loss of Certainty. Attempts to place axiomatic number theory and axiomatic set theory on firmer ground by proving its consistency starting with a much weaker system are doomed to failure by Goedel's Theorem, and thus the possibility that the whole thing is built on a house of cards becomes ever more reasonable. I agree with Kline that the strongest evidence for the consistency of these theories, or at least some weaker version which would still contain most of our important discoveries, is just how very well it seems to work in practice. Let me use an example. The mathematicians Wiener and Banach lived at about the same time. Banach chose a much more abstract direction to take his mathematics, looking at general properties of infinite dimensional normed vector spaces beyond Hilbert spaces and spaces of continuous or measurable function. Wiener looked more to real life applications, and building on the work of Einstein put the phenominum of Brownian motion onto a rigourous mathematical foundation. Many very talented mathematicians have studied Banach spaces, but it has proven to be a remarkably barren field. On the other hand Wiener's approach has led to the rich theory of stochastic calculus, which has proved extremely useful in studing the heat equation, harmonic functions, and even economics. Stephen === Subject: Re: Set Theory: Should you believe? reality is, in my opinion, not a crackpot position. It's a meaningless position unless you can give confine itself to physical reality in connection to a subject which is not *about* physical reality a meaning. > Let me use an example. The mathematicians Wiener and Banach lived at > about the same time. Banach chose a much more abstract direction to > take his mathematics, looking at general properties of infinite > dimensional normed vector spaces beyond Hilbert spaces and spaces of > continuous or measurable function. Wiener looked more to real life > applications, and building on the work of Einstein put the phenominum of > Brownian motion onto a rigourous mathematical foundation. Many very > talented mathematicians have studied Banach spaces, but it has proven to > be a remarkably barren field. On the other hand Wiener's approach has > led to the rich theory of stochastic calculus, which has proved > extremely useful in studing the heat equation, harmonic functions, and > even economics. Stephen, I suggest you confine yourself to topics you know something about, and cease spouting ignorant horse. A very famous incident--so famous it makes it dead obvious you don't know what the hell you are talking about--is Gelfand's proof of Wiener's theorem using Banach algebras. Wiener had proven that if Fourier series on a closed interval which converges absolutely to a function f such that f has no zero on the interval, then 1/f also has an absolutely convergent Fourier series on that interval. The proof was not easy, yet Gelfand proves it in a few lines using the theory of Banach algebras. === Subject: Re: Set Theory: Should you believe? > the ever more present wrong notion that mathematics is dependent on > computability and the expanding belief that mathematics is somehow > subjected to the constrains of physical reality. I know this was addressed to someone else, but I would also like to offer my thoughts on this matter. I contend that mathematics _is_ constrained by physical reality. The underlying logic which determines mathematics is a manifestation of physical reality. I believe what you are asserting is that mathematics should not be required to produce physically measurable results as a test of its validity. I really have to wonder if such a requirement is unrealistic. It's interesting to observe that some people are wont to point to the fact that formal proofs can be verified by computer programs. I note that you object to the idea that mathematics is dependent on computability. I don't know if you mean that exclusively in terms of solving equations and/or finding approximate numerical solutions, or if you also object to the idea the mathematical proofs should be machine verifiable. I curious to know what you think of this: http://metamath.org/ -- Nil conscire sibi === Subject: Re: Set Theory: Should you believe? like to offer my thoughts on this matter. I contend that > mathematics _is_ constrained by physical reality. The > underlying logic which determines mathematics is a > manifestation of physical reality. I believe what > you are asserting is that mathematics should not be > required to produce physically measurable results as > a test of its validity. I really have to wonder if > such a requirement is unrealistic. It's interesting > to observe that some people are wont to point to the > fact that formal proofs can be verified by computer programs. I've often wondered about similar issues myself. For one thing, we could argue that there is a difference between our interpretation of certain mathematical notions, such as completed infinite sets, and what we're actually doing, which is writing finitely many symbols down on paper in certain ways. There's no reason for which I can see that, because we can write certain symbols down in certain ways, that certain interpretations of what we're doing, above and beyond this, must follow. Add to this the fact that the act of writing down these symbols is only possible by the nature of our reality. Or at least, I don't see how we could prove its independence from our reality in a way that we would understand, because it seems to me that any such metaproof must also be within, and hence a feature of, our reality. L. Renfro === Subject: Re: Set Theory: Should you believe? the ever more present wrong notion that mathematics is dependent on > computability and the expanding belief that mathematics is somehow > subjected to the constrains of physical reality. > I know this was addressed to someone else, but I would also like to offer my > thoughts on this matter. I contend that mathematics _is_ constrained by > physical reality. The underlying logic which determines mathematics is a > manifestation of physical reality. I believe what you are asserting is That is an issue for Philosophers which I don't believe can be resolved definitely. I do not subscribe to it, but can not refute it either, so let us please agree to exclude it from this particular discussion.This is not howewer the flawed reasoning that I am refering to. I am talking much baser contentions made by the ill educated like denying the existence of certain mathematical objects on the ground that they can not be physically constructed. An example would be the set of Natural Numbers whichs existence is denied because there are not sufficient atoms in the universe to build a tangible physical representation of it, or specific irrational numbers on on the same grounds pertaining to their decimal base representation. > that mathematics should not be required to produce physically measurable > results as a test of its validity. I really have to wonder if such a > requirement is unrealistic. If you make such a requirement, then you will not get very far. What is a physically measurable result that gives validity to the number 2 ? Or to the law of distributivity? or to the differentiability of a given function? Mathematical concepts have no consequences in physical reality and the laws of the physical universe have no consequence for mathematical concepts. (as long as we keep to the agreement I requested above). > It's interesting to observe that some people > are wont to point to the fact that formal proofs can be verified by > computer programs. > I note that you object to the idea that mathematics is dependent on > computability. I don't know if you mean that exclusively in terms of > solving equations and/or finding approximate numerical solutions, or if you > also object to the idea the mathematical proofs should be machine > verifiable. I curious to know what you think of this: http://metamath.org/ I agree with the idea that mathematical proofs should be machine verifiable; as long as this is a statement about computer science and not about mathematics. === Subject: Re: Set Theory: Should you believe? >This > is not howewer the flawed reasoning that I am refering to. I am talking > much baser contentions made by the ill educated like denying the > existence of certain mathematical objects on the ground that they can > not be physically constructed. An example would be the set of Natural > Numbers whichs existence is denied because there are not sufficient > atoms in the universe to build a tangible physical representation of > it, or specific irrational numbers on on the same grounds pertaining to > their decimal base representation. I don't know that such an opinion is ill-educated or merely unpersuasive. OTOH, I don't believe that is the essence of most objections to the concept of infinite sets. For myself, I am willing to consider the existence of infinite sets without much reservation. At the same time, I believe that arguments and reasoning applicable to finite sets should be applied with caution to infinite sets, and carefully scrutinized to determine if it makes sense in any given instance. >> that mathematics should not be required to produce physically measurable >> results as a test of its validity. I really have to wonder if such a >> requirement is unrealistic. > If you make such a requirement, then you will not get very far. What is > a physically measurable result that gives validity to the number 2 ? Or > to the law of distributivity? or to the differentiability of a given > function? Mathematical concepts have no consequences in physical > reality and the laws of the physical universe have no consequence for > mathematical concepts. (as long as we keep to the agreement I requested > above). I am not stating this conclusively, but consider Cantor's transfinite induction, in comparison to the calculus of Newton and Leibnitz. We have ample evidence that the results of differentiation and integration correspond in many cases to physical experiments in convincing ways. Is there any result from Cantor's theory which leads to a verifiable prediction in terms of physical experiment? That is merely one aspect of the idea that mathematics might be physically testable. Another is whether the actual reasoning can be reproduced or verified using computers. >> It's interesting to observe that some people >> are wont to point to the fact that formal proofs can be verified by >> computer programs. >> I note that you object to the idea that mathematics is dependent on >> computability. I don't know if you mean that exclusively in terms of >> solving equations and/or finding approximate numerical solutions, or if >> you also object to the idea the mathematical proofs should be machine >> verifiable. I curious to know what you think of this: >> http://metamath.org/ > I agree with the idea that mathematical proofs should be machine > verifiable; as long as this is a statement about computer science and > not about mathematics. One might argue that this is merely using the computer as a tool. -- Nil conscire sibi === Subject: Re: Set Theory: Should you believe? This > is not howewer the flawed reasoning that I am refering to. I am talking > much baser contentions made by the ill educated like denying the > existence of certain mathematical objects on the ground that they can > not be physically constructed. An example would be the set of Natural > Numbers whichs existence is denied because there are not sufficient > atoms in the universe to build a tangible physical representation of > it, or specific irrational numbers on on the same grounds pertaining to > their decimal base representation. > I don't know that such an opinion is ill-educated or merely unpersuasive. It is not a matter of what the opinion itself is. The problem is with the person holding such an opinion. That person must have received a flawed or no mathematical education at all. > OTOH, I don't believe that is the essence of most objections to the concept > of infinite sets. For myself, I am willing to consider the existence of Of course it is not the essence of most objections to to the concept of infinite sets. It is an example of the flawed assumption that mathematics are constrained by physical reality in such a manner. > infinite sets without much reservation. At the same time, I believe that > arguments and reasoning applicable to finite sets should be applied with > caution to infinite sets, and carefully scrutinized to determine if it > makes sense in any given instance. That is not related to this discussion. >> that mathematics should not be required to produce physically measurable >> results as a test of its validity. I really have to wonder if such a >> requirement is unrealistic. > If you make such a requirement, then you will not get very far. What is > a physically measurable result that gives validity to the number 2 ? Or > to the law of distributivity? or to the differentiability of a given > function? Mathematical concepts have no consequences in physical > reality and the laws of the physical universe have no consequence for > mathematical concepts. (as long as we keep to the agreement I requested > above). > I am not stating this conclusively, but consider Cantor's transfinite > induction, in comparison to the calculus of Newton and Leibnitz. We have > ample evidence that the results of differentiation and integration > correspond in many cases to physical experiments in convincing ways. Is > there any result from Cantor's theory which leads to a verifiable > prediction in terms of physical experiment? You were the one arguing in favour of the requirement that mathematical concepts be subjected to verification by physical measurability. Obviously you have found a further difficulty in supporting that case. > That is merely one aspect of the idea that mathematics might be physically > testable. Another is whether the actual reasoning can be reproduced or > verified using computers. >> It's interesting to observe that some people >> are wont to point to the fact that formal proofs can be verified by >> computer programs. >> I note that you object to the idea that mathematics is dependent on >> computability. I don't know if you mean that exclusively in terms of >> solving equations and/or finding approximate numerical solutions, or if >> you also object to the idea the mathematical proofs should be machine >> verifiable. I curious to know what you think of this: >> http://metamath.org/ > I agree with the idea that mathematical proofs should be machine > verifiable; as long as this is a statement about computer science and > not about mathematics. > One might argue that this is merely using the computer as a tool. Your use of the word 'merely' suggests that you might be thinking of manners of using a computer trascending those of a tool. What would those be ? === Subject: Re: Set Theory: Should you believe? > I know this was addressed to someone else, but I would also like to offer my > thoughts on this matter. I contend that mathematics _is_ constrained by > physical reality. The underlying logic which determines mathematics is a > manifestation of physical reality. I believe what you are asserting is > that mathematics should not be required to produce physically measurable > results as a test of its validity. I really have to wonder if such a > requirement is unrealistic. It's interesting to observe that some people > are wont to point to the fact that formal proofs can be verified by > computer programs. But formal proofs generally cannot be discovered by finitary algorithmic means. We still need Inspiration. If you regard all, so-called mental processes, as really physical then your assertion may have some basis. Bob Kolker === Subject: Re: Set Theory: Should you believe? >> I know this was addressed to someone else, but I would also like to offer >> my >> thoughts on this matter. I contend that mathematics _is_ constrained by >> physical reality. The underlying logic which determines mathematics is a >> manifestation of physical reality. I believe what you are asserting is >> that mathematics should not be required to produce physically measurable >> results as a test of its validity. I really have to wonder if such a >> requirement is unrealistic. It's interesting to observe that some people >> are wont to point to the fact that formal proofs can be verified by >> computer programs. > But formal proofs generally cannot be discovered by finitary algorithmic > means. Isn't that Chruch's theorem? > We still need Inspiration. If you regard all, so-called mental > processes, as really physical then your assertion may have some basis. The only thing I am asserting with absolute conviction that will never be shaken is that the thought processes which we call mathematics are governed by the Laws of Nature. That is to say Physical Laws. Whether that amounts to finitary algorithmic means is less certain. Everything else was intended contingently. Note that my comment regarding proofs involved verification, not production. -- Nil conscire sibi === Subject: Re: Set Theory: Should you believe? axioms of modern set theory are too burdensome for actual > mathematicians, who in practice take a somewhat Platonic view of their > subject. But he is unsatisfied with the pragmatic Platonist approach > that we take, particular in the manner in which mathematics is taught at > pre-college levels. How do you get from there to his claim that you can't prove the fundamental theorem of arithmetic? He thinks we need to roll up our sleeves and > rethink foundations so that we get something that really is usuable, so > that we can finally truly rid ourselves of Platonism, superstition, > instinct, gut reaction, religion, etc, etc. It seems to me what he's saying is simply incoherent. He likes Lie groups, so it's OK to talk about them, so long as you cross your fingers and say you are really talking about constructable numbers. But integers, which are about as constructable as it gets, he is willing to blow off, apparently merely out of disinterest in number theory as opposed to Lie groups. > But on the whole, even if his tone was not exactly politic, I liked a > lot of what he said. But I don't think it is going to change the world. Let's make things way harder to prove for no reason, and do it in an incoherent way is a tough sell. === Subject: Re: Set Theory: Should you believe? 1) most of `elementary mathematics' is not sufficiently well understood by the mathematical establishment. Well, a simple example: 1x3(aaa,bbb,ccc) 3x6(aab,abb,aac,acc,bbc,bcc) 6x1(abc) Which column is the Newton's one? What meaning has the second column? There are two polynomial coefficients. try it wor any similar products. Did you studied combinatorics? Why you do not know it? If you know this fact, why you did not used it? kunzmilan === Subject: Re: Set Theory: Should you believe? > that has caused a bit of discussion in some logic circles. Here's a quote from the paper, which I think shows how confused Wildberger is about axioms: In ordinary mathematics, statements are either true, false, or they don't make sense. If you have an elaborate theory of 'hierarchies upon hierarchies of infinite sets', in which you cannot even in principle decide if there is something between the first and second 'infinity' on your list, there's a time to admit you are no longer doing mathematics. Of course, the statement about not being able to decide if there is a cardinal between aleph_0 and aleph_1 is absurd, but leave that aside and look instead at the statement that there is a cardinal between aleph_0 and 2^aleph_0. How does this differ from the statement that the sum of the angles of a triangle cannot be determined in absolute geometry? Suppose, when faced with the fact that not everyone found the parallel postulate to be intuituively true, the Greek geometers had simply removed it from consideration. They could then be attacked with the sneer that they were not doing mathematics at all, because they could not answer so basic a question as whether the sum of the angles of a triangle was less than, equal to, or greater than two right angles. Yet I think it is clear they *would* be doing geometry. For that matter a group theorist who cannot tell you if a generic group is abelian or nonabelian, since it might be either, is not failing to do mathematics. The difference is that we've given up on the idea that there is a single correct geometry, but still feel (and that's nothing but intuition speaking) that there is a single true set theory, just as we think there is a single true number theory. But our intution is not strong enough to settle all the questions as to what this true set theory actually is. This is really a meta problem for mathematics, and not a question which can allow a person to conclude that set theorists are not mathematicians, any more than the existence of nonstandard models for first-order arithmetic would prove number theorists are not mathematicians. As to rigor, the mechanically verified proofs of the Mizar project are more rigorous than mere mortals like you, me, or Wildberger do, and they are based on set theory with (if needed) inacessible cardinals. The rigor argument is clearly therefore baloney, and Wildberger cannot seem to separate mathematics from metaphysics for the rest of it. === Subject: Re: Set Theory: Should you believe? > My claims in short: 1) most of `elementary mathematics' is not sufficiently > well understood by the mathematical establishment, leading to weaknesses in > K12 and college curriculum, 2) the current theory of `real numbers' is a > joke, and sidesteps the crucial issue of understanding the computational > specification of the continuum, and 3) `infinite sets' are a metaphysical > concept, and unnecessary for correct mathematics. Here's two relevant quotes: ordinary mathematical practice does not require an enigmatic metaphysical universe of sets (Nik Weaver) the actual infinite is not required for the mathematics of the physical world (Feferman) Most mathematicians working in the field of foundations understand and accept those quotes, but the average mathematician may not. What I have suggested is that mathematics needs a reality check; mathematics can and should be treated as a science in which testable consequences are required. I suspect that is what you are getting at, although I don't think you've said it especially clearly. > Analysts and set theorists are welcome to send me reasoned responses. Send them to you? Why would anyone do that? === Subject: Re: Set Theory: Should you believe? [...] >> Analysts and set theorists are welcome to send me reasoned responses. >Send them to you? Why would anyone do that? Well, after the initial hit-and-run posting he doesn't appear to be replying to anything else in this thread. -- Richard Herring === Subject: Re: Set Theory: Should you believe? > Most mathematicians working in the field of foundations understand and > accept those quotes, but the average mathematician may not. Most mathematicians are not enthusiastic about being forced to work only in terms of first-order arithmetic, or even second-order arithmetic, either. Because it would be a big, fat pain and accomplish nothing they see as necessary. On the other hand, they might very well be interested in the question of whether, and to what extent, what they are doing can be done using weaker assumptions--so long as you don't try to *make* them do it that way. === Subject: Re: Set Theory: Should you believe? > that has caused a bit of discussion in some logic circles. Your paper starts out saying a lot of things that are blatanly false. This is not a good way to make a reasoned argument. You say, for instance, that the Academy has consistently refused to get serious about foundational questions, whereas doing that has been a major theme of twentieth century mathematics, and many great mathematicians have made it their life's work. You claim physicists have trouble with string theory because they make use of set theory, which is simply nonsense. I suspect you don't know much about string theory. You claim the axioms of set theory have not been questioned, which is drivel. You sneer at people who do what you claim you think should be done, namely study the foundations of mathematics, and put the word supposedly before difficult, indicating that you think set theory and logic are dead easy. But do you know any serious set theory? You claim that most mathematicians could not define a vector or a function, as if you possessed the secret decoder ring which made you smarter than the rest of us. In short, you sound, just in your opening few paragraphs, altogether too much like the kind of people we constantly come across here on sci.math. You sound like a crank. I think you *are* a crank. I suggest you stick to subjects you've studied, and not sound off on topics you don't understand. === Subject: Re: Set Theory: Should you believe? >> http://web.maths.unsw.edu.au/~norman/views.htm >> that has caused a bit of discussion in some logic circles. Well, the discussion that I have seen - on this newsgroup (sci.math or sci.logic, I can't remember) - is that it is bull. Much of it admittedly from me. >> My claims in short: 1) most of `elementary mathematics' is not >> sufficiently >> well understood by the mathematical establishment, leading to weaknesses >> in >> K12 and college curriculum, I don't know about the mathematical establishment (as a whole) not understanding elementary mathematics, but your own writings on set theory and the axiomatic method don't fill me with confidence. 2) the current theory of `real numbers' is a >> joke, and sidesteps the crucial issue of understanding the computational >> specification of the continuum, and This is pure crank stuff. Describing a huge and extremely rigorously defined area such as the construction of the Reals as a joke without any mathematical justification is flakey at best; the phrase computational specification of the continuum (a phrase that gets exactly zero matches on Google) is crank babble. 3) `infinite sets' are a metaphysical >> concept, and unnecessary for correct mathematics. No, infinite sets are a mathematical concept, not unlike perfect circles and the exact value of the sqrt(2). Tell me, is the set of all natural numbers finite or infinite? Or if you can't form the set, why not? What about the set of all points on a perfect circle (ie all solutions to x^2 + y^2 = 1). Finite or infinite? Or don't you believe that I can define a set as being all points on the unit circle. If not, why not? What is wrong with {(x,y) | x^2 + y^2 = 1} as a set? Infinite or finite? Your paper has no mathematical content, and is pure crank stuff. The stuff about Axioms somehow being irrelevant to mathematics is just your own philosophical ramblings. Surprising, since you seem to accept the axioms of group theory, but not set theory (because they are too complicated and too abstract for your liking)? Do groups with an infinite number of elements exist, by the way? I would have ignored this post - and previous posts on your mathematical insights - in much the same way that I ignore posts about Einstein was wrong or Cantor's diagonal proof is flawed - as pure crank material. The only reason I haven't, is that you are a mathematics teacher, and it worries me that somebody (eg your students) may be being taught this stuff about set theory. Tell me, do you accept that there are models for the ZF Axioms if we drop Axiom 6 ? Do you accept that there is a model for the ZF axioms if we include an additional axiom: Exists S such that { } is an element of S, and x elements of S implies x union {x) is an element of S ??? (informally known as the axiom of infinity) If you don't, I can certainly show you a model (the von-Neumann construction of N). How is this axiom fundamentally different from the other axioms? >> Analysts and set theorists are welcome to send me reasoned responses. >> Assoc Prof N J Wildberger >> School of Maths >> UNSW > I've added sci.logic to your posting. karl m === Subject: Re: Set Theory: Should you believe? >> http://web.maths.unsw.edu.au/~norman/views.htm >> that has caused a bit of discussion in some logic circles. > Well, the discussion that I have seen - on this newsgroup (sci.math or > sci.logic, I can't remember) - is that it is bull. Ad hominem. >> My claims in short: 1) most of `elementary mathematics' is not >> sufficiently well understood by the mathematical establishment, >> leading to weaknesses in K12 and college curriculum, > I don't know about the mathematical establishment (as a whole) not > understanding elementary mathematics, but your own writings on set theory > and the axiomatic method don't fill me with confidence. Ad hominem. > 2) the current theory of `real numbers' is a >> joke, and sidesteps the crucial issue of understanding the computational >> specification of the continuum, and > This is pure crank stuff. Describing a huge and extremely rigorously defined > area such as the construction of the Reals as a joke without any > mathematical justification is flakey at best; the phrase computational > specification of the continuum (a phrase that gets exactly zero matches on > Google) is crank babble. The mathematical justification for describing the current theory of real numbers as a joke is given in the paper. You may not find it convincing - I may not find it convincing - but it's there. Crank babble is ad hominem. > 3) `infinite sets' are a metaphysical >> concept, and unnecessary for correct mathematics. > No, infinite sets are a mathematical concept, not unlike perfect circles and > the exact value of the sqrt(2). > Tell me, is the set of all natural numbers finite or infinite? Or if you > can't form the set, why not? I think Norm would say, 1) you can't form the set (and Norm's reasons mathematics you can do with the completed infinite set that you can't do without it. > What about the set of all points on a perfect circle (ie all solutions to > x^2 + y^2 = 1). Finite or infinite? Or don't you believe that I can define a > set as being all points on the unit circle. If not, why not? What is wrong > with {(x,y) | x^2 + y^2 = 1} as a set? Infinite or finite? Again, I think Norm is arguing against the completed infinite. For sensibly write about the set of all natural numbers, or any of these other sets that most mathematicians are quite happy with - moreover, you don't lose anything valuable if you discard them. > Your paper has no mathematical content, and is pure crank stuff. The stuff > about Axioms somehow being irrelevant to mathematics is just your own > philosophical ramblings. Surprising, since you seem to accept the axioms of > group theory, but not set theory (because they are too complicated and too > abstract for your liking)? Do groups with an infinite number of elements > exist, by the way? More ad hominem. Norm accepts the axioms of group theory as the definition of what a group is, and has no problem with them because he can construct (finite) models of them. He argues that the axioms of set theory (in particular, ZFC) don't define what a set is and don't lead to sensible constructions of infinite sets. > I would have ignored this post - and previous posts on your mathematical > insights - in much the same way that I ignore posts about Einstein was > wrong or Cantor's diagonal proof is flawed - as pure crank material. The > only reason I haven't, is that you are a mathematics teacher, and it worries > me that somebody (eg your students) may be being taught this stuff about set > theory. I'm not sure how you think your method of not ignoring Norm's post will prevent his students from being taught his ideas. > Tell me, do you accept that there are models for the ZF Axioms if we drop > Axiom 6 ? Do you accept that there is a model for the ZF axioms if we > include an additional axiom: > Exists S such that { } is an element of S, and x elements of S implies x > union {x) is an element of S ??? > (informally known as the axiom of infinity) > If you don't, I can certainly show you a model (the von-Neumann construction > of N). > How is this axiom fundamentally different from the other axioms? It seems to me Norm is making several points, two of which are that ZFC sucks and that mathematics doesn't need axioms in the first place. Your suggestions may or may not have any bearing on the first point, but they don't address the second. I'm dismayed by the level of vituperation in some of the posts in this thread. Norm is not presenting a high-school algebra proof of Fermat's Last Theorem, nor is he insisting that the reals are countable because you can always take that real number that you left off your list and stick it on at the end. He's adopting a finitistic, or constructivist, or computational view of mathematics. It's an unpopular view, it doesn't particularly appeal to me, but I don't see the need to go ballistic in response. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Set Theory: Should you believe? <44b5abd7$0$1207$afc38c87@news.optusnet.com.au> Norm accepts the axioms of group theory as the > definition of what a group is, Shut up. You don't know Norm, either. Norm attacks logicians and the axiomatic method generally. IF he accepted axioms as legitimate for defining things then he wouldn't have anything to argue about. > and has no problem with them Wrong. > because he can construct (finite) models of them. Obviously, he AND EVERYBODY ELSE can ALSO construct INFINITE models of them -- AND HE DEALS with infinite groups ALL THE TIME in his work. So this is simply not a legitimate distinction. > He argues that the axioms > of set theory (in particular, ZFC) don't define what a set is That is completely incoherent. Since, in ZFC, EVERYthing is a set, the axioms of ZFC could NOT POSSIBLY DO ANYTHING BUT define what a set is. > and don't lead to sensible constructions of infinite sets. That is ridiculous too. He is not even saying that. You need to quote him. What he DOES say is that textbooks aren't much on defining what an infinite set is, and he has looked. That just makes HIM look stupid. Arthur Rubin and others have posted 5 different definitions of infinite set, in case he was too stupid to find one. More to the point, classical FOL IN GENERAL is NOT constructive and one NEVER looks to the axioms themselves to construct ANYthing -- the model construction language is ALWAYS something DIFFERENT. > It seems to me Norm is making several points, two of which are > that ZFC sucks and that mathematics doesn't need axioms in the first > place. Well, now you are a lot closer to reading him rightly, but you do need to understand that your 2nd point here (about math not needing axioms in the first place) contradicts your point above about Norm respecting the axioms of group theory because he can construct finite models of them. Group theory is in fact defined by its axioms. Norm canNOT simultaneously appreciate that AND think that math doesn't need axioms. Neither can you. > I'm dismayed by the level of vituperation in some of the posts in > this thread. Well, you shouldn't be. That's what you get for farting in church. NW is presuming to pontificate about something he has not studied. > Norm is not presenting a high-school algebra proof of > Fermat's Last Theorem, nor is he insisting that the reals are countable > because you can always take that real number that you left off your > list and stick it on at the end. He's adopting a finitistic, or > constructivist, or computational view of mathematics. He IS NOT, you IDIOT! IF he were doing that, he would've actually GOOGLED constructive mathematics before presuming NW * doesn't know * about constructive mathematics! > It's an unpopular view, And one that he has never heard of; the only thing that is even MOTIVATING him to LEAN in that direction are the things would've been WISE TO QUOTE, IF you were going to be stupid enough to be attributing views to another person, things like the failure of the profession at large to logically ground its concepts. If this was what was actually bothering him then (obviously) he should've been supporting axioms, not attacking them. > it doesn't particularly appeal to me, but I don't see the need > to go ballistic in response. The going ballistic is NOT in response to constructivism. NW didn't cite Bishop or anybody who is trying to go that way. He HIMSELF FIRST went ballistic vs. logicians by calling them a priesthood. He embarrassed himself by sounding like . === Subject: Re: Set Theory: Should you believe? <44b5abd7$0$1207$afc38c87@news.optusnet.com.au> Well, the discussion that I have seen - on this newsgroup (sci.math or > sci.logic, I can't remember) - is that it is bull. > Ad hominem. > I don't know about the mathematical establishment (as a whole) not > understanding elementary mathematics, but your own writings on set theory > and the axiomatic method don't fill me with confidence. > Ad hominem. > This is pure crank stuff. Describing a huge and extremely rigorously defined > area such as the construction of the Reals as a joke without any > mathematical justification is flakey at best; the phrase computational > specification of the continuum (a phrase that gets exactly zero matches on > Google) is crank babble. > The mathematical justification for describing the current theory of > real numbers as a joke is given in the paper. You may not find it > convincing - I may not find it convincing - but it's there. > Crank babble is ad hominem. > Your paper has no mathematical content, and is pure crank stuff. The stuff > about Axioms somehow being irrelevant to mathematics is just your own > philosophical ramblings. Surprising, since you seem to accept the axioms of > group theory, but not set theory (because they are too complicated and too > abstract for your liking)? Do groups with an infinite number of elements > exist, by the way? > More ad hominem. Norm accepts the axioms of group theory as the > definition of what a group is, and has no problem with them because > he can construct (finite) models of them. He argues that the axioms > of set theory (in particular, ZFC) don't define what a set is and > don't lead to sensible constructions of infinite sets. An ad hominem fallacy is of the form: Person P made argument A Person P is ignorant/poor/gay/female/stupid/handicapped/etc Therefore argument A is invalid. It is *not* ad hominem to say: Person Q made argument B Argument B is incoherent/lacks reasoning/lacks evidence/is purely assertion/is muddled/makes categorical errors/equivocates/is nonsense/is irrelevant/etc Therefore person Q is ignorant/poorly read/uninformed/unreasonable/stupid/irrelevant/etc Sure, such a statement may be insulting. But it isn't invalid. > -- > Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) -Tez === Subject: Re: Set Theory: Should you believe? Gerry Myerson says... >I'm dismayed by the level of vituperation in some of the posts in >this thread. Norm is not presenting a high-school algebra proof of >Fermat's Last Theorem, nor is he insisting that the reals are countable >because you can always take that real number that you left off your >list and stick it on at the end. He's adopting a finitistic, or >constructivist, or computational view of mathematics. It's an unpopular >view, it doesn't particularly appeal to me, but I don't see the need >to go ballistic in response. Norm's paper is not a discussion of finitistic methods. It's not a discussion of the computational view of mathematics. It is a mean-spirited, sneering attacking on a huge swath of modern mathematics and modern mathematicians. It is *full* of vituperation. When it comes to reasonableness, I don't see why Norm's equating set theory with a religious is *more* reasonable than Peter Webb's equating Norm's paper with crank babble. Norm's paper is not a mathematical paper, it is a polemic. There is really no way to give a reasoned, mathematical response to it. In contrast, a paper that starts off saying that the author is rejecting the axiom of infinity because he wishes to see how much of mathematics can be done with minimal ontological commitment could be the start of a reasonable mathematical paper. An exploration of finitistic mathematics could be interesting mathematics. Nobody would accuse Norm of being a crank for writing such a paper, or even for dedicating his life to the development of finitistic methods. People accuse him of being a crank when he says things such as comparing set theory with a religious cult. *That's* what's crank material, not finitistic methods, and not the computational view of mathematics. However, I understand why Norm might prefer to write the inflamatory type of paper: because at least it generates a response, while the more reasonable exploration of finitistic methods would generate polite indifference and yawns. -- Daryl McCullough Ithaca, NY === Subject: Re: Set Theory: Should you believe? >> http://web.maths.unsw.edu.au/~norman/views.htm >> that has caused a bit of discussion in some logic circles. >> Well, the discussion that I have seen - on this newsgroup (sci.math or >> sci.logic, I can't remember) - is that it is bull. > Ad hominem. Not ad hominem. I talk about his post, not him as a person. >> My claims in short: 1) most of `elementary mathematics' is not >> sufficiently well understood by the mathematical establishment, >> leading to weaknesses in K12 and college curriculum, >> I don't know about the mathematical establishment (as a whole) not >> understanding elementary mathematics, but your own writings on set >> theory >> and the axiomatic method don't fill me with confidence. > Ad hominem. His argument is ad hominem. Mine is just a cheap shot. >> 2) the current theory of `real numbers' is a >> joke, and sidesteps the crucial issue of understanding the >> computational >> specification of the continuum, and >> This is pure crank stuff. Describing a huge and extremely rigorously >> defined >> area such as the construction of the Reals as a joke without any >> mathematical justification is flakey at best; the phrase computational >> specification of the continuum (a phrase that gets exactly zero matches >> on >> Google) is crank babble. > The mathematical justification for describing the current theory of > real numbers as a joke is given in the paper. You may not find it > convincing - I may not find it convincing - but it's there. > Crank babble is ad hominem. Not ad hominem. I talk about his post, not him as a person. >> 3) `infinite sets' are a metaphysical >> concept, and unnecessary for correct mathematics. >> No, infinite sets are a mathematical concept, not unlike perfect circles >> and >> the exact value of the sqrt(2). >> Tell me, is the set of all natural numbers finite or infinite? Or if you >> can't form the set, why not? > I think Norm would say, 1) you can't form the set (and Norm's reasons > mathematics you can do with the completed infinite set that you can't > do without it. Ohh, so there's no good mathematics that can be done with the axiom of infinity that can't be done without. So infinite set theory is not good mathematics? How about computability theory and Turing Machines? Not good mathematics? >> What about the set of all points on a perfect circle (ie all solutions to >> x^2 + y^2 = 1). Finite or infinite? Or don't you believe that I can >> define a >> set as being all points on the unit circle. If not, why not? What is >> wrong >> with {(x,y) | x^2 + y^2 = 1} as a set? Infinite or finite? > Again, I think Norm is arguing against the completed infinite. For > sensibly write about the set of all natural numbers, or any of these > other sets that most mathematicians are quite happy with - moreover, > you don't lose anything valuable if you discard them. >> Your paper has no mathematical content, and is pure crank stuff. The >> stuff >> about Axioms somehow being irrelevant to mathematics is just your own >> philosophical ramblings. Surprising, since you seem to accept the axioms >> of >> group theory, but not set theory (because they are too complicated and >> too >> abstract for your liking)? Do groups with an infinite number of elements >> exist, by the way? > More ad hominem. Not ad hominem. I talk about his post, not him as a person. >Norm accepts the axioms of group theory as the > definition of what a group is, and has no problem with them because > he can construct (finite) models of them. He argues that the axioms > of set theory (in particular, ZFC) don't define what a set is and > don't lead to sensible constructions of infinite sets. >> I would have ignored this post - and previous posts on your mathematical >> insights - in much the same way that I ignore posts about Einstein was >> wrong or Cantor's diagonal proof is flawed - as pure crank material. >> The >> only reason I haven't, is that you are a mathematics teacher, and it >> worries >> me that somebody (eg your students) may be being taught this stuff about >> set >> theory. > I'm not sure how you think your method of not ignoring Norm's post > will prevent his students from being taught his ideas. Well, maybe if he attempted to justify his ideas, and decided he couldn't, maybe he would drop them. >> Tell me, do you accept that there are models for the ZF Axioms if we drop >> Axiom 6 ? Do you accept that there is a model for the ZF axioms if we >> include an additional axiom: >> Exists S such that { } is an element of S, and x elements of S implies x >> union {x) is an element of S ??? >> (informally known as the axiom of infinity) >> If you don't, I can certainly show you a model (the von-Neumann >> construction >> of N). >> How is this axiom fundamentally different from the other axioms? > It seems to me Norm is making several points, two of which are > that ZFC sucks and that mathematics doesn't need axioms in the first > place. Your suggestions may or may not have any bearing on the first > point, but they don't address the second. His paper is on ZFC. His philosophy that axioms are not needed is secondary. I also address this philosophy, by asking whether axioms are needed in group theory. > I'm dismayed by the level of vituperation in some of the posts in > this thread. Norm is not presenting a high-school algebra proof of > Fermat's Last Theorem, nor is he insisting that the reals are countable > because you can always take that real number that you left off your > list and stick it on at the end. He's adopting a finitistic, or > constructivist, or computational view of mathematics. It's an unpopular > view, it doesn't particularly appeal to me, but I don't see the need > to go ballistic in response. Because his papers that I have seen either display a profound lack of knowledge of set theory, or are deliberately false/misleading. The guy is a professor of mathematics. I want to know which it is. I know you must know him professionally, which is presumably why you are concerned. I would hope that he can defend his ideas and papers for himself. So lets go. These are neccessarily out of context, but I don't think this misrepresents his position: For each of these I would expect the statement is true, it is wrong out of Norman's ingorance (eg crank material), or it is wrong and he knows it is wrong (deliberately misleading): -------------------------- To get you used to the modern magic of Cauchy sequences, here is one I just made up: [2/3, 2/3, 2/3 ...] Anyone want to guess what the limit is? Oh, you want some more information first? The initial billion terms are all 2/3. Now would you like to guess? No, you want more information. All right, the billion and first term is 2/3 Now would you like to guess? No, you want more information. Fine, the next trillion terms are all 2/3 You are getting tired of asking for more information, so you want me to tell you the pattern once and for all? Ha Ha! Modern mathematics doesn't require it! There doesn't need to be a pattern, and in this case, there isn't, because I say so. You are getting tired of this game, so you guess Good effort, but sadly you are wrong. The actual answer is -17. That's right, after the first trillion and billion and one terms, the entries start doing really wild and crazy things, which I don't need to describe to you, and then `eventually' they start heading towards but how they do so and at what rate is not known by anyone. Isn't modern religion fun? --------------------------- Compare and contrast: To get you used to the modern magic of decimal notation, here is one I just made up: 0.666... Anyone want to guess what the limit is? Oh, you want some more information first? The initial billion digits are all 6. Now would you like to guess? No, you want more information. All right, the billion and first digit is 6. Now would you like to guess? No, you want more information. Fine, the next trillion terms are all 6. You are getting tired of asking for more information, so you want me to tell you the pattern once and for all? Ha Ha! Modern mathematics doesn't require it! There doesn't need to be a pattern, and in this case, there isn't, because I say so. You are getting tired of this game, so you guess Good effort, but sadly you are wrong. The actual answer is 2/3 - 10^billion. That's right, after the first trillion and billion and one terms, the entries start doing really wild and crazy things, which I don't need to describe to you. Isn't modern religion fun? These seem exactly the same argument to me. True, crank or deliberately misleading? How about: --------------------------- Now that you are comfortable with the definition of real numbers, perhaps you would like to know how to do arithmetic with them? How to add them, and multiply them? And perhaps you might want to check that once you have defined these operations, they obey the properties you would like, such as associativity etc. Well, all I can say is---good luck. If you write this all down coherently, you will certainly be the first to have done so. -------------------------- True, crank or deliberately misleading? On Cauchy sequences: -------------------- On top of the manifold ugliness and complexity of the situation, you will be continually dogged by the difficulty that in all these sequences there does not have to be a pattern---they are allowed to be completely `arbitrary'. That means you are unable to say when two given real numbers are the same, or when a particular arithmetical statement involving real numbers is correct. ---------------------- True, crank or deliberately misleading? ----------------------- A set of rational numbers is essentially a sequence of zeros and ones, and such a sequence is specified properly when you have a finite function or computer program which generates it. Otherwise `it' is not accessible in a finite universe. ------------------------- (My mind boggles at what he thinks he means here). True, crank or deliberately misleading? -------------------------- Even the `computable real numbers' are quite misunderstood. Most mathematicians reading this paper suffer from the impression that the `computable real numbers' are countable, and that they are not complete. As I mention in my recent book, this is quite wrong. Think clearly about the subject for a few days, and you will see that the computable real numbers are not countable , and are complete. ------------------------- Oh, my God. He has written a book on it. I hope its not part of the UNSW pure mathemtics syllabus. True, crank or deliberately misleading? He then goes on: ------------------------- Think for a few more days, and you will be able to see how to make these statements without any reference to `infinite sets', and that this suffices for Cantor's proof that not all irrational numbers are algebraic. ------------------------- Sorry, I am having a lot of trouble understanding how I can think about all irrational numbers without thinking about an infinite set. I have even more trouble comparing the cardinality of all computable numbers and all algebriac numbers without using set theoretic constructions. True, crank or deliberately misleading? ------------------------------------ In my studies of Lie theory, hypergroups and geometry, there has never been a point at which I have pondered---should I assume this postulate about the mathematical world, or that postulate? --------------------------------- Hmm. Never seen a Group theorem that assumes a Group is commutative? ----------------------------- but the nature of the mathematical world that I investigate appears to me to be absolutely fixed. Either G2 has an eleven dimensional irreducible representation or it doesn't (in fact it doesn't). ----------------------------- The guy is a group theory expert. Presumably he has heard of the Whitehead problem. http://en.wikipedia.org/wiki/Whitehead_problem True, crank or deliberately misleading? > -- > Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) I could go on, almost every sentence is a jewel of mininformation or ignorance. He walks like a crank, talks like a crank, and sqwaks like a crank. No problem, unless you also run around telling people you are a Professor of Mathematics. === Subject: Re: Set Theory: Should you believe? <44b5abd7$0$1207$afc38c87@news.optusnet.com.au> mathematics you can do with the completed infinite set that you can't > do without it. Norm is also opposed to axioms. Without axioms, how do we know when we are forming an infinte set? If I state Euclid's theorem on the infinitude of primes, am I forming a set? Am I forming a set just by referencing the integers at all? If Norm won't give a set of axioms he finds acceptable, we can't very well say that measureable cardinals contradict his foundations for mathematics, because he hasn't really given a foundation. He has, in fact, claimed that infinite sets are metaphysics; but if they are metaphysics, he's not talking mathematics at all, but metaphysics. In which case, so what? What do his metaphysical beliefs have to do with mathematics? > More ad hominem. Norm accepts the axioms of group theory as the > definition of what a group is, and has no problem with them because > he can construct (finite) models of them. In which case, he can hardly say he is rejecting axioms, and ought to step forward and say what his proposed axioms are. Would ditching the axiom of infinity do it? If not, what would? === Subject: Re: Set Theory: Should you believe? sender: DontBother@Nowhere.com On 12 Jul 2006 23:34:03 -0700, Gene Ward Smith >> I think Norm would say, 1) you can't form the set (and Norm's reasons >> mathematics you can do with the completed infinite set that you can't >> do without it. >Norm is also opposed to axioms. Isn't everyone? The only reason for axioms is that people are too lazy or stupid to demonstrate the truth of their assumptions. > Without axioms, how do we know when we >are forming an infinte set? If I state Euclid's theorem on the >infinitude of primes, am I forming a set? Am I forming a set just by >referencing the integers at all? If Norm won't give a set of axioms he >finds acceptable, we can't very well say that measureable cardinals >contradict his foundations for mathematics, because he hasn't really >given a foundation. He has, in fact, claimed that infinite sets are >metaphysics; but if they are metaphysics, he's not talking mathematics >at all, but metaphysics. In which case, so what? What do his >metaphysical beliefs have to do with mathematics? >> More ad hominem. Norm accepts the axioms of group theory as the >> definition of what a group is, and has no problem with them because >> he can construct (finite) models of them. >In which case, he can hardly say he is rejecting axioms, and ought to >step forward and say what his proposed axioms are. Would ditching the >axiom of infinity do it? If not, what would? ~v~~ === Subject: Re: Set Theory: Should you believe? > On 12 Jul 2006 23:34:03 -0700, Gene Ward Smith >> Norm is also opposed to axioms. > Isn't everyone? The only reason for axioms is that people are too lazy > or stupid to demonstrate the truth of their assumptions. How do you demonstrate that? -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Set Theory: Should you believe? On Thu, 13 Jul 2006 20:19:13 +0300, Aatu Koskensilta >> On 12 Jul 2006 23:34:03 -0700, Gene Ward Smith > Norm is also opposed to axioms. >> Isn't everyone? The only reason for axioms is that people are too lazy >> or stupid to demonstrate the truth of their assumptions >How do you demonstrate that? That people assume their assumptions? Axioms are assumed. Assumption is the alternative to demonstration. Assumption is not demonstration. If people are not too lazy or stupid to demonstrate theorems - since many theorems have been demonstrated - then the inference is they are too lazy or stupid to demonstrate axioms by conversion of not not too lazy or stupid. ~v~~ === Subject: Re: Set Theory: Should you believe? > On Thu, 13 Jul 2006 20:19:13 +0300, Aatu Koskensilta > Isn't everyone? The only reason for axioms is that people are too lazy > or stupid to demonstrate the truth of their assumptions >> How do you demonstrate that? > That people assume their assumptions? Axioms are assumed. I'm not convinced. Please demonstrate that. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Set Theory: Should you believe? > I think Norm would say, 1) you can't form the set (and Norm's reasons > mathematics you can do with the completed infinite set that you can't > do without it. > Norm is also opposed to axioms. Without axioms, how do we know when we > are forming an infinte set? If I state Euclid's theorem on the > infinitude of primes, am I forming a set? Am I forming a set just by > referencing the integers at all? If Norm won't give a set of axioms he > finds acceptable, we can't very well say that measureable cardinals > contradict his foundations for mathematics, because he hasn't really > given a foundation. He has, in fact, claimed that infinite sets are > metaphysics; but if they are metaphysics, he's not talking mathematics > at all, but metaphysics. In which case, so what? What do his > metaphysical beliefs have to do with mathematics? All these questions are better directed to Norm than to me, but I'll make believe I know what he is on about, and answer thus: If I remember right, Euclid never said there's an infinitude of primes. He just said, given any prime, there's a bigger one. You and I are accustomed to interpreting the second as meaning the same thing as the first, but I think that until quite recent times, mathematicians didn't. They didn't accept the completed infinity, and they were still able to develop the theory of numbers, prove the quadratic reciprocity theorem, the four squares theorem, etc. If you state the theorem the way Euclid did, you are not forming an infinite set, and you can get on perfectly well that way. As for measureable cardinals, I'm guessing the question of their existence doesn't interest Norm one way or the other, on the grounds that the stability or otherwise of the Sydney Harbor Bridge is unlikely to depend on the outcome. The real world questions mathematics sprang from do not depend on these abstractions, nor on the axiom systems in which they are debated. > More ad hominem. Norm accepts the axioms of group theory as the > definition of what a group is, and has no problem with them because > he can construct (finite) models of them. > In which case, he can hardly say he is rejecting axioms, and ought to > step forward and say what his proposed axioms are. Would ditching the > axiom of infinity do it? If not, what would? I don't know, and if you really want to know, you could try asking him. But perhaps he is only saying, the axioms of group theory define an interesting set of structures, which I can construct, and which help me answer questions about physics, while the axioms of set theory only help me study set theory, which is not where the real value of mathematics is. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Set Theory: Should you believe? <44b5abd7$0$1207$afc38c87@news.optusnet.com.au> I think Norm would say, 1) you can't form the set (and Norm's reasons >mathematics you can do with the completed infinite set that you can't >do without it. > Norm is also opposed to axioms. Without axioms, how do we know when we > are forming an infinte set? If I state Euclid's theorem on the > infinitude of primes, am I forming a set? Am I forming a set just by > referencing the integers at all? If Norm won't give a set of axioms he > finds acceptable, we can't very well say that measureable cardinals > contradict his foundations for mathematics, because he hasn't really > given a foundation. He has, in fact, claimed that infinite sets are > metaphysics; but if they are metaphysics, he's not talking mathematics > at all, but metaphysics. In which case, so what? What do his > metaphysical beliefs have to do with mathematics? > All these questions are better directed to Norm than to me, > but I'll make believe I know what he is on about, and answer > thus: > If I remember right, Euclid never said there's an infinitude of > primes. He just said, given any prime, there's a bigger one. > You and I are accustomed to interpreting the second as meaning > the same thing as the first, but I think that until quite recent > times, mathematicians didn't. They didn't accept the completed > infinity, and they were still able to develop the theory of > numbers, prove the quadratic reciprocity theorem, the four squares > theorem, etc. If you state the theorem the way Euclid did, > you are not forming an infinite set, and you can get on perfectly > well that way. > [...] Well, the Greeks were not able to resolve Zeno's paradoxes, were they? The point is that the Greek way of thinking leads to a paradox that Achilles can never catch up with a tortoise (moving ahead of him in a straight line at a slower velocity). This is the problem with the Greek concept of potential infinity -- which will not permit Achilles to complete the infinitely many conscious acts needed to catch up with the tortoise (first Achilles has to reach a point where the tortoise was previuously located, but now has moved ahead -- this step repeats itself ad infinitum). I don't think classical real analysis resolves Zeno's paradoxes with the concept of limits either -- this leads to the issue of how infinitely many finite, non-zero and non-infinitesimal intervals of reals can sum to a finite interval. Norman Wildberger, like the Greeks, has the problem of explaining what he means by an arbitrary natural number or prime number, etc. For example, when NW says that some proposition holds for an arbitrary natural number, is that not an unverifiable assertion about infinitely many naturals that cannot be proven unless one makes postulates (or axioms) to which he is so opposed? And such axioms *must* take for granted the existence of an infinite class of natural numbers for which the said proposition holds -- otherwise we don't have a proof of this proposition. And NW does oppose the concept of an arbitrary real number-- here is a quote from his paper Set theory: should you believe? at : But here is a very important point: we are not obliged, in modern mathematics, to actually have a rule or algorithm that specifies the sequence r1, r2, r3, .87 .87 .87 . In other words, 'arbitrary' sequences are allowed, as long as they have the Cauchy convergence property. This removes the obligation to specify concretely the objects which you are talking about. Sequences generated by algorithms can be specified by those algorithms, but what possibly could it mean to discuss a 'sequence' which is not generated by such a finite rule? Such an object would contain an 'infinite amount' of information, and there are no concrete examples of such things in the known universe. This is metaphysics masquerading as mathematics. I agree with NW's conclusion that an arbitrary real number, in the classical sense, is problematic. But definitely not with his stated reason for this conclusion -- that such an arbitrary real number (not specified by a finite rule) contains an infinite amount of information. The fact is that even, say, a Cauchy sequence converging to Pi, specified by a finite rule, does contain an infinite amount of information -- no human mind can complete the task of running through all the terms of this sequence generated by this finite rule. And NW will have to explain how he can accept statements like For any real number x>0, (1/x)>0, if he does not accept the concept of an arbitrary Cauchy sequence of rationals. For x is an arbitrary real number, is it not? My point of view, explained in my arxiv paper , is that the problem is not with the existence of an inifnite class of naturals -- one can accept that an arbitrary prime can only be defined as that belonging to an infinite class of primes, for example. What consititutes infinitary reasoning (in my view) is quantifying over these infinite classes, i.e., formally referring to infinitely many such infinite classes in a single formula. Thus an arbitrary real number x, in the classical sense, is not allowed in my proposed logic NAFL-- because x can only be defined as belonging to an infinite class of reals, which requires quantification over reals -- not permitted in NAFL because each real is an infinite object. Infinite sets do not exist in NAFL, but infinite classes can exist (in fact MUST exist whenever the infinitely many finite objects belonging to that class exist). Infinite classes, like real numbers, are proper classes-- so the reals do not consititute a class and you can't quantify over reals in NAFL. I believe that the correct resolution of Zeno's paradoxes is in the version of real analysis proposed in the above-cited arxiv paper . In particular, see Sec. 4, which is more or less self-contained. === Subject: Re: Set Theory: Should you believe? <44b5abd7$0$1207$afc38c87@news.optusnet.com.au> asking him. But perhaps he is only saying, the axioms > of group theory define an interesting set of structures, which > I can construct, and which help me answer questions about physics, > while the axioms of set theory only help me study set theory, > which is not where the real value of mathematics is. The Heine-Borel theorem is awfully useful for developing real analysis. Is it a good thing or a bad one in Norm's view to have a system strong enough to prove it? How the hell can anyone know, unless he bites the bullet and says what he thinks you can assume? === Subject: Re: Set Theory: Should you believe? <44b5abd7$0$1207$afc38c87@news.optusnet.com.au> this thread. Norm starts out his paper, which I didn't read because the beginning was so extremely unpromising, in what seems to me to be a very insulting way. If he has ideas he wants to be taken seriously I suggest he remove the sneers directed at set theorists, who apparently are beneath contempt, and wild remarks about physics and the like. Present a reasoned argument in a reasonable way and people are likely to react more positively, and less likely to conclude that you are an idiot and simply quit reading. > He's adopting a finitistic, or > constructivist, or computational view of mathematics. He's also spitting on people who don't. I think it is terribly arrogant to dismiss people like Shelah or Woodin with such utter contempt like this, and I didn't see any signs, as far as I had gotten, that he even knows anything about modern set theory. Does he? === Subject: Re: Set Theory: Should you believe? <44b5abd7$0$1207$afc38c87@news.optusnet.com.au> I'm dismayed by the level of vituperation in some of the posts in > this thread. > Norm starts out his paper, which I didn't read because the beginning > was so extremely unpromising, in what seems to me to be a very > insulting way. If he has ideas he wants to be taken seriously I suggest > he remove the sneers directed at set theorists, who apparently are > beneath contempt, and wild remarks about physics and the like. Present > a reasoned argument in a reasonable way and people are likely to react > more positively, and less likely to conclude that you are an idiot and > simply quit reading. > He's adopting a finitistic, or > constructivist, or computational view of mathematics. > He's also spitting on people who don't. I think it is terribly arrogant > to dismiss people like > Shelah or Woodin with such utter contempt like this, Hugh Woodin is a con-artist/leech who needs to get a real job. Ideally, what we need to do with people like Woodin, is haul them into the dock for public hearings. Where did that $1 million in federal grants you sucked down in the last 15 years go, Mr. Woodin? What were the practical spin-offs? Why should we fund you, as opposed to someone working on real-world problems, like bird flu etc.? What practical benefit does your research have? We're going to need an explanation, Mr. Woodin, otherwise we can't sign the check. We simply can't fund research which has NO practical applications. It's worse than the toilet seat scandals. When the government buys a toilet seat for $1 million, at least you get the toilet seat. When you give Hugh Woodin $1 million, you get nothing of practical value, not even a toilet seat. That public money should be rerouted to people doing work which actually benefits society. If you wanna do theology, do it on your own dime. === Subject: Re: Set Theory: Should you believe? >> I'm dismayed by the level of vituperation in some of the posts in >> this thread. > Norm starts out his paper, which I didn't read because the beginning > was so extremely unpromising, in what seems to me to be a very > insulting way. If he has ideas he wants to be taken seriously I suggest > he remove the sneers directed at set theorists, who apparently are > beneath contempt, and wild remarks about physics and the like. Present > a reasoned argument in a reasonable way and people are likely to react > more positively, and less likely to conclude that you are an idiot and > simply quit reading. >> He's adopting a finitistic, or >> constructivist, or computational view of mathematics. > He's also spitting on people who don't. I think it is terribly arrogant > to dismiss people like > Shelah or Woodin with such utter contempt like this, and I didn't see > any signs, as far as I had gotten, that he even knows anything about > modern set theory. Does he? I think he is all too familiar with modern set theorists. Set theorists have written the book on how to treat others with contempt. Most branches of mathematics will accept any reasonable proof. Set theorists demand proofs in set theory. This is like the Catholic Church requiring Mass be given in Latin. It is a method of guaranteeing only the priests (the true believers) know the Church's doctrines. It is designed to prevent skeptics (non-believers) from being able to question Church doctrine, since you need to know a dead language to have any idea what that doctrine is. Imagine if Einstein had been told he had to prove relativity in Newtonian physics before anyone would consider his ideas. Russell - 2 many 2 count === Subject: Re: Set Theory: Should you believe? >> I'm dismayed by the level of vituperation in some of the posts in >> this thread. > Norm starts out his paper, which I didn't read because the beginning > was so extremely unpromising, in what seems to me to be a very > insulting way. If he has ideas he wants to be taken seriously I suggest > he remove the sneers directed at set theorists, who apparently are > beneath contempt, and wild remarks about physics and the like. Present > a reasoned argument in a reasonable way and people are likely to react > more positively, and less likely to conclude that you are an idiot and > simply quit reading. >> He's adopting a finitistic, or >> constructivist, or computational view of mathematics. > He's also spitting on people who don't. I think it is terribly arrogant > to dismiss people like > Shelah or Woodin with such utter contempt like this, and I didn't see > any signs, as far as I had gotten, that he even knows anything about > modern set theory. Does he? > I think he is all too familiar with modern set theorists. > Set theorists have written the book on how to treat > others with contempt. Most of the contempt is reserved for those whose criticisms exhibit profound ignorance of what they are criticizing and those whose criticisms are so contemptuous as to inspire countercontempt. > Most branches of mathematics will accept any reasonable proof. > Set theorists demand proofs in set theory. They will accept reasonable proofs of reasonable claims but require extraordinary, or at least rigorous, proofs of extraordinary claims. And their standard of judging other's proof is no stricter than that they apply to their own proofs. > This is like the Catholic Church requiring Mass be given in Latin. > It is a method of guaranteeing only the priests (the true > believers) know the Church's doctrines. It may seem like that to those unfamiliar with set theory. Any speciality tends to look arcane to those outside it. > It is designed to prevent skeptics (non-believers) from being > able to question Church doctrine, since you need to know > a dead language to have any idea what that doctrine is. This is precisely the attitude of those like Russell who dump on set theorists which inspires the set theorists to dump back. If the technicalities of set theory were as easy to learn as Russell seems to think it ought to be then everyone would learn it in grade school. In fact, it, like many specialities, usually takes years of study for one to become really good at it. Russell seems to expect a Set Theory for Dummies short course which will bring him up to PhD levels in an afternoons reading. === Subject: Re: Set Theory: Should you believe? > If the technicalities of set theory were as easy to learn as Russell > seems to think it ought to be then everyone would learn it in grade > school. In fact, it, like many specialities, usually takes years of > study for one to become really good at it. A good number of people can master concepts of mathematics sufficiently to solve difficult problems in, say, fluid mechanics without much grasp of set theory. That makes me wonder if set theory really is fundamental to mathematics. I ask what it is that underlies the pragmatic application of sophisticated mathematics. What are the intuitive assumptions these people have made, and how are they manipulating ideas? I believe what I'm asking is, what are the anthropological foundations of mathematics? -- Nil conscire sibi === Subject: Re: Set Theory: Should you believe? sender: Lester Zick On Thu, 13 Jul 2006 09:33:58 -0400, Hatto von Aquitanien >> If the technicalities of set theory were as easy to learn as Russell >> seems to think it ought to be then everyone would learn it in grade >> school. In fact, it, like many specialities, usually takes years of >> study for one to become really good at it. >A good number of people can master concepts of mathematics sufficiently to >solve difficult problems in, say, fluid mechanics without much grasp of set >theory. That makes me wonder if set theory really is fundamental to >mathematics. I believe set theory is essential to modern math. > I ask what it is that underlies the pragmatic application of >sophisticated mathematics. What are the intuitive assumptions these people >have made, and how are they manipulating ideas? I believe what I'm asking >is, what are the anthropological foundations of mathematics? Not exactly the same question. The historical or anthropological foundations of math are certainly interesting. But the more interesting question is why people are using intuitive assumptions in math at all? ~v~~ === Subject: Re: Set Theory: Should you believe? > On Thu, 13 Jul 2006 09:33:58 -0400, Hatto von Aquitanien >>A good number of people can master concepts of mathematics sufficiently to >>solve difficult problems in, say, fluid mechanics without much grasp of >>set >>theory. That makes me wonder if set theory really is fundamental to >>mathematics. > I believe set theory is essential to modern math. What can you do with set theory which you can't do with symbolic logic and/or some kind of BNF-like grammar? > Not exactly the same question. The historical or anthropological > foundations of math are certainly interesting. But the more > interesting question is why people are using intuitive assumptions in > math at all? > ~v~~ Because without intuitive assumptions they would have absolutely not concept of existence. That would make doing mathematics rather difficult. My point is that if one were able to identify and codify these assumptions, it may be possible to establish a good formal foundation that way. -- Nil conscire sibi === Subject: Re: Set Theory: Should you believe? >>If the technicalities of set theory were as easy to learn as Russell >>seems to think it ought to be then everyone would learn it in grade >>school. In fact, it, like many specialities, usually takes years of >>study for one to become really good at it. > A good number of people can master concepts of mathematics sufficiently to > solve difficult problems in, say, fluid mechanics without much grasp of set > theory. That makes me wonder if set theory really is fundamental to > mathematics. I ask what it is that underlies the pragmatic application of > sophisticated mathematics. What are the intuitive assumptions these people > have made, and how are they manipulating ideas? I believe what I'm asking > is, what are the anthropological foundations of mathematics? I have (tried) working in fluid dynamics for a number of years, and certainly some sort of naive set theory is indispensible. I already indicated this in another post where I talked about problems in showing which differential equations have unique solutions and which don't. Basically anything beyond numerical simulations or solving simple laminar flows requires sophisticated techniques that involve objects much more abstract than the real numbers (namely things like measure spaces and Besov spaces and things like that). A good example is the result of Caffarelli, Kohn and Nirenberg which looks at the maximal possible Hausdorff dimension of the set of singularities in the solution to the Navier-Stokes equation. Stephen === Subject: Re: Set Theory: Should you believe? <44b5abd7$0$1207$afc38c87@news.optusnet.com.au> laminar flows requires sophisticated techniques that involve objects > much more abstract than the real numbers (namely things like measure > spaces and Besov spaces and things like that). If you want a really scary thought, try this one--the large cardinal axioms set theorists like to ponder have implications for measure theory and descriptive set theory. === Subject: Re: Set Theory: Should you believe? <44b5abd7$0$1207$afc38c87@news.optusnet.com.au> Set theorists demand proofs in set theory. What the hell does this mean, if anything? === Subject: Re: Set Theory: Should you believe? >> Most branches of mathematics will accept any reasonable proof. >> Set theorists demand proofs in set theory. > What the hell does this mean, if anything? It means Russell has repeatedly failed to prove anything. :) Stephen === Subject: Re: Set Theory: Should you believe? > I didn't see any signs, as far as I had gotten, that he even knows > anything about modern set theory. Does he? I don't know. I reject astrology, even though I don't know anything about modern astrology (I don't even know if there is such a thing). I reject creation science and intelligent design, even though I haven't read any recent writings of their advocates. I don't have to; I know where they're going, and I know they're never going to get anywhere useful, going in that direction. I personally don't put set theory in the same category as astrology or creation science. Maybe Norm does. I don't know. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Set Theory: Should you believe? >> I didn't see any signs, as far as I had gotten, that he even knows >> anything about modern set theory. Does he? > I don't know. > I reject astrology, even though I don't know anything about modern > astrology (I don't even know if there is such a thing). I reject > creation science and intelligent design, even though I haven't > read any recent writings of their advocates. I don't have to; I > know where they're going, and I know they're never going to get > anywhere useful, going in that direction. > I personally don't put set theory in the same category as astrology > or creation science. Doesn't this undermine your whole analogy? Why didn't you pick an orthodox theory like Evolution, Special Realtivity or Plate Techtonics as being the theory he is attacking? (Set theory is every bit as well accepted as any of these other topics). Because he looks less of a crank if you compare him to attacking astrology than him attacking (say) the Theory of Evolution, even though this is a much closer analogy? >Maybe Norm does. I don't know. > -- > Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) I don't see you publishing any papers on astrology. Nor do you sign your posts as a Professor of Astronomy. And finally, I don't see you saying Astronomers and cosmologists are welcome to send me reasoned responses., as if your level of knowledge of astrology was so advanced you didn't think non-specialists should be able to respond. Doesn't it worry you that a professional mathematician can write a paper on set theory (that has caused a bit of discussion in some logic circles) and you can't tell from the paper if he actually knows anything about modern set theory ? === Subject: Re: Set Theory: Should you believe? <44b5abd7$0$1207$afc38c87@news.optusnet.com.au> or creation science. Maybe Norm does. I don't know. Norm apparently puts number theory in that category. === Subject: Re: Set Theory: Should you believe? >> I personally don't put set theory in the same category as astrology >> or creation science. Maybe Norm does. I don't know. >Norm apparently puts number theory in that category. Quite right, too. We have, on at least as good authority as that which either the astrolgers or creation scientists have for their belief systems, that God created both the firmament (stars included) and the integers. Then there's that bit where Jesus is asserted to be both a generic cardinal *and* the first infinite ordinal. Lee Rudolph === Subject: Re: Set Theory: Should you believe? > Then there's that bit where Jesus is asserted to be both a generic cardinal > *and* the first infinite ordinal. But most Christians don't find him quite inaccessible, right? Remarkable, subtle and strong, perhaps, somewhat ineffable and indescribable at times - though probably not totally indescribable. I think I can see a set theoretical proof of the non-existence of Jesus. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Set Theory: Should you believe? sender: Lester Zick On Thu, 13 Jul 2006 12:11:34 +1000, Peter Webb > http://web.maths.unsw.edu.au/~norman/views.htm > that has caused a bit of discussion in some logic circles. >Well, the discussion that I have seen - on this newsgroup (sci.math or >sci.logic, I can't remember) - is that it is bull. Much of it admittedly >from me. > My claims in short: 1) most of `elementary mathematics' is not > sufficiently > well understood by the mathematical establishment, leading to weaknesses > in > K12 and college curriculum, >I don't know about the mathematical establishment (as a whole) not >understanding elementary mathematics, but your own writings on set theory >and the axiomatic method don't fill me with confidence. >2) the current theory of `real numbers' is a > joke, and sidesteps the crucial issue of understanding the computational > specification of the continuum, and >This is pure crank stuff. Describing a huge and extremely rigorously defined >area such as the construction of the Reals as a joke without any >mathematical justification is flakey at best; the phrase computational >specification of the continuum (a phrase that gets exactly zero matches on >Google) is crank babble. >3) `infinite sets' are a metaphysical > concept, and unnecessary for correct mathematics. >No, infinite sets are a mathematical concept, not unlike perfect circles and >the exact value of the sqrt(2). >Tell me, is the set of all natural numbers finite or infinite? Or if you >can't form the set, why not? >What about the set of all points on a perfect circle (ie all solutions to >x^2 + y^2 = 1). Well technically of course all solutions you point out define a perfect sphere not a perfect circle. > Finite or infinite? Or don't you believe that I can define a >set as being all points on the unit circle. If not, why not? What is wrong >with {(x,y) | x^2 + y^2 = 1} as a set? Infinite or finite? >Your paper has no mathematical content, and is pure crank stuff. The stuff >about Axioms somehow being irrelevant to mathematics is just your own >philosophical ramblings. Surprising, since you seem to accept the axioms of >group theory, but not set theory (because they are too complicated and too >abstract for your liking)? Do groups with an infinite number of elements >exist, by the way? >I would have ignored this post - and previous posts on your mathematical >insights - in much the same way that I ignore posts about Einstein was >wrong or Cantor's diagonal proof is flawed - as pure crank material. The >only reason I haven't, is that you are a mathematics teacher, and it worries >me that somebody (eg your students) may be being taught this stuff about set >theory. >Tell me, do you accept that there are models for the ZF Axioms if we drop >Axiom 6 ? Do you accept that there is a model for the ZF axioms if we >include an additional axiom: >Exists S such that { } is an element of S, and x elements of S implies x >union {x) is an element of S ??? >(informally known as the axiom of infinity) >If you don't, I can certainly show you a model (the von-Neumann construction >of N). >How is this axiom fundamentally different from the other axioms? > Analysts and set theorists are welcome to send me reasoned responses. > Assoc Prof N J Wildberger > School of Maths > UNSW >> I've added sci.logic to your posting. karl m ~v~~ === Subject: Re: Set Theory: Should you believe? >What about the set of all points on a perfect circle (ie all solutions to >x^2 + y^2 = 1). > Well technically of course all solutions you point out define a > perfect sphere not a perfect circle. With only two variables, x and y, there is no need to presume three dimensions, but if one does, one gets a right circular cylinder, not a sphere. If one doesn't assume three (or more) dimensions, one does get a circle. === Subject: Re: Set Theory: Should you believe? sender: Lester Zick >What about the set of all points on a perfect circle (ie all solutions to >x^2 + y^2 = 1). >> Well technically of course all solutions you point out define a >> perfect sphere not a perfect circle. >With only two variables, x and y, there is no need to presume three >dimensions, but if one does, one gets a right circular cylinder, not a >sphere. And as long as one gets to assume whatever one wants one gets whatever one wants to assume. There is no unambiguous right angle dimension to any other when fewer than three dimensions are involved. If you're talking about the set of all points equidistant from any other the figure is a sphere. >If one doesn't assume three (or more) dimensions, one does get a circle. Well the basic problem here is that it's a little difficult to discern what the author is complaining about in set theory. Sets of properties are perfectly useful. However typical set theory definitions which run along the lines of a set of all points which . . . do turn out to be a joke because they invariably rely on various geometric assumptions regarding figures such as planes, lines, etc. ~v~~ === Subject: Re: Set Theory: Should you believe? > Well the basic problem here is that it's a little difficult to discern > what the author is complaining about in set theory. Sets of properties > are perfectly useful. However typical set theory definitions which run > along the lines of a set of all points which . . . do turn out to be > a joke because they invariably rely on various geometric assumptions > regarding figures such as planes, lines, etc. > ~v~~ My suspicion is that the objection is to the idea of applying operations which are valid for finite sets to infinite sets and arriving at conclusions which challenge conventional wisdom. For example it was recently pointed out that permutations of N result in a demonstration that there exist uncountable bijections. Another example is Cantor's theory of transfinite induction. One can reasonably ask if any of these findings are meaningful. I'm told by some that the L.9awenheim-Skolem theorem will do all kinds of wonderful things for me. The theorem depends upon Cantor's findings. Now, if I take the time to truly understand this stuff, will I simply be led down the primrose path? -- Nil conscire sibi === Subject: Re: Set Theory: Should you believe? > For example it was recently pointed out that permutations of N result in a demonstration that > there exist uncountable bijections. Really? Where can I find out more about this startling discovery? -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Set Theory: Should you believe? <44b5abd7$0$1207$afc38c87@news.optusnet.com.au> For example it was recently pointed out that >> permutations of N result in a demonstration that >> there exist uncountable bijections. > Really? Where can I find out more about this > startling discovery? The identity map on the set of all permutations of N is an example of an uncountable permutation, but I'm not sure if this is what Hatto von Aquitanien had in mind. L. Renfro === Subject: Re: Set Theory: Should you believe? >> For example it was recently pointed out that permutations of N result in a demonstration that >> there exist uncountable bijections. > Really? Where can I find out more about this startling discovery? The above is Aquitanien's garbled interpretation of the fact that there exist countable sets that are not recursively enumerable. Stephen === Subject: Re: Set Theory: Should you believe? > For example it was recently pointed out that permutations of N result in > a demonstration that there exist uncountable bijections. >> Really? Where can I find out more about this startling discovery? > The above is Aquitanien's garbled interpretation of the fact > that there exist countable sets that are not recursively > enumerable. > Stephen If you cannot recursively enumerate it, you can't count it. -- Nil conscire sibi === Subject: Re: Set Theory: Should you believe? >> For example it was recently pointed out that permutations of N result >> in a demonstration that there exist uncountable bijections. > Really? Where can I find out more about this startling discovery? >> The above is Aquitanien's garbled interpretation of the fact >> that there exist countable sets that are not recursively >> enumerable. >> Stephen > If you cannot recursively enumerate it, you can't count it. I guess that may not be completely accurate. As long as I know that every location has a unique element I can count the elements and not know which one I am counting. I stand corrected. -- Nil conscire sibi === Subject: Re: Set Theory: Should you believe? On Thu, 13 Jul 2006 08:37:51 -0700, Norman Wildberger fed this fish to the penguins: >that has caused a bit of discussion in some logic circles. >My claims in short: 1) most of `elementary mathematics' is not sufficiently >well understood by the mathematical establishment, leading to weaknesses in >K12 and college curriculum, 2) the current theory of `real numbers' is a >joke, and sidesteps the crucial issue of understanding the computational >specification of the continuum, and 3) `infinite sets' are a metaphysical >concept, and unnecessary for correct mathematics. >Analysts and set theorists are welcome to send me reasoned responses. I have read it all and I have survived. I will not offer any reasoned some points your position is tenable and understandable, there are just too many blunders, confusions and errors to try to answer it. G. Rodrigues === Subject: Re: Absolutely continuous, L^2 question >> <5036829.1152728209310.JavaMail.jakarta@nitrogen.mathf >> orum.org>, > <1820258.1152720081570.JavaMail.jakarta@nitrogen.mathf > orum.org>, > >On Wed, 12 Jul 2006 09:00:56 EDT, James >This question has been killing me. Please >> help > with >any insight: >Let f be absolutely continuous on [0,x] for >> all > x >> 0. Let f, f ' be in L^2[0,oo). Let f(0) = >> 0. >Prove that >lim f(x) = 0 >>x --> oo >There was a part (a) to this problem that I >> got > : >Prove that int_[0,x] |f f '| <= .5* >> (int_[0,x] |f > '| >)^2. But I don't see how this helps for part > (b). >I have tried several ways to prove that lim >> f(x) > = 0 >as x --> oo. >1) I have proven that if f is in L^1[0,oo) >> and f > is >uniformly continuous, then lim f(x) = 0 as x >> -->oo. Well, that fact that f' is in L^2 shows that >> f is >uniformly >continuous: |f(x+h) - f(x)| = ______ <= _______, which tends to 0 as h -> 0, uniformly in x. Are you asserting that f is uniformly >> continuous on > all of [0,oo)? Your >argument shows that f is uniformly continuous >> on > [0,x] since f is AC on >[0,x]. > > I don't think you know what his argument is. >> (Yes, f > is UC on > [0,oo).) > > > I don't see it. You say there is a universal delta >> on all of [0,oo)? > > |f(x+h) - f(x)| = |int_[x,x+h] f' | <= ||f'||_2 * >> h^(1/2) ---> 0 as h goes to > 0. But to justify |f(x+h) - f(x)| = |int_[x,x+h] >> f' | you need to say that f > is AC on [0,a] where a is greater than x+h. So you >> have that f is uniformly > continuous on [0,a]. The fact that |f(x+h) - f(x)| >> --> 0 as h ---> 0 for all > x in [0,a] doesn't give you what you want. If you >> want to say that f is > uniformly continuous beyond a, then when you write >> |f(x+h) - f(x)| ---> 0 as > h --> 0 (uniformly in x), this means that for all >> eps > 0 there is an s > 0 > (i.e. delta > 0) with |f(x+h) - f(x)| < eps for all >> h < s for all x in [0,a]. > But what I am saying is that if you want to check >> uniform continuity at y > > a, I think that this delta (or s) changes. > > In any case, whether or not my babbling above makes >> any sense, please share > with me how you are getting your universal delta >> for the uniform continuity > of f on all of[0,oo). >> If 0 <= x < y, you have |f(y) - f(x)| = |int_[x,y] f' >> | <= >> [int_[x,y] |f'|^2)^(1/2)]*|y-x|^(1/2) <= [int_[0,oo) >> |f'|^2)^(1/2)]*|y-x|^(1/2). So there is a constant C >> such that >> |f(y) - f(x)| <= C*|y-x|^(1/2) for all x, y in >> [0,oo). >It's really the first part of your response that I am having trouble with. You say >If 0 <= x < y, you have |f(y) - f(x)| = |int_[x,y] f' | >It seems to me that what you are basically saying is that f is absolutely continuous on all [0,oo). In order to say that |f(y) - f(x)| = |int_[x,y] f' |, you need to first say that f(y) = int_[0,y] f' and f(x) = int[0,x] f'. In order to say those two things you need to say that f is absolutely continuous on [0,z] where z is greater than y and x. So ok, given x and y, there is a z that makes this work. Right. So what the heck is the problem? > If you pick x' and y', then there is a z' that makes this work. So it changes. So what? It's still true that for any x and y f(y) - f(x) = int_x^y f'(t) dt, because f is AC on [x,y]. How does the fact that if you change x and y then you change x and y affect this? >I am missing a logical step here. Side question : If f is absolutely continuous on [0,x] for all x > 0, then does this imply that f is absolutely continuous on [0,oo)? No. And that doesn't matter one bit - all we use above is the fact that it's AC on [x,y]. >James ************************ David C. Ullrich === Subject: Re: Infinite Induction and the Limits of Curves > Mike Kelly said: >Then you'll >agree with this logic : >> T_n = { naturals up to n } >S_n = {even naturals up to n } >> T_oo = { natural numbers } >S_oo = { even natural numbers } >> For all n, the number of elements in T_n is exactly equal to the number >of elements in S_n, and hence, taking the limit as n -> oo, we conclude >that the number of elements in T_oo equals the number of elements in >S_oo. Thus the naturals and the evens are equinumerous. > The first sentence starts out with a false premise, yet you arrive at the >> standard conclusion. >> For all n, the number of elements in T_n is exactly equal to the >number of elements in S_n? >> Choose any standard natural number n. >The number of elements in T_n is n. >The number of elements in S_n is n. >n = n >So the number of elements in T_n is exactly equal to the number of >elements in S_n. > What are you on? For any n in N T_n has n elements and S_n has >floor(n/2) elements. You think there are as many even naturals between 1 >and 10 as there are naturals between 1 and 10? Go back to kindergarten. > >What is wrong with that? Which steps do you agree with and which do you >disagree with? >> Maybe the problem was that when I said for all n I should have >explicitly stated for all standard natural numbers n? > Maybe you need to define what you mean by up to n. That would >generally be read as some type of value range. Did you mean it as an >index? If so, say the first n of the... and it will be more clear. > I should have said Up to the nth. Sorry if that was unclear. You seem > to have figured out what I intended to say, though... > Let's start again : > T_n = { naturals up to the nth natural } > S_n = { even naturals up to the nth even natural } > T_3 = {1, 2, 3} - 3 elements > S_3 = {2, 4, 6} - 3 elements > T_n = {1, 2, 3, 4, ..., n} - n elements > S_n = {2, 4, 6, 8, ..., 2n} - n elements > T_oo = { All naturals } > S_oo = { All even naturals } > Note that the value range for every S_n is twice that for the corresponding > T_n. When you include ALL of them, to the elements of S_n still have twice the > value range of T_n, or do they both now share the same value range, that of N? > This is where standard theory errs. What definition of value range are you using? To what sequences can it be applied? > Now do you agree that the inductively proven equality between element > count of the set T and the element count of the set S which holds for > all n also holds in the infinite case? If not, why not? > If it does, then the value ranges are also different for T_oo and S_oo, but > both are taken to cover the range of all naturals. If you measure the sets > within any value range for the elements, T_n will have twice as many elements > as S_n. What does it mean to measure a set within a value range for the elements? > For all n, the number of elements in T_n is exactly equal to the number > of elements in S_n, and hence, taking the limit as n -> oo, we conclude > that the number of elements in T_oo equals the number of elements in > S_oo. Thus the naturals and the evens are equinumerous. > That would be a nice proof if you also included the part about S_oo having > twice the range of T_oo, but then you'd run into the contradiction that T_oo is > only half complete. What is the range of S_oo? What is the range of T_oo? In what sense is one twice the other? What does half complete mean? > ---- > What about the sets > X_n = { 2 * 1, 2 * 2, 2 * 3, 2 * 4, ..., 2 * n } - n elements > Y_n = { 17 * 1, 17 * 2, 17 * 3, 17 * 4, .... 17 * n } - n elements > A_n = { 1, 2, 3, 4, ..., n } - n elements > B_n = { 2, 4, 6, 8, ..., 2n } - n elements > X_3 = { 2, 4, 6 } - 3 elements > Y_n = { 17, 34, 51 } - 3 elements > A_n = { 1, 2, 3 } - 3 elements > B_n = { 2, 4, 6, } - 3 elements > X_oo = { All even naturals } > Y_oo = { All naturals divisible by 17 } > A_oo = { All strings which represent naturals } > B_oo = { All strings which represent even naturals } > Do these all have the same element count up to the nth element? What > about in the infinite case? > Up to the nth element, yes, but over the same entire value range, no. What does the same entire value range mean? Consider sequences of the form S_n = { 2, 4, 6, 8, ...., 2n} R_n = { 1/3, 1/6, 1/9, 1/12, ... 1/3n } Do these have the same element count up to the nth element? What about in the infinite case? What does value range mean here? -- mike. === Subject: Re: Is {} and element of all sets? (some helpful stuff that understood) Did the OP find my reference to the power set of any use? (and) > -- > It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) > Arturo Magidin > magidin@math.berkeley.edu I don't know about Hobbes, but Calvin is clearly a minimifidianist, as am I. The Zen folks remind us that suffering is optional, and to Believe Nothing. Why believe if you can know instead? Good question, lost on many folks. The other two points of the Zen book I am studying are Practice Awareness and Don't Take It Personally. These three points alone can alleviate suffering. Any one of them can relieve suffering. It's amazing how they do that. Doug === Subject: Newton's formula 1x3(aaa,bbb,ccc) 3x6(aab,abb,aac,acc,bbc,bcc) 6x1(abc) Which column is the Newton's one? === Subject: Re: B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2) > This may be an old problem. > Let B(x) be a function such that > If x is not an integer > B(x) = 0 > If x is an integer > B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2) > Prove (or disprove) that > B(2) = 1 > James Dow Allen B(0) = 2^(-0) + B(0) + B(-1/3) + B(2) which, since B(-1/3) = 0, implies B(2) = -1, as far as I can see... Is the problem supposed to be more complicated than this? === Subject: Re: B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2) Was my question too well-known to be interesting? Or just uninteresting? > Let B(x) be a function such that > If x is not an integer > B(x) = 0 > If x is an integer > B(x) = 2^-x + B(4x/3) + B((2x-1)/3) + B(4x+2) > Prove (or disprove) that > B(2) = 1 > ... What does B(0) = ? It doesn't matter: When x is a positive integer, the rightside of the equation will never lead to anything but non-integers or positive integers. But let's sidestep this bugaboo, by stipulating B(x) = 0 whenever x is not a positive integer. > Need you ask? ... Can you clarify this question? I wasn't *compelled* to post on Usenet, if that's what you ask. James === Subject: Re: Why you can not count real number? > Is there any problem if I make a hypothesis that real > number is > countable? > Just the probability that you can make more and more > number as you go? Perhaps the following can help you. Kant notably holds that arithmetic is synthetic a priori and has to do with the pure intuition of time. The fundamental idea is that in the process of counting we freeze time, nothing to do with the continuum. Independently of this, the concept of bijectivity solve your question in negative sense. === Subject: Re: Why you can not count real number? The list of all ratios 0/inf, 1/inf, 2/inf, till inf/inf couns all real numbers in the range 0-1, regardless of definition of infinity. Between them [square root from 2]/inf is not included, since [square root from 2] was not in the original list of the natural numbers. Similar arguments are true for any [square root from 2]/k, k =/> 2. kunzmilan === Subject: Re: precedence of negation and times I would say -xy means -(xy) , but as noted in this thread it is equal to (-x)y . === Subject: quasiconvex functions I would like to know if the sum of a quasiconvex function and an affine function is always/sometimes/never a quasiconvex function? I thank you in advance, John === Subject: Re: JSH: Surprised even me > Did you know that for a brief period last week, if you searched > Google images for jstevh, the first hit was to a page called > Company sells real dog poop.? > It still does. (And before JSH calls anyone a liar, he should try it > out. The link is directly follow your second link above), all I get is the two of clubs. Ten days ago I also got the dog poop link, but I don't anymore. (I do still *think* of dog poop when I think of James though.) Why would Google images behave differently for you than for me? === Subject: Re: JSH: Surprised even me >You people are totally incompetent. > All these years I thought that at some level maybe some of you knew >mathematics, but now I see you don't. > I am actually silly enough to still be further disappointed. > You people have no real interest in mathematics. > You are play-acting a role of what you think a mathematician would be >like, but if Gauss were alive today, he would not talk to any of you, I >am sure. > I was working at hard problems, actually discovering mathematics, so my >attention was always somewhat diverted, and the energy I directed >against any of you was always muted. > You do not know me, or what I can do. You clearly have no clue about >what it is like to face a real mathematician in his own domain. > And now I get to concentrate on taking you apart versus dealing with >the hard world of discovery. > And I have no compunctions against repeatedly stripping you all of your >illusions. > Remember, part of my goal is to shut down entire math departments. > I need you help to accomplish it. I need you to behave as you are now, >as the anti-mathematicians you are. > Without your help in avoiding powerful and simple mathematics, I can't >toss out people like Barry Mazur or even Andrew Wiles. > I need your simple faith in social ways against mathematical proof to >build the energy. > I need what you are doing now. > > > Harris, aren't you afraid they will put you away in the funny house? I > mean, not because you are a Math Crank, but because of the threats you > are making ever more often. If you slip out there in the real world > this is going to stop being some weird situation where you are just > some guy posting on usenet and you are going to become a guy > considered to be a danger to others and himself by the pertinent > authorities. Stop this madness Harris, you are no mathematician. Put > your abilities to work, get a grip on your life, it's not too late. > You're wasting your breath, he's not listening. > But look on the bright side. I'm sure that once they put him in the > looney bin and find out he has no insurance, they'll turn him out > on the street cured. Did wonders for what's-his-name, the guy > with the Scientific Proof of God. > So we'll have to suffer for a year or so with no JSH, but think how > much funnier he'll be when they throw his ass out on the street. Darn, for a while I thought we were going to be the only to repliers in the shortest JSH thread ever. We might have gotten the Fields Medal or something. Now the place is crowded. I guess you are right, he is not listening. But there are times when it is hard to oversee that there is a real and really sad side to all this story. Harris is a real person suffering from real psychological impairment. He also really is an asshole, but that doesn't prevent me from feeling compassion sometimes. Anyway I couldn't picture a worse scenario than Harris being put away for a year! what would we do with all of our time? Especially now that the Soccer World Championship is over. God forbid. === Subject: Re: JSH: Surprised even me >You people are totally incompetent. >> All these years I thought that at some level maybe some of you knew >mathematics, but now I see you don't. >> I am actually silly enough to still be further disappointed. >Did you know that for a brief period last week, if you searched >Google images for jstevh, the first hit was to a page called >Company sells real dog poop.? > I am not making this up. > It still does. (And before JSH calls anyone a liar, he should try it > out. The link is > Hey! Who are you calling dog poop? > And that's not even a picture of jstevh, just has that word on the web > page. > This is the link you want: > Are there any more cards? (Archimedes Plutonium should be another card.) Don't worry if you can't come up with 52 cards. They aren't playing with a full deck anyway. 8-) --- Christopher Heckman === Subject: Re: JSH: Surprised even me >> You people are totally incompetent. >> All these years I thought that at some level maybe some of you knew >> mathematics, but now I see you don't. >> I am actually silly enough to still be further disappointed. >> Did you know that for a brief period last week, if you searched >Google images for jstevh, the first hit was to a page called >Company sells real dog poop.? >> I am not making this up. > It still does. (And before JSH calls anyone a liar, he should try it >out. The link is > Hey! Who are you calling dog poop? > And that's not even a picture of jstevh, just has that word on the web > page. > This is the link you want: > (Archimedes Plutonium should be another card.) The reason there's a JSH card is because some joke opportunity presented itself which I couldn't resist (although I've forgotten actually what it was). And someone actually thought there was some racial undertone by making JSH a spade! Luckily, my secret identity had not yet been comprimised. > Don't worry if you can't come up with 52 cards. They aren't playing > with a full deck anyway. 8-) I suppose I could start with crank.net, but the joke's over and it's certainly not worth wasting any more time. > --- Christopher Heckman === Subject: Re: JSH: Surprised even me <44b587c0$1@news.auckland.ac.nz >And I have no compunctions against repeatedly stripping you all of your >illusions. >>Yes! We have no compunctions. >>We have no >>Compunctions. >>Today. > I once stripped an executable. But careful with that compunctuation, > Eugene! > Rest in peace: Roger Syd Barrett, January 6, 1946 - July 7, 2006 > --c > Syd died? At least we still have Astronomy Domine. Yep. Tuesday. The music industry claims another victim. --- Christopher Heckman === Subject: Re: Bunch of actors Mail-To-News-Contact: abuse@dizum.com >> It took years for me to realize it, as I kept thinking that there was >> some substance to modern math society somewhere, but it finally settled >> in, there was not. >> So now, you think you can just talk your way out of a solution to the >> factoring problem. >> Fine. Try it. See what happens. >Huh? 2*2 = 4? I don't believe it!! Your just wrong! It's the distributive property! You can't argue with the distributive property. The 2 just multiplies through, forcing you into the object ring. Geez. -- Michael F. Stemper #include No animals were harmed in the composition of this message. === Subject: Re: An exact simplification challenge - 8 > Hello the computer algebra fans, > Is there a person who can show us how to simplify > using some CAS operations, the following expression > a := 1+I: > b := conjugate(a): > c := abs(a): > a*c*EllipticPi(b/2, 1/c) - > a*c*EllipticPi(c, b/2, 1/c) + > b*c*EllipticPi(c, a/2, 1/c) - > b*c*EllipticPi(a/2, 1/c); > It is a funny way to write Pi. Maple (version V R4) computes the value numerically as -3.15538673832555336190522019368 Chris > Rephrase the expression in terms of integrals, use the > range of the integrand to 'reduce' to the real case and > after changing variables (i did it not the elegant way) > arrive at Int(4*x^(1/2)/(1+x^2)/(1-x^2)^(1/2),x=0..1). > And now use: > changevar(x=(xi)^2,%,xi): subs(xi=x,%): #simplify(%); > changevar(1+x^4=(xi)^1,%,xi): subs(xi=x,%); #simplify(%); > value(%); > -- > please use mail at ... instead of test3 at ... to send me a mail === Subject: Re: An exact simplification challenge - 8 > Hello the computer algebra fans, > Is there a person who can show us how to simplify > using some CAS operations, the following expression > a := 1+I: > b := conjugate(a): > c := abs(a): > a*c*EllipticPi(b/2, 1/c) - > a*c*EllipticPi(c, b/2, 1/c) + > b*c*EllipticPi(c, a/2, 1/c) - > b*c*EllipticPi(a/2, 1/c); > It is a funny way to write Pi. >Maple (version V R4) computes the value numerically as >-3.15538673832555336190522019368 Maple 10.04 (using Digits:= 50) computes it as 3.1415926535897932384626419083070865997433942587131+.3e-48*I which agrees pretty well with Pi (an error of about 1.5*10^(-24)). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: An exact simplification challenge - 8 > Is there a person who can show us how to simplify > using some CAS operations, the following expression > a := 1+I: > b := conjugate(a): > c := abs(a): > a*c*EllipticPi(b/2, 1/c) - > a*c*EllipticPi(c, b/2, 1/c) + > b*c*EllipticPi(c, a/2, 1/c) - > b*c*EllipticPi(a/2, 1/c); >> It is a funny way to write Pi. > Maple (version V R4) computes the value numerically as > -3.15538673832555336190522019368 So they will tell you to upgrade :) But wait a moment! Let's see: Maple V Release 5, IBM INTEL NT, Jun 16 1998 -3.1553867383255533619052201936751 Hmm, this will not help ;-) However you can simplify the expression -- via EllipticK for example. Peter === Subject: Re: An exact simplification challenge - 8 Transforming into integral, changing variables and valuating gives : 8*2^(1/2)*EllipticPi(-3+2*2^(1/2),3-2*2^(1/2))/(2^(1/2)+1)+2*(2*2^(1/2)-4)*E llipticK(3-2*2^(1/2)); ,which appears to be Pi. Chris > Is there a person who can show us how to simplify > using some CAS operations, the following expression > a := 1+I: > b := conjugate(a): > c := abs(a): > a*c*EllipticPi(b/2, 1/c) - > a*c*EllipticPi(c, b/2, 1/c) + > b*c*EllipticPi(c, a/2, 1/c) - > b*c*EllipticPi(a/2, 1/c); >> It is a funny way to write Pi. > Maple (version V R4) computes the value numerically as > -3.15538673832555336190522019368 > So they will tell you to upgrade :) > But wait a moment! Let's see: > Maple V Release 5, IBM INTEL NT, Jun 16 1998 > -3.1553867383255533619052201936751 > Hmm, this will not help ;-) > However you can simplify the expression -- > via EllipticK for example. > Peter === Subject: Re: An exact simplification challenge - 8 > Transforming into integral, changing variables and valuating gives : > 8*2^(1/2)*EllipticPi(-3+2*2^(1/2),3-2*2^(1/2))/(2^(1/2)+1)+2*(2*2^(1/2)-4)*E l lipticK(3-2*2^(1/2)); > ,which appears to be Pi. > Chris >Is there a person who can show us how to simplify >using some CAS operations, the following expression >a := 1+I: >b := conjugate(a): >c := abs(a): >a*c*EllipticPi(b/2, 1/c) - >a*c*EllipticPi(c, b/2, 1/c) + >b*c*EllipticPi(c, a/2, 1/c) - >b*c*EllipticPi(a/2, 1/c); >> It is a funny way to write Pi. >Maple (version V R4) computes the value numerically as >-3.15538673832555336190522019368 > So they will tell you to upgrade :) > But wait a moment! Let's see: > Maple V Release 5, IBM INTEL NT, Jun 16 1998 > -3.1553867383255533619052201936751 > Hmm, this will not help ;-) > However you can simplify the expression -- > via EllipticK for example. > Peter Here is my hack (I should have changed the int bounds at the end): restart; interface(version); Classic Worksheet Interface, Maple 10.04, Windows, May 30 2006 Build ID 233114 a := 1+I: b := conjugate(a): c := abs(a): a*c*EllipticPi(b/2, 1/c) - a*c*EllipticPi(c, b/2, 1/c) + b*c*EllipticPi(c, a/2, 1/c) - b*c*EllipticPi(a/2, 1/c): S8:=%: evalf(S8,40); 3.141592653589793238462643383279502884195-.121857397313417128471e-18*I 'convert(S8,Int)'; combine(%): subs(_alpha1=x, %): normal(%); J:=integrand(%); J := 32*I*(-1+x^2)/(4-2*x^2)^(1/2)/(2-x^2+x^2*I)/(2-2*x^2)^(1/2)/(-2+x^2+x^2*I) assume( 1 < x, x < sqrt(2)); getassumptions(x); evalc(J): simplify(%): Int(%,x=1..sqrt(2)); Int(-16*(-1+x^2)/(2-2*x^2+x^4)/(-2+2*x^2)^(1/2)/(4-2*x^2)^(1/2),x = 1 .. 2^(1/2)) changevar(1-x^2=eta,%,eta); Int(8*eta/(2*eta+(1-eta)^2)/(-2*eta)^(1/2)/(2+2*eta)^(1/2)/(1-eta)^(1/2),eta = -1 .. 0) changevar(1-eta=xi+1,%,xi); Int(-4*xi^(1/2)/(xi^2+1)*2^(1/2)/(2-2*xi)^(1/2)/(xi+1)^(1/2),xi = 0 .. 1) evalf(%,40); -3.141592653589793238462643383279502884197 === Subject: Re: The Fundamental Theorem of Calculus > On Tue, 11 Jul 2006 18:00:43 -0700, The World Wide > Wade ><18595607.1152655602764.JavaMail.jakarta@nitrogen.mat > hforum.org>, >> let X be a bounded Lebesgue measurable subset of R >> (a bounded Jordan measurable subset of R) and >> f:X->R a function such that there exists F:X->R, > with >> F'(x)=f(x) for every x in X. >> Is f Lebesgue (respectively Riemann) integrable? >No. Let X = [0,1], F(x) = x^2*sin(1/x^2), x > 0, > F(0) = 0. Then F >is differentiable on [0,1], but F' is not Lebesgue > integrable on >[0,1]. >> If X=[a,b], does the integral of f over [a,b] > equal >> F(b)-F(a)? >If F' = f is in L^1[a,b], then the Lebesgue integral > of f over >[a,b] equals F(b)-F(a). > True. But it's worth pointing out that one needs to > be > very careful here: This is true if F is > differentiable > at _every_ point and the derivative is integrable. > Seems like it should be pointed out, since F' in > L^1 > is often taken to mean something very different from > that. Out of curiosity: what is this alternative meaning? >However, even if f is bounded on [a,b], f >need not be Riemann integrable on any subinterval of > [a,b]. > ************************ > David C. Ullrich Maury === Subject: Re: The Fundamental Theorem of Calculus <16447624.1152798568161.JavaMail.jakarta@nitrogen.mathforum.org> True. But it's worth pointing out that one needs to >> be very careful here: This is true if F is differentiable >> at _every_ point and the derivative is integrable.> Seems like it should be pointed out, since F' in >> L^1 is often taken to mean something very different >> from that. > Out of curiosity: what is this alternative meaning? I think he's talking about elements of L^1 being certain equivalence classes of functions, where two functions are considered equivalent if they differ on a set of measure zero. L. Renfro === Subject: Re: The Fundamental Theorem of Calculus On Wed, 12 Jul 2006 13:51:54 EDT, Maury Barbato >> On Wed, 12 Jul 2006 07:25:13 EDT, Maury Barbato >> <18595607.1152655602764.JavaMail.jakarta@nitrogen.math >> forum.org>, >> >> let X be a bounded Lebesgue measurable subset of >> R >> (a bounded Jordan measurable subset of R) and >> f:X->R a function such that there exists F:X->R, >> with >> F'(x)=f(x) for every x in X. >> Is f Lebesgue (respectively Riemann) integrable? >> >> No. Let X = [0,1], F(x) = x^2*sin(1/x^2), x > 0, >> F(0) >> = 0. Then F >> is differentiable on [0,1], but F' is not Lebesgue >> integrable on >> [0,1]. >> >> If X=[a,b], does the integral of f over [a,b] >> equal >> F(b)-F(a)? >> >> If F' = f is in L^1[a,b], then the Lebesgue >> integral >> of f over >> [a,b] equals F(b)-F(a). However, even if f is >> bounded >> on [a,b], f >> need not be Riemann integrable on any subinterval >> of >> [a,b]. >Forgive my ignorance: if F':[a,b]->R is bounded, >> then >does it belong to L^1[a,b]? >I don't know Lebesgue Integral Theory, but, as I >suspected, my problem has a natural placing in it. >> F' is a pointwise limit of continuous functions, >> hence measurable. So yes. >Can you write explicetly the sequence of functions, >please (I think you refer to functions constructed using >difference quotient)? Yes: F_n -> F', if F_n(x) = (F(x+1/n) - F(x)) / (1/n). (Or if we're being vary careful: If F is only defined on [a,b] then F_n as above is not defined on all of [a,b]. Define G(x) = F(x) for x in [a,b], G(x) = 0 otherwise. Then G_n(x) -> F'(x) for all x in (a,b), if G_n is as above with G in place of F.) >> You should really learn about the Lebesgue integral! >You'll right, surely!!! I'll do it in my next year >of my mathematical studies. >> All these questions you've asked over the last few >> months or so, people have assumed you knew about that >> and were wondering what was true in other situations. >> Yes, the Lebesgue integral does give very nice >> answers >> to many questions - that's why it's so popular. >Maury >> ************************ >> David C. Ullrich >Maury ************************ David C. Ullrich === Subject: Re: The Fundamental Theorem of Calculus <22353065.1152726744074.JavaMail.jakarta@nitrogen.mathforum.org> > On Wed, 12 Jul 2006 07:25:13 EDT, Maury Barbato > <18595607.1152655602764.JavaMail.jakarta@nitrogen.math >> forum.org>, >> let X be a bounded Lebesgue measurable subset of >> R >> (a bounded Jordan measurable subset of R) and >> f:X->R a function such that there exists F:X->R, >> with >> F'(x)=f(x) for every x in X. >> Is f Lebesgue (respectively Riemann) integrable? >> No. Let X = [0,1], F(x) = x^2*sin(1/x^2), x > 0, >> F(0) >> = 0. Then F >> is differentiable on [0,1], but F' is not Lebesgue >> integrable on >> [0,1]. >> If X=[a,b], does the integral of f over [a,b] >> equal >> F(b)-F(a)? >> If F' = f is in L^1[a,b], then the Lebesgue >> integral >> of f over >> [a,b] equals F(b)-F(a). However, even if f is >> bounded >> on [a,b], f >> need not be Riemann integrable on any subinterval >> of >> [a,b]. >>Forgive my ignorance: if F':[a,b]->R is bounded, >> then >does it belong to L^1[a,b]? >I don't know Lebesgue Integral Theory, but, as I >suspected, my problem has a natural placing in it. >> F' is a pointwise limit of continuous functions, >> hence measurable. So yes. >Can you write explicetly the sequence of functions, >please (I think you refer to functions constructed using >difference quotient)? > Yes: F_n -> F', if F_n(x) = (F(x+1/n) - F(x)) / (1/n). > (Or if we're being vary careful: If F is only defined > on [a,b] then F_n as above is not defined on all of > [a,b]. Define G(x) = F(x) for x in [a,b], G(x) = 0 > otherwise. Then G_n(x) -> F'(x) for all x in (a,b), > if G_n is as above with G in place of F.) If you're being very careful, you should be more careful: G_n is not continuous at b - 1/n. Try G(x) = F(b) for x > b. An alternative that doesn't require looking outside the interval [a,b] is to take F_n(x) = n (F(x + (x-a)/(n(b-a))) - F(x + (x-b)/(n(b-a)))) Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Integrals > Looks right. Integral over X can be divided into two > parts, integral over S and integral over X-S. The > sum of the two will be less than meas(X)*sup(f), > unless meas(S)=0. Not so simple: what assures you that S is Jordan (or Lebesgue) measuranble?!? Maury === Subject: platonic solids hello everybody always wants to prove there are no more than five of these. how do you prove that the five actually exist? that's all i had to say: what i'm doing now is padding the post with extra text because in my experience these messages aren't taken seriously if they're too short. for good measure i will protest my serious intentions: i may be overlooking something obvious, but i have in fact tried to answer this question for myself, and i am really interested in knowing the answer. thank you for your attention. peace, stm === Subject: Re: platonic solids > hello > everybody always wants to prove there are no more than five of these. > how do you prove that the five actually exist? Construction of the five is in Euclid's Elements. I believe Book XIII. > that's all i had to say: > what i'm doing now is padding the post with extra text because in my > experience these messages aren't taken seriously if they're too short. > for good measure i will protest my serious intentions: i may be > overlooking something obvious, but i have in fact tried to answer this > question for myself, and i am really interested in knowing the answer. > thank you for your attention. So first consult Euclid, then come back if you did not understand something... http://www.perseus.tufts.edu/cgi-bin/ptext?doc=Perseus%3Aabo%3Atlg%2C179 9%2C001&query=init. http://aleph0.clarku.edu/~djoyce/java/elements/elements.html === Subject: Re: platonic solids > everybody always wants to prove there are no more than five of these. > how do you prove that the five actually exist? By actually exhibiting them. Jose CArlos Santos === Subject: Re: A scientific approach to proving whether man landed on the moon - photogrammetric rectification > You are wrong--film *does not* have sufficient DR to capture sun lit > landscapes and star fields simultaneously... can you show me *one* > example? No you can't! I might be able to, but it sure won't be Ektachrome. It might be something like a super long toe B&W film (say Verichrome Pan) in a highly compensating pyrocatchin developer. It used to be possible to get outrageously nonlinear characteristics with that stuff. Ektachrome doesn't even have the range to capture sun-lit landscapes all by themselves... -- C'est un Nagra. C'est suisse, et tres, tres precis. === Subject: Re: A scientific approach to proving whether man landed on the moon - photogrammetric rectification > The Command/Service module was part of a single moonship design and > sat on-top of a landing lower stage making an enormous stack nearly 90 > feet tall. The whole thing would land and the CSM would later act as > the launcher for the return, leaving the lower stage used in the > landing in-place. This is one reason why the service module had such > a big main engine. Here's a 1961 drawing of that concept: http://content.answers.com/main/content/wp/en/1/17/Apollo_Direct_Ascent.png Pat === Subject: Re: A scientific approach to proving whether man landed on the moon - photogrammetric rectification *TROLL BAIT* The trolls go in. The trolls go out. The trolls really love it when people freak out. They love to post their silly troll bait, to catch some more people with some of their hate. Some troll-like nuts are kooks with some luck, while others try harder to get you all fluffed! They know just how to pull your chain, and get you to give them the fame that they crave. They flood the groups with all of their stuff, then gleefully sit back and watch threads amuck. The Facts are not what they wish to discuss. They just want to see you in a bit of a fuss! Trolls do not bother to listen at all. You'd have better luck with a solid brick wall! The more you try to counter their hoots, the more they will post crap to mess up *our* groups. So just ignore them. They aren't worth the time. Or put them in killfiles so you won't see their whines. D. Knisely === Subject: Re: A scientific approach to proving whether man landed on the moon - photogrammetric rectification Good stuff! Also see http://www.trolls.com . They have some really good Trolls.. > *TROLL BAIT* > The trolls go in. > The trolls go out. > The trolls really love it > when people freak out. > They love to post > their silly troll bait, > to catch some more people > with some of their hate. > Some troll-like nuts > are kooks with some luck, > while others try harder > to get you all fluffed! > They know just how > to pull your chain, > and get you to give them > the fame that they crave. > They flood the groups > with all of their stuff, > then gleefully sit back > and watch threads amuck. > The Facts are not what > they wish to discuss. > They just want to see you > in a bit of a fuss! > Trolls do not bother > to listen at all. > You'd have better luck > with a solid brick wall! > The more you try > to counter their hoots, > the more they will post crap > to mess up *our* groups. > So just ignore them. > They aren't worth the time. > Or put them in killfiles > so you won't see their whines. > D. Knisely === Subject: Re: A scientific approach to proving whether man landed on the moon - photogrammetric rectification <1CG9805P38906.4996180556@twistycreek.com> <3idtg.32465$FQ1.3554@attbi_s71> <9609b$44b5ea7e$471d40b4$11781@ALLTEL.NET *TROLL BAIT* > The trolls go in. > The trolls go out. > The trolls really love it > when people freak out. > They love to post > their silly troll bait, > to catch some more people > with some of their hate. > Some troll-like nuts > are kooks with some luck, > while others try harder > to get you all fluffed! > They know just how > to pull your chain, > and get you to give them > the fame that they crave. > They flood the groups > with all of their stuff, > then gleefully sit back > and watch threads amuck. > The Facts are not what > they wish to discuss. > They just want to see you > in a bit of a fuss! > Trolls do not bother > to listen at all. > You'd have better luck > with a solid brick wall! > The more you try > to counter their hoots, > the more they will post crap > to mess up *our* groups. > So just ignore them. > They aren't worth the time. > Or put them in killfiles > so you won't see their whines. > D. Knisely But you are all such pagan liars, and of the worse possible born-again kind of incest collaborators to your Third Reich at that. - Brad Guth === Subject: Re: A scientific approach to proving whether man landed on the moon - photogrammetric rectification >>*TROLL BAIT* >>The trolls go in. >>The trolls go out. >>The trolls really love it >>when people freak out. >>They love to post >>their silly troll bait, >>to catch some more people >>with some of their hate. >>Some troll-like nuts >>are kooks with some luck, >>while others try harder >>to get you all fluffed! >>They know just how >>to pull your chain, >>and get you to give them >>the fame that they crave. >>They flood the groups >>with all of their stuff, >>then gleefully sit back >>and watch threads amuck. >>The Facts are not what >>they wish to discuss. >>They just want to see you >>in a bit of a fuss! >>Trolls do not bother >>to listen at all. >>You'd have better luck >>with a solid brick wall! >>The more you try >>to counter their hoots, >>the more they will post crap >>to mess up *our* groups. >>So just ignore them. >>They aren't worth the time. >>Or put them in killfiles >>so you won't see their whines. >>D. Knisely > But you are all such pagan liars, and of the worse possible born-again > kind of incest collaborators to your Third Reich at that. > Brad Guth Sorry I have to quit polluting sci.astro.amateur, among other, replying to your troll... we can continue in sci.physics if you like... but here you are going into my killfile... bye! *Plonk* === Subject: Re: A scientific approach to proving whether man landed on the moon - photogrammetric rectification >> You have to read documents from that period carefully. Apollo *was* seen >> as the first step toward an eventual lunar landing; statements that it was >> intended to directly support a landing mission don't mean that it was >> meant to *fly* that mission. >Actually, at one point, it was intended to do just that. While >originally conceived as a merely a follow-on to project Mercury for >Earth-orbital missons, once Kennedy's mandate for getting men on the >moon was stated, that changed dramatically... Yes, but we're talking about Apollo as it was originally conceived, *before* that mandate appeared. -- spsystems.net is temporarily off the air; | Henry Spencer mail to henry at zoo.utoronto.ca instead. | henry@spsystems.net === Subject: Re: A scientific approach to proving whether man landed on the moon - photogrammetric rectification >> Brad, tell us the exposure for 1) a star field; 2) a landscape at >> the distance of the earth or moon illuminated by the sun; 3) the >> dynamic range (density) in film required to capture both simultaneously; >> and 4) the actual dynamic (density) range of film. > That's been done a thousand times over. What's the big freaking deal? The big deal is that if *you* do it you will realize something about photographs taken on the moon that you currently don't understand or choose to ignore! === Subject: Re: A scientific approach to proving whether man landed on the moon - photogrammetric rectification <1CG9805P38906.4996180556@twistycreek.com> <3idtg.32465$FQ1.3554@attbi_s71> <7Igtg.32709$FQ1.10364@attbi_s71 >> Brad, tell us the exposure for 1) a star field; 2) a landscape at >> the distance of the earth or moon illuminated by the sun; 3) the >> dynamic range (density) in film required to capture both simultaneously; >> and 4) the actual dynamic (density) range of film. > That's been done a thousand times over. What's the big freaking deal? > The big deal is that if *you* do it you will realize something about > photographs taken on the moon that you currently don't understand or > choose to ignore! You're actually an incest cloned borg of the Third Reich, arnt you. What's to ignore? What's not to easily understand about being snookered? Am I supposed to ignore of whatever's officially NASA/Apollo, as having been obtained from orbit? (would you like to see those examples?) So many others have taken photos and even gotten them published of our moon along with accommodating multiple other items within the very same frame of exposure, including those of the NASA/Apollo missions having accomplished while their robotic instruments and cameras were in orbit and having such a really good look-see at the entire Earth depicted with it's good range of albedo of less than 0.1 to greater than 0.7 as being rather nicely included along with the extremely dark and golden sooty brown surface of our mostly basalt with titanium, iron and otherwise being a salty moon in the foreground. So, what exactly is your pathetic point as to why Venus remained as invisible as WMD, and why other planets and of a few sufficiently bright stars (especially to that of an unfiltered Kodak moment) were playing such a game of stealth? What's you're best brown-nosed excuse for all of the artificial xenon lamp spectrum worth of the somewhat spotty illumination? As I'd said before, this has all been done at least a thousand times over. What's the big freaking deal? If you're supposedly capable of having photographed the physically dark (0.07 albedo) moon (somewhat darker yet because of the low solar angle), as such you're unavoidably photographing any other available planet and even a few of those brighter stars that so happen to be within frame, as in no matters what. You can't have one without the other because, film has sufficient DR, and there have been countless numbers of amateur photographs and even a few of those pesky NASA official images that totally proves I'm right. In fact, there's even a good number of those official NASA/Apollo shots as having been obtained from orbit that proves I'm right. How the hell can you argue against NASA? Yourself and the likes of good old David Knisely are two incest cloned status quo farts of a kind. And, you are each such totally born-again liars, arnt you. Do tell, where the heck was Venus in relationship the lunar horizon, especially on two of their missions where the vibrance of nearby Venus had to have been burning holes in that film? Besides, what's the point of folks taking those unfiltered Kodak moments when being so nicely surrounded by so much gamma and of those unavoidable hard-X-rays? Why don't you folks show us any one good example of those NASA/Apollo EVA obtained pictures that's not being xenon lamp illuminated? Then please show us village idiots, on a map of the moon, exactly where it's 55+% (guano island like) albedo for as far as the human eye can see and of the unfiltered Kodak eye can record. I believe we're talking about a zone that represents at least that of a lunar near-white-out zone, that's good for at least 10 km in all directions. By the way; some significant portions of those moonpans reproductions look as though we're talking damn near 75% reflective. Then show us a working R&D scaled prototype of those fly-by-rocket landers, as in proof-testing any part of itself. Then show us an actual frame of any film or even that of any leader/trailer portion of their film. Please explain the nearly 30% inert GLOW and otherwise the nearly 60:1 ratio of accomplishing those missions. Oops, I forgot that you're an incest cloned borg. Sorry about that. - Brad Guth === Subject: Re: A scientific approach to proving whether man landed on the moon - photogrammetric rectification >> Brad, tell us the exposure for 1) a star field; 2) a landscape at >> the distance of the earth or moon illuminated by the sun; 3) the >> dynamic range (density) in film required to capture both simultaneously; >> and 4) the actual dynamic (density) range of film. >That's been done a thousand times over. What's the big freaking deal? >> The big deal is that if *you* do it you will realize something about >> photographs taken on the moon that you currently don't understand or >> choose to ignore! > You're actually an incest cloned borg of the Third Reich, arnt you. > What's to ignore? What's not to easily understand about being > snookered? > Am I supposed to ignore of whatever's officially NASA/Apollo, as having > been obtained from orbit? (would you like to see those examples?) > So many others have taken photos and even gotten them published of our > moon along with accommodating multiple other items within the very same > frame of exposure, including those of the NASA/Apollo missions having > accomplished while their robotic instruments and cameras were in orbit > and having such a really good look-see at the entire Earth depicted > with it's good range of albedo of less than 0.1 to greater than 0.7 as > being rather nicely included along with the extremely dark and golden > sooty brown surface of our mostly basalt with titanium, iron and > otherwise being a salty moon in the foreground. So, what exactly is > your pathetic point as to why Venus remained as invisible as WMD, and > why other planets and of a few sufficiently bright stars (especially to > that of an unfiltered Kodak moment) were playing such a game of > stealth? > What's you're best brown-nosed excuse for all of the artificial xenon > lamp spectrum worth of the somewhat spotty illumination? > As I'd said before, this has all been done at least a thousand times > over. What's the big freaking deal? If you're supposedly capable of > having photographed the physically dark (0.07 albedo) moon (somewhat > darker yet because of the low solar angle), as such you're unavoidably > photographing any other available planet and even a few of those > brighter stars that so happen to be within frame, as in no matters > what. You can't have one without the other because, film has > sufficient DR, and there have been countless numbers of amateur > photographs and even a few of those pesky NASA official images that > totally proves I'm right. In fact, there's even a good number of those > official NASA/Apollo shots as having been obtained from orbit that > proves I'm right. How the hell can you argue against NASA? You are wrong--film *does not* have sufficient DR to capture sun lit landscapes and star fields simultaneously... can you show me *one* example? No you can't! === Subject: Re: A scientific approach to proving whether man landed on the moon - photogrammetric rectification <1CG9805P38906.4996180556@twistycreek.com> <3idtg.32465$FQ1.3554@attbi_s71> <7Igtg.32709$FQ1.10364@attbi_s71> <6ditg.32814$FQ1.12433@attbi_s71 >> Brad, tell us the exposure for 1) a star field; 2) a landscape at >> the distance of the earth or moon illuminated by the sun; 3) the >> dynamic range (density) in film required to capture both simultaneously; >> and 4) the actual dynamic (density) range of film. >That's been done a thousand times over. What's the big freaking deal? >> The big deal is that if *you* do it you will realize something about >> photographs taken on the moon that you currently don't understand or >> choose to ignore! > You're actually an incest cloned borg of the Third Reich, arnt you. > What's to ignore? What's not to easily understand about being > snookered? > Am I supposed to ignore of whatever's officially NASA/Apollo, as having > been obtained from orbit? (would you like to see those examples?) > So many others have taken photos and even gotten them published of our > moon along with accommodating multiple other items within the very same > frame of exposure, including those of the NASA/Apollo missions having > accomplished while their robotic instruments and cameras were in orbit > and having such a really good look-see at the entire Earth depicted > with it's good range of albedo of less than 0.1 to greater than 0.7 as > being rather nicely included along with the extremely dark and golden > sooty brown surface of our mostly basalt with titanium, iron and > otherwise being a salty moon in the foreground. So, what exactly is > your pathetic point as to why Venus remained as invisible as WMD, and > why other planets and of a few sufficiently bright stars (especially to > that of an unfiltered Kodak moment) were playing such a game of > stealth? > What's you're best brown-nosed excuse for all of the artificial xenon > lamp spectrum worth of the somewhat spotty illumination? > As I'd said before, this has all been done at least a thousand times > over. What's the big freaking deal? If you're supposedly capable of > having photographed the physically dark (0.07 albedo) moon (somewhat > darker yet because of the low solar angle), as such you're unavoidably > photographing any other available planet and even a few of those > brighter stars that so happen to be within frame, as in no matters > what. You can't have one without the other because, film has > sufficient DR, and there have been countless numbers of amateur > photographs and even a few of those pesky NASA official images that > totally proves I'm right. In fact, there's even a good number of those > official NASA/Apollo shots as having been obtained from orbit that > proves I'm right. How the hell can you argue against NASA? > You are wrong--film *does not* have sufficient DR to capture sun lit > landscapes and star fields simultaneously... can you show me *one* > example? No you can't! Forget it. The likes of David Knisely, Pat Flannery, Dale, Mary Pegg, Jud McCranie, Secret237 and yourself; You're all nothing but a serious incest collective of such born-again liars, and none the less of such liars with each of your pants on fire. Why is it that you folks simply don't like to see or otherwise talk about any of those fully certified NASA/Apollo images as having been obtained from lunar orbit, clearly showing that of our extremely dark and nasty moon, along with that of mother Earth having loads of photons to spare, as having shown us exactly what a 0.05 albedo portion of Earth looks like, that which would otherwise prove as within other frames as having to include the likes of a few other planets and even being capable of having recorded a few of those pesky bright stars within any number of those very same frames as obtained from the lunar deck? You simply can not possibly have photographed an average unfiltered worth of a 0.07 albedo lunar terrain and having continually excluded all else, other than via being a certified liar, and not to mention your otherwise having rad-hard DNA to boot. If so much as one those EVA Kodak moments is hocus-pocus, much less all of them as having been xenon lamp illuminated, then what's to believe about any staged hammer/feather sequence that could have been easily accomplished without their being on the moon. It has been explained multiple times as to how to go about physically and otherwise having photographically and/or video simulated whatever gravity, but you folks are stuck in auto-butt-save naysayism as well as auto-exclude whatever evidence that doesn't suit mode from the very get go. Obviously you've never been to the movies in the past few decades, thus you'll believe in absolutely anything thing that comes along with your NASA good house keeping stamp of approval. You wont even admit that upon at least two of those Missions is where Venus was simply unavoidably situated above the lunar horizon, yet never once having been included within any given frame that had more than sufficient camera/lens scope and the unfiltered sensitive worth of film DR to have recorded such. Therefore, you're all nothing but the worse possible pagan SOB liars on this Earth, and your God(NASA) is what seriously sucks and blows almost as bad off as did your perpetrated cold-war, and now we have that of what your resident LLPOF warlord(GW Bush) that represents as just more of the same status quo that suits your skewed mindset, the likes of what give your Hitler a good shot and otherwise put Christ on a stick in the first place. Does this mess get any worse; you bet it does. - Brad Guth === Subject: Re: A scientific approach to proving whether man landed on the moon - photogrammetric rectification > Why is it that you folks simply don't like to see or otherwise talk > about any of those fully certified NASA/Apollo images as having been > obtained from lunar orbit, clearly showing that of our extremely dark > and nasty moon, along with that of mother Earth having loads of photons > to spare, as having shown us exactly what a 0.05 albedo portion of > Earth looks like, that which would otherwise prove as within other > frames as having to include the likes of a few other planets and even > being capable of having recorded a few of those pesky bright stars > within any number of those very same frames as obtained from the lunar > deck? > You simply can not possibly have photographed an average unfiltered > worth of a 0.07 albedo lunar terrain and having continually excluded > all else, other than via being a certified liar, and not to mention > your otherwise having rad-hard DNA to boot. > If so much as one those EVA Kodak moments is hocus-pocus, much less all > of them as having been xenon lamp illuminated, then what's to believe > about any staged hammer/feather sequence that could have been easily > accomplished without their being on the moon. It has been explained > multiple times as to how to go about physically and otherwise having > photographically and/or video simulated whatever gravity, but you folks > are stuck in auto-butt-save naysayism as well as auto-exclude whatever > evidence that doesn't suit mode from the very get go. > Obviously you've never been to the movies in the past few decades, thus > you'll believe in absolutely anything thing that comes along with your > NASA good house keeping stamp of approval. > You wont even admit that upon at least two of those Missions is where > Venus was simply unavoidably situated above the lunar horizon, yet > never once having been included within any given frame that had more > than sufficient camera/lens scope and the unfiltered sensitive worth of > film DR to have recorded such. > Therefore, you're all nothing but the worse possible pagan SOB liars on > this Earth, and your God(NASA) is what seriously sucks and blows almost > as bad off as did your perpetrated cold-war, and now we have that of > what your resident LLPOF warlord(GW Bush) that represents as just more > of the same status quo that suits your skewed mindset, the likes of > what give your Hitler a good shot and otherwise put Christ on a stick > in the first place. > Does this mess get any worse; you bet it does. > Brad Guth You are wrong--film *does not* have sufficient DR to capture sun lit landscapes and star fields simultaneously... can you show me *one* example? No you can't! === Subject: Re: A scientific approach to proving whether man landed on the moon - photogrammetric rectification <3idtg.32465$FQ1.3554@attbi_s71> <7Igtg.32709$FQ1.10364@attbi_s71> <6ditg.32814$FQ1.12433@attbi_s71> landscapes and star fields simultaneously... can you show me *one* > example? No you can't! Over the many years, I've shown you folks dozens of such examples, and I've proven by the hard-scientific matter of replicated Kodak facts and numbers, that other planets and even a few of the brighter stars had to have been unavoidably recorded. You're the ones without so much as a clue nor an honest speck of evidence (other than the usual butt-loads of your infomercial-science) to boot. What sort of need of saturated star fields are you talking about? You do realize that Venus is obviously that of a much smaller than Earth item, but otherwise simply so much brighter than Earth? What sort of pathetic film with such absolutely piss-poor DR are you folks talking about? What sort of worse possible scanning and/or moderation of that film are you into using? What sort of crapy lens having applied such horrific optical spectrum cut-off and/or of such narrow band-pass filters are you using? I'll have to say this one again and again: Forget it, folks, there's absolutely no freaking question that we haven't yet walked upon that gamma/hard-X-ray and otherwise salty and physically dark moon of ours. The likes of David Knisely, Pat Flannery, Dale, Mary Pegg, Jud McCranie, Secret237 and most likely yourself; You're all nothing but a serious incest collective, of such borg like born-again liars, and none the less of such pagan liars with each of your pants on fire. Why is it that you folks simply don't like to see or otherwise talk about any of those fully certified NASA/Apollo images as having been obtained from lunar orbit, clearly showing that of our extremely dark and nasty moon, along with that of mother Earth having loads of photons to spare, as having shown us exactly what a 0.05 albedo portion of Earth looks like, that which would otherwise prove as within other frames as having to include the likes of a few other planets and even being capable of having recorded a few of those pesky bright stars within any number of those very same frames as obtained from the lunar deck? You simply can not possibly have photographed an average unfiltered worth of any 0.07 albedo lunar terrain and then having continually managed to have excluded all else, other than via being a certified liar, and not to mention your otherwise having loads of rad-hard DNA to boot. If so much as one those EVA Kodak moments is hocus-pocus, much less all of them as having been xenon lamp illuminated, then what's to believe about any staged hammer/feather sequence that could have been easily accomplished without their being on the moon. It has been explained multiple times as to how to go about physically and otherwise having photographically and/or video simulated whatever gravity, but you folks are stuck in auto-butt-save naysayism as well as auto-exclude whatever evidence that doesn't suit mode from the very get go. Obviously you've never been to the movies in the past few decades, thus you'll believe in absolutely anything thing that comes along with your NASA good house keeping stamp of approval. You wont even admit that upon at least two of those Missions is where Venus was simply unavoidably situated above that lunar horizon, yet never once having been included within any given frame that had more than sufficient camera/lens scope and the unfiltered sensitive worth of Kodak film DR to have recorded such. Therefore, because you're so all-knowing means that you're all representing nothing but the absolute worse possible pagan SOB liars upon this Earth, and your God(NASA) is what seriously sucks and blows almost as bad off as did your decades of perpetrated cold-wars, and now we have to survive that of what your resident LLPOF warlord(GW Bush) created, as representing more of the same status quo that suits your skewed mindset, the likes of what give your Hitler a good shot and having otherwise put Christ on a stick in the first place. Does this NASA/Apollo fiasco from hell get itself any worse; you bet it does. - Brad Guth === Subject: Re: A scientific approach to proving whether man landed on the moon - photogrammetric rectification <3idtg.32465$FQ1.3554@attbi_s71> <7Igtg.32709$FQ1.10364@attbi_s71> <6ditg.32814$FQ1.12433@attbi_s71> Sam Wormley and Scott Dorsey can't deal with the truth, can't even deal with their own NASA/Apollo facts that proves we've been snookered and dumbfounded ever since. >Scott Dorsey; Ektachrome doesn't even have the range to >capture sun-lit landscapes all by themselves... Liar! Besides the same arguments can be said of their B&W film, that proves I'm right, and proves that you're a liar. There are loads of even better photographic examples to go by, whereas Spaz and many others have been totally right, and the likes of George Evans is not exactly right (in fact he's somewhat of a soft-core liar, like GW Bush), as well as are a good many others that'll claim whatever it takes. Besides, there's more than a couple of unavoidably bright stars that are going to be a rather neat trick to avoid, as well as we have a few of those pesky planets to contend with. Over many years, I've shown these folks dozens of such photographic examples hosting a few other than moon/Earth items, and I've otherwise proven by the hard-scientific matter of replicated Kodak facts and numbers, that other planets and even a few of the brighter stars had to have been unavoidably recorded. However, in spite of the truth, they're the ones without so much as a clue nor an honest speck of evidence (other than the usual disinformation butt-loads of their infomercial-science) to boot. That freaking American flag alone is most often the photographic spectrum proof-positive that we haven't set a a rad-hard moonboot upon that physically dark and nasty moon of ours, and it has absolutely nothing to do with whatever's solar or otherwise physically windy or not about our moon. What sort of need of our having to prove saturated star fields are they talking about? All you have to accomplish is having to basically realize that Venus is obviously that of a much smaller than Earth item, but otherwise remains as photographically (especially to the unfiltered Kodak eye) of what's simply so much brighter than Earth? What sort of pathetic film with such absolutely piss-poor DR are these folks talking about? What sort of worse possible scanning and/or moderation of that film are they into using? What sort of crapy lens having applied such horrific optical spectrum cut-off and/or of such narrow band-pass filters are they using? I'll have to say this one again and again: Forget it, folks, there's absolutely no freaking question that we haven't yet walked upon that gamma/hard-X-ray and otherwise salty and physically dark moon of ours. The likes of David Knisely, Pat Flannery, Dale, Mary Pegg, Jud McCranie, Secret237 and of most likely others the likes of anyone that's pro-NASA; They're all nothing but a serious incest collective, of such borg like born-again liars, and none the less of such pagan liars with each of their pants on fire. Why is it that such folks simply don't like to see or otherwise talk about any of those fully certified NASA/Apollo images as having been obtained from lunar orbit, clearly showing that of our extremely dark and nasty moon, along with that of mother Earth having loads of photons to spare, as having shown us exactly what a 0.05 albedo portion of Earth looks like, that which would otherwise prove as within other frames as having to include the likes of a few other planets and even being capable of having recorded a few of those pesky bright stars within any number of those very same frames as obtained from the lunar deck? You simply can not possibly have photographed an average unfiltered worth of any 0.07 albedo lunar terrain and then having continually managed to have excluded all else, other than by way of being a certified liar, and not to mention your otherwise having butt-loads of rad-hard DNA to boot. If so much as one those EVA Kodak moments is hocus-pocus, much less all of them as having been xenon lamp illuminated, then what's there to believe about any staged hammer/feather sequence that could have been easily accomplished without their being on the moon. It has been explained multiple times as to how to go about physically and otherwise having photographically and/or video simulated whatever gravity, and there are a few honest mistakes worth arguing about, but you folks are stuck in auto-butt-save naysayism as well as auto-exclude as to whatever evidence that doesn't suit your all-or-nothin mindset from the very get go. Obviously they've never been to the movies in the past few decades, thus they'll most likely believe in absolutely anything that comes along with their NASA good house keeping stamp of approval. These folks wont even admit, that upon at least two of those Missions is where Venus was simply unavoidably situated above that lunar horizon, yet never once having been included within any given frame that had more than sufficient camera/lens scope (as field of view) and the unfiltered sensitive worth of Kodak film DR to have recorded such. Therefore, because they're so all-knowing means that they're all of what's representing nothing but the absolute worse possible pagan SOB liars upon this Earth, and their God(NASA) is what seriously sucks and blows almost as bad off as did their decades of perpetrated cold-wars, and now we have to survive that of whatever their resident LLPOF warlord(GW Bush) created, as representing more of the same status quo that suits their skewed and obviously bigoted mindset, the likes of what give their Hitler a good shot and having otherwise put Christ on a stick in the first place. Does this NASA/Apollo fiasco from hell get itself any worse; you bet it does. - Brad Guth === Subject: Re: A scientific approach to proving whether man landed on the moon - photogrammetric rectification You guys have all been baited by this Troll and you swallowed it hook, line, and sinker. Will this thread ever end? > Brad, tell us the exposure for 1) a star field; 2) a landscape at > the distance of the earth or moon illuminated by the sun; 3) the > dynamic range (density) in film required to capture both > simultaneously; > and 4) the actual dynamic (density) range of film. >>That's been done a thousand times over. What's the big freaking deal? > The big deal is that if *you* do it you will realize something about > photographs taken on the moon that you currently don't understand or > choose to ignore! >> You're actually an incest cloned borg of the Third Reich, arnt you. >> What's to ignore? What's not to easily understand about being >> snookered? >> Am I supposed to ignore of whatever's officially NASA/Apollo, as having >> been obtained from orbit? (would you like to see those examples?) >> So many others have taken photos and even gotten them published of our >> moon along with accommodating multiple other items within the very same >> frame of exposure, including those of the NASA/Apollo missions having >> accomplished while their robotic instruments and cameras were in orbit >> and having such a really good look-see at the entire Earth depicted >> with it's good range of albedo of less than 0.1 to greater than 0.7 as >> being rather nicely included along with the extremely dark and golden >> sooty brown surface of our mostly basalt with titanium, iron and >> otherwise being a salty moon in the foreground. So, what exactly is >> your pathetic point as to why Venus remained as invisible as WMD, and >> why other planets and of a few sufficiently bright stars (especially to >> that of an unfiltered Kodak moment) were playing such a game of >> stealth? >> What's you're best brown-nosed excuse for all of the artificial xenon >> lamp spectrum worth of the somewhat spotty illumination? >> As I'd said before, this has all been done at least a thousand times >> over. What's the big freaking deal? If you're supposedly capable of >> having photographed the physically dark (0.07 albedo) moon (somewhat >> darker yet because of the low solar angle), as such you're unavoidably >> photographing any other available planet and even a few of those >> brighter stars that so happen to be within frame, as in no matters >> what. You can't have one without the other because, film has >> sufficient DR, and there have been countless numbers of amateur >> photographs and even a few of those pesky NASA official images that >> totally proves I'm right. In fact, there's even a good number of those >> official NASA/Apollo shots as having been obtained from orbit that >> proves I'm right. How the hell can you argue against NASA? > You are wrong--film *does not* have sufficient DR to capture sun lit > landscapes and star fields simultaneously... can you show me *one* > example? No you can't! === Subject: Re: A scientific approach to proving whether man landed on the moon - photogrammetric rectification <1CG9805P38906.4996180556@twistycreek.com> <12b57tugjqh504f@corp.supernews.com > setting (NB: Bucky calls daytime sunsight, and nighttime, > sunclipse). > It is not possible to see the earth rise and set from the places, where > apollos landed. how is that? thus: tension is only and always deployed unto compression; --it takes some to jitterbug! http://members.tripod.com/~american_almanac http://www.rwgrayprojects.com/synergetics/plates/figs/plate01.html http://larouchepub.com/other/2006/3322_ethanol_no_science.html http://www.wlym.com/pdf/iclc/howthenation.pdf === Subject: Re: A scientific approach to proving whether man landed on the moon - photogrammetric rectification > setting (NB: Bucky calls daytime sunsight, and nighttime, > sunclipse). >> It is not possible to see the earth rise and set from the places, where >> apollos landed. > how is that? Because the moon is tidally locked to the earth. This means that a lunar day is equal to a lunar month, which in turn means that the moon presents the same face (the near side) to the earth at all times. (You actually can see the earth rise and set from some places on the moon, those on the edges of the near-side. Due to libation, the earth will move up and down a couple of degrees during the month.) === Subject: A=B*C how to get B? hello, I have a system of equations that can be written in matrix vector notation as follows: A(1*n)=B(1*m)*C(m*n) In a book i saw that B = A* C' * (C*C')^(-1) However, i do not recall how to get to that solution. can anybody help me with this? olea === Subject: Re: A=B*C how to get B? > hello, > I have a system of equations that can be written in matrix vector > notation as follows: > A(1*n)=B(1*m)*C(m*n) > In a book i saw that B = A* C' * (C*C')^(-1) > However, i do not recall how to get to that solution. can anybody help > me with this? It's the least-squares solution. I'm used to working with column vectors but let me see if I can do this right. Look for the B that minimizes || A - BC ||^2 = (A-BC)(A-BC)' = AA' - AC'B' - BCA' + BCC'B' Note: AC'B' is a scalar so equals its transpose BCA', || A - BC ||^2 = A'A - 2AC'B' + BCC'B' Now take the gradient of this with respect to B and set it =0. -2AC' + 2BCC' B(CC') = (AC') This is a full-rank linear system, and its solution is B = (AC')(CC')^-1 Note that there may be no B which exactly solves your original problem A = BC. This B is the one which minimizes the difference between left and right hand sides in the sense of the Euclidean norm. - Randy === Subject: Re: A=B*C how to get B? >I have a system of equations that can be written in matrix vector >notation as follows: >A(1*n)=B(1*m)*C(m*n) >In a book i saw that B = A* C' * (C*C')^(-1) C'(CC')^(-1) is called the pseudoinverse of C. >However, i do not recall how to get to that solution. can anybody help >me with this? AC' = BCC' => B = AC'(CC')^-1 provided that the inverse exists. It does when C has rank n. === Subject: Military logistic problem First of all I'm not sure if this belongs here, if not, could you suggest a more appropriate group. I'm having some problems researching the possibility of explicit solutions to military logistics problems. Although, I think the actual problem is more general. I have a time-step simulation tool that simulates strategic-lift (aircraft, ships, etc...) moving stuff from a start port to an end port. Obviously aircraft make many flights back and forth. While a time-step simulation suits this problem, it is quite expensive in time. The question I need answering is whether there exists an explicit solution to such a problem. I'm pretty sure it's not linear, and perhaps chaotic behaviour prevents such a solution. By the way, when I say solution I mean things like finding: a) How long it takes to move everything with various amounts of start-lift, port infrastructure (#berths, #AC handlers, etc...) and so on; b) The optimum number of the above to produce the quickest time; etc... I would be very grateful if you could point me in a direction to find if such a thing is possible and if so, whether the complexity of such a solution would make it impractical. confusing!) Chris Riddle === Subject: Re: Military logistic problem Chris === Subject: Full body massages for stress and muscle tension relief. Cleveland Ohio (Serious) My name is Robert, I am giving massages to relieve stress, muscle tension and sores in upper and lower body. My massages last 90-95 minutes and I am focusing mainly on so-called tension-accumulated stress points, such as shoulders, upper/mid and lower back, hips, back and front of legs and feet. Stress, backache and muscle tension in extremities are major trouble areas I am working on. I just charge $10 (ten) per session, but I do in-calls only at my location in Cleveland Heights, Ohio, using massage table. I have been giving massages on regular basis ten years now and have quite a few regular customers. Please get in touch if you have further questions or would like to set up your massage with me. Robert Website http://www.geocities.com/massotherapybyrob === Subject: DFT of a Gaussian I would like to know the side effects of discretisation, i.e. I've got a Gaussian function of the following form g_a(n) = |a|/(sqrt(2pi)M)*exp(-n^2a^2/2) and I want to know what is its discrete Fourier transform, n being the time variable. In clear, what is the result of sum_{0}^{M-1} g_a(n) exp(-2i Pi nm/M) ? I guess it depends on the variance of g_a and on M but I can't quantify the aliasing. Carine. === Subject: How to write a normalisation depending on the domain of a function I've got a function of the form I(x) = 1/F int_{-F/2}^{F/2} g(x,f) df I would like the 1/F to be dependent on the limits of the integral, i.e. if I apply a rectangle filter which is 1_[F1,F2](f) = 1 if fin [F1,F2]; = 0 otherwise on g(x,f), I want to have the following filtered function: I_f(x) = 1/F int_{-F/2}^{F/2} g(x,f) 1_[F1,F2](f)df = 1/(F2-F1)int_F1^F2 g(x,f) df And I can't find out how to write that function I(x) in a clever way. It sounds stupid, hey? Carine. === Subject: distance between sets Hello Suppose X and Y are two closed, bounded and convex sets. I need a metric such that d(X,Y) = 0 iff X=Y. for example the Hausdorff metric (http://planetmath.org/encyclopedia/HausdorffMetric.html) can help me. Is there any other popular metric that has that property? Diego === Subject: Re: distance between sets > Suppose X and Y are two closed, bounded and convex sets. > I need a metric such that d(X,Y) = 0 iff X=Y. > for example the Hausdorff metric > (http://planetmath.org/encyclopedia/HausdorffMetric.html) > can help me. > Is there any other popular metric that has that property? ?? *Every* metric satisfies d(X,Y) = 0 iff X=Y. (Well, maybe not in general relativity ...) Nevertheless, perhaps some of the following could be useful: http://www.fas.org/sgp/othergov/doe/lanl/pubs/00285752.pdf http://tinyurl.com/hdqzf http://www.maths.warwick.ac.uk/~jcr/re/usc.pdf [see p. 3] http://www.sztaki.hu/~szatmari/thesis.pdf http://cadpipe.vtt.fi/documents/hardwbasrenddataredu.pdf [pp. 36-38?] L. Renfro === Subject: Re: distance between sets sure sounds pretty interesting steppy the gamer Games I play: http://www.gamestotal.com http://uc.gamestotal.com http://gc.gamestotal.com http://gc.gamestotal.com/manual/ http://gc.gamestotal.com/hef.cfm http://uc.gamestotal.com/freegames/ http://uc.gamestotal.com/manual/ http://uc.gamestotal.com/hef.cfm http://3700ad.gamestotal.com http://aw.gamestotal.com http://www.stephenyong.com > Hello > Suppose X and Y are two closed, bounded and convex sets. > I need a metric such that d(X,Y) = 0 iff X=Y. > for example the Hausdorff metric > (http://planetmath.org/encyclopedia/HausdorffMetric.html) can help > me. > Is there any other popular metric that has that property? > Diego === Subject: Re: Turing vs. Godel (Newbie Question) (A simple example) > What is the relationship between Godel's 1st. > Incompleteness Theorem > and Turing's Proof of the Unsolvability of the > Halting Problem? I > mean the exact Mathematical relationship between > these two theorems. > C-B Since people have been on this topic recently, I have a quick (and hopefully simple) question. Does anyone have a simple example of a turing machine (or computer program) which (we assume) doesn't halt, but is unprovable. Or does anyone have an simple example of a (presumed) true but unprovable proposition? Or does anyone have a simple example of a number which is inconstructible? I'm also new to this area so hopefully the question has meaning.. I'm simply interested in an easy to understand example of one of these. Tom. === Subject: Re: Turing vs. Godel (Newbie Question) What is the relationship between Godel's 1st. Incompleteness Theorem >and Turing's Proof of the Unsolvability of the Halting Problem? I >mean the exact Mathematical relationship between these two theorems. >> C-B > There is a nice discussion of the relationship in Section 2 of this > http://www.calculemus.org/MathUniversalis/NS/10/04feferman.html > Yes, it is a very popular question for discussion. But in a nut shell, > from what you have seen, what would you say is the answer to this > question? > C-B > Why do you want an answer in a nutshell? I would have thought > Feferman's discussion of the issue was very clear and very useful, and > not too long. > In a nutshell, Goedel's 1st incompleteness theorem is *approximately* a > restatement of Turing's theorem, given Church's thesis (which appeared > after the two theorems). But Godel concludes that a particular object (sentence) is not in either of 2 sets (provable and refutable sentences) while Turing proved that a particular problem (set) is not solvable (recursive.) They're not of the same form. How do you take these two different structures and show they are the same? BTW There are many other theorems (even similar ones) in Proof Theory and in the Theory of Computation. Are you saying that by coincidence the first one discovered in each of these two theories happened to be the same? C-B === Subject: Re: Turing vs. Godel (Newbie Question) > BTW There are many other theorems (even similar ones) in Proof Theory > and in the Theory of Computation. Are you saying that by coincidence > the first one discovered in each of these two theories happened to be > the same? are you aware of the curry-howard isomorphism? -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: Turing vs. Godel (Newbie Question) > BTW There are many other theorems (even similar ones) in Proof Theory > and in the Theory of Computation. Are you saying that by coincidence > the first one discovered in each of these two theories happened to be > the same? > Are you aware of the curry-howard isomorphism? Of course. http://en.wikipedia.org/wiki/Curly_Howard C-B > -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar === Subject: Re: Turing vs. Godel (Newbie Question) > BTW There are many other theorems (even similar ones) in Proof Theory > and in the Theory of Computation. Are you saying that by coincidence > the first one discovered in each of these two theories happened to be > the same? > are you aware of the curry-howard isomorphism? I asked (you) first. C-B > -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar === Subject: Re: Turing vs. Godel (Newbie Question) >> What is the relationship between Godel's 1st. Incompleteness Theorem >> and Turing's Proof of the Unsolvability of the Halting Problem? I >> mean the exact Mathematical relationship between these two theorems. >> C-B >> There is a nice discussion of the relationship in Section 2 of this >> http://www.calculemus.org/MathUniversalis/NS/10/04feferman.html > Yes, it is a very popular question for discussion. But in a nut shell, >from what you have seen, what would you say is the answer to this >question? > C-B > Why do you want an answer in a nutshell? I would have thought > Feferman's discussion of the issue was very clear and very useful, and > not too long. > In a nutshell, Goedel's 1st incompleteness theorem is *approximately* a > restatement of Turing's theorem, given Church's thesis (which appeared > after the two theorems). > But Godel concludes that a particular object (sentence) is not in > either of 2 sets (provable and refutable sentences) while Turing proved > that a particular problem (set) is not solvable (recursive.) They're > not of the same form. How do you take these two different structures > and show they are the same? to. Have you looked at it yet? Are there are parts you're having trouble understanding? Maybe I can help you. > BTW There are many other theorems (even similar ones) in Proof Theory > and in the Theory of Computation. Are you saying that by coincidence > the first one discovered in each of these two theories happened to be > the same? > C-B === Subject: Re: Turing vs. Godel (Newbie Question) >> What is the relationship between Godel's 1st. Incompleteness Theorem >> and Turing's Proof of the Unsolvability of the Halting Problem? I >> mean the exact Mathematical relationship between these two theorems. > C-B >> There is a nice discussion of the relationship in Section 2 of this >> http://www.calculemus.org/MathUniversalis/NS/10/04feferman.html >> Yes, it is a very popular question for discussion. But in a nut shell, >from what you have seen, what would you say is the answer to this >question? >> C-B > Why do you want an answer in a nutshell? I would have thought >Feferman's discussion of the issue was very clear and very useful, and >not too long. > In a nutshell, Goedel's 1st incompleteness theorem is *approximately* a >restatement of Turing's theorem, given Church's thesis (which appeared >after the two theorems). > But Godel concludes that a particular object (sentence) is not in > either of 2 sets (provable and refutable sentences) while Turing proved > that a particular problem (set) is not solvable (recursive.) They're > not of the same form. How do you take these two different structures > and show they are the same? > to. Are there are parts you're having trouble understanding? Maybe I can help you. really is the relationship between Godel and Turing's theorems? How can they be the same - don't they have a different structure - an undecidable wff (individual object) vs. a non-recursive set? How do you equate them? C-B > BTW There are many other theorems (even similar ones) in Proof Theory > and in the Theory of Computation. Are you saying that by coincidence > the first one discovered in each of these two theories happened to be > the same? > > C-B === Subject: Re: Turing vs. Godel (Newbie Question) >>What is the relationship between Godel's 1st. Incompleteness Theorem >>and Turing's Proof of the Unsolvability of the Halting Problem? I >>mean the exact Mathematical relationship between these two theorems. > C-B > There is a nice discussion of the relationship in Section 2 of this > http://www.calculemus.org/MathUniversalis/NS/10/04feferman.html >> Yes, it is a very popular question for discussion. But in a nut shell, >> from what you have seen, what would you say is the answer to this >> question? >> C-B >> Why do you want an answer in a nutshell? I would have thought >Feferman's discussion of the issue was very clear and very useful, and >not too long. >> In a nutshell, Goedel's 1st incompleteness theorem is *approximately* a >restatement of Turing's theorem, given Church's thesis (which appeared >after the two theorems). > But Godel concludes that a particular object (sentence) is not in >either of 2 sets (provable and refutable sentences) while Turing proved >that a particular problem (set) is not solvable (recursive.) They're >not of the same form. How do you take these two different structures >and show they are the same? > to. Are there are parts you're having trouble understanding? Maybe I can help you. > really is the relationship between Godel and Turing's theorems? How > can they be the same - don't they have a different structure - an > undecidable wff (individual object) vs. a non-recursive set? How do > you equate them? Well, in Section 2 he begins by stating two theorems. They are really essentially the same theorem. The first one is about a Turing Machine A which indicates that C_q(n) does not halt by halting on the pair (q,n). The theorem says that if A is sound, it is not complete. The second theorem says that if a formal system is sound for a certain predicate P, it is not complete. This says essentially the same thing. (The second theorem is a weak form of Goedel's first incompleteness theorem. It does not imply the existence of an undecidable formula. However, it is closely related to Goedel's first incompleteness theorem in the form it is usually given and the arguments are very similar). Saying that if A is sound it is not complete implies that the set of (q,n) for which C_q(n) does not halt is not recursively enumerable. Since the set of (q,n) for which C_q(n) does halt clearly is recursively enumerable, this is equivalent to saying that the set of (q,n) for which C_q(n) halts is not recursive. So you can see how a theorem which asserts the existence of a true but unprovable proposition can be essentially the same as one which asserts that a certain set is not recursive. > C-B >BTW There are many other theorems (even similar ones) in Proof Theory >and in the Theory of Computation. Are you saying that by coincidence >the first one discovered in each of these two theories happened to be >the same? C-B === Subject: Re: Turing vs. Godel (Newbie Question) > Which form of the first incompleteness theorem do you have in mind? [1] > For the purposes of THIS question, that certainly CANNOT > matter. > There are many formulations of the first incompleteness theorem; in some > forms it follows rather directly from the unsolvability of the halting > problem while in others subtler considerations are needed. [2] Actually, your first comment [1] seems to be referring to which theorem, while your latter [2] refers to which formalization. [1] doesn't make much sense if you are referring to which formalization (the question is about theorems, not formalizations), and there certainly aren't many 1st. Incompleteness Theorems. as far as [2] goes. (Which did you mean in each case?) C-B > -- > Aatu Koskensilta (aatu.koskensilta@xortec.fi) > Wovon man nicht sprechen kann, daruber muss man schweigen > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Turing vs. Godel (Newbie Question) There are many formulations of the first incompleteness theorem; in some > forms it follows rather directly from the unsolvability of the halting > problem while in others subtler considerations are needed. > All that is needed is a framework rich enough to define what > it means for a set to be recursively enumerable or semi- > decidable. That just says that they are both theorems of Recursion Theory, but doesn't say how they are related. Is that the only connection? > PA turns out to be a framework rich enough to do that. > TMs in general do as well. No subtlety is needed in any case. > Furthermore, the unidirectionality in what you write above is > troubling, as though the question were only the degree of difficulty > in showing that G1 follows from HP. Obviously, HP follows from G1 > as well. Please substantiate both assertions (like any good scientist would.) That would be a pretty reasonable answer to the question. > There is no obvious reason why either direction should > be more or less difficult than the other. The core difficulty is that > they are usually phrased in completely different axiomatic frameworks > to begin with; Who has axiomatized Turing's proof (present company excluded)? What are the axioms and rules? > the possibility of ANY relationship between them is > by DEFAULT *linguistically* precluded. One has to wax translatory > and analogous BEFORE one can even get the two problems into > the same arena. The difference between Turing Machines and Peano Arithmetic wffs is syntactic only (not semantic.) > Which of course C-B didn't know. I have to admit you're right on that one. I wouldn't have said much of what you just did. What's the answer to the OP BTW? C-B === Subject: Re: Turing vs. Godel (Newbie Question) > What is the relationship between Godel's 1st. Incompleteness Theorem > and Turing's Proof of the Unsolvability of the Halting Problem? I > mean the exact Mathematical relationship between these two theorems. > they both embed the diagonal lemma > in a very formal sense > they categorically express the same thing > the meaning assigned to terms differ > but the nature of the self-reference > is essentially the same > a somewhat classical explanation of this > can be found in bools + jefferey (computability and logic) > in my opinion > category theory is the modern method to > describe similarities across term meanings > and the standard reference for this is > lawvere (diagonal arguments and cartesian closed categories) > reprinted at How does that answer the question? What is the exact relationship? C-B > -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar === Subject: Re: Turing vs. Godel (Newbie Question) > Which form of the first incompleteness theorem do you have in mind? > For the purposes of THIS question, that certainly CANNOT > matter. > There are many formulations of the first incompleteness theorem; in some > forms it follows rather directly from the unsolvability of the halting > problem while in others subtler considerations are needed. I would be interested in those various formulations, including a proof that Godel's result(s) follows directly from Turing's. (And what better impetus would one ask for?) C-B > -- > Aatu Koskensilta (aatu.koskensilta@xortec.fi) > Wovon man nicht sprechen kann, daruber muss man schweigen > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Turing vs. Godel (Newbie Question) >> There are many formulations of the first incompleteness theorem; in some >> forms it follows rather directly from the unsolvability of the halting >> problem while in others subtler considerations are needed. > I would be interested in those various formulations, including a proof > that Godel's result(s) follows directly from Turing's. The first incompleteness theorem is stated in many inequivalent ways in the literature ranging from Peano arithmetic is incomplete to for any theory in the language of arithmetic which contains Q we can mechanically find a sentence which is undecidable in the theory if the theory is consistent. The first incompleteness theorem in the form any sound axiomatizable extension of Q is incomplete follows from the existence of an r.e. but non-recursive set expressible in the language of arithmetic, and the recursive enumerability of theorems of an axiomatizable theory. This can be strengthened to require only 1-consistency of the theory instead of soundness by examining the expressing formulas and noting that they can be taken to be Sigma-1. G.9adel's original proof can be seen as a proof of the productivity of the set of arithmetical truths, which is a stronger recursion theoretic result than this set being just non-r.e. Rosser's strengthening of G.9adel's result shows essentially that we can take the expressible formula for a r.e. set to be such that it also represents the set in any consistent, not necessarily 1-consistent, extension of Q. This observation can be applied to either the weaker or stronger version (just incompleteness or productivity). -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Turing vs. Godel (Newbie Question) <44B2B85A.2080209@yahoo.com > Yes, it is a very popular question for discussion. But in a nut shell, > from what you have seen, what would you say is the answer to this > question? > The word in English is nutshell. > A nut shell would be something that you personally > held up to your ear for inspiration after finding it on the beach. Whatever it takes to inspire you. (I thought it was where most posters here live - at least my political foes, anyway.) > There are a lot of ways to answer this question, Ok, how about summarizing a few? > but there really is ONLY ONE important thing going on here. > That thing is the concept of a set being recursively enumerable > but NOT just plain recursive. I.e., of being recursively enumerable > but having a complement that is NOT recursively enumerable. I take it you're assuming it's expressible. But actually it (provability) need not be r.e. Even if both provability and nonprovability are not r.e. Godel's theorem still holds. > Both Godel's theorem and Turing's theorem about the semi- > decidability of the Halting problem are proofs that some set > has this particular level of complexity. > In the case of Godel's thm it is the set of theorems following > from the inference rules of first-order logic. > In the case of Turing's it is the set of (TM,input) descriptions > describing a TM that will eventually halt on the input. > In both of these cases there is an algorithm for confirming > membership in the set (there is a theorem-proving TM and > a universal TM), but there is no TM or algorithm for confirming > non-membership in the set. But in Turing's case we have a problem that is not solvable (set not recursive) vs. in Godel's case an individual object (wff) that is not in either of two sets (provable or refutable.) You describe it as just referring to two different sets, but the results are not the same except for the sets. The final results (theorems) are structurally different. How do you relate the actual theorems? > Your question, as you originally phrased it, was about > the precise mathematical relationship between. > That is just a stupid way of phrasing it. It is NOT > about a relationship between two DIFFERENT things. It's two different theorems, silly. > It is about two examples of ONE concept (namely, being > recursively enumerable but not recursive/decidable). That's what I am looking for (a relationship), except your claim that it's just two different sets (as well as the need for the complement being r.e.) is not so. C-B === Subject: Re: Turing vs. Godel (Newbie Question) What is the relationship between Godel's 1st. Incompleteness Theorem > and Turing's Proof of the Unsolvability of the Halting Problem? I > mean the exact Mathematical relationship between these two theorems. > Which form of the first incompleteness theorem do you have in mind? Good question. In fact, I purposely worded it that way (which is how most of texts word it), rather than differentiating the weaker short (based on w-consistency) to see who would bite and show their lack of detailed understanding of Godel's results. You were the only one to point out that distinction. Congratulations and a C-B Brownie Point to you (to show off to all of your friends*.) You can wimp out like most authors and consider only the shorter proof if you'd like. Apparently nobody has ever reduced the longer proof much conceptually (only leaving out details) - other than my own formal proof which still takes about 10 steps vs. about 3 for the shorter proof. Of course, in the final analysis we want to relate all incompleteness results (Godel, Rosser, Turing, Smullyan - maybe even Lob et. al.). I just quoted the two most famous, most talked-about and simplest ones, to see if anybody has gotten even that far. (Speaking of which, Rosser's 1936 result is generally described as being stronger because it uses a more complex sentence than Godel's, but formalized it is actually simpler than Godel's latter result and is not stronger, but simply uses different assumptions.) C-B *your 18 cohorts? > -- > Aatu Koskensilta (aatu.koskensilta@xortec.fi) > Wovon man nicht sprechen kann, daruber muss man schweigen > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: The list of all natural numbers don't exist Mail-To-News-Contact: abuse@dizum.com >> Except that in standard set theory, every set has a cardinality. >> Standard theory claims so. Prove it. >I can't prove a definition. Others can state this better than me, >but... I'm not sure if this will help or not, but I'd just like to point out that the definition does not claim that the cardinalities of any pair of sets can be compared. The notion that the cardinalities of *any* two sets can be compared is trichotomy, which requires that one also assume the Axiom of Choice. So, if TO doesn't like being able to compare the cardinalities of any two sets, all that he has to do is work without Choice. (Not intended to contradict you, just piggybacking on your comments.) -- Michael F. Stemper #include Always remember that you are unique. Just like everyone else. === Subject: Re: The list of all natural numbers don't exist > There is a difference between assuming > all natural are finite and proving all naturals > are finite. > Let's consider a different axiom system. How about the axioms for a > group? > Question: Does a group have more than one element? > None of the group axioms requires an element other than the identity to > exist, but there is likewise no axiom to forbid it. > So, can we assume that there are no groups with more than one element? > Or would this require an additional axiom? >> Is there a proof in group theory that groups with more than >> one element can't exist? Is the statement groups can have >> only one member a theorem of group theory? > Funny you should ask that. > Is there such a proof? You ignorance is showing again, Russell. Do a little research on groups on your own. >> Each and every ordinal has a largest element is a theorem >> of ZF-I. > You have not proved that, any more than you have proved any of the > corresponding statements about groups. That was precisely my point. > I am not saying there is such a proof in group theory. You are saying there is a proof for ZF-I, but you have been shown that your alleged proof is not a valid proof enough times for it to soak in to any head capable of learning. > You have repeatedly stated that infinite sets are consistent > with the axioms of ZF-I. My proof makes no assumptions > other than the axioms of ZF-I. > How can you claim I assume infinite sets don't exist? While your false assumptions in your several attempts to prove your falsehood have been varied, they never have been absent. And in each one so far presented, someone has pointed out at least one fatal flaw. Challenge to Russell: Present us here with any alleged proof that all ordinals in ZF-I MUST be finite, and someone here will find its flaw. Possibly several of us will find at least one flaw and some of us may find several flaws, but at least one of us will find at least one flaw. I know this because I know that Russell is incapable of proving ZF-I to be inconsistent, even if it actually were to be. === Subject: Re: The list of all natural numbers don't exist > Suppose x is a limit ordinal and x is a member of x. > Since membership is a well ordering on ordinals, we can write x < x to > mean x is a member of itself. We also have x = Ux, because that's what > it means to be a limit ordinal. It means that the members of x are the > same as the members of the members of x. (Notice that statement does not > hold for successor ordinals, but only for limit ordinals.) > Depending on how you define a well ordering, x < x might already be a > contradiction. But we can also deduce that since x < x = Ux, there must > be some y < x such that x < y. This certainly contradicts the > requirement that < must be a well ordering. > The contradiction means there can't be a limit ordinal that belongs to > itself, Q.E.D. You prove that if x is an element of itself then x can't be well ordered. So x is a member of itself and x is not well ordered. How have you proven x is not a member of itself? Russell - 2 many 2 count === Subject: Re: The list of all natural numbers don't exist <2aKdneku1LY8VC7ZnZ2dnUVZ_tqdnZ2d@comcast.com> <_L2dnWc6LJ7GeC7ZnZ2dnUVZ_oednZ2d@comcast.com> <6qOdnXUopr1J5ynZnZ2dnUVZ_q6dnZ2d@comcast.com> <4aCdnV7Aj707KSjZnZ2dnUVZ_oadnZ2d@comcast.com > Suppose x is a limit ordinal and x is a member of x. > Since membership is a well ordering on ordinals, we can write x < x to > mean x is a member of itself. We also have x = Ux, because that's what > it means to be a limit ordinal. It means that the members of x are the > same as the members of the members of x. (Notice that statement does not > hold for successor ordinals, but only for limit ordinals.) > Depending on how you define a well ordering, x < x might already be a > contradiction. But we can also deduce that since x < x = Ux, there must > be some y < x such that x < y. This certainly contradicts the > requirement that < must be a well ordering. > The contradiction means there can't be a limit ordinal that belongs to > itself, Q.E.D. > You prove that if x is an element of itself then x can't be well ordered. > So x is a member of itself and x is not well ordered. > How have you proven x is not a member of itself? If x is a member of itself then it is not well ordered. Consequently, if x is well ordered then it is not a member of itself. Part of the definition of limit ordinal is the property of being well ordered. Thus IF x is a member of itself THEN it is not a limit ordinal. IF x is a limit ordinal THEN it is not a member of itself. (This is pretty basic logic. Do you really not follow it?) This is *exactly* what you asked to prove in your previous post. What is the confusion here? Is it that he used the letter x to mean something different from what someone previously refered to as x? Russell: > Let x be the set of all sets that are not members of themselves. > Is x a member of x? Dave: >> Show me your proof in ZF-R that such a set exists. Even if there are >> some sets that are members of themselves, it doesn't follow that there is >> a single set that contains all those sets. Russell: >Why am I the one that provides all the proofs? >If I said the empty set is empty, you would demand a proof. >It appears that in ZF one can't prove anything without >an axiom that already assumes what is to be proven. >I will prove there is such a set as soon as you prove >a non-empty limit ordinal doesn't contain itself as as >element in ZF-R. has proved that a non-empty limit ordinal does not contain itself as an element in ZF-R. You should now present your proof that a set that contains all sets that are not members of themselves exists in ZF-R. -- mike. > === Subject: Re: The list of all natural numbers don't exist > Suppose x is a limit ordinal and x is a member of x. > Since membership is a well ordering on ordinals, we can write x < x > to > mean x is a member of itself. We also have x = Ux, because that's what > it means to be a limit ordinal. It means that the members of x are the > same as the members of the members of x. (Notice that statement does > not > hold for successor ordinals, but only for limit ordinals.) > Depending on how you define a well ordering, x < x might already be a > contradiction. But we can also deduce that since x < x = Ux, there > must > be some y < x such that x < y. This certainly contradicts the > requirement that < must be a well ordering. > The contradiction means there can't be a limit ordinal that belongs to > itself, Q.E.D. >> You prove that if x is an element of itself then x can't be well ordered. >> So x is a member of itself and x is not well ordered. >> How have you proven x is not a member of itself? > Your supposition is about a limit ordinal being a member of itself. > The response deals with that issue completely, so why are you trying to > change your supposition now? Can't stand being shown up as the idiot you > are? He assumed what he was trying to prove. Why would I assume a limit ordinal is well ordered? Russell - 2 many 2 count === Subject: Re: The list of all natural numbers don't exist > Suppose x is a limit ordinal and x is a member of x. > Since membership is a well ordering on ordinals, we can write x < x > to > mean x is a member of itself. We also have x = Ux, because that's what > it means to be a limit ordinal. It means that the members of x are the > same as the members of the members of x. (Notice that statement does > not > hold for successor ordinals, but only for limit ordinals.) > Depending on how you define a well ordering, x < x might already be a > contradiction. But we can also deduce that since x < x = Ux, there > must > be some y < x such that x < y. This certainly contradicts the > requirement that < must be a well ordering. > The contradiction means there can't be a limit ordinal that belongs to > itself, Q.E.D. > You prove that if x is an element of itself then x can't be well ordered. > So x is a member of itself and x is not well ordered. > How have you proven x is not a member of itself? >> Your supposition is about a limit ordinal being a member of itself. >> The response deals with that issue completely, so why are you trying to >> change your supposition now? Can't stand being shown up as the idiot you >> are? > He assumed what he was trying to prove. No. He assumed the opposite of what he wanted to prove. He assumed that there existed an ordinal that was a member of itself and showed that this led to a contradiction. > Why would I assume a limit ordinal is well ordered? Because all ordinals are well-ordered by definition. Do you have a similar trouble assuming that an even number is divisible by 2? Stephen === Subject: Re: The list of all natural numbers don't exist <6qOdnXUopr1J5ynZnZ2dnUVZ_q6dnZ2d@comcast.com> <4aCdnV7Aj707KSjZnZ2dnUVZ_oadnZ2d@comcast.com> <_8CdnWWzGY2rfijZnZ2dnUVZ_oednZ2d@comcast.com> Suppose x is a limit ordinal and x is a member of x. > Since membership is a well ordering on ordinals, we can write x < x >to >mean x is a member of itself. We also have x = Ux, because that's what >it means to be a limit ordinal. It means that the members of x are the >same as the members of the members of x. (Notice that statement does >not >hold for successor ordinals, but only for limit ordinals.) > Depending on how you define a well ordering, x < x might already be a >contradiction. But we can also deduce that since x < x = Ux, there >must >be some y < x such that x < y. This certainly contradicts the >requirement that < must be a well ordering. > The contradiction means there can't be a limit ordinal that belongs to >itself, Q.E.D. > You prove that if x is an element of itself then x can't be well ordered. > So x is a member of itself and x is not well ordered. >How have you proven x is not a member of itself? >> Your supposition is about a limit ordinal being a member of itself. >> The response deals with that issue completely, so why are you trying to >> change your supposition now? Can't stand being shown up as the idiot you >> are? > He assumed what he was trying to prove. > No. He assumed the opposite of what he wanted to prove. > He assumed that there existed an ordinal that was a member > of itself and showed that this led to a contradiction. > Why would I assume a limit ordinal is well ordered? > Because all ordinals are well-ordered by definition. > Do you have a similar trouble assuming that an even number is > divisible by 2? > Stephen That's a great example of the fallacy here. Russell asks how we can assume that a limit ordinal is well ordered in the proof that no limit ordinal is a member of itself. That is analogous to responding to the proof : No even number greater than two can be a prime number. Proof : any even number greater than two is divisible by two and hence is a composite number. with Why should we assume an even number is divisible by 2? I'll have to remember that one :-) I'm always amazed by how people can ask the question How can we assume an object with property P has property P? I think in Russell's case it has something to do with him conflating definition with assertion of existence. He seems to think that defining limit ordinal as transitive well ordered set with no largest element is assuming that such an object exists, when really it is just shorthand. In the same way as even number is shorthand for divisible by 2. -- mike. === Subject: Re: The list of all natural numbers don't exist > Why would I assume a limit ordinal is well ordered? >> Because all ordinals are well-ordered by definition. >> Do you have a similar trouble assuming that an even number is >> divisible by 2? >> Stephen > That's a great example of the fallacy here. Russell asks how we can > assume that a limit ordinal is well ordered in the proof that no limit > ordinal is a member of itself. That is analogous to responding to the > proof : > No even number greater than two can be a prime number. Proof : any > even number greater than two is divisible by two and hence is a > composite number. > with > Why should we assume an even number is divisible by 2? > I'll have to remember that one :-) I'm always amazed by how people can > ask the question How can we assume an object with property P has > property P? Following your analogy, Russell's conclusion appears to be that even numbers do not exist. If the existence of an even prime number greater than 2 leads to a contradiction, it must be the case that even numbers do not exist. > I think in Russell's case it has something to do with him conflating > definition with assertion of existence. He seems to think that defining > limit ordinal as transitive well ordered set with no largest > element is assuming that such an object exists, when really it is just > shorthand. In the same way as even number is shorthand for divisible > by 2. Russell's problem is that he has decided what the answer should be and does not really care what the answer really is. He knows that set theory is inconsistent. It really does not matter to him that he cannot prove it. He just sees that as another fault of set theory. Apparently in a good theory it is easy to prove whatever you want (or probably whatever Russell wants), including proving the theory is inconsistent. The fact that it has been so hard to prove set theory inconsistent is sure evidence that there is something wrong with it. :) Stephen === Subject: Re: The list of all natural numbers don't exist >Suppose x is a limit ordinal and x is a member of x. >> Since membership is a well ordering on ordinals, we can write x < x > to > mean x is a member of itself. We also have x = Ux, because that's what > it means to be a limit ordinal. It means that the members of x are the > same as the members of the members of x. (Notice that statement does > not > hold for successor ordinals, but only for limit ordinals.) >> Depending on how you define a well ordering, x < x might already be a > contradiction. But we can also deduce that since x < x = Ux, there > must > be some y < x such that x < y. This certainly contradicts the > requirement that < must be a well ordering. >> The contradiction means there can't be a limit ordinal that belongs to > itself, Q.E.D. >> You prove that if x is an element of itself then x can't be well ordered. >> So x is a member of itself and x is not well ordered. >> How have you proven x is not a member of itself? > Your supposition is about a limit ordinal being a member of itself. > The response deals with that issue completely, so why are you trying to > change your supposition now? Can't stand being shown up as the idiot you > are? > He assumed what he was trying to prove. He assumed no more that your statement of the problem required him to assume. Suppose x is a limit ordinal and x is a member of x. requires assumption that x is a limit ordinal, with all that the definitin of being a limit ordinal implies. > Why would I assume a limit ordinal is well ordered? Possibly because being well ordered is a necessary property of ALL ordinals, both limit and successor. Do a little research before asking such stupid questions. For exxample see http://en.wikipedia.org/wiki/Ordinal_number The ingenious definition suggested by John von Neumann, and which is now taken as standard, is this: define each ordinal as a special well-ordered set, namely that of all ordinals before it. Formally: A set S is an ordinal if and only if S is totally ordered with respect to set containment and every element of S is also a subset of S. (Here, set containment is another name for the subset relationship.) Such a set S is automatically well-ordered with respect to set containment. === Subject: Re: The list of all natural numbers don't exist <2aKdneku1LY8VC7ZnZ2dnUVZ_tqdnZ2d@comcast.com> <_L2dnWc6LJ7GeC7ZnZ2dnUVZ_oednZ2d@comcast.com> <6qOdnXUopr1J5ynZnZ2dnUVZ_q6dnZ2d@comcast.com> <4aCdnV7Aj707KSjZnZ2dnUVZ_oadnZ2d@comcast.com> <_8CdnWWzGY2rfijZnZ2dnUVZ_oednZ2d@comcast.com > Suppose x is a limit ordinal and x is a member of x. >> Since membership is a well ordering on ordinals, we can write x < x > to > mean x is a member of itself. We also have x = Ux, because that's what > it means to be a limit ordinal. It means that the members of x are the > same as the members of the members of x. (Notice that statement does > not > hold for successor ordinals, but only for limit ordinals.) >> Depending on how you define a well ordering, x < x might already be a > contradiction. But we can also deduce that since x < x = Ux, there > must > be some y < x such that x < y. This certainly contradicts the > requirement that < must be a well ordering. >> The contradiction means there can't be a limit ordinal that belongs to > itself, Q.E.D. >> You prove that if x is an element of itself then x can't be well ordered. >> So x is a member of itself and x is not well ordered. >> How have you proven x is not a member of itself? > Your supposition is about a limit ordinal being a member of itself. > The response deals with that issue completely, so why are you trying to > change your supposition now? Can't stand being shown up as the idiot you > are? > He assumed what he was trying to prove. > Why would I assume a limit ordinal is well ordered? Uh, because that's part of the *definition* of a limit ordinal? Why would you assume a triangle has three sides or that a car has an engine and wheels? Because that's what those words *mean*... Seaman is using here a proof method called Proof by contradiction which you seem completely uncomfortable with. An outline of how it works : - Make an assumption (there exists a limit ordinal which is a member of itself) - Derive a contradiction (such an object would be both a limit ordinal and not be a limit ordinal) - Conclude that your original assumption was false (there does not exist a limit ordinal which is a member of itself) Are you genuinely unfamiliar with this type of argument? -- mike. === Subject: Re: The list of all natural numbers don't exist > Uh, because that's part of the *definition* of a limit ordinal? Why > would you assume a triangle has three sides or that a car has an > engine and wheels? Because that's what those words *mean*... So you just assume limit ordinals exist and they are transitive and well ordered and have no largest element. I assume you are wrong. Since we can assume anything we want in set theory, I guess that means we can assume set theory is inconsistent. > Seaman is using here a proof method called Proof by > contradiction which you seem completely uncomfortable with. It's not a proof by contradiction. It's a proof of contradiction. proves limit ordinals are not well ordered. > An outline > of how it works : > - Make an assumption (there exists a limit ordinal which is a member of > itself) There is a limit ordinal that is transitive, well ordered and doesn't have a largest element. > - Derive a contradiction (such an object would be both a limit ordinal > and not be a limit ordinal) Such an ordinal can't be well ordered. > - Conclude that your original assumption was false (there does not > exist a limit ordinal which is a member of itself) There is no limit ordinal that is transitive, well ordered and has no largest element. > Are you genuinely unfamiliar with this type of argument? Sounds like a great argument to me. Russell - 2 many 2 count === Subject: Re: The list of all natural numbers don't exist > Uh, because that's part of the *definition* of a limit ordinal? Why > would you assume a triangle has three sides or that a car has an > engine and wheels? Because that's what those words *mean*... > So you just assume limit ordinals exist and they are transitive > and well ordered and have no largest element. > I assume you are wrong. Since we can exhibit an obvious example of a limit ordinal which is known to exist in ZF and NBG, and have frequently done so, why should we be wrong? > Since we can assume anything we want in set theory, > I guess that means we can assume set theory is inconsistent. Russell seems to think HE can assume what he wants but honest set theorists are constrained to assume no more than is stated in the axiom systems they use. > Seaman is using here a proof method called Proof by > contradiction which you seem completely uncomfortable with. > It's not a proof by contradiction. > It's a proof of contradiction. > proves limit ordinals are not well ordered. knows better than that, though Russell apparently does no know that he is claiming white is black. > An outline > of how it works : > - Make an assumption (there exists a limit ordinal which is a member of > itself) > There is a limit ordinal that is transitive, well ordered and doesn't > have a largest element. Since by definition ordinals are transistive and well ordered, that part is ok, and since the definition of *limit* ordinals requires that they not have a largest element, that part is ok, too. > - Derive a contradiction (such an object would be both a limit ordinal > and not be a limit ordinal) WRONG! Such an object would be a normal everyday limit ordinal of which N is the prime example, being transitive, well ordered, and with no largest element. > Such an ordinal can't be well ordered. Except that N certainly is well ordered. So Russell is wrong again. > - Conclude that your original assumption was false (there does not > exist a limit ordinal which is a member of itself) Russell has, as usual, missed the point. A limit ordinal such as N is by definition well ordered and has no largest member by inclusion or by membership. A set which is a member of itself HAS a largest member by membership, namely itself as a member of itself So that Russell is trying to say that a set which has no largest member does have a largest member. Everyone else realizes that is not possible, so that sets like N (limit ordinals) cannot be members of themselves. > There is no limit ordinal that is transitive, well ordered > and has no largest element. EVERY limit ordinal is transitive, well ordered and has no largest element. Including N. What Russell denies is precisely what is required for a set to be a limit ordinal. Russell has it exactly backwards. One of the things one must learn in mathematics, in particular in set theory, is that one must respect definitions to the letter. Russell's casual way of ignoring or even negating standard definitions or axioms of set theory prevents him from getting anything right at all in set theory. === Subject: Re: The list of all natural numbers don't exist <2aKdneku1LY8VC7ZnZ2dnUVZ_tqdnZ2d@comcast.com> <_L2dnWc6LJ7GeC7ZnZ2dnUVZ_oednZ2d@comcast.com> <6qOdnXUopr1J5ynZnZ2dnUVZ_q6dnZ2d@comcast.com> <4aCdnV7Aj707KSjZnZ2dnUVZ_oadnZ2d@comcast.com> <_8CdnWWzGY2rfijZnZ2dnUVZ_oednZ2d@comcast.com> Uh, because that's part of the *definition* of a limit ordinal? Why > would you assume a triangle has three sides or that a car has an > engine and wheels? Because that's what those words *mean*... > So you just assume limit ordinals exist and they are transitive > and well ordered and have no largest element. > I assume you are wrong. > Since we can assume anything we want in set theory, You completely miss the point. Dave's argument does *not* assert the existence of limit ordinals. It shows that no limit ordinal which is a member of itself can exist. His argument assumes that limit ordinals which contain themselves as elements exists and derives a contradiction. This proves that limit ordinals which contain themsleves do not exist. This says *nothing whatsoever* about the existence of limit ordinals in general. If you can't follow this then you should really stop trying to debunk the foundations of mathematics and try reading an introductory text on proof methods instead. > Seaman is using here a proof method called Proof by > contradiction which you seem completely uncomfortable with. > It's not a proof by contradiction. > It's a proof of contradiction. > proves limit ordinals are not well ordered. No, he proves that no limit ordinal which is a member of itself exists. You're totally incompetent in this area so I'm not surprised you can't follow his proof, or even correctly state what it is proving. > An outline > of how it works : > - Make an assumption (there exists a limit ordinal which is a member of > itself) > There is a limit ordinal that is transitive, well ordered and doesn't > have a largest element. OK. Now ***IF*** you can derive a contradiction from this, then you've proved that all limit ordinals have a largest element. > - Derive a contradiction (such an object would be both a limit ordinal > and not be a limit ordinal) > Such an ordinal can't be well ordered. This is the step that you have failed to prove, failed consistently and repeatedly. In fact, I think this whole branch of the thread is a result of your attempts to change the subject from complaints about your failed proof of this step. Your earlier proof ran something like... -Successor ordinals have property P -Limit ordinals do not have property P -Therefore, limit ordinals are not ordinals. This is illogical garbage. You have not presented a proof that a limit ordinal cannot be well ordered. If you have something other than the rubbish you tried to pass off as a proof before, present it here. Prove that limit ordinals cannot be well ordered. > - Conclude that your original assumption was false (there does not > exist a limit ordinal which is a member of itself) > There is no limit ordinal that is transitive, well ordered > and has no largest element. > Are you genuinely unfamiliar with this type of argument? > Sounds like a great argument to me. Just a shame you are either too dim or too obstinate to understand that each step of a prood has to be justified, and that you have ***FAILED*** to justify the crucial step in your proof. -- mike. === Subject: Re: The list of all natural numbers don't exist > Suppose x is a limit ordinal and x is a member of x. > Since membership is a well ordering on ordinals, we can write x < x to > mean x is a member of itself. We also have x = Ux, because that's what > it means to be a limit ordinal. It means that the members of x are the > same as the members of the members of x. (Notice that statement does not > hold for successor ordinals, but only for limit ordinals.) > Depending on how you define a well ordering, x < x might already be a > contradiction. But we can also deduce that since x < x = Ux, there must > be some y < x such that x < y. This certainly contradicts the > requirement that < must be a well ordering. > The contradiction means there can't be a limit ordinal that belongs to > itself, Q.E.D. > You prove that if x is an element of itself then x can't be well ordered. > So x is a member of itself and x is not well ordered. > How have you proven x is not a member of itself? Your supposition is about a limit ordinal being a member of itself. The response deals with that issue completely, so why are you trying to change your supposition now? Can't stand being shown up as the idiot you are? === Subject: Re: The list of all natural numbers don't exist >> Suppose x is a limit ordinal and x is a member of x. >> Since membership is a well ordering on ordinals, we can write x < x to >> mean x is a member of itself. We also have x = Ux, because that's what >> it means to be a limit ordinal. It means that the members of x are the >> same as the members of the members of x. (Notice that statement does not >> hold for successor ordinals, but only for limit ordinals.) >> Depending on how you define a well ordering, x < x might already be a >> contradiction. But we can also deduce that since x < x = Ux, there must >> be some y < x such that x < y. This certainly contradicts the >> requirement that < must be a well ordering. >> The contradiction means there can't be a limit ordinal that belongs to >> itself, Q.E.D. > You prove that if x is an element of itself then x can't be well ordered. > So x is a member of itself and x is not well ordered. > How have you proven x is not a member of itself? The proof assumes that x is an ordinal, which means that x is well ordered. x cannot be both well ordered and be a member of itself, and given that it is well ordered, then it cannot be a member itself. Stephen === Subject: Re: The list of all natural numbers don't exist <48OdnclzAZ2pGjPZnZ2dnUVZ_uidnZ2d@comcast.com> <2aKdneku1LY8VC7ZnZ2dnUVZ_tqdnZ2d@comcast.com> <_L2dnWc6LJ7GeC7ZnZ2dnUVZ_oednZ2d@comcast.com> We agreed on a definition of ordinal. >> Now you are trying to redefine the defintion of some ordinals >> by calling them successor ordinals. > Some ordinals are successors, and some are not. The ones that are not > successors are called limit ordinals. How is this changing the > definition of ordinal? Everything that we agreed is an ordinal is > still an ordinal, and everything that we agreed is not an ordinal is > still not an ordinal. I think that's Russell's problem up to this point - he does not know about limit ordinals. Hence his reliance on successor to define (and understand) ordinals. To be fair, that may be the fault of most of the mathematicians posting in this thread, that they simply assumed he knew there was a type of ordinal other than successor ordinal, since it's taken this long before anyone explicitly mentioned limit ordinals. === Subject: Re: The list of all natural numbers don't exist > We agreed on a definition of ordinal. > Now you are trying to redefine the defintion of some ordinals > by calling them successor ordinals. >> Some ordinals are successors, and some are not. The ones that are not >> successors are called limit ordinals. How is this changing the >> definition of ordinal? Everything that we agreed is an ordinal is >> still an ordinal, and everything that we agreed is not an ordinal is >> still not an ordinal. > I think that's Russell's problem up to this point - he does not know > about limit ordinals. Hence his reliance on successor to define > (and understand) ordinals. > To be fair, that may be the fault of most of the mathematicians > posting in this thread, that they simply assumed he knew there > was a type of ordinal other than successor ordinal, since it's > taken this long before anyone explicitly mentioned limit ordinals. Limit ordinals were explictly mentioned over a week ago. Russell himself used the words 'limit ordinal' in his responses. Of course this does not mean that he knows what they are. Stephen === Subject: Re: The list of all natural numbers don't exist If we want a definition of Cardinality then it should apply on both >> finite and infinite sets, and clearly what your set theorist say is not >> applicable for finite sets. since in finite sets , if there exist an >> injective function that is not surjective then this is quite enough to >> determine strict inquality between them. >> This would mean that Z-cardinality (careful which terms you use here!) >> would entirely depend on which functions we could construct or imagine. >> Cantor's cardinality, on the other hand, is dependent on *all* >> functions between the 2 compared sets. > I don't think so, Cantors cardinality is biased to the bijective > functions. let me tell you how. > for sets A and B if there exist a bijective function then card A = card > B, this is cantors definition of equality of cardinality, now even if > there exist another injective function between A and B that is > injective but not surjective ( ie not bijective ) then this is not > considered to mean anything, since there exist another function that is > bijective, that's why | say that Cantor's cardinality is biased, while > my concept of the generational function being the cardinality > determiner is not biased. Cantor says that sets A and B have the same cardinality if there exists a bijection between the two sets. So it is sufficient to show the existence of only one bijection to demonstrate |A| = |B|. The existence of other injections from A into other sets is irrelevant, and does not change the fact that |A| = |B| as long as set B can be found. >> My new concept of Cardinality doesn't say that Cantors cardinality is >> unvalid. It only states that Cantor's cardinality is a poor >> discriminator of set multiplicity. But of course it can discrimiante >> between large difference in multiplicities of members in sets , like >> between the multiplicity of members in a set and that in its power set. >> What exactly is multiplicity? If you tried to formalise that, you >> might find that we could actually help you find a proper abstraction. You appear to be defining a particular Z-cardinality of set A based on an injection between A and some set C, which only demonstrates |A| <= |C|. You then seem to say that a different Z-cardinality exists for every injection that can be defined between A and some other set Ci. If you can find several different sets with an injection from A, then you say you have a multiplicity of Z-cardinalities for set A. Is this a fair interpretation? If so, how is Z-cardinality more useful than Cantor's cardinality? What does it tell us that cardinality does not? === Subject: Factoring c-1 No, I am not factoring a polynomial in the speed of light. Ask JSH how to factor (c-1), if you are interested in such questions. In my consideration of random number generators, which are useful in Monte Carlo methods and cryptography, I keep seeing phi(c) as the maximum period of the popular random number generator ( (a^n) mod c) with (a,c) = 1. For prime c, phi(c) = c-1. Why? Why is the count of numbers that don't divide p equal to the numbers below p or p-1? I say it is a partitioning problem with two partitions: the numbers that divide p (or in general c), and those that don't. What I am seeing now is that any number has at least two factors in the first partition: itself and 1. c and 1 in set F, for Factor. Everything below c not a factor of c goes in set PHI. phi(c) is just the size of set PHI. Well, if you look at periods, that is, mulitples of primes factoring composite integer c, you see that every prime period hits on c, and the power of each prime factor is the number of periods before that hit. Simple. For p, every prime, and so every number, less than p is a non-hit, and so belongs in set PHI. So it seems to me, that for any number c coprime to a in a random number generator ( ( a^n ) mod c ) the factors of c-1 will be precisely the set PHI because they'll all be non-hits for c. Did I get that right? That is, F union PHI = all the numbers below and including c, down to 1. F intersection PHI = {}. This seems to have application in the factoring problem and I am probably just rediscovering it. I mean, if you could factor x >> 1 by factoring x-1, you'd have an algorithm for factoring, it seems to me, because then you could factor x-2 etc. and the factoring problems is essential to knowing the period of a random number generator. This gets real interesting for that other generator (a^n + b^n) mod c, with a,b,c pairwise coprime. The sets F.a and F.b have {} intersection, etc... Doug Goncz Replikon Research Seven Corners, VA 22044-0394 === Subject: Re: Factoring c-1 [Doug Goncz] > No, I am not factoring a polynomial in the speed of light. LOL! I'm very glad of that. > Ask JSH how to factor (c-1), if you are interested in such questions. > In my consideration of random number generators, which are useful in > Monte Carlo methods and cryptography, I keep seeing phi(c) as the > maximum period of the popular random number generator ( (a^n) mod c) > with (a,c) = 1. > For prime c, phi(c) = c-1. > Why? It's immediate from the definition of phi; e.g., see: http://mathworld.wolfram.com/TotientFunction.html > Why is the count of numbers that don't divide p equal to the numbers > below p or p-1? It isn't, but that's not what phi means. The count of integers i with 1 <= i <= p and i doesn't divide prime p is p-2. phi(p) is the count of integers i with 1 <= i <= p and gcd(p, i) = 1; it's the count of strictly positive integers <= p with no (non-unit) factors in common with p. 1|p so 1 is not counted in the former, but gcd(1, p)=1 so 1 is counted in the latter. When p is prime, gcd(i, p) = 1 for all i in [1, p), but gcd(p, p) = p =/= 1, so phi(p) = p-1. > I say it is a partitioning problem with two partitions: the numbers > that divide p (or in general c), and those that don't. It's the integers > 0 and <= c that share a non-unit factor with c, versus those that don't. The integers <= c that divide c includes 1 but is otherwise a subset of those sharing a non-unit factor -- it's quite possible for i to share a non-unit factor with c but not divide c (e.g., 4 shares a factor of 2 with 6, but 4 doesn't divide 6). If c's prime factorization is the product of p_i^e_i, the number of c's divisors is the product of (e_i + 1) (proof sketch: you get a divisor of c by picking 0 through e_1 of p_1, 0 through e_2 of p_2, etc). For example, 50 = 2^1*5^2, so 50 has (1+1)*(2+1) = 6 divisors (2^0*5^0, 2^0*5^1, 2^0*5^2, 2^1*5^0, 2^1*5^1, 2^1*5^2). But there are 50-phi(50) = 50 - 20 = 30 integers <= 50 that have a non-unit factor in common with 50 (= |{i: 1 <= i <= 50 and gcd(i, 50) > 1}|). > ... === Subject: Re: Factoring c-1 > No, I am not factoring a polynomial in the speed of light. Ask JSH how > to factor (c-1), if you are interested in such questions. > In my consideration of random number generators, which are useful in > Monte Carlo methods and cryptography, I keep seeing phi(c) as the > maximum period of the popular random number generator ( (a^n) mod c) > with (a,c) = 1. > For prime c, phi(c) = c-1. > Why? > Why is the count of numbers that don't divide p equal to the numbers > below p or p-1? > I say it is a partitioning problem with two partitions: the numbers > that divide p (or in general c), and those that don't. > What I am seeing now is that any number has at least two factors in the > first partition: itself and 1. c and 1 in set F, for Factor. Everything > below c not a factor of c goes in set PHI. phi(c) is just the size of > set PHI. > Well, if you look at periods, that is, mulitples of primes factoring > composite integer c, you see that every prime period hits on c, and > the power of each prime factor is the number of periods before that > hit. Simple. > For p, every prime, and so every number, less than p is a non-hit, and > so belongs in set PHI. > So it seems to me, that for any number c coprime to a in a random > number generator > ( ( a^n ) mod c ) > the factors of c-1 will be precisely the set PHI because they'll all be > non-hits for c. > Did I get that right? > That is, > F union PHI = all the numbers below and including c, down to 1. > F intersection PHI = {}. > This seems to have application in the factoring problem and I am > probably just rediscovering it. I mean, if you could factor x >> 1 by > factoring x-1, you'd have an algorithm for factoring, it seems to me, > because then you could factor x-2 etc. and the factoring problems is > essential to knowing the period of a random number generator. > This gets real interesting for that other generator (a^n + b^n) mod c, > with a,b,c pairwise coprime. The sets F.a and F.b have {} intersection, > etc... For prime p, since all a with 1 <= a < p are 1 or relatively prime, then phi(p) = p-1. But for non-prime q, phi(q) < q-1. For gcd(a,n) = 1, the period of a^k (mod n) divides phi(n). This is because a^phi(n) (mod n) is 1. Keep in mind phi(n) is around the same size as n, so unless you can find an a that produceds a small period, this would not be very useful. === Subject: Re: Factoring c-1 > . . . if you could factor x > 1 by factoring x-1, > you'd have an algorithm for factoring, it seems > to me, because then you could factor x-2 etc. . . . That would be exponential in the size of x. Make it (x-1)/2, ((x-1)/2-1)/2, etc. to be polynomial. -- === Subject: Re: An uncountable countable set > What we have said, and what is quite true, is that for every natural > there is a larger natural. > > Of course. And therefore it is impossible to exhaust all of them or to > find a set which is larger than all naturals together. > But nowhere an attempt is made to exhaust them. There is only an axiom > from which it can be derived that there is a set that contains them all. > With that axiom, Cantor's argument is a proof. Without that axiom, > Cantor's argument is meaningless. > When Cantor's proof was published, there was not such an axiom. > You are not arguing against Cantor's > argument, you are arguing against that axiom. > If an axiom states the existence of 100 natural numbers below 20, it > has to be abolished in order to save mathematics.The axiom of > infinity, interpreted as you do, is such an axiom There is a greater need to save mathematics from the incompetence of such as mueckenh that to save it from ZF and the axiom of infinity. === Subject: Re: An uncountable countable set > Moreover, > different orderings of the same subset of transpositions will yield > different results. In the formula I gave for the diagonal number, > calculation of one digit does *not* depend on the calculation of > another digit. > > And what is the consequence of this? > That the calculation of the digits can be done in parallel? > It is done in zero time. mueckenh seems unaware of the difference between a requirement of serial processing as exemplified in his algorithm and the option of parallel processing which Cantor's diagonal rule achieves. > How do you define the limit? And if you define that, is that limit > also well-ordered? Those are things you have to prove. > I define limit by: *Using all finite natural numbers* just as like as > Cantor does. Cantor does not need any limit process because his rule does a parallel processing for all naturals at one go. === Subject: Re: An uncountable countable set <44b11de9$0$17957$892e7fe2@authen.yellow.readfreenews.net> <44b18705$0$17996$892e7fe2@authen.yellow.readfreenews.net> <44b290b2$0$17984$892e7fe2@authen.yellow.readfreenews.net> <44b3eb8f$0$17998$892e7fe2@authen.yellow.readfreenews.net> You are using to index simultaneously for two different meanings. The >first is to index a specific position the second is undefined. What >does 0.1111 is indexed by the list number 4 = 0.1111 exaclty mean? > The unary number n indexes every position 1,2,3,...,n. > Actually a unary numeral, or any other numeral for a natural number, can > only index one position. The MEMBERS of a natural, however expressed, > can index up to that natural. The members of N can index all natural > positions. Why then is 0.111... not in the sequence 0.1, 0.11, 0.111, ...? 0.111... cannot be indexed by the members of any natural. Does 0.111... have more 1's than can be indexed by any natural? Or what prevents it from being indexed by any of the naturals? === Subject: Re: An uncountable countable set But it does not guarantee what is existing there. In order to find out > you must count. > I do not know what axiom of infinity, 'mueckenh is referring to , but > the ones in ZF, or ZFC or NBG guarantee existence of all of the > naturals ( as finite ordinals) without having to count anything. Perhaps those naturals guaranteed by the axiom even cannot be used for counting? They are, after all, no numbers. === Subject: Re: An uncountable countable set The binary tree: All paths, not yet disperged, start through the root > edge a. > There is no need for a root edge at all. the tree can quite comforably > be rooted in a rot node from which two edges branch out to two cild > nodes, and so on. > (Let us denote them as a single path as long as they are > together.) > As paths are sets of edges, this does not work. Call it bunch of paths cross sections, if you like. I abbreviate that by path. > Map this root edge a on this single path. In the next level > the path splits in two paths. Map half of edge a on each of them. The > right one passing through edge b, gets b mapped on it and it inherits > half of a. After splitting again, each of the paths gets the next edge, > say c, half of b and quarter of a. > | a > o > / b > o o > / / c > This tree contains 4 paths indicated by sequences of two branchings, > left or right right, so that {LL,LR, RL, RR} represents the set of all > paths for the tree as shown. > For larger binary trees one gets more and longer strings of left to > right branchings. > For infinite binary trees each such list of branchings is an infinite > sequence. And there are uncountably many possible such infinite > sequences. But we know that there are nearly twice as much edges. There are infinitely many of them too. And the the set of edges is countable. === Subject: Re: An uncountable countable set The axiom says there is an infinite set. It does not say that > 0.111... does belong to that set, in particular because all numbers > which in fact do belong to the set are different from 0.111... . > But 0.111... in effect IS that set to which it does not belong. Yes. 111... represents aleph_0 or omega. But we see that it is impossible to have aleph_0 or omega natural numbers. Every smaller number is finite. If aleph_0 is the first transfinite cardinal, then the set of natural numbers alone is not actually infinite. === Subject: Re: An uncountable countable set <44b119fc$0$17975$892e7fe2@authen.yellow.readfreenews.net> <44b1845b$0$17984$892e7fe2@authen.yellow.readfreenews.net> <44b28a08$0$17992$892e7fe2@authen.yellow.readfreenews.net> <44b3bd33$0$17981$892e7fe2@authen.yellow.readfreenews.net> <44b4f4f0$0$17943$892e7fe2@authen.yellow.readfreenews.net> That the transpositions are > conditional does not disturb the proof. > It does mean that they must be performed in order though. > And that does distrub the proof. Why should it? Do you dislike order? Cantor's list is ordered, the set N is ordered. Do they not exist for this sake? is there any hint why order should destroy infinity? > Cantor's diagonal argument is > also conditional. > But there are no conditions on any digit which depend on any other digit > having been calculated previously, i.e., no sequential conditions. Thus > one can have a general rule that does eachdigit without reference to any > others. so is valid for all in one rule and one step. In order to identify any digit, you must find it. That requires counting. That requires order. > That the conditions have to be executed one after the > other does not disturb the proof. > It means that themueckenh process, as described, can never end. This is a set and it exists as well as any other infinite set. Not more and not less. > Cantor's diagonal argument requires > counting which is a sequential act too. > Not at all. Did you ever count? Or do you mathematics without numbers but with letters only? > Getting the list in the first place may require counting, > though that is dubious, but once it is presented, no further counting is > required. Finding the n-th line of the list, and within this line the n-th digit requires hard counting, if n is large. === Subject: Re: An uncountable countable set yet found it. I am starting to doubt that. I doubt that too, because neither in his collected works nor in his correspondence I have found any hint on that. > What I find is the following: > 1. Cantor's first diagonal argument: proves (amongst others) that the > rational numbers are countable. > 2. Cantor's second diagonal argument: proves that there are sets with > cardinality greater than the cardinality of the natural numbers (the > proof we are discussing here). > 3. His proof that the reals are not countable (published in 1874). > Of course, Zermelo did show that (2) can be transformed to a proof that > the reals are not countable, but I do not think that Cantor did do that > transformation. Cantor himself generalized his proof (in the same paper) to all well-defined sets like the real numbers (= Linearcontinuum) and functions which only can take the values 0 and 1. (Therefore I said one could safely interpret his list as a list of binary numbers. Zermelo only remarked the problem with double representation of binary fractions.) Cantor concluded: da¤ die M.8achtigkeit wohldefinierter Mannigfaltigkeiten kein Maximum haben oder, was dasselbe ist, da¤ jeder gegebenen Mannigfaltigkeit L eine andere M an die Seite gestellt werden kann, welche von st.8arkerer M.8achtigkeit ist als L. For non-German-readers: He recognized that cardinalities do not have a maximum. === Subject: Re: An uncountable countable set <44b119fc$0$17975$892e7fe2@authen.yellow.readfreenews.net> <44b1845b$0$17984$892e7fe2@authen.yellow.readfreenews.net> <44b28a08$0$17992$892e7fe2@authen.yellow.readfreenews.net> Cantor was wrong (or using a different idea of the limit), or WM is > misinterpreting him, my German is not good enough to tell. Daraus folgt, da¤ solche Umformungen einer wohlgeordneten Menge die Anzahl derselben unge.8andert lassen, welche sich auf eine endliche oder unendliche Folge von Transpositionen von je zwei Elementen zur.9fckf.9fhren lassen, ... It follows that only such transformations of a well-ordered set leave its (ordinal) number unchanged, which can be derived from a finite or infinite sequence of transpositions of each two elements, ... > In any case, > it's a proposition WM needs to prove to complete his argument, and he > hasn't done so. What should that be good for? In the binary tree everyone can see that no path can split without two additional edges. So it is by no means possible that there were more paths than edges. Nevertheless the set theorists go through eyes wide shut and do not care. Who would accept logic arguments if set theory was concerned? === Subject: Re: An uncountable countable set > that a sequence of transpositions can not change the order-type. Either > Cantor was wrong (or using a different idea of the limit), or WM is > misinterpreting him, my German is not good enough to tell. > Daraus folgt, da¤ solche Umformungen einer wohlgeordneten Menge die > Anzahl derselben unge.8andert lassen, welche sich auf eine endliche oder > unendliche Folge von Transpositionen von je zwei Elementen > zur.9fckf.9fhren lassen, ... > It follows that only such transformations of a well-ordered set leave > its (ordinal) number unchanged, which can be derived from a finite or > infinite sequence of transpositions of each two elements, ... ordered set into a set which is simultaneously well ordered and not well ordered. > In any case, > it's a proposition WM needs to prove to complete his argument, and he > hasn't done so. > What should that be good for? > In the binary tree everyone can see that no path can split without two > additional edges. So it is by no means possible that there were more > paths than edges. As long as paths all end. > Nevertheless the set theorists go through eyes wide > shut and do not care. Set theorists see that the set of infinite binary paths can be bijected with the power set of the set of edges of those paths. That this result differs from that for finite trees may seem anomalous, but it is in no way contradictory. Infinite is different than finite in many surprising ways. > Who would accept logic arguments if set theory was concerned? === Subject: Re: An uncountable countable set <44b119fc$0$17975$892e7fe2@authen.yellow.readfreenews.net> <44b1845b$0$17984$892e7fe2@authen.yellow.readfreenews.net> <44b28a08$0$17992$892e7fe2@authen.yellow.readfreenews.net> that a sequence of transpositions can not change the order-type. Either >> Cantor was wrong (or using a different idea of the limit), or WM is >> misinterpreting him, my German is not good enough to tell. >Daraus folgt, da¤ solche Umformungen einer wohlgeordneten Menge die >Anzahl derselben unge.8andert lassen, welche sich auf eine endliche oder >unendliche Folge von Transpositionen von je zwei Elementen >zur.9fckf.9fhren lassen, ... >It follows that only such transformations of a well-ordered set leave >its (ordinal) number unchanged, which can be derived from a finite or >infinite sequence of transpositions of each two elements, ... of each two elements. That is a strange expression in this context. ,each of two elements would make more sense, although redundant unless he hadn't previously defined transposition. If that is the case, it would seem Cantor was wrong, you should have no trouble believing that. >> In any case, >> it's a proposition WM needs to prove to complete his argument, and he >> hasn't done so. >What should that be good for? You would have finally proven a contradiction within standard set theory. You've been trying to do that for many years, why give up now, when you're so close? All you need to do is prove that, in this quotation, Cantor was right! >In the binary tree everyone can see that no path can split without two >additional edges. So it is by no means possible that there were more >paths than edges. Nevertheless the set theorists go through eyes wide >shut and do not care. >Who would accept logic arguments if set theory was concerned? Most of us, if you would just provide one. -- David Hartley === Subject: Re: An uncountable countable set <44b11de9$0$17957$892e7fe2@authen.yellow.readfreenews.net> <44b18705$0$17996$892e7fe2@authen.yellow.readfreenews.net> <44b290b2$0$17984$892e7fe2@authen.yellow.readfreenews.net> <44b3eb8f$0$17998$892e7fe2@authen.yellow.readfreenews.net> <44b52c43$0$17938$892e7fe2@authen.yellow.readfreenews.net >> You are using to index simultaneously for two different meanings. >> The first is to index a specific position the second is undefined. >> What does 0.1111 is indexed by the list number 4 = 0.1111 exaclty >> mean? > The unary number n indexes every position 1,2,3,...,n. > How exactly is that achieved? The unary number n provides the existence of every m with 0 < m =< n in my list. All these m have been used to index the digits of n, so these indexes, now in n, can serve to index further numbers. >> What does completely indexed mean? > All digits are indexed. > All positions *are* indexed. There is a bijection between the index set > N and the figures a_i: > ... i ... > ^ > | > v > ... a_i ... That is correct for all a_i of list numbers = naturla numbers. That is not possible for 0.111... . By the way, this is the very reason why 0.111... is not in the list. === Subject: Re: An uncountable countable set >> You are using to index simultaneously for two different meanings. >> The first is to index a specific position the second is undefined. >> What does 0.1111 is indexed by the list number 4 = 0.1111 exaclty >> mean? > The unary number n indexes every position 1,2,3,...,n. > How exactly is that achieved? > The unary number n provides the existence of every m with 0 < m =< n in > my list. All these m have been used to index the digits of n, so these > indexes, now in n, can serve to index further numbers. >> What does completely indexed mean? > All digits are indexed. > All positions *are* indexed. There is a bijection between the index set > N and the figures a_i: > ... i ... > ^ > | > v > ... a_i ... > That is correct for all a_i of list numbers = naturla numbers. That is > not possible for 0.111... . By the way, this is the very reason why > 0.111... is not in the list. In a sense, it IS the list. === Subject: Re: An uncountable countable set >> You are using to index simultaneously for two different >> meanings. The first is to index a specific position the second >> is undefined. What does 0.1111 is indexed by the list number 4 = >> 0.1111 exaclty mean? > The unary number n indexes every position 1,2,3,...,n. >> How exactly is that achieved? > The unary number n provides the existence of every m with 0 < m =< n > in my list. All these m have been used to index the digits of n, so > these indexes, now in n, can serve to index further numbers. Could you reword the last paragraph by means of coherent language? >> What does completely indexed mean? > All digits are indexed. >> All positions *are* indexed. There is a bijection between the index >> set N and the figures a_i: >> ... i ... >> ^ >> | >> v >> ... a_i ... > That is correct for all a_i of list numbers = naturla numbers. That is > not possible for 0.111... . Your presumption that infiite things are not possible has no meaning in contemporary mathematics. Do you still want to show a contradiction that is not based on your absurd presumptions but within i.e ZFC? F. N. -- xyz === Subject: Re: An uncountable countable set > Except that after each (1,2), there will only be a finite initial > subsequence in numerical order and an infinite terminal sequence not yet > ordered. > Using 1 origin indexing, the nth occurrence of (1,2) occurs at the > (n^2 + n)/2 th position in the list of transpostions. > At which (n^2 + n)/2 th operation is the entire list ordered? > At which number can I find the last element of Cantor's diagonal? The successor to the one which finishes your ordering by magnitude. Which, since neither exists, works correctly === Subject: Re: An uncountable countable set <44b119fc$0$17975$892e7fe2@authen.yellow.readfreenews.net> <44b1845b$0$17984$892e7fe2@authen.yellow.readfreenews.net> Except that after each (1,2), there will only be a finite initial >subsequence in numerical order and an infinite terminal sequence not yet >ordered. > Using 1 origin indexing, the nth occurrence of (1,2) occurs at the >(n^2 + n)/2 th position in the list of transpostions. > At which (n^2 + n)/2 th operation is the entire list ordered? > At which number can I find the last element of Cantor's diagonal? > The successor to the one which finishes your ordering by magnitude. > Which, since neither exists, works correctly Both do no exist. That is the reason why Cantor's arguing is correct and mine is not. I see. === Subject: Re: An uncountable countable set > But not all unary sequences are representations of natural numbers. > 0,111... is a sequence which does not represent a natural. > And it cannot be completely indexed by natural numbers. That is precisely wrong, it is only the set of all natural numbers which CAN index it. === Subject: Re: An uncountable countable set <44b11de9$0$17957$892e7fe2@authen.yellow.readfreenews.net> <44b18705$0$17996$892e7fe2@authen.yellow.readfreenews.net> <44b290b2$0$17984$892e7fe2@authen.yellow.readfreenews.net> But not all unary sequences are representations of natural numbers. >0,111... is a sequence which does not represent a natural. > And it cannot be completely indexed by natural numbers. > That is precisely wrong, it is only the set of all natural numbers which > CAN index it. It cannot be completely indexed by all natural numbers. That is no question. Therefore the set of all does not exists. === Subject: Re: An uncountable countable set > The representation 0.111... *does* exist. There is a specific definition > for it (as I have already given), and an axiom through which you may > prove that it does exist. > The axiom says there is an infinite set. It does not say that > 0.111... does belong to that set, in particular because all numbers > which in fact do belong to the set are different from 0.111... . You are using the word set with two different meanings in the same sentence. *And* you do not answer my primary comment. There is a specific definition for a notation like 0.111... Without such a definition it is just a string of eight symbols. > What is wrong about it? Take some particular p and take n any value larger > than or equal to p. We see that An[p] = 1 = K[p]. So what is wrong? > If you take some particular p, this p is a natural, otherwise you could > not take it. Therefore you cannot show that there are digits in > 0.111... which are not indexed by naturals. > For all p there is an n such that An[p] = K[p]; > But as K is not in the list > of natural numbers, it must differ from the numbers in the list. And > this difference cannot be accomplished other than by K having more > digits than any number in the list. Again, this is impossible. Hence we > have a contradiction. Pray *prove* why that is impossible. You always only state it but provide not prove. Each An has finitely many digits. K has infinitely many digits. If you disagree with the second you disagree with the axiom of infinity, but that is philosophical. > 0.111... - 0.111...1 = 0.000...0111... > > such that there are more 1's in K than in any An. > > The last is true. But it is also true that for all p there is an An. > Or please exhibit a p for which it is not wrong. > This equation > 0.111... - 0.111...1 = 0.000...0111... > says that for any An = 0.111...1 there is a p = n+1 which cannot be > indexed. This makes absolutely no ense to me. And again is no answer to the question I posed. > All digits which are indexed by smaller list numbers can be indexed by > one larger list number. Therefore always only one is required. > > Care to explain the implication you mention above? > How does K differ from all natural unary numbers together? Again you fail to answer a question, Your implication was: > If you think the sentence all positions of K = 0.111...are indexed by > list numbers is not equivalent to the sentence K > is in the list, then you seem to imply that more than one list number > is required to index the digits of K. Care to explain it? > > You do not need the first two list numbers 0.1 and 0.11, because > > the third alone is sufficient: 0.111 does index the first three > > digits p = 1 to 3 of 0.111... > > I am a bit at a loss here what you mean with indexing. > > Again care to explain? How can a single number index three digits? > instead of to indeex you can also say to cover, i.e. to have more > 1's than (or at least as many as) the number indexed or covered. Well, to index can be stated as to cover. But not the other way around. > So let us find a resolution of this dilemma by mathematics. > > Define *- by > a_i *- b_i = a_i - b_i if a_i > b_i else a_i *- b_i = 0 > (subtraction down to zero but without negative numbers) > > Consider 0.111... and the list numbers as sequences or as vectors, such > that *- can be applied to each term separately. > > If you are right, you must maintain the result: > > 0.111... *- (0.1 + 0.11 + 0.111 + ...) = 0.000... > > I can make no sense of the second term. What do you *mean* with > (0.1 + 0.11 + 0.111 + ...) > The sum of all list numbers by digit, as shown in the following > example: > 0.1 > +0.11 > +0.111 > _____ > =0.321 Ok. What about + ...? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: An uncountable countable set The axiom says there is an infinite set. It does not say that > 0.111... does belong to that set, in particular because all numbers > which in fact do belong to the set are different from 0.111... . > You are using the word set with two different meanings in the same sentence. The set of naturals shold be the same set as the set of indices of 0.111... . But it cannot. > *And* you do not answer my primary comment. There is a specific definition > for a notation like 0.111... Without such a definition it is just a string > of eight symbols. There is a specific definition that the set of all naturals has a cardinal number larger than any element of the set. This definiion is wrong. I show this by the fact that 0.111... is not in the set of unary reprsetations of all natural numbers. aleph_0, the number of 1's in 0.111... is *not* the cardinal number of |N. > For all p there is an n such that An[p] = K[p]; Te assumption that all digits of 0.111... can be inexed by natural numbers while 0.111... has the property to to be outside of the list. > But as K is not in the list > of natural numbers, it must differ from the numbers in the list. And > this difference cannot be accomplished other than by K having more > digits than any number in the list. Again, this is impossible. Hence we > have a contradiction. > Pray *prove* why that is impossible. You always only state it but provide > not prove. Each An has finitely many digits. K has infinitely many digits. > If you disagree with the second you disagree with the axiom of infinity, > but that is philosophical. K cannot have infinitely many digits if all can be indexed by finite numbers. Infinite is larger than any finite number. Therefore every list number taken will not be capably of indexing K. 0.111... - 0.111...1 = 0.000...0111... > > such that there are more 1's in K than in any An. > > The last is true. But it is also true that for all p there is an An. > Or please exhibit a p for which it is not wrong. > > This equation > 0.111... - 0.111...1 = 0.000...0111... > says that for any An = 0.111...1 there is a p = n+1 which cannot be > indexed. > This makes absolutely no ense to me. And again is no answer to the > question I posed. Whatever p you choose, there is always a place in 0.111... which cannot be indexed. Therefore there is a place which cannot be indexed in principle. > Again you fail to answer a question, Your implication was: > > If you think the sentence all positions of K = 0.111...are indexed by > > list numbers is not equivalent to the sentence K > > is in the list, then you seem to imply that more than one list number > > is required to index the digits of K. > Care to explain it? You know that one list number is not capable of indexing 0.111... , bu you assume that 0.111... can be indexed completely. Therefore more than one list number should be required. But this would not help. > Well, to index can be stated as to cover. But not the other way around. Therefore never more than one number is required for indexing purposes. > > 0.111... *- (0.1 + 0.11 + 0.111 + ...) = 0.000... > > I can make no sense of the second term. What do you *mean* with > (0.1 + 0.11 + 0.111 + ...) > > The sum of all list numbers by digit, as shown in the following > example: > > 0.1 > +0.11 > +0.111 > _____ > =0.321 > Ok. What about + ...? Definition: 1+1+1+... := 1 === Subject: Re: An uncountable countable set > The axiom says there is an infinite set. It does not say that > 0.111... does belong to that set, in particular because all numbers > which in fact do belong to the set are different from 0.111... . > You are using the word set with two different meanings in the same > sentence. > The set of naturals shold be the same set as the set of indices of > 0.111... . But it cannot. Maybe not in mueckenh's world but it works fine in mine. if mueckenh chooses to exile himself in such a way from the mainstream, he should not be surprised when the mainstream elects not to join him in his self-exile. > There is a specific definition that the set of all naturals has a > cardinal number larger than any element of the set. This definiion is > wrong. Not in Zf or ZFC or NBG. Mueckkenh must make some assumptions or he cannot conclude anything. But absent a statement of anything of his assumptions except for his distaste for infinite sets, we have no way to test the validity of his conclusions. > I show this by the fact that 0.111... is not in the set of unary > reprsetations of all natural numbers. Neither is 2. So what? > Aleph_0, the number of 1's in > 0.111... is *not* the cardinal number of |N. Really? I seem to see a trivial bijection between the numbers of 1's preceding any of the 1's in 0.111... and the members of N, which bijextion disproves mueckenh's claim. > Te assumption that all digits of 0.111... can be inexed by natural > numbers while 0.111... has the property to to be outside of the list. But none of 0.111,,,'s digits need go unindexed. One can, in fact, use the members of the endless well ordered set {0.1, 0.11, 0.111, ...} to index the digits of 0.111... . > K cannot have infinitely many digits if all can be indexed by finite > numbers. Assumptions made without proof, such as the above, must either be axioms or are of no consquence. > Whatever p you choose, there is always a place in 0.111... which cannot > be indexed. Therefore there is a place which cannot be indexed in > principle. Whatever finite natural one picks, there is always a larger one, but how does that prohibit existence of a totality of all of them? Only if one ->assumes<- some axiom or set of axioms justifying such a prohibition. So that mueckenh's views are quite as artificial as ours, in that he makes unjustifiable asssumptions about what one may imagine and then claims they are justifiable. When we make assumptions, we merely call them axioms and, until someone deduces a conflict between axioms, we go on from there. > You know that one list number is not capable of indexing 0.111... , bu > you assume that 0.111... can be indexed completely. Our axiom systems provide adequate mechanisms. Mueckenh's objections are only valid in a different axiom system which has no more ultimate justification than our own. === Subject: Re: An uncountable countable set > > I proved in my special list even that the diagonal number is a > > rational. > > I wonder whether it was a proof or just some handwaving. > > 0.0 > 0.1 > 0.11 > 0.111 > ... > replace 0 by 1. > > As far as I see the diagonal starts with 1.000... Am I right? > I use only the digits behind the point: So the diagonal is 0.111... = > 1/9. So you imply additional 0's after your notation. I was not sure. > If there is no other outcome possible, I don't need a further > definition. 0.999... = 1 follows from the definition of (+,-,*,/) in > the real numbers. Pray tell me how. Somewhere else you stated that the representation 0.111... did not exist. What are you arguing? Either the representation 0.111... does exist or not. And if it does exist the definitions of the mathematical operations are not sufficient to give a meaning to it. Anyhow, how do you show that 1.000... - 0.999... = 0 with the definitions you are using? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: An uncountable countable set > I proved in my special list even that the diagonal number is a > > rational. > > > I wonder whether it was a proof or just some handwaving. > > 0.0 > > 0.1 > > 0.11 > > 0.111 > > ... > > replace 0 by 1. > > As far as I see the diagonal starts with 1.000... Am I right? > > I use only the digits behind the point: So the diagonal is 0.111... = > 1/9. > So you imply additional 0's after your notation. I was not sure. > If there is no other outcome possible, I don't need a further > definition. 0.999... = 1 follows from the definition of (+,-,*,/) in > the real numbers. > Pray tell me how. Somewhere else you stated that the representation > 0.111... did not exist. What are you arguing? Either the representation > 0.111... does exist or not. And if it does exist the definitions of the > mathematical operations are not sufficient to give a meaning to it. > Anyhow, how do you show that 1.000... - 0.999... = 0 with the definitions > you are using? In usual mathematics we have 10 * 0.999... = 9.999... and from that we get easily 0.999... = 1. I argue that 0.111... does not exist. When I will have succeded, also 0.999... will be abolished. But that has not yet been generally accepted. OK? === Subject: Re: An uncountable countable set > In usual mathematics we have 10 * 0.999... = 9.999... and from that > we get easily 0.999... = 1. > I argue that 0.111... does not exist. When I will have succeded, also > 0.999... will be abolished. But that has not yet been generally > accepted. OK? As mueckenh will never succeed in convincing the majority of mathematicians to believe what they do not believe, 0.111... and 0.999... will both exist forever. === Subject: Re: An uncountable countable set > > > I proved in my special list even that the diagonal number is a > > rational. > > > I wonder whether it was a proof or just some handwaving. > > > 0.0 > > 0.1 > > 0.11 > > 0.111 > > ... > > replace 0 by 1. > > As far as I see the diagonal starts with 1.000... Am I right? > > I use only the digits behind the point: So the diagonal is 0.111... = > 1/9. > So you imply additional 0's after your notation. I was not sure. > If there is no other outcome possible, I don't need a further > definition. 0.999... = 1 follows from the definition of (+,-,*,/) in > the real numbers. > Pray tell me how. Somewhere else you stated that the representation > 0.111... did not exist. What are you arguing? Either the representation > 0.111... does exist or not. And if it does exist the definitions of the > mathematical operations are not sufficient to give a meaning to it. > Anyhow, how do you show that 1.000... - 0.999... = 0 with the definitions > you are using? > In usual mathematics we have 10 * 0.999... = 9.999... and from that > we get easily 0.999... = 1. > I argue that 0.111... does not exist. If by usual mathematics you mean a mathematics in which the meaning of infinite decimals have not been defined, then we don't have 0.999... or 9.999... either. When you say 0.111... does not exist, do you mean (a) the notation is undefined, or (b) when you define it, you get a number that doesn't exist? If (a), then that is fixed by defining it, which can be done with limits. If (b), then you are arguing that 1/9 doesn't exist? - Randy === Subject: Re: An uncountable countable set ... > Indeed, at that time it was not yet an axiom, but the thought was > present that infinite sets did exist. > > There was not such an assumption in general. Infinite sets did not > exist (as they do not exist yet), but only the potential infinite was > accepted. Cantor was little understood and was blamed to do philosophy > or theology but not mathematics. > > ... die Gesamtheit aller endlichen Zahlen 1, 2, 3, ..., v, ..., Cantor. > just because of views like that. present that infinite sets did exist. And indeed, Cantor had not yet formulated it as an axiom, but nevertheless did state it. What you are arguing against is *not* against the proof, but against the axiom. > It leads to the inconsistency that you must demand 0.111... *- (0.1 + > 0.11 + 0.111 + ...) = 0.000... but 0.111... not being in the sum. > > As you have not defined the meaning of the second term, this makes no > sense. Offhand I would say it goes 0.1, 0.21, 0.321, 0.4321, 0.54321, > 0.654321, 0.7654321, 0.87654321, 0.987654321, 1.0987654321, and I see > no obvious limit emerging when we do the infinite sum. > The numbers are to be understood as sequences or vectors, so your last > sum is 0.(10),9,8,7,6,5,4,3,2,1. The next would be 0. > (11),(10),9,8,7,6,5,4,3,2,1 and so on. The magnitudes of the numbers do > not matter if they are larger than zero, because 1 *- omega = 1 *- 1 = > 0. My question is whether > 0.111... *- (0.1 + 0.11 + 0.111 + ...) = 0.000...? Next question. What is the meaning of + ...? How do you define that? I remind you that you are trying to show an inconsistency with the axiom of infinity, so you should define it using that axiom in order to be able to show an inconsistency. And so there is no largest natural number. > And you must > refuse to calculate with fractions in mappings. > > Here I have no idea what this means. > The binary tree: Your binary tree again. What is the relation with refuse to calculate with franction in mappings? Why do you never give direct answers to questions? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: An uncountable countable set ... > The first well-order taken from Cantor's. The others remain well-orders > because: wenn in einer wohlgeordneten Menge irgend zwei Elemente m und > m' ihre Pl.8atze in der gesamten Rangordnung wechseln, so wird dadurch > der Typus nicht ver.8andert, also auch nicht die Anzahl oder > Ordnungszahl. Daraus folgt, da¤ solche Umformungen einer > wohlgeordneten Menge die Anzahl derselben unge.8andert lassen, welche > sich auf eine endliche oder unendliche Folge von Transpositionen von je > zwei Elementen zur.9fckf.9fhren lassen, d. h. alle solche Änderungen, > welche durch Permutation der Elemente entstehen. I translate (not needed for Franziska, but needed for others): When in a well-ordered set any two elements m and m' change place in the common order, the order-type will not change, so also the ordinal of a well-ordered set leave the ordinal number unchanged, that can be written as a finite or infinite sequence of transpositions of two elements, that is, all such changes that emerge through permutations of elements. I see a problem here: It is not clear what the meaning is of the words transformations (Umformungen) and permutations (Permutation). I think, it is clear that it does not include *any* re-ordering, although many of them can be written as an infinite sequence of transpositions, while some those *do* change the ordinal number. The reordering of the naturals (0, 1, 2, ...) to (1, 2, 3, ..., 0) can be written as an infinite sequence of transpositions, but it changes the ordinal number. But is it a permutation? On the other hand, there are many infinite sequences of transpositions that do not change the ordinal number. Consider a well-ordered set of type w * w. Interchange the first two elements of each maximal subset of order type w. Anyhow, either WM's interpretation is wrong, and so his conclusion is wrong, or WM's interpretation is right and Cantor's statement is wrong, and so (again) WM's conclusion is wrong. I think the interpretation by WM is wrong. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: An uncountable countable set The first well-order taken from Cantor's. The others remain well-orders > because: wenn in einer wohlgeordneten Menge irgend zwei Elemente m und > m' ihre Pl.8atze in der gesamten Rangordnung wechseln, so wird dadurch > der Typus nicht ver.8andert, also auch nicht die Anzahl oder > Ordnungszahl. Daraus folgt, da¤ solche Umformungen einer > wohlgeordneten Menge die Anzahl derselben unge.8andert lassen, welche > sich auf eine endliche oder unendliche Folge von Transpositionen von je > zwei Elementen zur.9fckf.9fhren lassen, d. h. alle solche Änderungen, > welche durch Permutation der Elemente entstehen. > I translate (not needed for Franziska, but needed for others): > When in a well-ordered set any two elements m and m' change place in > the common order, the order-type will not change, so also the ordinal > of a well-ordered set leave the ordinal number unchanged, that can be > written as a finite or infinite sequence of transpositions of two > elements, that is, all such changes that emerge through permutations > of elements. > I see a problem here: > It is not clear what the meaning is of the words transformations > (Umformungen) and permutations (Permutation). I think, it is clear > that it does not include *any* re-ordering, although many of them can be > written as an infinite sequence of transpositions, while some those *do* > change the ordinal number. These words are uninteresting. Transpositions or exchanges of two elements are clearly defined. Infinitely many are admitted by Cantor. > The reordering of the naturals (0, 1, 2, ...) to (1, 2, 3, ..., 0) can > be written as an infinite sequence of transpositions, but it changes the > ordinal number. But is it a permutation? Unimportant for Cantor's statement. There is no transposition which could change the ordinal number. Only the infinite process does what cannot be done by any finite set of transpositions. This shows that infinite processes and infinity is doubtful. My example with |Q shows that infinity is impossible. > On the other hand, there are many infinite sequences of transpositions > that do not change the ordinal number. Consider a well-ordered set of > type w * w. Interchange the first two elements of each maximal subset of > order type w. > Anyhow, either WM's interpretation is wrong, and so his conclusion is wrong, > or WM's interpretation is right and Cantor's statement is wrong, and so > (again) WM's conclusion is wrong. > I think the interpretation by WM is wrong. I know your interpretation is wrong. But that does not at all affect my proof, because, as Virgil, emphasized frequently, there is no transposition which changes the well-order to normal order while maintaining the initial enumeration (= well-order). Would infinitely many transpositions be possible, this would necessarily happen. === Subject: Re: An uncountable countable set > ... > The first well-order taken from Cantor's. The others remain well-orders > because: wenn in einer wohlgeordneten Menge irgend zwei Elemente m und > m' ihre Pl.8atze in der gesamten Rangordnung wechseln, so wird dadurch > der Typus nicht ver.8andert, also auch nicht die Anzahl oder > Ordnungszahl. Daraus folgt, da¤ solche Umformungen einer > wohlgeordneten Menge die Anzahl derselben unge.8andert lassen, welche > sich auf eine endliche oder unendliche Folge von Transpositionen von je > zwei Elementen zur.9fckf.9fhren lassen, d. h. alle solche Änderungen, > welche durch Permutation der Elemente entstehen. > I translate (not needed for Franziska, but needed for others): > When in a well-ordered set any two elements m and m' change place in > the common order, the order-type will not change, so also the ordinal > of a well-ordered set leave the ordinal number unchanged, that can be > written as a finite or infinite sequence of transpositions of two > elements, that is, all such changes that emerge through permutations > of elements. > I see a problem here: > It is not clear what the meaning is of the words transformations > (Umformungen) and permutations (Permutation). I think, it is clear > that it does not include *any* re-ordering, although many of them can be > written as an infinite sequence of transpositions, while some those *do* > change the ordinal number. > These words are uninteresting. Transpositions or exchanges of two > elements are clearly defined. Infinitely many are admitted by Cantor. > The reordering of the naturals (0, 1, 2, ...) to (1, 2, 3, ..., 0) can > be written as an infinite sequence of transpositions, but it changes the > ordinal number. But is it a permutation? > Unimportant for Cantor's statement. That must mean that meuckenh does not see any sequence of permutataions that can produce that result. but even if one finds such a sequence of transpostions, the result is still well ordered, though of a different order type, that of the successor of N. > There is no transposition which > could change the ordinal number. Only the infinite process does what > cannot be done by any finite set of transpositions. This shows that > infinite processes and infinity is doubtful. My example with |Q shows > that infinity is impossible. To those who believe it is impossible, perhaps it is, but to those who believe otherwise, it is not. > I think the interpretation by WM is wrong. > I know your interpretation is wrong. It ain't what you don't know that harms you, it's what you know that ain't so. Mark Twain. Received-SPF: None; receiver=nym.alias.net; client-ip=70.89.231.62; envelope-from=; helo=mail.twistycreek.com === Subject: Re: Republican *Database* Against Liberals `` Comments: This message did not originate from the above address. It was automatically remailed by an anonymous re-mail service. Visit http://www.twistycreek.com/ for details. Contact admin^at^twistycreek^dot^com Mail-To-News-Contact: postmaster@nym.alias.net -----BEGIN PGP SIGNED MESSAGE----- AS REPUBLICANS, WE already know to aggressively & proactively boycott Liberals across-the-board. But the recent aiding-and- abetting-terrorists scandal at the New York Times serves as a stark reminder that we Republicans need to put these traitors out-of-business, and do everything we can to force them into bankruptcy; help them to lose their jobs, their homes, their families, and eventually lose their miserable inhuman lives whereupon their diseased, unrepentant, rabidly hateful souls will have Hell to pay--then Judgment, the karmic books opened... We already know to boycott every so-called Buy-Blue Liberal- owned business, and we already know to boycott Liberals on a personal level, e.g. passing them over for promotion at work, getting them fired from their jobs, making their workload so difficult that they quit their jobs. And having them globally blacklisted from working at *any* Republican-owned businesses nationwide, and eventually worldwide. Make them go flat broke. Any Liberals who fill out job applications, or submit their resumes, seeking employment at any Republican-owned business should have their dossiers filed in a *DO-NOT-HIRE* database that we can share with all other Republican-owned businesses across the country. If a Liberal wants to do business with you, whether it's buy, sell or trade, always make sure they pay a premium to do so. Conversely, reward Republicans with discounts, and do a better job for them, whatever it is you do for a living. At least NEVER give a Liberal an even break! ^^^^^ We already know that our Republican-dominated United States Federal Government intelligence agencies, notably the N.S.A., have everybody's number. They know exactly who is Republican, meaning those who are loyal, law-abiding, tax-paying American patriots who want to help our government fight against these despicable terrorist-supporting Liberals in every way we can... And the N.S.A. know exactly who is Liberal, meaning those who *HATE* God, *HATE* Family, and *HATE* Country, and are doing everything in their Liberal power to compromise our National Security, to support and defend terrorists, and surrender our great nation to viciously Anti-American, Communistic-Liberal Atheist militants like the A.C.L.U. (i.e. the most dangerous organization in the world!), the U.N. (second most dangerous!), and other known genocidal enemies of United States of America, extremely dangerous Islamo-Fascists like Al-Qaeda, Hammas, and criminally-insane dictators of countries like Iran, North Korea, Venezuela, etc. The Liberals support these genocidal criminals! No matter what relatively minor disagreements some Republicans may have with some of our other fellow Republicans, hot-button issues like illegal immigration, states-rights, etc., we need to stand faithfully 100% UNITED against the Liberal Terrorist enemy. We need to identify every Liberal in the country, then we need to build a Republican-moderated public database on the Internet, which anyone can search on-line in the event someone suspected of being a Liberal applies for a job, or submits bids on contracts, etc. I've already posted helpful instructions for boycotting Liberals, But even small businesses not listed on the Buy-Blue website need to be blacklisted on a nationwide Internet database which is owned and controlled by Republicans. This will help prevent the anarchist Liberal disinformation agents from posting false reports, false names, and other vicious slander that Liberals are so well known for fabricating and spreading around like a virus (e.g. look at all the Antichrist Liberal Atheist garbage infesting these public newsgroup forums). Also, every Church, Synagogue, Mosque, Ashram, Classroom, etc. should be actively encouraging their flocks to vote Republican across-the-board in every election. Every business owner should do the same with their employees. Conspicuous notices should be posted to remind everyone from the executive suites down to the mailroom to register & vote Republican in the November elections. Children, and young adults not yet old enough to vote, should be reminded how much the Liberal A.C.L.U. Democrats *HATE* the Boy Scouts, and *HATE* Jesus Christ & His ekklesia most especially! (Remind children that Liberals are the enemy of the human race.) Teachers in public schools and university classrooms should be encouraging their students to support our great and devoutly- faithful President George W. Bush and our Republican-dominated U.S. Federal Government. Any teachers who don't should be fired, like that rabidly-insane inhuman animal Ward Churchill deserves to be dropped off buck-naked in the heart of the Sunni Triangle. _______________________________________________________________ Again, please forward this message to everyone you know, and to every newsgroup you are subscribed to. We are a nation at WAR, and we need to destroy the Liberal enemy using every weapon at our disposal. The pen being mightier than the sword, we need to get the word out to boycott all Liberals until they go bankrupt. We need to vote every Liberal out of public office, especially on the federal level. We must purge our government of Liberals across-the-board. Appoint all nine U.S. Supreme Court Justices as patriotic, God-fearing, traditional Family-values oriented Republican moderate-to-conservatives. We need more than a 5-4 vote. We need a Nine-to-Zero vote effectively rubberstamping our Republican agenda. Our President needs the line-item veto, so we can pass Republican legislation and defeat the Liberals on all fronts. We need to **de-Liberalize** America, and soon... _____________________________________________________ My suggestion is for a comprehensive, Republican-made list of *every* Liberal in the United States, so that Republican-owned businesses will know not to hire or otherwise conduct any business with them *whatsoever*, any more than we Republicans would want to do business directly with al-Qaeda, etc. Seeing that the Liberals openly support, aid and abet Anti-American Terrorists, I see no reason why any Republican in America should do business with any Liberal--even if it's just hiring them for chinaman's wages to flip burgers, clean toilets, or do other menial grunt work. I wouldn't hire them period. Of course, practically all employers, large and small, run routine background checks on potential employees, including searching the web and Google Groups archives for possible clues as to the prospect's own political, religious and other affiliations, their biases, sexual preference, criminal record, financial status, etc, as experienced skiptracers can find information on anyone. Beyond this, all Republican-owned businesses should join forces in blacklisting all known Liberal applicants from employment at any of the aforesaid businesses anywhere in the United States of America, and abroad where applicable; any job from slave-class labor, to CEO of Fortune 500 Co's. The idea is to bankrupt the Liberal enemy of the U.S.A., in much the same way as we are fighting the global war against the Liberals' dear friends, the islamo-Fascist Terrorists, around the world. I just heard Bill O'Reilly say the NY Times has lost thousands of subscribers since their latest, treasonous, terrorist-supporting escapade. Meanwhile, Ann Coulter's *excellent* new book 'Godless: The Church of Liberalism', keeps selling like hotcakes! http://www.newsmax.com/adv/godless/ I believe we need to step up the war on Liberals here on the home front, much as Israel just stepped up the war on Hammas in the Gaza Strip. As they say on HBO's 'The Sopranos', the Liberal Atheists have gotta go... VOTE Republican! Daniel Joseph Min http://pgp.mit.edu:11371/pks/lookup?op=get&search=0x2B1CCFE7 *Download Min's Banned (Freeware) Books: http://www.2hot2cool.com/11/danieljosephmin/ *Min's Spiritual I.Q. Test (how smart are you, really): *Min's Google-Archived Home Page On The WWW: -----BEGIN PGP SIGNATURE----- iQA/AwUBRLZcaZljD7YrHM/nEQJ4SwCgov7eTdZRF28pMHho5cyoCBq6/F4AoOpm IgN+c4Pl3J4hhfnS3rAgndL0 =bZgL -----END PGP SIGNATURE----- === Subject: Re: Republican *Database* Against Liberals `` bad filter. bad bad bad. let min through. bad filter. i'll fix you so you recognize this min. then good filter. nice filter. thank you, filter. === Subject: Re: Question about Cantor's proof <%uatg.17857$1G3.17305@reader1.news.jippii.net> Cantor seems to have suffered of severe clinical depression, but there >> is no reason to suppose his illness had anything much to do with his >> mathematics or the opposition it met. >> The irrational backlash from many of the mainstream mathematicians >> of the time and their efforts to slander him could not have helped his >> depression. > What efforts to slander Cantor are you talking about? My mistake. The two signs of old age are loss of memory. It was once thought that Cantor's recurring bouts of depression were triggered by the opposition his work met at the hands of Kronecker. While Cantor's mathematical worries and his difficulties dealing with certain people were greatly magnified by his depression, it is doubtful whether they were its cause, which was probably bipolar disorder. === Subject: Re: Question about Cantor's proof cardinality of integers is fine, it's a nice mathematical result which > has no application in the physical world. What UTTER bull. For the nth time: Cantor's Theorem DOES NOT MENTION infinity. The theorem is that the powerset is bigger than the set, and the proof is by diagonalization. The powerset of a set is the set of all its subsets. If the set is countable to begin with then you can make a LIST of (at least some of) the subsets. Then you can ask how long this list is. The theorem says that NO list whose length is the same as the number of elements in the set (i.e. no SQUARE array of bits, where the side-length is the size of the set) contains ALL the subsets. We can prove this simply by taking the anti-diagonal of the square. It is guaranteed to disagree in (at least) one place, with EVERY row (and EVERY column) of the square, so it is NOT one of the subsets listed. That is all. This DOES SO TOO have physical consequences; you can take 30 coins (Heads or Tails) and arrange 25 of them in a 5x5 square and use this theorem to prove that there has to be an arrangement of them that does NOT occur as a row or column of your square. You can then take the remaining 5 coins and arrange THEM in this HT pattern that did NOT appear in the square. Knowing that you can always do this is nice to know. Thinking that somehow infinity makes it different is, well, crazy. > To feed people or send > spacecraft to Mars you need the cybernetic interpretation of infinity: This is just idiotic. However these finally wind up getting done, I am quite confident it will NOT involve ANYbody asking YOU about cybernetic infinity. > The potential infinity has real applications, such as programming a > spacecraft to report data forever, knowing that some physical failure > will evertually stop it. The actualized infinity is a mathematical > concept only. A mathematical concept that was absolutely necessary to the discorver of the math that was used to compute the trajectory of the data-reporter. And just for the record, it is NOT known that some physical failure will eventually stop it. Strange encounters may be possible out there. Even if they are not, YOU don't know it. > Cantor's nightmarish infinity of infinities must have been infinitey > stressful for him, for he committed suicide. The crank-posters here are slowly doing the same thing to themselves; life is short, and they are wasting it, one post at a time. I can only wonder whether we are not dragging ourselves down with them. === Subject: Re: Question about Cantor's proof > As such, mathematics is scientific in nature -- it has testable > consequences. But Cantorian set theory, which was created to provide a > formal basis for proving Cantor's theorem, leads to mathematics with > no testable consequences. You have absolutely no idea what testable consequences means. You have tried to allege that |a^2 -2b^2|>=1 is a hypothesis that has testable consequences, while the square root of 2 is irrational is not one. How can either of these have more or fewer testable consequences than the powerset of a set is bigger than the set, EVEN when the set is infinite? How can you expect to be taken seriously? === Subject: Re: Question about Cantor's proof > The other theory I have is that this is related to the prevalence of >bull in current society. Jiri You mean like equating opponenets of Cantor's proof with trisectors and circle squarers? === Subject: Re: Question about Cantor's proof >> Cantor seems to have suffered of severe clinical depression, but there >> is no reason to suppose his illness had anything much to do with his >> mathematics or the opposition it met. > > The irrational backlash from many of the mainstream mathematicians > of the time and their efforts to slander him could not have helped his > depression. > What efforts to slander Cantor are you talking about? Think about Kronecker. The dictator of German mathematics at that time. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Question about Cantor's proof > The irrational backlash from many of the mainstream mathematicians > of the time and their efforts to slander him could not have helped his > depression. > > What efforts to slander Cantor are you talking about? > Think about Kronecker. The dictator of German mathematics at that time. Kronecker told deliberate falsehoods about Cantor to defame him? -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Question about Cantor's proof > My observation is that most of the unbelievers have problems > accepting that an infinite sequence of digits can be completed > as a real value. This usually indicates a deeper misunderstanding > or unacceptance of infinite sequences, actual infinities, or just > plain problems dealing with the concept of infinity. Those unbelievers mostly come from Germany. At least, that is my impression. What bothers me more is that some of those are teaching mathematics at universities. > Unbelievers at the other end of the spectrum accept the concept > of infinity, but don't accept that there is more than one size of > infinity, so all infinities are equivalent. Countable and > uncountable are false distinctions to them. I think that is the same group. When you accept a completed infinity, there is only one infinity. At least, that appears to be the reasoning. But in most cases the reasoning relies on the incomprehension of infinite sets. > Somewhere in between are the cranks who think that infinity is > just another number, so that oo/2, for example, has meaning. > Since Cantor's theorems don't embrace such intuitive notions > of infinite values, he must be wrong according to them. That is the section where I find computer scientists only. > And then there are those who can't deal with decimal fractions > for representing reals. They don't accept Cantor's diagonal > because the decimal fractions can't really represent irrational > values. These guys usually also believe that 0.999... is less > than 1 by some infinitesimal quantity. Again, this is generally > indicative of a deeper misunderstanding of infinity. This is a much larger group. That is the group that thinks that something like 0.999... has meaning (as a number) without proper definitions. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Question about Cantor's proof ... > I notice that many of the unbelievers and cranks come from > a computer programming background, and as such their > ideas about binary arithmetic, fixed-length registers, finite > Turing machines, computability, and so forth, somehow must > govern the rules of abstract mathematics. Not only there. You will also find them amongst the physicists. Especially those that are technically inclined. David Petry is one of that class. They think that mathematics is a science that should help them to develop their theories. That the theories (string theory, Ensteinian stuff) do not conform to what they think should be right is blamed on mathematics. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Question about Cantor's proof > Now uncountability of the reals is untestable by some > experiment and is definately not the result of common sense. > That is ENTIRELY wrong. > There IS a relevant experiment and it happens > EVERY time you try to COUNT the reals: you conduct > the OPPOSITE experiment. You say, Hmmm, let me COUNT > the reals. And you hypothesize a counter. And you conduct > your counting experiment. And it VERY quickly FAILS: USING > COMMON SENSE. Everytime I try to count the reals I run out of paper to write on. I make a proof so that I don't have to run experiments. That was my point. Further about common sense: Common sense is what is common among the population. Try going to a local bar and see how many people you convince with your proof. I had hard time explaining this proof to my dad and he's a phd (in chemistry). > These are people who think that mathematical truth > is guided by their intuition and their imagination. > Those are precisely the two things you need to use > to show that your attempt to count the reals has failed. No, I use formal logic. Intuition only gets me so far. If this was an intuitive result it would have been known far before Cantor. A mathematician should obviously not be guided only by thinking this is the way it SHOULD be which is what intuition gives you. Mathematical truth doesn't change while my intuition does (it gets refined or perhaps changed entierly). Jiri === Subject: Re: Question about Cantor's proof > Try going to a local bar and see how many people you > convince with your proof. Hmph -- try going to a bar and seeing how many people remember what the concept of an infinite decimal is. Most people never understood infinity right THE FIRST time. > I had hard time explaining this proof to my > dad and he's a phd (in chemistry). That's your fault, not his. >These are people who think that mathematical truth >is guided by their intuition and their imagination. > Those are precisely the two things you need to use > to show that your attempt to count the reals has failed. > No, I use formal logic. Intuition only gets me so far. If this was an > intuitive result it would have been known far before Cantor. What utter bull. It probably WAS known, actually. You should see if Franz Fritsche will accept your patronage for a search back through the 18th and 19th century literature at large for examples of diagonal arguments. The important point about the Cantorian argument/result is that it actually has NOTHING to do with infinity. As the proof occurs in ZF, the axiom of infinity IS NOT used. This is because the proof applies EQUALLY well TO FINITE sets. No square array contains its own anti-diagonal, as a row or as a column. That REALLY IS common sense. It remains common-sensical even if you allow the side-length of the square to be omega (or any other countable ordinal) as opposed to a finite ordinal -- NOT that there is any common sense relevant to understanding what a countable ordinal (in general) is. But there is some relevant to understanding what the simplest/smallest one (omega) is, since it is just the collection of all those finite things, all of which WERE available to common sense. === Subject: Re: Question about Cantor's proof > Try going to a local bar and see how many people you > convince with your proof. > Hmph -- try going to a bar and seeing how many people remember > what the concept of an infinite decimal is. Most people never > understood infinity right THE FIRST time. original point). I see no point in further discussing what you or I mean by common sense. We obviously have a very different definition of those two words. Jiri === Subject: Re: Question about Cantor's proof Ha ha! But anyway Cantor did have nervous breakdowns and was in and > out of mental institutions from 1884 on. > Cantor seems to have suffered of severe clinical depression, but there > is no reason to suppose his illness had anything much to do with his > mathematics or the opposition it met. Some of us have the nagging suspicion (with no direct supporting evidence) that Cantor knew he had done something sneaky and wrong, without knowing exactly what, and the burden of guilt about this may have contributed to his illness. === Subject: Re: Question about Cantor's proof >Ha ha! But anyway Cantor did have nervous breakdowns and was in and >out of mental institutions from 1884 on. > Cantor seems to have suffered of severe clinical depression, but there > is no reason to suppose his illness had anything much to do with his > mathematics or the opposition it met. > Some of us have the nagging suspicion (with no direct supporting > evidence) that Cantor knew he had done something sneaky and wrong, > without knowing exactly what, and the burden of guilt about this may > have contributed to his illness. Some of us have the nagging suspicion that the many unprincipled personal attacks on Cantor by Kronecker had a good deal to do with it. === Subject: Re: Question about Cantor's proof evidence) that Cantor knew he had done something sneaky and wrong, > without knowing exactly what, and the burden of guilt about this may > have contributed to his illness. Some of us have the nagging suspicion that you wash your hands compulsively and check under the bed each night to see if burglars are hiding there. === Subject: Re: Question about Cantor's proof = 1, as a mathematician we know that is > a true statement, because there is a proof. The physicist (at least David > Petry) does not accept the proof, and states it has testable consequences. > Making it truer everytime he tests it. FWIW, my opinion of you has just dropped precipitously. You are a liar. === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? >> There's only one theory with no axioms. >> Correct. It is the one with no theorems. > No that's not what it is. > Consider Goedel, vis-a-vis incompleteness, and the physicists' notion > of a Theory of Everything. Apparently, those people never heard of > Goedel, or didn't agree that his results about incompleteness hold in > their case, because they talk about a Theory of Everything. You're confused. The physicists' TOE delas with unifying gravity and quantum physics, and while it uses a lot of complex math, it has nothing to do with set theory. > There is > no Theory of Everything in ZF or other regular set theories. There > is no universe in ZF. Quantify over sets, in ZF: it's not a set. So, > it's a non-sets theory. If you mean there can be no set of all sets, yes, that is well known. > There's only one theory with no axioms. It has all the theorems. Any > other is inconsistent or incomplete, just ask Goedel. Your regular set > theory is incomplete, via Goedel, and inconsistent, via universal > quantiification, not to mention paradoxes in them, generally paradoxes > of unrestricted comprehension or the Liar, or about > symmetry/antisymmetry. Incomplete means inconsistent, of a universal > theory. > There's only one theory with no axioms, the null axiom theory. > I refute. It still sounds like gibberish. Could you show us a theorem in this theory with no axioms? === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? Mail-To-News-Contact: abuse@dizum.com >> One can establish a bijection between the points of R^2 and the interior >> of a circle in R^2. >As a side note. I had a friend raise a very simple objection to the notion >that every finite region of R^2 has the same number of points. His >argument was that if we draw a circle and confine a bunch of points in it, >then draw a larger circle around it. there are points in the larger circle >which are not in the smaller one. It would seem to follow that there are >more points in the larger circle than in the smaller one. Let's define two disks. (A disk is the set of points enclosed by a circle.) Disk 1 is {(x,y) | x^2 + y^2 < a^2} and disk 2 is {(x,y) | x^2 + y^2 < b^2} . (a, b, x, and y are all elements of the reals.) I can define a function that maps disk 1 to disk 2 as follows: f(x,y) = ( x*(b/a), y*(b/a) ) One can easily show that for any point (w,z) in disk 2, the point ( w*(a/b), z*(a/b) ) in disk 1 will map to it: f( w*(a/b), z*(a/b) ) = ( w*(a/b)*(b/a), z*(a/b)*(b/a) ) = (w,z) Therefore, every point in disk 2 has a corresponding point in disk 1 and vice versa. Therefore, there is a bijection between them. This is true any time that both a and b are non-zero. (I was surprised to realize that a and b do not need to be positive. Making one negative merely flips the mapping.) There's no requirement that a be equal to b for this to work. > Yes, I know I'm >not supposed to entertain such doubts, but.... Sure you are. Not only are you supposed to entertain them, you're supposed to investigate them. Try writing down a function or two. Try proving that a function can't exist. -- Michael F. Stemper #include Always remember that you are unique. Just like everyone else. === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? > One can establish a bijection between the points of R^2 and the interior > of a circle in R^2. >>As a side note. I had a friend raise a very simple objection to the >>notion >>that every finite region of R^2 has the same number of points. His >>argument was that if we draw a circle and confine a bunch of points in it, >>then draw a larger circle around it. there are points in the larger >>circle >>which are not in the smaller one. It would seem to follow that there are >>more points in the larger circle than in the smaller one. > Let's define two disks. (A disk is the set of points enclosed by a > circle.) > Disk 1 is {(x,y) | x^2 + y^2 < a^2} and disk 2 is {(x,y) | x^2 + y^2 < > b^2} . (a, b, x, and y are all elements of the reals.) > I can define a function that maps disk 1 to disk 2 as follows: > f(x,y) = ( x*(b/a), y*(b/a) ) > One can easily show that for any point (w,z) in disk 2, the point > ( w*(a/b), z*(a/b) ) in disk 1 will map to it: > f( w*(a/b), z*(a/b) ) = ( w*(a/b)*(b/a), z*(a/b)*(b/a) ) = (w,z) > Therefore, every point in disk 2 has a corresponding point in disk 1 > and vice versa. Therefore, there is a bijection between them. This is > true any time that both a and b are non-zero. (I was surprised to > realize that a and b do not need to be positive. Making one negative > merely flips the mapping.) There's no requirement that a be equal > to b for this to work. >> Yes, I know I'm >>not supposed to entertain such doubts, but.... > Sure you are. Not only are you supposed to entertain them, you're > supposed to investigate them. Try writing down a function or two. > Try proving that a function can't exist. I understand the math. That is not the issue. There are some real issues related to concepts such as geometrodynamics and the integral of the stress energy tensor over a region of space which does cause a bit of intellectual vertigo. But my point really was much simpler. You can show me all the mathematical reasoning you want, and at the end of the day, I am still going to *feel* less than 100% convinced. -- Nil conscire sibi === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? <8rCdnaTl0cQNei7ZnZ2dnUVZ_vidnZ2d@speakeasy.net> <5NKdnWOLUfwA-CnZnZ2dnUVZ_vydnZ2d@speakeasy.net> <-YSdnf1j6fygkyjZnZ2dnUVZ_o-dnZ2d@speakeasy.net> <200607131722.k6DHMOM44738@walkabout.empros.com> mathematical reasoning you want, and at the end of the day, I am still > going to *feel* less than 100% convinced. But your personality quirks aren't really on topic. === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? <13229727.1152625775854.JavaMail.jakarta@nitrogen.mathforum.org> We are using Arab writing and reading of numbers, from left (ones) to the right (the largest decimal). What would happen with some arguments used in this discussion, if the ordering were inversed, and the last digit in the notation of a number were read as the largest decimal place, etc.? kunzmilan === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? > One can establish a bijection between the points of R^2 and the > interior of a circle in R^2. Such bijections cannot be recursive > enumerations. >> Are you suggesting these points are countable? I note that you did not address this question. -- Nil conscire sibi === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? >> When dealing with a finite subset M of N, all bijections >> of M onto M can be deterministically produced. >> Do you agree? > I need to know what you mean by deterministically produced. > This sounds non-mathematical to me, but maybe you have some > mathematical definition in mind. >> I really mean that you can compute all bijections of M onto M. > What does compute a bijection mean? For example, let > M = {2, 5 ,8} and define f:M --> M to be the bijection > {(2,5), (5,8), (8,2)}. Please give the details of > verifying that f can be computed. What you have provided is a permutation. And that is my contention. All bijections from M->M are permutations. > Moreover, in order > to show that the concept is not vacuuous, please give > an example of a bijection that one can't compute. Uh...???..., which concept? I just asserted that all finite bijections are permutations and can thus be computed. That really is what I meant by deterministic. The term has intuitive meaning in computer science. It basically means that for a finite system same garbage in -> same garbage out. > Or are you talking about a property of the *set* of all > bijections, rather than a property of individual bijections? > If this is the case, then below is the collection of all > bijections of M. Please give the details of verifying > that it can be computed. Moreover, give an example of > a collection of bijections that can't be computed. > { {(2,2), (5,5), (8,8)}, {(2,2), (5,8), (8,5)}, > {(2,5), (5,2), (8,8)}, {(2,5), (5,8), (8,2)}, > {(2,8), (5,2), (8,5)}, {(2,8), (5,5), (8,2)} } I don't know what it means to prove that I can produce a permutation. I can show that there are n! permutations for every ordered set of n elements, but I hold that the notion of permutation is proved by definition. You have asked me to prove that a commonly used concept in mathematics is not vacuous. I believe that burden is upon /you/. -- Nil conscire sibi === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? <8rCdnaTl0cQNei7ZnZ2dnUVZ_vidnZ2d@speakeasy.net> <5NKdnWOLUfwA-CnZnZ2dnUVZ_vydnZ2d@speakeasy.net> <-YSdnf1j6fygkyjZnZ2dnUVZ_o-dnZ2d@speakeasy.net> <5ZednTBYEc7g0CjZnZ2dnUVZ_t-dnZ2d@speakeasy.net> > What does compute a bijection mean? For example, let >> M = {2, 5 ,8} and define f:M --> M to be the bijection >> {(2,5), (5,8), (8,2)}. Please give the details of >> verifying that f can be computed. > What you have provided is a permutation. And that is my > contention. All bijections from M->M are permutations. Isn't this true by definition? The term permutation, when used in this way, simply means a bijection from the set onto itself. This is the definition of permutation that you'll find in most any elementary abstract algebra text and combinatorics text. Exactly what does permutation mean that makes your statement All bijections from M->M are permutation say anything more than All objects satisfying property P satisfy property P? In mathematics, especially in abstract algebra and related areas, a permutation is a bijection from a finite set X onto itself. http://en.wikipedia.org/wiki/Permutation A bijective function is also called a bijection or permutation. The latter is more commonly used when X = Y. http://en.wikipedia.org/wiki/Bijection (Books 1 - 10 with 2040 pages on permutation bijection), but this is so trivial to find that I'll leave it to you if you're still unconvinced. L. Renfro === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? >> I really mean that you can compute all bijections of M onto M. > What does compute a bijection mean? For example, let > M = {2, 5 ,8} and define f:M --> M to be the bijection > {(2,5), (5,8), (8,2)}. Please give the details of > verifying that f can be computed. >> What you have provided is a permutation. And that is my >> contention. All bijections from M->M are permutations. > Isn't this true by definition? The term permutation, > when used in this way, simply means a bijection from > the set onto itself. This is the definition of permutation > that you'll find in most any elementary abstract algebra > text and combinatorics text. Exactly what does permutation > mean that makes your statement All bijections from M->M > are permutation say anything more than All objects > satisfying property P satisfy property P? > In mathematics, especially in abstract algebra and related > areas, a permutation is a bijection from a finite set X > onto itself. > http://en.wikipedia.org/wiki/Permutation Quod Erat Demonstrandum -- Nil conscire sibi === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? <8rCdnaTl0cQNei7ZnZ2dnUVZ_vidnZ2d@speakeasy.net> <5NKdnWOLUfwA-CnZnZ2dnUVZ_vydnZ2d@speakeasy.net> <-YSdnf1j6fygkyjZnZ2dnUVZ_o-dnZ2d@speakeasy.net> that every finite region of R^2 has the same number of points. His > argument was that if we draw a circle and confine a bunch of points in it, > then draw a larger circle around it. there are points in the larger circle > which are not in the smaller one. It would seem to follow that there are > more points in the larger circle than in the smaller one. I have never > fully convinced myself that his reasoning is lacks merit. Yes, I know I'm > not supposed to entertain such doubts, but.... Take a sheet of rubber, stretch it, and draw a circle on it (that's your larger circle). Then let the rubber contract so that the larger circle becomes the smaller circle. Are there any points in the larger circle that don't become points in the smaller circle? === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? >> As a side note. I had a friend raise a very simple objection to the >> notion >> that every finite region of R^2 has the same number of points. His >> argument was that if we draw a circle and confine a bunch of points in >> it, >> then draw a larger circle around it. there are points in the larger >> circle >> which are not in the smaller one. It would seem to follow that there are >> more points in the larger circle than in the smaller one. I have never >> fully convinced myself that his reasoning is lacks merit. Yes, I know >> I'm not supposed to entertain such doubts, but.... > Take a sheet of rubber, stretch it, and draw a circle on it (that's > your larger circle). Then let the rubber contract so that the larger > circle becomes the smaller circle. Are there any points in the larger > circle that don't become points in the smaller circle? Take a sheet of rubber and paint a circle on it, and then stretch it. Either the paint will fragment because it doesn't stretch with the rubber, or it will become thinner and its color will therefore be less intense. I fully understand the reasoning that says there are as many points in the smaller circle as there are inside the larger circle, but that doesn't quell some intuitive reluctance I have to completely dismiss my friends counter argument. -- Nil conscire sibi === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? <8rCdnaTl0cQNei7ZnZ2dnUVZ_vidnZ2d@speakeasy.net> <5NKdnWOLUfwA-CnZnZ2dnUVZ_vydnZ2d@speakeasy.net> <-YSdnf1j6fygkyjZnZ2dnUVZ_o-dnZ2d@speakeasy.net> > As a side note. I had a friend raise a very simple objection to the >> notion >> that every finite region of R^2 has the same number of points. His >> argument was that if we draw a circle and confine a bunch of points in >> it, >> then draw a larger circle around it. there are points in the larger >> circle >> which are not in the smaller one. It would seem to follow that there are >> more points in the larger circle than in the smaller one. I have never >> fully convinced myself that his reasoning is lacks merit. Yes, I know >> I'm not supposed to entertain such doubts, but.... > Take a sheet of rubber, stretch it, and draw a circle on it (that's > your larger circle). Then let the rubber contract so that the larger > circle becomes the smaller circle. Are there any points in the larger > circle that don't become points in the smaller circle? > Take a sheet of rubber and paint a circle on it, and then stretch it. > Either the paint will fragment because it doesn't stretch with the rubber, > or it will become thinner and its color will therefore be less intense. I > fully understand the reasoning that says there are as many points in the > smaller circle as there are inside the larger circle, but that doesn't > quell some intuitive reluctance I have to completely dismiss my friends > counter argument. I guess it comes down to how you make sense of terms like more (in this case, more points) in the case of infinite sets. If you accept that two infinite sets have the same number of elements when their members can be put into one-to-one correspondence, then you inevitably have the paradox that a set (points in larger circle) can have the same number of elements as a subset of itself (points in smaller circle). If not then you need another method of comparison. (I haven't been following this thread in detail, so I may not be on the same page as everyone else...) === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? >> >This really is exactly what you are doing when you say that the >cardinality of the natural numbers is identical to the cardinality of >the multiples of 10. >> http://www.ifi.unizh.ch/math/bmwcs/section_9_3.xhtml#ex:cauchy > I don't see how your reply relates to my comment. Did you run the applet? -- Nil conscire sibi === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? > >>This really is exactly what you are doing when you say that the >>cardinality of the natural numbers is identical to the cardinality of >>the multiples of 10. >http://www.ifi.unizh.ch/math/bmwcs/section_9_3.xhtml#ex:cauchy >>I don't see how your reply relates to my comment. > Did you run the applet? No. One of the disadvantages of using unix over windows is that these things (java, flash, etc) don't come automatically. Could you tell me what the applet does? Stephen === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? On Wed, 12 Jul 2006 15:14:20 -0400, Hatto von Aquitanien >> Peano's Axioms are the paradigm of recursive enumeration. > > Nonresponsive to my point. Not every bijection is a recursive > enumeration. > > I am not persuaded that it is meaningful to both describe a set as > countable and to say that it cannot be recursively enumerated. >> One can establish a bijection between the points of R^2 and the interior >> of a circle in R^2. Such bijections cannot be recursive enumerations. >Are you suggesting these points are countable? >As a side note. I had a friend raise a very simple objection to the notion >that every finite region of R^2 has the same number of points. His >argument was that if we draw a circle and confine a bunch of points in it, >then draw a larger circle around it. there are points in the larger circle >which are not in the smaller one. It would seem to follow that there are >more points in the larger circle than in the smaller one. I have never >fully convinced myself that his reasoning is lacks merit. Whether the reasoning is correct or not depends on exactly what we mean by has the same number of points. One could possibly give a definition of that phrase so that the reasoning above would be correct. But if A has the same number of points as B is, as usual, taken as an informal version of A has the same cardinality as B, ie there is a bijection from A onto B then there's simply no reasoning at all here - so there are points in B which are not in A, that's no evidence at all that there is no bijection from A onto B. > Yes, I know I'm >not supposed to entertain such doubts, but.... ************************ David C. Ullrich === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? >> Basically, you've given an algorithm for generating the members of S and >> turned it into a predicate. For the sake of keeping track of ideas which >> underly the conclusion, should we not consider this concept >> of /doing/ /something/ /repeatedly/ as one of the premises? > Why should we? > If I say that y=z+1 and x=y+1, does that somehow imply that the > value of x exists one step later than the value of z? If z=3, > doesn't x equal 5 immediately? I go back to where numbers come from. As I've said, in one sense I can understand the notion of using something such as Peano's axioms as a predicate to immediately construct as set. OTOH, numbers are a result of our ancestors (clearly it was a member of my family who first discovered this ;) ) coming to realize that keeping track of things was easier if there were some general abstraction which could be applied to collections of units. They therefore came up with names for these abstractions beginning with the smallest quantities and then extending the idea. First they counted digits. (Fingers and thumbs). When they ran out of digits, they held up on fist and said something like elv. For the next number they held up a second fist and uttered two-elv. We humans have amazing abilities to abstract and extend upon these basic ideas. The concept of natural numbers is a good example. One one hand, I can understand N as falling out of Plato's Heaven in one thunk. On the other hand, I see it as an extention of our notion of a finite ordered set. In any finite set of integers it is meaningful to talk about counting backwards from the largest to the smallest, or dividing the set in half, etc. Even these ideas are extentions of the original concept of enumerating items. Some of these ideas applicable to enumerating the elements of finite collections fail to be applicable when discussing infinite sets. One example which seems most fundamental is that we are not able to start from the top and count down. Our notion of N appears to be an extension of the most primitive essence of counting. When carefully analyzing Cantor's reasoning, it seems important that we examine exactly what we mean by infinity, infinite sets, N, etc. When I think about decimal point notation, I think about some abstraction of the process of cutting something in tenths, and then cutting the remainder in tenths, etc... That is how I built up that idea when I first learned it. That idea seems implicit in Cantor's reasoning, even if he didn't explicitly use decimal point notation. Whether or not the notion of recursively subdividing a remainder is essential to understanding Cantor's reasoning, it seem clear that there is some notion of extrapolating, ad infinitum in two directions (I intentionally avoided the term dimensions). The individual sequences extend to the right, and the collection of sequences extends vertically. It might even be said that he is attempting to form a bijection between NxN and N. -- Nil conscire sibi === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? > Whether or not the notion of recursively subdividing a remainder is > essential to understanding Cantor's reasoning, it seem clear that there is > some notion of extrapolating, ad infinitum in two directions (I > intentionally avoided the term dimensions). The individual sequences > extend to the right, and the collection of sequences extends vertically. > It might even be said that he is attempting to form a bijection between NxN > and N. That has already ben done in other contexts, like ennumerating the rationals, and was not needed in the context of Cantor's 2nd uncountability proof. === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? >> Whether or not the notion of recursively subdividing a remainder is >> essential to understanding Cantor's reasoning, it seem clear that there >> is some notion of extrapolating, ad infinitum in two directions (I >> intentionally avoided the term dimensions). The individual sequences >> extend to the right, and the collection of sequences extends vertically. >> It might even be said that he is attempting to form a bijection between >> NxN and N. > That has already ben done in other contexts, like ennumerating the > rationals, Indeed it has. I have been trying to figure out was the essential difference is between the enumeration of the rational numbers and the diagonal contradiction proof. > and was not needed in the context of Cantor's 2nd > uncountability proof. I looked at that, but I have yet to fully grasp the concept. -- Nil conscire sibi === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? That's not really true. In order to give a constructive proof of the > existence of transcendental numbers, you must show that the digits of > algebraic numbers can actually be computed... > But that step is trivial. If it really is trivial, then you ought to be able to answer the following question with great ease. Suppose I compute the value of a root of a seventh degree equation with all coefficients less than 100, and the value I get is 1.39999999999999999... Can I conclude that the root is rational? === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? That's not really true. In order to give a constructive proof of the >existence of transcendental numbers, you must show that the digits of >algebraic numbers can actually be computed... > But that step is trivial. > If it really is trivial, then you ought to be able to answer the > following question with great ease. Suppose I compute the value of a > root of a seventh degree equation with all coefficients less than 100, > and the value I get is 1.39999999999999999... Can > I conclude that the root is rational? This has nothing to do with the question of showing the digits are computable; it's a question of diophantine approximation. There is an effective version of Louiville's theorem, so presumably I could look this up, but really it is *not* relevant. === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? proof of the existence of transcendental numbers, you must > show that the digits of algebraic numbers can actually > be computed... >> But that step is trivial. > If it really is trivial, then you ought to be able to answer > the following question with great ease. Suppose I compute > the value of a root of a seventh degree equation with all > coefficients less than 100, and the value I get is > 1.39999999999999999... Can I conclude > that the root is rational? What does proving a given algebraic number is rational or irrational have to do with proving that algebraic numbers for the life of me I can't figure out anything close to to Gene Ward Smith's remark. L. Renfro === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? That's not really true. In order to give a constructive > proof of the existence of transcendental numbers, you must > show that the digits of algebraic numbers can actually > be computed... >> But that step is trivial. > If it really is trivial, then you ought to be able to answer > the following question with great ease. Suppose I compute > the value of a root of a seventh degree equation with all > coefficients less than 100, and the value I get is > 1.39999999999999999... Can I conclude > that the root is rational? > What does proving a given algebraic number is rational or > irrational have to do with proving that algebraic numbers > for the life of me I can't figure out anything close to > to Gene Ward Smith's remark. It helps one to compute that digit right before the string of nines. === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? the following question with great ease. Suppose I compute > the value of a root of a seventh degree equation with all > coefficients less than 100, and the value I get is > 1.39999999999999999... Can I conclude > that the root is rational? >> What does proving a given algebraic number is rational or >> irrational have to do with proving that algebraic numbers >> for the life of me I can't figure out anything close to >> to Gene Ward Smith's remark. > It helps one to compute that digit right before the > string of nines. This still makes no sense. Look, I don't care whether the number is rational or irrational, let alone whether it's easy or difficult to prove this. What does this have to do with how difficult it is to prove that each algebraic number is computable? The only difficulty is proving the computability of the various operations involved in one of the many ways of obtaining arbitrary precise approximations to an algebraic number. Get a numerical analysis text, get a book on recursive function theory, look up one of the numerical methods for solving algebraic equations, and (assuming all the coefficients are integers) verify with the recursive function theory book that all the procedures are computable functions. By the way, the (complex) computable numbers also form an algebraically closed field. This takes a little more work, but it essentially comes down to coming up with a sufficiently constructive proof of the fundamental theorem of algebra. An example of such a proof is given in the following paper. [See also Theorem 6 (pp. 788-790) in H. Gorgon Rice, Recursive real numbers, Proceedings of the American Mathematical Society 5 (1954), 784-791.] P. C. Rosenbloom, An elementary constructive proof of the fundamental theorem of algebra, American Mathematical Monthly 52 #10 (December 1945), 562-570. L. Renfro === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? > ... >Cantor's method uses decimal representation, not binary. There's a >reason for this, related to the fact that different digit strings can >represent the same real number (such as 0.011111... = 0.100000... in base >2, or 0.0999999... = 0.100000... in base 10). > Do you know where and when this argument by Cantor was published? I can not > find it in the Gesammelte Abhandlungen (Collected Works) edited by > Zermelo, but perhaps I am not looking good enough. > Unfortunately, no. I have searched for this before, with no results. I nearly come to the conclusion that that argument is not from Cantor. As far as I can see Cantor himself has only stated one proof of the uncountability of the reals. That one uses two subsequences of the given list: one increasing and one decreasing. It goes about densely ordered sets without gaps. There are two diagonal proofs by Cantor, the first goes about the countability of the rationals (using the well-known diagonal method), and extends to other sets, like the algebraic numbers (also in his paper). The second is his famous diagonal proof, where he finds that there is a set of sequences for which no bijection with the naturals exists. In addition, that paper shows that there is no bijection between a set and its powerset (not really, but it is an easy conclusion *). In a note Zermelo states how the first part can be reformulated to a proof that shows how there is no bijection between the naturals and the reals in binary notation (using the fact that pairs of countable sets can be mapped to a single countable set). I think it is somebody else who came to the proof using decimal expansion of the reals. I was wondering because there is only one place where Cantor considers decimal expansions. That is in the paper with his first diagonal argument, where he shows a mapping from R to RxR, and uses decimal expansions. And where he explicitly takes care of the possible dual representations. (To parafrase: when there is a choice between a representation that terminates with all nines or all zeros, the first one should be taken, except if the number is 0.) (*) The proof shows that the set of binary functions f(x) on a set does not have the same cardinality as the given set. But there is an obvious bijection betwen the set of binary functions on a set and its subsets. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? > Back to Goedel and statements about completeness of theories of, say > the natural numbers, around here it is thought that Presburger > arithmetic is complete where Peano arithmetic is incomplete. Yet, a > finite proof in one is a finite proof in the other, the objects within > the two are the same. Does not that strike you as inconsistent? > No, although your confusion is understandable. Goedel's Incompleteness > Theorem proves that all systems *which are omega-consistent* are not > decidable nor can they prove their own conssistencies. As it turns > out, Presburger Arithmetic is too weak to have Goedel apply to it. On > the positive side, Presburger Arithmetic is completely decidable; on > the negative, it is too weak to do very much (for example, it cannot > even do multiplication, only basic addition). > The two are not the same, and a finite proof in Peano is not > necessarily a finite proof in Presburger. For example, 2 x 2 = 4 is > not a theorem in Pressburger. > More about that, it was noted here that a result of Paris and Kirby is > that for there to be made all true statements of the natural integers, > there are statements about infinite natural integers, infinite in size, > not quantity. Do not you agree in mutual implication? If those > statements may only hold true when they do, and they do, then there > are. > I'm not quite sure what you are saying here. Paris & Kirby simply > proved that Goodstein's Theorem is undecidable in Peano Airthmetic, but > is decidable in stronger systems (such as those which contain the Axiom > of Infinity). This proof makes use of Nonstandard Analysis (which is > cool). However, there are no contradictions here. It is merely > stating that with fewer axioms, the weaker the system, and fewer > theorems can be proven. > There's only one theory with no axioms. It has all the theorems. > It, in fact, has no theorems. Give me an example of a proof which has > no axioms. (You can't even put a first line down, since you must > assume something to move forward.) > Any other is inconsistent or incomplete, just ask Goedel. > No, any sufficiently strong system is. Presburger Arithmetic, as you > correctly point out, is complete. > Your regular set set theory is incomplete, via Goedel... > As is virtually any set theory > ...and inconsistent, via universal quantiification... > Huh? Where did you get THAT from??? > not to mention paradoxes in them, generally paradoxes > of unrestricted comprehension or the Liar, or about > symmetry/antisymmetry. > I don't know what you are talking about here. > Incomplete means inconsistent, of a universal theory. > This is provably false. An inconsistent theory is always complete > (since every statement is a theorem). For it to be incomplete, it must > therefore be consistent. > There's only one theory with no axioms, the null axiom theory. > Sounds more like a null-theorem theory. > Jonathan Hoyle I'm talking about the Goedelian incompleteness result (references are to generally two of them, in various orders) that a strong theory is unable to prove true statements about objects defined in the system. There, there is no theory of those things, because there are true statements that are not inferable from any theory of them. That's so in theories with proper/non-logical axioms, and not all of them. Does that makes sense? If there is no complete theory of anything, there is no theory of anything. Peano arithmetic can not be called a theory of the natural numbers if it is incomplete, in its incompleteness it is only a fragment of the complete theory of natural numbers. By accepting incompleteness as some kind of truism, there is sacrificed any sense of comprehension. Compare PA and PA, Presburger and Peano. Each has the same objects, zero and successors. In Peano, 2 x 2 = 4 is the statement in Presburger that 2 + 2 = 4, more clearly that 3 x 2 = 6 is the statement that 2 + 2 + 2 = 6. They're the same statement. A finite proof of either exists, in Peano basically with accessory of notation. So, I am talking about a different result of Goedel. When I say ZF is inconsistent, that's about basically the universal quantification over sets. It's either possible or not. In a similar sense as aboe about there being no complete theory of anything, thus no theory of anything, where there is no complete collection of anything, there is no collection of anything. So, as it is thus impossible to quantify over sets in ZF, then none exist. In terms of the null axiom theory, and an object of theory as basically an/the ur-element, and thus a continuum of individua, via tautology and Janus' introspection and the irrelevant liar, yes those are most comprehensively addressed in philosophy. Infinite sets are equivalent. Ross (Points are polydimensional.) === Subject: Re: Attempts to Refute Cantor's Uncountability Proof? > I'm talking about the Goedelian incompleteness result (references are > to generally two of them, in various orders) that a strong theory is > unable to prove true statements about objects defined in the system. > There, there is no theory of those things, because there are true > statements that are not inferable from any theory of them. That's so > in theories with proper/non-logical axioms, and not all of them. Regardless, Presburger arithmetic is too weak to have either Goedel theorem applied. > Does that makes sense? If there is no complete theory of anything, > there is no theory of anything. Peano arithmetic can not be called a > theory of the natural numbers if it is incomplete, in its > incompleteness it is only a fragment of the complete theory of > natural numbers. By accepting incompleteness as some kind of truism, > there is sacrificed any sense of comprehension. >Whether you accept it or not, any consistent theory is going to be incomplete. > Compare PA and PA, Presburger and Peano. Each has the same objects, > zero and successors. In Peano, 2 x 2 = 4 is the statement in > Presburger that 2 + 2 = 4, more clearly that 3 x 2 = 6 is the statement > that 2 + 2 + 2 = 6. They're the same statement. A finite proof of > either exists, in Peano basically with accessory of notation. Of those statements, perhaps, but not of all of them. For example, you cannot encode into Presburger the following Peano theorem: For all x, there exists a y such that for all z > 0: yz > x. > So, I am talking about a different result of Goedel. Doesn't matter. > When I say ZF is inconsistent, that's about basically the universal > quantification over sets. It's either possible or not. I don't follow. Both universal quantification and existential quantification are part of ZF. There is no inconsistency here. > In a similar sense as aboe about there being no complete theory > of anything, thus no theory of anything, where there is no complete > collection of anything,there is no collection of anything. How do you go from no complete theory of everything to no theory of everything? Just because the theory is incomplete doesn't mean it stops existing. It exists, it's just incomplete. > So, as it is thus impossible to quantify over sets in ZF, then none exist. Again, you are making sweeping false statements. You can quantify over sets in ZF. And even if you removed universal quantification from ZF, that doesn't mean that the sets stop existing...it just means you can't make universal statements. > In terms of the null axiom theory, and an object of theory as basically > an/the ur-element, and thus a continuum of individua, via tautology and > Janus' introspection and the irrelevant liar, yes those are most > comprehensively addressed in philosophy. These are words in search of an idea and failing to find one. Please give me an example of a theorem you can prove without using any axioms. (And I in turn will show you the axioms you are using to make that proof happen.) > Infinite sets are equivalent. Provably false. Jonathan Hoyle === Subject: The White Box INTRODUCING THE WHITE BOX Brought To You By: CoreyWhite@gmail.com Alt.Magick Introduction: So, you are paranoid enough to wonder if the police have been tapping your line? Maybe it is the 8 month long drug probe you heard about that led to 19 arrests in your area? Or not, but that's what tipped me off to inventing the White Box. And if you are like me then you will appreciate the peace of mind that comes with knowing that the Department of Homeland Security isn't reading your e-mails to Grandma. What is the white box? It is a device you can attach to your own phone line, which will generate a moderate amount of line noise that makes it impossible for the government to sniff traffic between you and your ISP. But it doesn't generate enough line noise to interfere with your connection. The device is usually called a noise box, and can also be used to add an abnormal amount of noise on the line, which would make data transmissions almost impossible and voice communications annoying. I discovered this Box by accident because one of the silver wires in my phone cords was exposed for most of the year. Whenever anything touched it, the cord would generate a small amount of noise on the line. And because my phone was being tapped as part of a drug investigation, my ISP would *drop my call*, but because someone else was listening I couldn't get a dial-tone until I manually forced my modem to hang-up. That may not happen with all wire-taps, but you can be sure that a little line noise makes it really hard for them to know what's happening during a data transmission. So, now that you know what it is, let us go about the steps of building one. Building/Installing A Noise-Box: Materials: 1 RJ11 Coupler 1 5k Potentiometer/10k Potentiometer 15 turn 1 1.0-1.5 uf Capacitor, non-polarized, 100vdc+ 1 100 ohm, 1/2 or 1/4 watt resistor Procedure: 1) Solder one end of the capacitor to the middle lug of the pot. 2) Solder one end of the resistor to either remaining end on the pot. 3) Open up the RJ11 Coupler and expose expose some copper on the red and green wires. 4) Take your white box and hook the capacitor to the green wire. 9) Hook the resistor to the red wire. 10) Plug a phone line into the white box. 11) You can now adjust the pot to add noise to the line. 11) Remember to remove the white box when you have finished testing. Closing: Pretty nifty, huh? Well, I know there are easier ways of doing this, and I will experiment a bit to find the best combination, so you won't have to go around adjusting pots and because this guide was created to interfere with making calls. And not blocking out Wire taps. It may not be perfectly suited as a tool to make it harder to log your traffic, but just a minimal amount of line noise will cause problems while you are being tapped. While it would take a lot more to do it when your line is clean.