mm-2409 === Subject: Integral again again I have to calculate an integral. I suppose this one should be easier, perhaps you can help: int_{-infinity}^{u} 1 / (g^2 + u^2 + c/a^2 * [1-Cos(a*u)]) du I could not find it in any table. Mathematica was also unable to solve it. Can you do the special case g=0, c=a=1 ? === Subject: Re: Integral again > again I have to calculate an integral. I suppose this one should be > easier, perhaps you can help: > > int_{-infinity}^{u} 1 / (g^2 + u^2 + c/a^2 * [1-Cos(a*u)]) du > integral 1/(a + b.cos x) dx = 2/sqr(a^2 - b^2) * arctan[(sqr(a^2 - b^2) tan x/2)/(a + b)] === Subject: Re: Integral again > > again I have to calculate an integral. I suppose this one should be > easier, perhaps you can help: > > int_{-infinity}^{u} 1 / (g^2 + u^2 + c/a^2 * [1-Cos(a*u)]) du > > integral 1/(a + b.cos x) dx > = 2/sqr(a^2 - b^2) * arctan[(sqr(a^2 - b^2) tan x/2)/(a + b)] > As OP responded, the original problem contains an extra u^2 term, which makes it likely to be non-elementary. But there is another issue, even if the u^2 term were absent. (1) (The case with non-vanishing denominator:) The expression involving tan(x/2), while revered by most Calculus textbook writers, and also by the writers of many symbolic integration programs, has severely restricted validity: as x passes through odd multiples of pi, the answer has jump discontinuities to stay periodic, while the desired result ought to be continuous and monotone. (2) (The case of denominator with simple roots:) The answer has its own singularities (of logarithmic nature), and the tan(x/2) type answer typically creates extra jump discontinuities, as above. If you want an honest answer for case (1), valid for the whole real line, here is a formula - and I assume without loss of generality that A>0. Denote R := (A^2 - B^2 - C^2)^(1/2): then integral(dx/(A + B * cos(x) + C * sin(x)) = (1/R) * (x - 2 * arctan ((B * sin(x) - C * cos(x)) / (R + A + B * cos(x) + C * sin(x)))) (for safety's sake, checked using a symbolic algebra program). === Subject: Re: Integral again Well this does not seem to be the integral I need. You do not have x^2, dont you? And I do not see a way to write my intgegral in your form. Can you tell me about your idea? === Subject: Re: masses vs, weights > Its > its Yes nature's like a cook; everything's put together with a little of this, and a little of that, to make up good and bad things. Humans have to invent ways to preserve their well being, and of course it requires utilizing and understanding nature. Don === Subject: Re: Some Weyl group automorphisms On 23-Jul-2005, mareg@mimosa.csv.warwick.ac.uk () > > >What are the outer automorphisms of the Weyl group B_n, and how can >it be proved? (B_n is the wreath product C_2 wr_n S_n of the cyclic >group of order 2 with the symmetric group of degree n; if the C_2's >are identified with the reflections along n mutually orthogonal >directions in n-dimensional Euclidean space, then B_n is the >isometry group of the n-dimensional square/(hyper)cube.) > >It's easy to see that for n > 2 there's always an outer >automorphism that takes the set of odd permutations of the S_n to >their negatives (i.e., their products with Z(B_n) =~ C_2), and for >n > 2 and even, there's also an outer automorphism that takes the >set of generators of the C_2's to *their* negatives. > >I believe these are the only outer automorphisms (this is certainly >true for n = 3 and n = 4), but I'm having trouble proving it. One >approach I've tried is to show that for n =/= 4 all subgroups of >B_n that are isomorphic to the alternating group A_n are conjugate >to each other (they're certainly 'automorphic' to each other), but >I haven't yet succeeded. [...] > I can prove that it is correct, but I am not sure whether I will be able > to find an elementary proof. It goes roughly like this. > > The group G has an elementary abelian normal subgroup N of order 2^n > with G/N = S_n. You can think of N as being an n-dimensional module > for G/N over the field of order 2, where the module action is defined > by conjugation in G. > > The automorphism group A of G has a chain of normal subgroups > 1 < C < B < A, where C consists of thsoe automorphisms that induce the > identity on both N and on G/N, and B consists of those that induce the > identity on G/N. > > Automorphisms in C permute the complements of N in G, where inner > automorphisms map conjugates to conjugate complements. So roughly, > outer automorphisms interchange conjugacy classes of complements. > It can be shown that there are two such classes. The classes also > correspond to the cohomology group H^1(G/N,N), and there is a result > in cohomology called Shapiro's Lemma, which enables you to deduce that > this has order 2. Of course this is just your outer automorphism that > maps odd permutations to their negatives. Since N contains Z(G) > with |Z(G)| = 2, we have |C| = |N| = 2^n. > > B/C is isomorphic to the endomorphism ring of N regarded as a G/N-module > over the field of order 2. For n odd, this module is a direct sum of > two irreducibles, of dimensions 1 and n-1, so the endomorphism ring is > trivial. For n even, the module is uniserial, with submodules of > dimensions > 1 and n-1. It is not hard to see that the endomorphism ring has order > at most 2, and your outer automorphism mapping elements in N to their > negatives shows that it has order exactly 2. > > A/B is isomorphic to a subgroup of Aut(G/N). For n != 6, Aut(G/N) = G/N = > S_n, so the whole of Aut(G/N) occurs as inner automorphisms, and we get > |A/B| = n!. In fact, for n=6, the outer automorphism of S_6 does > not induce an automorphism of G (because ther eis no way it could > act on N), and so |A/B| = n! is true also for n=6. > > Hope this sheds a little light on things! It is certainly possible that > there is a more elementary proof than this. this; in particular I know almost nothing about cohomology. Maybe this will motivate me to learn something about it. -- Jim Heckman === Subject: Re: Spherical geometry On 23-Jul-2005, gregegan@netspace.zebra.net.au (Greg Egan) : [...] > Suppose we have a triangle on the unit sphere, with A, B, C the angles at > the three vertices, and a, b, c the angles subtended at the centre of the > sphere by the sides opposite those vertices. > > We can always choose a coordinate system such that the vertices are the > following unit vectors in R^3: > > vA = (0, 0, 1) > vB = (sin c, 0, cos c) > vC = (cos A sin b, sin A sin b, cos b) > > In terms of spherical polar coordinates, we put vertex A at the north > pole, vertex B on the prime meridian, and vertex C on the meridian with > longitude A. The co-latitudes (angles from the pole) of vertices B and C > are c and b respectively. > > Cosine Laws [...] > Substituting the geometry of the second triangle into the First Cosine Law > gives the Second Cosine Law: cos A = cos a sin B sin C - cos B cos C I haven't checked the steps in your derivation, but there must be a sign error somewhere. It should be: cos A = cos a sin B sin C + cos B cos C [...] -- Jim Heckman === Subject: Re: Spherical geometry >On 23-Jul-2005, gregegan@netspace.zebra.net.au (Greg Egan) >: > >[...] > >> Suppose we have a triangle on the unit sphere, with A, B, C the angles at >> the three vertices, and a, b, c the angles subtended at the centre of the >> sphere by the sides opposite those vertices. >> >> We can always choose a coordinate system such that the vertices are the >> following unit vectors in R^3: >> >> vA = (0, 0, 1) >> vB = (sin c, 0, cos c) >> vC = (cos A sin b, sin A sin b, cos b) >> >> In terms of spherical polar coordinates, we put vertex A at the north >> pole, vertex B on the prime meridian, and vertex C on the meridian with >> longitude A. The co-latitudes (angles from the pole) of vertices B and C >> are c and b respectively. >> >> Cosine Laws > >[...] > >> Substituting the geometry of the second triangle into the First Cosine Law >> gives the Second Cosine Law: >> >> cos A = cos a sin B sin C - cos B cos C > >I haven't checked the steps in your derivation, but there must be a >sign error somewhere. It should be: > >cos A = cos a sin B sin C + cos B cos C > >[...] Nope. As Greg said, for every spherical triangle, there is a dual triangle where each side is replaced by a supplementary angle and each angle is replaced by a supplementary side. Since cos(pi-x) = -cos(x), and sin(pi-x) = sin(x), the dual of the formula cos(a) = cos(A) sin(b) sin(c) + cos(b) cos(c) is -cos(A) = -cos(a) sin(B) sin(C) + cos(B) cos(C) which is the same as cos(A) = cos(a) sin(B) sin(C) - cos(B) cos(C) Rob Johnson take out the trash before replying === Subject: Re: one math exercise >> >>function langauge so as to apply to this fourm. it looks like I did not >>do a very good job at it. >> >>X is a random variable with ensemble {-n,-n+1,...,0,1,...,n}. suppose we >>want P(X=-k) = P(X=k), k=1,2,...,n and P(|X|=k+1)=2/3P(|X|=k). >>find the probabilty mass distribution for X > > > You could have saved us a lot of work if you had said that in the first > place! > > But I still recommend you follow the steps I've outlined. > Suppose n = 7, and P(X = 0) = a; work out P(X = 1) and > P(X = 2) and so on, calculate P(X = -7) + ... + P(X = 7), > and see what a has to be to make this 1; then, generalize. > great; I was able to come up with the prob. mass distro. P(X=k) = (2/3)^k P(X=k), let P(X=k)=a then we have P(X=k) = ((2/3)^k)a according to what you recommended, now I am trying to figure out what a has to be to make sum of P(X=n)=1 for n=-7,...,7 a = 1/( (2/3)^7 + (2/3)^6 + ... + 2/3 + 1 + 2/3 + (2/3)^2 + ...+ (2/3)^7) I hope I am correct, but how can you get (5 - 6(2/3)^n+1 )^-1 out of my solution. t === Subject: Re: one math exercise >> >>function langauge so as to apply to this fourm. it looks like I did not >>do a very good job at it. >> >>X is a random variable with ensemble {-n,-n+1,...,0,1,...,n}. suppose we >>want P(X=-k) = P(X=k), k=1,2,...,n and P(|X|=k+1)=2/3P(|X|=k). >>find the probabilty mass distribution for X > > > You could have saved us a lot of work if you had said that in the first > place! > > But I still recommend you follow the steps I've outlined. > Suppose n = 7, and P(X = 0) = a; work out P(X = 1) and > P(X = 2) and so on, calculate P(X = -7) + ... + P(X = 7), > and see what a has to be to make this 1; then, generalize. > great; I was able to come up with the prob. mass distro. P(X=k) = (2/3)^k P(X=k), let P(X=k)=a then we have P(X=k) = ((2/3)^k)a according to what you recommended, now I am trying to figure out what a has to be to make sum of P(X=n)=1 for n=-7,...,7 a = 1/( (3/2)^7 + (3/2)^6 + ... + 3/2 + 1 + 2/3 + (2/3)^2 + ...+ (2/3)^7) I hope I am correct, but how can you get (5 - 6(2/3)^n+1 )^-1 out of my solution. === Subject: mean-value theorem? f is continuous on [0,1] and differentiable in (0,1). f(1) = 0. Show that, for m>0, there exists c in (0,1), such that c^2 f'(c) + m f(c) = 0. === Subject: Re: mean-value theorem? > f is continuous on [0,1] and differentiable in (0,1). f(1) = 0. > Show that, for m>0, there exists c in (0,1), such that > c^2 f'(c) + m f(c) = 0. That's a nasty one, provided I read it thus: c^2 * f'(c) + m * f(c) = 0 The trick is to take the expression x^2 * f'(x) + m * f(x), multiply it by an appropriate function g(x) so that (x^2 * f'(x) + m * f(x)) * g(x) becomes an exact derivative (did they tell you about the integrating factor of a differential equation?) - then manipulate some more to fit it into MVT, or actually into Rolle's Theorem. === Subject: regular languages behaviour at infinity let R be a regular language over the alphabet Sigma. (w.l.o.g. |Sigma|>=2). Prove or give a counterexample: lim_{n -> infty} |{w in R | |w|<=n }| / |{w in Sigma | |w|<=n}| exists. ( Equivalently up to a multiplicative constant: lim_{n -> infty} |{w in R | |w|<=n }| / |Sigma|^n exists. ) Any ideas or litrature hints? Sasha. === Subject: Re: regular languages behaviour at infinity > > let R be a regular language over the alphabet Sigma. (w.l.o.g. |Sigma|>=2). > Prove or give a counterexample: > > lim_{n -> infty} |{w in R | |w|<=n }| / |{w in Sigma | |w|<=n}| > exists. > > ( > Equivalently up to a multiplicative constant: > lim_{n -> infty} |{w in R | |w|<=n }| / |Sigma|^n > exists. > ) > > Any ideas or litrature hints? The limit need not exist. Consider the language R consisting of all even-length strings over Sigma. Rick === Subject: Re: regular languages behaviour at infinity > > let R be a regular language over the alphabet Sigma. (w.l.o.g. |Sigma|>=2). > Prove or give a counterexample: > > lim_{n -> infty} |{w in R | |w|<=n }| / |{w in Sigma | |w|<=n}| > exists. > > ( > Equivalently up to a multiplicative constant: > lim_{n -> infty} |{w in R | |w|<=n }| / |Sigma|^n > exists. > ) > > Any ideas or litrature hints? > > Sasha. My initial hunch is that the the limit exists and is equal to zero, and that the pumping lemma is the key to showing this. Try to compute an upper bound on the number of strings which satisfy the conclusion of the pumping lemma (with fixed pumping length). Hope that helps. -John Coleman === Subject: Re: regular languages behaviour at infinity > My initial hunch is that the the limit exists and is equal to zero, and > that the pumping lemma is the key to showing this. Try to compute an > upper bound on the number of strings which satisfy the conclusion of > the pumping lemma (with fixed pumping length). Zero is often the case, but not always. Let Gamma be any subset of Sigma with |Gamma|>=2. Then lim |{w in Gamma^* | |w|<=n}| / |{w in Sigma^* | |w|<=n}| = ... = |Gamma|(|Sigma|-1) / (|Sigma| (|Gamma|-1)) Sasha. === Subject: Re: amicable & sociable numbers > >> >> > >> Ignacio Larrosa Ca.96estro >> >> >JoeS escribi.97: > >>This method was first discovered by >>Abu-l-Hasan Thabit ben Korrah around the ninth century... > >A little extension: > >p = (2^s + 1)2^(n - s) - 1 > >q = (2^s + 1)2^n - 1 > >r = (2^s + 1)^2*2^(2n - s) - 1) > >(the Thabit ibn Kurrah formula is for s = 1). >> >>My understanding is that ibn is Arabic, ben is Hebrew. > > >Ahmed ben Bella was the leader of the movement >against the French in Algeria(1962). >The Arabic variants ibn, ben, bin are either >variants in transcription rules or reflect >dialectic variations. I don't know which. >> >>So you're quite confident that he wasn't the son of Bella Abzug. > > > Or Bela Lugosi? Bela Lugosi Jr. is now a successful Lawyer. If you thought his father was scary... === Subject: a limit involving Gamma function Show that lim_{s -> inf} Gamma(s+1/2) / (sqrt(s) Gamma(s)) = 1. === Subject: Re: a limit involving Gamma function > > Show that > lim_{s -> inf} Gamma(s+1/2) / (sqrt(s) Gamma(s)) = 1. This is essentially the same as showing that lim[Gamma(s + 1/2)] = lim[s^(1/2) Gamma(s)] as s --> oo. I haven't tried to do this on paper, but a few months ago when I was playing around with the gamma function I noticed that Gamma[s] = k * Integral[n^(ks - 1)e^(-(n^k))dn] for any k > 0 and the integral is over n from 0 --> oo. This can be accomplished by starting with the regular gamma function (where the integral is over t) and u-substiuting t = n^k. Anyways, maybe you should work with two different limits lim[Gamma(s)] as s --> oo and lim[x^(1/2) Gamma(x)] as x --> oo and try to show that the latter can be transformed into the former via that u-sub trick I mentioned earlier (by letting k = x^(1/2)). If the square root can be absorbed into the integarl this way (I haven't tried it so I'm not sure) then obviously they're one in the same and the original limit you mentioned would be unity. Please let me know if that works out for you. Kyle === Subject: Re: a limit involving Gamma function > Show that > lim_{s -> inf} Gamma(s+1/2) / (sqrt(s) Gamma(s)) = 1. Maybe you could show something like Gamma(s) is asymptotic to int_{s-e}^{s+e} x^s e^{-x} dx for any positive e>0. (I must admit that I haven't even tried, so this could be completely wrong. But the idea would be to see where x^s e^{-x} attains in maximum as a function of x, and then show that this maximum heavily dominates the other values.) === Subject: Re: a limit involving Gamma function > Show that > lim_{s -> inf} Gamma(s+1/2) / (sqrt(s) Gamma(s)) = 1. > Why not use Sterling's formula? === Subject: Re: a limit involving Gamma function > >> Show that >> lim_{s -> inf} Gamma(s+1/2) / (sqrt(s) Gamma(s)) = 1. >> > >Why not use Sterling's formula? Or we can use the fact/definition which says that the Gamma function is log-convex and Gamma(s+1) = s Gamma(s) for all s: log(Gamma(s)) <= (log(Gamma(s-1/2)) + log(Gamma(s+1/2)))/2 = log(Gamma(s+1/2)) - log(s-1/2)/2 log(Gamma(s+1/2)) <= (log(Gamma(s)) + log(Gamma(s+1)))/2 = log(Gamma(s)) + log(s)/2 Putting these last two inequalities together we get log(s-1/2)/2 <= log(Gamma(s+1/2)) - log(Gamma(s)) <= log(s)/2 sqrt(1-1/(2s)) <= Gamma(s+1/2)/(sqrt(s)Gamma(s)) <= 1 Letting s -> oo, we get the desired result. Rob Johnson take out the trash before replying === Subject: Re: a limit involving Gamma function <250720050915131488%anniel@nym.alias.net.invalid> What about if we don't use Stirling's formula? === Subject: Re: a limit involving Gamma function > What about if we don't use Stirling's formula? > Why not? === Subject: Re: Conformal Gravity? crank === Subject: Re: Conformal Gravity? On 25 Jul 2005 06:11:30 -0700, brian0918 >crank jingleballs === Subject: Re: Conformal Gravity? > On 25 Jul 2005 06:11:30 -0700, brian0918 > >>crank > > jingleballs > In your defence, George, your post in this thread was probably the most coherent, reasoned message I have ever seen from you. I hope a decent debate ensues and you find some answers. === Subject: Re: Conformal Gravity? <9f8ae1lm2o4oivcnjrs0ra5m2i5qji8shi@4ax.com> <1OqdnYDwPPIBrXjfRVnyrw@pipex.net> > On 25 Jul 2005 06:11:30 -0700, brian0918 > >>crank > > jingleballs > > > In your defence, George, your post in this thread was probably the most > coherent, reasoned message I have ever seen from you. I hope a decent debate > ensues and you find some answers. He'll never get a response back. Sarfatti's posts are actually communiques between him & his close associates. But, he also has them delivered to Usenet for posterity. -Mark Martin === Subject: Re: Conformal Gravity? > He'll never get a response back. Sarfatti's posts are actually > communiques between him & his close associates. But, he also has them > delivered to Usenet for posterity. > This newsgroup does attract more than its fair share of nutcases doesn't it. === Subject: Re: Conformal Gravity? <9f8ae1lm2o4oivcnjrs0ra5m2i5qji8shi@4ax.com> <1OqdnYDwPPIBrXjfRVnyrw@pipex.net> He'll never get a response back. Sarfatti's posts are actually > communiques between him & his close associates. But, he also has them > delivered to Usenet for posterity. > > > This newsgroup does attract more than its fair share of nutcases doesn't it. -Mark Martin === Subject: Re: Damned if you do and damned if you don't > I am not going to comment on the theoretical physics. > I just want to state that there are non exotic explanations > for. > > 1) Pioneer anomoly. Recently Sedna has been discovered > and there is the possibility that there is a clould of > snowballs in the outer solar system. This would slighly > increase the gravity experienced by Pioneer. No, actually. Not until the satellite went outside their orbits. Until then, they would have littel or no effect. David A. Smith === Subject: Re: Damned if you do and damned if you don't No, actually. Not until the satellite went outside their orbits. Until then, they would have littel or no effect. The effect only took place at long distances from the Sun. That is to say outside the orbits of Sedna and its smaller companions. The real question is the nature of missing mass. In the outer solar system I am convinced that missing mass is in the form of ice balls. Jupiters in the galaxy? === Subject: Re: Damned if you do and damned if you don't I have had some thoughts about how effects actually occur. The electromagnetic device described is clearly a kind of rocket. Ion propusion is very much mainsream propulsion and promises to extend the life of satellites. To make a claim for exotic physics you neeed to exclude more basic physics. For example some of the atoms becoming ionised and being accelerated to high speed. It seems crystal clear to me that this is what is hapenning. There is one other point which I feel should be made. That is that if people he really was meddling with zero point energy it would be rather dangerous. If lamda could be varied by some means wouldn't it be rather dangerous. I have a vision of lamda becoming rather larger than anyone would want. Inflation ?! The fact that ions are being accelerated to high speed is irrelevant. The INTENTION is there and he does not seem to know what he is doing. The question of the risks taken for our so called benefit is a far wider one. The Pentagon had an antigravity project which was carried out in great secrecy. In the 18th century Jenner carried out vaccinations with cowpox with is an attenuated form of smallpox. To make vaccines you do not need a full strength version contrary to what the Pentagon would lead us to believe. Similarly the Americans have only themselves for the Antrax attacks. They in their wisdom stored a weaponized version. === Subject: Re: Difficult Integral - who can help? ... >during my calculations for a physical problem I derived the following >integral that must be solved: > >Integral[z]:= > >int_{-infinity}^{z} dz' Exp[-i pi (z'^2 - i alpha z')] * >int_{-infinity}^{z'} dz'' Exp[+i pi (z''^2 - i alpha z'')] > >As you see its a double-integral. > >Please note: > >z is real >i = sqrt(-1) >alpha is real and also alpha > 0 > >I solved the integral for z=infinity. So I have Integral[infinity] >which gives 1/2. > >But I also need Integral[z]. I asked lots of people and tried some >tricks, but nothing worked for this integral. >... Correction to my notation of Friday: At least one of the integrals should be solvable with the substitution z'+z'' = t z'-z'' = s The Jacobian is 1/2 -> (1/2) Int(t=-infty..2z) Int(s=2z-t ..0) exp[-i*pi*t*s - pi*alpha*s] Since the original (z',z'') area of integration was some sort of infinite triangle with the upper right corner at (z'=z''=z), one must write down how the new variables (t,s) (diagonals and antidiagonals) are limited by the old main diagonal and the old upper ceiling at z'=z. -> (1/2) Int(t=-infty..2z) {1-exp[-i*pi*t*s - pi*alpha*s]}/[-i*pi*t-pi*alpha] -> (1/2) Int(t=-infty..2z) {1-exp[-i*pi*t*(2z-t) - pi*alpha*(2z-t)]} /[-i*pi*t-pi*alpha] Another approach would remove the linear term in the inner integral int_{-infinity}^{z'} dz'' Exp[+i pi (z''^2 - i alpha z'')] = int_{-infinity}^{z'} dz'' Exp[+i pi (z'' - i/2 alpha)^2+const] = const* int_{-infinity}^{z'+i/2*alpha} dw Exp[+i pi w^2] which is some form of error function. Anyway, I do not see any closed form for the result. === Subject: Re: Very basic topology question > >> I am approached with some language and pictures in my textbook (Bredon's >> Topology and Geometry) and they are never defined or explained. What is >> the >> rigorous definition of a square with opposite edges identified? There >> are >> two pictures in my book : >> (1) A square whose right and left sides have arrows pointing upwards, and >> whose top and bottom sides have arrows pointing rightwards >> (2) A square whose right and left sides have arrows pointing upwards, and >> whose top and bottom sides have arrows pointing leftwards >> >> The first one is supposed to be the torus, the second the Klein bottle. > > Someone has made a mistake. (2) is also a torus. To get a Klein bottle, > one pair of opposite sides must have arrows pointing in opposite > directions. > You're right, I copied it down wrong. >> The trouble is I don't know what these arrows are supposed to mean. If I >> take a square sheet of paper, roll it up to get a cylinder, then attach >> the >> ends of the cylinder to get a torus, I can see that I am identifying >> points, >> but what does that have to do with the arrows I mention above? I guess >> I'm just not sure what the arrows mean. What if the right and left sides >> have arrows pointing downwards instead of upwards? What if one is up and >> one is down? > > Let's start with something a little simpler. Suppose the top & bottom > have arrows left (and no arrows on the other two sides). Then you see > how to get a cylinder. But now suppose top has arrow left & bottom has > arrow right. Then before you roll up you have to twist so the arrows > go the same way; this give you a Mobius strip. > > In other words, before you can attach edges to each other, you have to > get the arrows aligned, and that may mean twisting. If one pair of > arrows is parallel and the other anti-parallel, you can't actually do > the required twisting in 3-dimensional space. That's why people use > the square with opposite edges identified model of the Klein model; > it has all the topological attributes of the 4-dimensional Klein bottle > in a convenient two-dimensional representation. > So say we have a square whose right and left sides have arrows pointing upwards, and whose top and bottom sides have arrows pointing rightwards. What is a mathematical definition of this? Is it the square (where by square I mean the square as a subspace of R^2 using the subspace topology) modulo the equivalence relation (0,y) ~ (1,y) and (x,0) ~ (x,1)? A square modulo the equivalence relation > -- > Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Very basic topology question > >> I am approached with some language and pictures in my textbook (Bredon's >> Topology and Geometry) and they are never defined or explained. What is >> the rigorous definition of a square with opposite edges identified? >> There are two pictures in my book : >> (1) A square whose right and left sides have arrows pointing upwards, and >> whose top and bottom sides have arrows pointing rightwards >> (2) A square whose right and left sides have arrows pointing upwards, and >> whose top and bottom sides have arrows pointing leftwards >> >> The first one is supposed to be the torus, the second the Klein bottle. >> >> The trouble is I don't know what these arrows are supposed to mean. If I >> take a square sheet of paper, roll it up to get a cylinder, then attach >> the ends of the cylinder to get a torus, I can see that I am identifying >> points, but what does that have to do with the arrows I mention above? >> I guess I'm just not sure what the arrows mean. What if the right and >> left sides have arrows pointing downwards instead of upwards? What if >> one is up and one is down? > > The idea is this: if both arrows point upwards, you're identifying the > bottom point of the left side of the square with the bottom point of the > right side, you're identifying the top point of the left side with the > topo point of the right side and, more generally, you identify each > point of the left side with the point of the right side such that the > line that unites both points is parallel to the sides that have not > been mentioned yet. > Ok. I now get the idea. But what would be a rigorous definition of these arrows? I am trying to do some problems related to this, and I understand that what you state above is correct, but if I want to solve a problem, I should use a mathematical definition. For example as above, say we have a square whose right and left sides have arrows pointing upwards, and whose top and bottom sides have arrows pointing rightwards. What is a mathematical definition of this? Is it the square (where by square I mean the square as a subspace of R^2 using the subspace topology) modulo the equivalence relation (0,y) ~ (1,y) and (x,0) ~ (x,1)? > If both arrows were pointing downwards, it would be the same thing. > > If the left arrow points upwards and the right arrow points downwards, > then you're identifying each point of the left side with the point of > the right side which is obtained if you apply a rotation of pi radians > of the original point around the center of the square. In particular > the bottom point (respectively the top point) of the left side is > identified with the top point (resp. the bottom point) of the right > side. > Is your pi radians rotation for reverse arrows a definition? Would it still hold if we had a hexagon? That is, starting from the top side of a hexagon and going clockwise, we have the arrows pointing clockwise clockwise clockwise counterclockwise counterclockwise counterclockwise. Then the top and bottom sides are opposite, so we identify each point on the top side with a point on the bottom side which is obtained if you apply a rotation of pi radians of the original point around the center of the hexagon? > > Jose Carlos Santos === Subject: Re: Euclidean postulates > Perhaps I don't understand your question. > > Under the assumptions of Euclidean geometry, four > non-coplanar points > determine a sphere. You need the assumptions of > Euclidean geometry. > If you omit the Euclidean parallel postulate, then > there are models for > the geometry that admit surfaces such as horospheres > and hyperspheres > that contain four non-coplanar points. of course,for points lie on a surface of any shape,but,not,iff, I take them arbitrary > > This is pretty well-known stuff. For modern authors, > consult > Greenburg's Euclidean and Non-Euclidean Geometries, > Hartshorne's > Geometry: Euclid and Beyond, Henderson's > Experiencing Geometry -- or > many others. For historical sources: Hilbert's > Foundations of > Geometry, Bonola's Non-Euclidean Geometry (which > includes Bolyai's and > Lobachevsky's original papers) -- not too mention > Euclid's Elements. > > I hope this is helpful. > > --charlie > I should like to thank you for the information you provide But,I should like to explain the following point: There is an Euclidean postulate state that for any three points not on a straight line there exist a unique circle line with finite radius that passes through them. And,if this is a postulate,doesn't the bigger fact: mentioned here(there exist a unique spherical surface with finite radius in space for any four points in space (that are not on the same plan) and passes through them) Deserves to be anther POSTULATE Bassam Karzeddin Al-Hussein Bin Talal University Jordan === Subject: Re: Euclidean postulates <28918747.1122302632621.JavaMail.jakarta@nitrogen.mathforum.org>, > > I should like to thank you for the information you provide > > But,I should like to explain the following point: > > There is an Euclidean postulate state that for any three points not on a > straight line there exist a unique circle line with finite radius that passes > through them. > It is not a postulate, it is a proposition. Euclid: Book IV, Proposition 5: To circumscribe a circle about a given triangle. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Is symmetry Enough ? Please tell me, here is an old Olympiads problem: Prove that if BC+AC=tan(C/2)*( BC*tan(A)+AC*tan(B) ) (1) then ABC is isosceles ; we may see that exchanging letters A and B inside (1) let the equation inchanged ;may we infer points A and B have same 'roles and ABC isosceles ? Alain. === Subject: Re: Is symmetry Enough ? >Please tell me, > >here is an old Olympiads problem: > >Prove that if BC+AC=tan(C/2)*( BC*tan(A)+AC*tan(B) ) (1) >then ABC is isosceles ; >we may see that exchanging letters A and B inside (1) let the equation >inchanged ;may we infer points A and B have same 'roles and >ABC isosceles ? > >Alain. No, the symmetry of the identity only proves that if triangle ABC satisfies the the identity then so does triangle BAC. But this proves nothing since triangles ABC and BAC are the same triangle. quasi === Subject: Re: Is symmetry Enough ? > >>Please tell me, >> >>here is an old Olympiads problem: >> >>Prove that if BC+AC=tan(C/2)*( BC*tan(A)+AC*tan(B) ) (1) >>then ABC is isosceles ; >>we may see that exchanging letters A and B inside (1) let the equation >>inchanged ;may we infer points A and B have same 'roles and >>ABC isosceles ? >> >>Alain. > >No, the symmetry of the identity only proves that if triangle ABC >satisfies the the identity then so does triangle BAC. But this proves >nothing since triangles ABC and BAC are the same triangle. > >quasi To see the logical flaw exposed, use the same reasoning to prove this statement: If for triangle ABC, a^2+b^2=c^2 then triangle ABC is isosceles. proof: Assume a^2+b^2=c^2. Interchange a, b in the identity, and note that the equation is unchanged, hence is still true. Therefore a=b, hence triangle ABC is isosceles. Obviously this proof is flawed since otherwise we have proved that all right triangles are isosceles. quasi === Subject: The Sum of Its Parts The Sum of Its Parts ------------- Matter and material interactions are exactly the sum of their parts. Intelligence, the mind, and mental effects are much more than the sum of their parts; they are the difference of their parts. If you doubt this just run the numbers. In the history of the earth probably no more than 10 billion members of homo s. have ever existed. At 1/10th ton each, the cumulative weight would run to some 1 billion tons or probably less than the weight of all the water in the Boxing managed to generate vastly more activity than most small planets and probably hold the potential to generate more activity than most small stars. Contrast this with materialist/physicalist fantasies. Taken literally they would have you believe that the behavior of the 1 billion tons of humanity ever extant should be analyzed according to f=ma. Behaviorist dogma is even more uncritically juvenile. They would have you believe the psychology of humanity is no more than the circumstantial material consequences analogous to those geological events giving rise to the Boxing Day tsunami. Yet they worship on because no one has been able to explain the mechanics of differences underlying sentient behavior. === Subject: Re: The Sum of Its Parts > The Sum of Its Parts > ------------- > > Matter and material interactions are exactly the sum of their parts. > Intelligence, the mind, and mental effects are much more than the sum > of their parts; they are the difference of their parts. > > If you doubt this just run the numbers. In the history of the earth > probably no more than 10 billion members of homo s. have ever existed. > At 1/10th ton each, the cumulative weight would run to some 1 billion > tons or probably less than the weight of all the water in the Boxing > managed to generate vastly more activity than most small planets and > probably hold the potential to generate more activity than most small > stars. > > Contrast this with materialist/physicalist fantasies. Taken literally > they would have you believe that the behavior of the 1 billion tons of > humanity ever extant should be analyzed according to f=ma. Behaviorist > dogma is even more uncritically juvenile. They would have you believe > the psychology of humanity is no more than the circumstantial material > consequences analogous to those geological events giving rise to the > Boxing Day tsunami. Yet they worship on because no one has been able > to explain the mechanics of differences underlying sentient behavior. > It is very unclear to me what you are talking about. The small actions of the brain are converted to large actions via amplification not by F = m * a JC === Subject: Re: The Sum of Its Parts > The Sum of Its Parts > ------------- > > Matter and material interactions are exactly the sum of their parts. > Intelligence, the mind, and mental effects are much more than the sum > of their parts; they are the difference of their parts. What a strange statement. What that difference of their parts means? Could it mean something like this: If we have 3 sets {abcd}, {bcde} and {cdef}, abcd bcde cdef then the difference of their parts would be a and f, or what? When you say part, could you mean software component of strong AI / real AI? ------------------------------------------------------- mikk2lnx@yahoo.co.uk === Subject: Re: The Sum of Its Parts %IW48mQf3K=Ci&gZ7]]aazx@]Y-nq!r5{yH/#,?@lDdUDvOfByB2hVW0.@OM%{l/{cT'{w > The Sum of Its Parts > ------------- > > Matter and material interactions are exactly the sum of their parts. > Intelligence, the mind, and mental effects are much more than the sum > of their parts; they are the difference of their parts. > > If you doubt this just run the numbers. In the history of the earth > probably no more than 10 billion members of homo s. have ever existed. > At 1/10th ton each, the cumulative weight would run to some 1 billion > tons or probably less than the weight of all the water in the Boxing > managed to generate vastly more activity than most small planets and > probably hold the potential to generate more activity than most small > stars. Ok, so where's the logic here? How does the activty of 1 billion tons of biomass show the idea that mental effects are the difference of their parts? I see no connection here. > Contrast this with materialist/physicalist fantasies. Taken literally > they would have you believe that the behavior of the 1 billion tons of > humanity ever extant should be analyzed according to f=ma. Ok, you believe that the 1 billion tons of biomass created all the activity (motion), by violating f=ma? How did it do that? How did all that material get moved without there being a force? At what point in all that activity was f=ma violated? Buildings are built, mountans moved, and usenet messages are written, all beause energy is flowing from the sun, through the earth, like a river. And like the water in a river, the energy doesn't tend to flow evenly. It tends to form into complex streams of energy flow. Humans, and all life, and all the things we say that humans do, are just a natural side effect of the complex patterns the energy flows create. > Behaviorist > dogma is even more uncritically juvenile. They would have you believe > the psychology of humanity is no more than the circumstantial material > consequences analogous to those geological events giving rise to the > Boxing Day tsunami. Yeah, that's the way it works. > Yet they worship on because no one has been able > to explain the mechanics of differences underlying sentient behavior. Then why don't you explain it to us? How did the mechanics of differences make this usenet message happen? How did it do it without following f=ma? -- Curt Welch http://CurtWelch.Com/ curt@kcwc.com http://NewsReader.Com/ === Subject: Re: The Sum of Its Parts > > The Sum of Its Parts > ------------- > > Matter and material interactions are exactly the sum of their parts. [snip crap] General Relativity is non-linear. You are WRONG. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: Epistemology 201: The Science of Science An unedited copy of the first month of possibly the most successful thread in the history of the science and math usenet (~ 8000 posts over four months) is now available. Some of the luminaries who participated in the discussion, both pro and con, were Thomas Davidson, Bob Kolker, Jason Stevens, Aaron Boyden, Seymour Metz, Neil Rickert, Allan Cybulskie, Stephen@nomail.com, Albert Wagner, Richard Hall, Richard Herring, Wolf Kirchmeir, Osher Doctorow, Tony Orlow, Mitch Perkins, Mitch Harris, Mati Meron, Will Twentyman, and others without whose input, often acrimonious yet occasionally informative, this thread could never have become an overwhelming popular success. Anyone interested in the history of this thread may obtain a copy of my end (that is posts to which I replied) via email. Please email a request under the subject Science and I will forward a Microsoft Wordpad copy (approximately 2mb) via return attachment to you. - LZ Epistemology 201: The Science of Science > --------- > (Scientific Reduction) > >If asked what science is, most people would reply that science is >empirical in nature and conducts experiments to falsify unsound >hypotheses. Some might also vouchsafe that mathematics is a >scientific discipline to the extent that unsound hypotheses can be >falsified through contradictions with foundational axioms. But why >do we need science at all and how is it used? > >Let's look at reality in general through unscientific eyes. What we >see is a collation of events in historical terms. We see them succeed >one another and opine that various events cause one another. There is >no way to determine whether this is true in any fundamental sense, >only that history documents that various sequences of events have in >fact succeeded one another. > >So, what is science is expected to do? What is apparent to everyone is >that we have one historical tapestry of events and science is expected >to make sense of that tapestry. But how to do this? Basically science >can only make sense of the tapestry by reducing the number and >complexity of causes evident for events. Science must take the run of >events evident to everyone and show which characteristics and >properties govern the emergence of certain events as manifested in the >characteristics and properties of other events. > >In so doing science regresses consideration of events to properties of >events and shows how the emergence of one event is implied in the >emergence of other events. And thereby science reduces the panoply >of history to manifold considerations evident in all or most events. > Methodology > --------- > >Initially at least empirical sciences approach this reduction in the >same way mathematics does through finite tautological regression. >However, whereas mathematics regresses its observations through to >consistency with foundational axioms, through the vagaries of history >empirical sciences are left only with contradiction between empirical >observations as the basis for its regressive foundation. > >Consequently empirical science has been left with no understanding of >its own intellectual mechanics. It pretends to be different from logic >and mathematics and claims no finite tautological regressions limit >its empiricism. However, this is only partly true just as it is only >partly true for mathematics and logic. None of the three have finite >ending points that limit application of the respective disciplines. >But all three have finite tautological regressions which define and >limit their starting points, what I refer to as ur regressions. > Ur Regressions for True, False, and Not > -------------- > >True, False, and Not are defined in reciprocal terms in the following >way. For any empirical observation [subject] the proposition > >p:[subject][not subject] is always true. And the proposition > >p:[subject not subject] is always false. And the empirical observation > >P:[not] is always true because the proposition > >P:[not not] is always false. -------------- > >These seem to be the only reducible definitions for true, false, and >not. The problem is analogous to the definition of factorability in >mathematics where given i=j*k we have for any number, i, two factors, >j and k but only one equation, which means there is no general >solution possible for factors of i lying between 1 and i. -------------- > >In other words every empirical observation is regressable through >tautologies or it cannot be true because tautologies in the formal >sense are always true. Tautologies are not perfect, however, because >even though they account for everything true they do not account for >everything. In order to do that they would also have to account not >only for everything true but everything false as well. And we find >that perfecting ordinary tautologies requires the addition of some >component which is always false. For example, for > >t:[subject][not subject] > >t:[subject][not subject][subject not subject] > >wherein the self contradictory alternative [subject not subject] is >appended to an ordinary tautology to form a comprehensive or perfect >tautology inclusive of all possibilities. > >However such a regression through to self contradiction is not >possible in the case of one empirical observation [not] which forms >an irreducible regression directly in tautuological terms: > >T:[not][not not] > >contradiction forms the basis of all tautological regressions in >general, which in fact is exactly consistent with the form of the >tautology itself. > === Subject: Re: Legendre Transform > > > > This is all very interesting, but may I note: > 1. the Legendre transform is not a Laplace transform, or vice versa. > No-one has yet come up with a sensible explanation of Wikipedia's claim I had never seen that before, and I still think it is odd. Cute if true. > 2. Yes, the Laplace transform is closely connected to the Fourier > transform, and so is the Hankel transform, which is the 2D Fourier > transform reexpressed in terms of rotational eigenfunctions. At what point is all this stuff explained? The key ideas seem to be hiding in the land of differential geometry, if my weekend reading is any indication. Which would neatly explain why the base concepts are so hard to find. > > > One of these days I really want to play with numerically inverting the > Laplace transform, among others. It would probably help if I had an > idea how to invert a contour integral, though. That would help a lot. > > > Don't do it. The matrix of the discrete Laplace transform always has an > inverse, but AFAIK there is no simple expression for it corresponding > to the inverse of the DFT, and it's horrendously ill-conditioned. So the Laplace transform isn't terribly useful if you can't analyticaly invert what you find? This is why im curious to see what makes them work. Perhaps find something better...or, if I actually understood what was going on, find the limited range in which you can numerically invert the transform without much grief. > > > Zigoteau. === Subject: Re: Legendre Transform > 2. Yes, the Laplace transform is closely connected to the Fourier > transform, and so is the Hankel transform, which is the 2D Fourier > transform reexpressed in terms of rotational eigenfunctions. > > At what point is all this stuff explained? It was probably all explained in detail a century ago, and people have moved on to more exciting things. If you rediscover it, you can't get it published in a journal that people read because it's already known. functional analysis, and you can still find them on library shelves. They're still useful, but they are long out of print and no-one is going to republish them because the market for them is not enormous. I in fact rediscovered it when thinking about the eigenvalue problem for the Laplace operator del^2 in two dimensions with circular boundary conditions (including the case of no boundary conditions). Since del^2 commutes with rotation, the angular part must be an eigenvalue of d/dtheta, i.e. exp(i*n*theta). The radial part is then the Bessel function J_n(r). If you Fourier transform to momentum space, del^2 becomes r^2, so the eigenfunctions are nonzero only on a circular ring.. del^2 also commutes with d/dx and d/dy, so it is also possible to express the solution as a sum of plane waves, i.e. a 2D Fourier transform. Skipping a few steps, it follows that the 2D Fourier transform of J_n(r)*exp(i*n*theta) has the form a*delta(r-b)*exp(i*n*theta). > The key ideas seem to be > hiding in the land of differential geometry, if my weekend reading is > any indication. Which would neatly explain why the base concepts are so > hard to find. ?? Not sure about that. The operator del^2 is relevant to spaces that are as flat as a pancake. > One of these days I really want to play with numerically inverting the > Laplace transform, > Don't do it. The matrix of the discrete Laplace transform always has an > inverse, but AFAIK there is no simple expression for it corresponding > to the inverse of the DFT, and it's horrendously ill-conditioned. > > So the Laplace transform isn't terribly useful if you can't analyticaly > invert what you find? Make up your mind - you just said you wanted to invert it numerically. There is a list of analytical functions for which you can write down the analytical inverse Laplace transform. It's the numerical inverse which is ill-conditioned. The analytical inverse Laplace transform is involved in one of the methods for solving linear differential equations with constant coefficients, such as are encountered in electric wave filters . There are a number of other methods, but some people prefer Laplace transforms. > This is why im curious to see what makes them work. Perhaps find > something better...or, if I actually understood what was going on, find > the limited range in which you can numerically invert the transform > without much grief. I'm all in favor of understanding what is going on. Zigoteau. === Subject: Re: automorphisms of subspaces of the reals Mail-To-News-Contact: abuse@dizum.com >Let's see if I can get it right this time. Let R be the real line, let >c = |R| be the cardinal of the continuum, and let Q be the set of all >rational numbers. >> >Now consider a continuous injection f:Q-->R, not the identity. >> >>Maybe I'm just dense, but I don't see how a function that's only defined >>on Q can be continuous. > >What's the definition of the word continuous? Apparently it's not quite what I thought that it was. Here's what I learned, decades back: 1. A function, f, is continuous at a point, P, iff Lim(x->P) f(x) = f(P) 2. A function, f, is continuous over an interval, (a,b), iff it is continuous for all x such that a Writing about jazz is like dancing about architecture - Thelonious Monk === Subject: Re: Invariant Galilean Transformations (FAQ) On All Laws [snip lies] > > Invariant Galilean Transformations (FAQ) On All Laws > (c) Eleaticus/Oren C. Webster > Thnktank@concentric.net [snip 1300 lines of trolled garbage] eleaticus, Oren Webster, is a despised and stooopid troll, http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Crimes.html Several crimes against logic and science Ha ha ha! Originally trolled across sci.physics sci.physics.relativity alt.physics sci.math sci.answers alt.answers news.answers Psychotic ineducable boring troll Eleaticus, Were there to be internal inconsistencies in SR (meaning inconsistencies of a purely mathematical logical nature) that would automatically lead to contradictions in number theory, itself, and arithmetic, since the mathematics of Minkowski geometry is equiconsistent with the theory of real numbers and with arithmetic. Eleaticus explicitly demonstrates that he is completely ignorant of multivariable calculus. He has no concept of the Chain Rule in multivariable calculus. Consider his Galilean Transformation goo and dribble: t' = t, x' = x - vt, y' = y, z' = z. His refusal to accept that t' must be introduced as a separate variable springs from a massive emprical stupidity re space and time are described as a four-dimensional manifold, with four coordinates instead of a time evolution of a three-dimensional manifold, and that the change of coordinate system should be a change of four coordinates, and not a time-dependent change of three coordinates. This is particularly vital when it comes to fields over space and time (electric and magnetic fields for example). The transformation law for the differential operators under the Galilean transformation is given by: d/dt' = d/dt + v d/dx, d/dx' = d/dx, d/dy' = d/dy, d/dz' = d/dz. This shows the necessity of introducing a new variable t', since partial differentiation with respect to t' (constant x', y', z') is a different operation to partial differentiation with respect to t (constant x, y, z). The above transformation law is determined by the Chain Rule: d/dt' = dt/dt' d/dt + dx/dt' d/dx + dy/dt' d/dy + dz/dt' d/dz, d/dx' = dt/dx' d/dt + dx/dx' d/dx + dy/dx' d/dy + dz/dx' d/dz, d/dy' = dt/dy' d/dt + dx/dy' d/dx + dy/dy' d/dy + dz/dy' d/dz, d/dz' = dt/dz' d/dt + dx/dz' d/dx + dy/dz' d/dy + dz/dz' d/dz. The presence of the term involving d/dx in the expression for d/dt' is indicative of the fact that x depends on t' (x', y', z', being held constant), as can be seen from the fact that the coefficient of d/dx in the expression for d/dt' is dx/dt'. Because of the now demonstrated fact that Eleaticus has no formal education in multivariable calculus, he has managed, somehow, to get it into his head that the presence of the term involving d/dx in the expression for d/dt' is indicative of t' depending on x (t, y, z, being held constant). Because of his stupidty Eleaticus cannot get the correct transformation law for the differential operators under the Galilean Transformation, and he cannot determine the invariance or otherwise of Maxwell's Equations under the Galilean Transformation. The first advice to Eleaticus is to learn multivariable calculus. Eleaticus should not pretend that he can understand how to determine invariance or otherwise of Maxwell's Equations under the Galilean Transformation, or under the Lorentz Transformation, until he understands the multivariable calculus which underlies such considerations. Eleaticus is a loud idiot. The homogeneous Maxwell equations are invariant under the Galilean Transformation, with transformation laws: E_x' = E_x, E_y' = E_y - v B_z, E_z' = E_z + v B_y, B_x' = B_x, B_y' = B_y, B_z' = B_z. The derivation of these transformation laws was determined using the transformation laws for the differential operators given above. These transformation laws have the additional advantage that they determine the correct transformation for the force law, thus providing further evidence in favour of the transformation law for the differential operators, as above. The inhomogeneous Maxwell equations are also invariant under the Galilean transformation, with transformation laws: E_x' = E_x, E_y' = E_y, E_z' = E_z, B_x' = B_x, B_y' = B_y + v/c^2 E_z, B_z' = B_z - v/c^2 E_y, rho' = rho, J_x' = J_x - v rho, J_y' = J_y, J_z' = J_z. Note the the transformation laws for the charge density and current density are as they should be under the Galilean transformation. Homogeneous equations are invariant under the Galilean Transformation, and inhomogeneous equations are invariant under the Galilean Transformation, but Maxwell's Equations as a whole are NOT invariant under the Galilean Transformation, since the transformation laws required for the EM field for the two cases are inconsistent with each other. The transformation law for the EM field which makes the homogeneous equations invariant will not also make the inhomogeneous equations invariant. The transformation law for the EM field which makes the inhomogeneous equations invariant will not also make the homogeneous equations invariant. On the other hand, all of Maxwell's equations are invariant under the Lorentz Transformation, with transformation laws: E_x' = E_x, E_y' = gamma (E_y - v B_z), E_z' = gamma (E_z + v B_y), B_x' = B_x, B_y' = gamma (B_y + v/c^2 E_z), B_z' = gamma (B_z - v/c^2 E_y), rho' = gamma (rho - v/c^2 J_x), J_x' = gamma (J_x - v rho), J_y' = J_y, J_z' = J_z, where gamma = 1/sqrt(1 - v^2/c^2). Idiot Oren Webster sees himself this way, http://www.mazepath.com/uncleal/effete6.jpg The entire remainder of the planet sees him this way, http://www.mazepath.com/uncleal/effete3.png http://www.mazepath.com/uncleal/sunshine.jpg http://www.apa.org/journals/psp/psp7761121.html http://insti.physics.sunysb.edu/~siegel/quack.html Hey, stooopid troll Eleaticus - Do you want EVIDENCE? Each of the 24 GPS satellites carries either four cesium atomic clocks or three rubidum atomic clocks in orbit, with full relativistic corrections being applied. Internal inconsistencies in SR (meaning inconsistencies of a purely mathematical logical nature) automatically lead to contradictions in number theory, itself, and arithmetic, since the mathematics of Minkowski geometry is equiconsistent with the theory of real numbers and with arithmetic. Mathematics of gravitation Equivalence Principle testing http://arXiv.org/abs/hep-th/0111236 Geometric structure of reality http://arxiv.org/abs/gr-qc/0103044 http://arXiv.org/abs/hep-th/0307140 GR structure, especially Part 4/p. 7 http://arXiv.org/abs/gr-qc/0311039 Experimental constraints on General Relativity http://www.eftaylor.com/pub/projecta.pdf Relativity in the GPS system http://arXiv.org/abs/gr-qc/9909014 falling light http://metrologyforum.tm.agilent.com/cesium.shtml http://arxiv.org/abs/physics/0008012 Hafele-Keating Experiment http://www.hawaii.edu/suremath/SRtwinParadox.html Twin Paradox http://arXiv.org/abs/astro-ph/0401086 http://arxiv.org/abs/astro-ph/0312071 Deeply relativistic neutron star binaries http://arxiv.org/abs/hep-th/0405160 Black hole evaporation http://physicstoday.org/vol-57/iss-7/p40.shtml No aether http://fsweb.berry.edu/academic/mans/clane/ No Lorentz violation http://arXiv.org/abs/gr-qc/0409089 Spin-2 gravitons have problems (so does the proposal) http://arXiv.org/abs/gr-qc/0411113 http://arXiv.org/abs/gr-qc/0301024 Nordtvedt Effect http://map.gsfc.nasa.gov/ http://arxiv.org/abs/astro-ph/0403292 http://arXiv.org/abs/astro-ph/0310723 WMAP + Sloane Digital Sky Survey http://arxiv.org/abs/hep-ph/0404175 Dark matter candidates Carroll on what it all means. Special Relativity is physics on a topologically trivial Lorentzian manifold with a metric whose curvature tensor is zero. This is a perfectly diffeomorphism-invariant condition and does not require any particular coordinate choice. It is invariant under the full group of diffeomorphisms. The Poincare group is the group of *isometries* of the metric in special relativity. The Special Relativity metric is *non-dynamical* (unlike GR). It defines the coupling *constants* of your theory. If you change the metric in any nontrivial way you are changing your theory. An operation can only be called a symmetry of a special-relativistic (non-gravitational) theory if it preserves the metric, and therefore the symmetry of special-relativistic theories is the Poincare group only. General Relativity (gravitation) has a dynamic metric. NIM A 355 537 (1995) Physics Letters B 328 103 (1994) Physical Review Letters 64 1697 (1990) Physical Review Letters 39 1051 (1977) Physical Review 135 B1071 (1964) Physics Letters 12 260 (1964) Europhysics Letters 56(2) 170-174 (2001) General Relativity and Gravitation 34(9) 1371 (2002) http://fourmilab.to/etexts/einstein/specrel/specrel.pdf Longitudinal and transverse mass Physics Today 58(3) 34 (2005) Time passage, equator vs. poles http://arxiv.org/abs/gr-qc/0306076.pdf http://www.navcen.uscg.gov/pubs/gps/gpsuser/gpsuser.pdf http://www.navcen.uscg.gov/pubs/gps/sigspec/default.htm http://www.navcen.uscg.gov/pubs/gps/icd200/default.htm http://www.trimble.com/gps/index.html http://sirius.chinalake.navy.mil/satpred/ http://www.phys.lsu.edu/mog/mog9/node9.html http://egtphysics.net/GPS/RelGPS.htm http://www.schriever.af.mil/gps/Current/current.oa1 http://edu-observatory.org/gps/gps_books.html -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: Unknown Functions & Einstein's Incompetence [snip lies] > Unknown Functions & Einstein's Incompetence (FAQ) > (c) Eleaticus/Oren C. Webster > Thnktank@concentric.net [snip 1300 lines of trolled garbage] eleaticus, Oren Webster, is a despised and stooopid troll, http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Crimes.html Several crimes against logic and science Ha ha ha! Originally trolled across sci.physics sci.physics.relativity alt.physics sci.math sci.answers alt.answers news.answers Psychotic ineducable boring troll Eleaticus, Were there to be internal inconsistencies in SR (meaning inconsistencies of a purely mathematical logical nature) that would automatically lead to contradictions in number theory, itself, and arithmetic, since the mathematics of Minkowski geometry is equiconsistent with the theory of real numbers and with arithmetic. Eleaticus explicitly demonstrates that he is completely ignorant of multivariable calculus. He has no concept of the Chain Rule in multivariable calculus. Consider his Galilean Transformation goo and dribble: t' = t, x' = x - vt, y' = y, z' = z. His refusal to accept that t' must be introduced as a separate variable springs from a massive emprical stupidity re space and time are described as a four-dimensional manifold, with four coordinates instead of a time evolution of a three-dimensional manifold, and that the change of coordinate system should be a change of four coordinates, and not a time-dependent change of three coordinates. This is particularly vital when it comes to fields over space and time (electric and magnetic fields for example). The transformation law for the differential operators under the Galilean transformation is given by: d/dt' = d/dt + v d/dx, d/dx' = d/dx, d/dy' = d/dy, d/dz' = d/dz. This shows the necessity of introducing a new variable t', since partial differentiation with respect to t' (constant x', y', z') is a different operation to partial differentiation with respect to t (constant x, y, z). The above transformation law is determined by the Chain Rule: d/dt' = dt/dt' d/dt + dx/dt' d/dx + dy/dt' d/dy + dz/dt' d/dz, d/dx' = dt/dx' d/dt + dx/dx' d/dx + dy/dx' d/dy + dz/dx' d/dz, d/dy' = dt/dy' d/dt + dx/dy' d/dx + dy/dy' d/dy + dz/dy' d/dz, d/dz' = dt/dz' d/dt + dx/dz' d/dx + dy/dz' d/dy + dz/dz' d/dz. The presence of the term involving d/dx in the expression for d/dt' is indicative of the fact that x depends on t' (x', y', z', being held constant), as can be seen from the fact that the coefficient of d/dx in the expression for d/dt' is dx/dt'. Because of the now demonstrated fact that Eleaticus has no formal education in multivariable calculus, he has managed, somehow, to get it into his head that the presence of the term involving d/dx in the expression for d/dt' is indicative of t' depending on x (t, y, z, being held constant). Because of his stupidty Eleaticus cannot get the correct transformation law for the differential operators under the Galilean Transformation, and he cannot determine the invariance or otherwise of Maxwell's Equations under the Galilean Transformation. The first advice to Eleaticus is to learn multivariable calculus. Eleaticus should not pretend that he can understand how to determine invariance or otherwise of Maxwell's Equations under the Galilean Transformation, or under the Lorentz Transformation, until he understands the multivariable calculus which underlies such considerations. Eleaticus is a loud idiot. The homogeneous Maxwell equations are invariant under the Galilean Transformation, with transformation laws: E_x' = E_x, E_y' = E_y - v B_z, E_z' = E_z + v B_y, B_x' = B_x, B_y' = B_y, B_z' = B_z. The derivation of these transformation laws was determined using the transformation laws for the differential operators given above. These transformation laws have the additional advantage that they determine the correct transformation for the force law, thus providing further evidence in favour of the transformation law for the differential operators, as above. The inhomogeneous Maxwell equations are also invariant under the Galilean transformation, with transformation laws: E_x' = E_x, E_y' = E_y, E_z' = E_z, B_x' = B_x, B_y' = B_y + v/c^2 E_z, B_z' = B_z - v/c^2 E_y, rho' = rho, J_x' = J_x - v rho, J_y' = J_y, J_z' = J_z. Note the the transformation laws for the charge density and current density are as they should be under the Galilean transformation. Homogeneous equations are invariant under the Galilean Transformation, and inhomogeneous equations are invariant under the Galilean Transformation, but Maxwell's Equations as a whole are NOT invariant under the Galilean Transformation, since the transformation laws required for the EM field for the two cases are inconsistent with each other. The transformation law for the EM field which makes the homogeneous equations invariant will not also make the inhomogeneous equations invariant. The transformation law for the EM field which makes the inhomogeneous equations invariant will not also make the homogeneous equations invariant. On the other hand, all of Maxwell's equations are invariant under the Lorentz Transformation, with transformation laws: E_x' = E_x, E_y' = gamma (E_y - v B_z), E_z' = gamma (E_z + v B_y), B_x' = B_x, B_y' = gamma (B_y + v/c^2 E_z), B_z' = gamma (B_z - v/c^2 E_y), rho' = gamma (rho - v/c^2 J_x), J_x' = gamma (J_x - v rho), J_y' = J_y, J_z' = J_z, where gamma = 1/sqrt(1 - v^2/c^2). Idiot Oren Webster sees himself this way, http://www.mazepath.com/uncleal/effete6.jpg The entire remainder of the planet sees him this way, http://www.mazepath.com/uncleal/effete3.png http://www.mazepath.com/uncleal/sunshine.jpg http://www.apa.org/journals/psp/psp7761121.html http://insti.physics.sunysb.edu/~siegel/quack.html Hey, stooopid troll Eleaticus - Do you want EVIDENCE? Each of the 24 GPS satellites carries either four cesium atomic clocks or three rubidum atomic clocks in orbit, with full relativistic corrections being applied. Internal inconsistencies in SR (meaning inconsistencies of a purely mathematical logical nature) automatically lead to contradictions in number theory, itself, and arithmetic, since the mathematics of Minkowski geometry is equiconsistent with the theory of real numbers and with arithmetic. Mathematics of gravitation Equivalence Principle testing http://arXiv.org/abs/hep-th/0111236 Geometric structure of reality http://arxiv.org/abs/gr-qc/0103044 http://arXiv.org/abs/hep-th/0307140 GR structure, especially Part 4/p. 7 http://arXiv.org/abs/gr-qc/0311039 Experimental constraints on General Relativity http://www.eftaylor.com/pub/projecta.pdf Relativity in the GPS system http://arXiv.org/abs/gr-qc/9909014 falling light http://metrologyforum.tm.agilent.com/cesium.shtml http://arxiv.org/abs/physics/0008012 Hafele-Keating Experiment http://www.hawaii.edu/suremath/SRtwinParadox.html Twin Paradox http://arXiv.org/abs/astro-ph/0401086 http://arxiv.org/abs/astro-ph/0312071 Deeply relativistic neutron star binaries http://arxiv.org/abs/hep-th/0405160 Black hole evaporation http://physicstoday.org/vol-57/iss-7/p40.shtml No aether http://fsweb.berry.edu/academic/mans/clane/ No Lorentz violation http://arXiv.org/abs/gr-qc/0409089 Spin-2 gravitons have problems (so does the proposal) http://arXiv.org/abs/gr-qc/0411113 http://arXiv.org/abs/gr-qc/0301024 Nordtvedt Effect http://map.gsfc.nasa.gov/ http://arxiv.org/abs/astro-ph/0403292 http://arXiv.org/abs/astro-ph/0310723 WMAP + Sloane Digital Sky Survey http://arxiv.org/abs/hep-ph/0404175 Dark matter candidates Carroll on what it all means. Special Relativity is physics on a topologically trivial Lorentzian manifold with a metric whose curvature tensor is zero. This is a perfectly diffeomorphism-invariant condition and does not require any particular coordinate choice. It is invariant under the full group of diffeomorphisms. The Poincare group is the group of *isometries* of the metric in special relativity. The Special Relativity metric is *non-dynamical* (unlike GR). It defines the coupling *constants* of your theory. If you change the metric in any nontrivial way you are changing your theory. An operation can only be called a symmetry of a special-relativistic (non-gravitational) theory if it preserves the metric, and therefore the symmetry of special-relativistic theories is the Poincare group only. General Relativity (gravitation) has a dynamic metric. NIM A 355 537 (1995) Physics Letters B 328 103 (1994) Physical Review Letters 64 1697 (1990) Physical Review Letters 39 1051 (1977) Physical Review 135 B1071 (1964) Physics Letters 12 260 (1964) Europhysics Letters 56(2) 170-174 (2001) General Relativity and Gravitation 34(9) 1371 (2002) http://fourmilab.to/etexts/einstein/specrel/specrel.pdf Longitudinal and transverse mass Physics Today 58(3) 34 (2005) Time passage, equator vs. poles http://arxiv.org/abs/gr-qc/0306076.pdf http://www.navcen.uscg.gov/pubs/gps/gpsuser/gpsuser.pdf http://www.navcen.uscg.gov/pubs/gps/sigspec/default.htm http://www.navcen.uscg.gov/pubs/gps/icd200/default.htm http://www.trimble.com/gps/index.html http://sirius.chinalake.navy.mil/satpred/ http://www.phys.lsu.edu/mog/mog9/node9.html http://egtphysics.net/GPS/RelGPS.htm http://www.schriever.af.mil/gps/Current/current.oa1 http://edu-observatory.org/gps/gps_books.html -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: Einstein (1905) Absurdities [snip lies] > > Einstein (1905) Absurdities > (c) Eleaticus/Oren C. Webster > Thnktank@concentric.net [snip 1300 lines of trolled garbage] eleaticus, Oren Webster, is a despised and stooopid troll, http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Crimes.html Several crimes against logic and science Ha ha ha! Originally trolled across sci.physics sci.physics.relativity alt.physics sci.math sci.answers alt.answers news.answers Psychotic ineducable boring troll Eleaticus, Were there to be internal inconsistencies in SR (meaning inconsistencies of a purely mathematical logical nature) that would automatically lead to contradictions in number theory, itself, and arithmetic, since the mathematics of Minkowski geometry is equiconsistent with the theory of real numbers and with arithmetic. Eleaticus explicitly demonstrates that he is completely ignorant of multivariable calculus. He has no concept of the Chain Rule in multivariable calculus. Consider his Galilean Transformation goo and dribble: t' = t, x' = x - vt, y' = y, z' = z. His refusal to accept that t' must be introduced as a separate variable springs from a massive emprical stupidity re space and time are described as a four-dimensional manifold, with four coordinates instead of a time evolution of a three-dimensional manifold, and that the change of coordinate system should be a change of four coordinates, and not a time-dependent change of three coordinates. This is particularly vital when it comes to fields over space and time (electric and magnetic fields for example). The transformation law for the differential operators under the Galilean transformation is given by: d/dt' = d/dt + v d/dx, d/dx' = d/dx, d/dy' = d/dy, d/dz' = d/dz. This shows the necessity of introducing a new variable t', since partial differentiation with respect to t' (constant x', y', z') is a different operation to partial differentiation with respect to t (constant x, y, z). The above transformation law is determined by the Chain Rule: d/dt' = dt/dt' d/dt + dx/dt' d/dx + dy/dt' d/dy + dz/dt' d/dz, d/dx' = dt/dx' d/dt + dx/dx' d/dx + dy/dx' d/dy + dz/dx' d/dz, d/dy' = dt/dy' d/dt + dx/dy' d/dx + dy/dy' d/dy + dz/dy' d/dz, d/dz' = dt/dz' d/dt + dx/dz' d/dx + dy/dz' d/dy + dz/dz' d/dz. The presence of the term involving d/dx in the expression for d/dt' is indicative of the fact that x depends on t' (x', y', z', being held constant), as can be seen from the fact that the coefficient of d/dx in the expression for d/dt' is dx/dt'. Because of the now demonstrated fact that Eleaticus has no formal education in multivariable calculus, he has managed, somehow, to get it into his head that the presence of the term involving d/dx in the expression for d/dt' is indicative of t' depending on x (t, y, z, being held constant). Because of his stupidty Eleaticus cannot get the correct transformation law for the differential operators under the Galilean Transformation, and he cannot determine the invariance or otherwise of Maxwell's Equations under the Galilean Transformation. The first advice to Eleaticus is to learn multivariable calculus. Eleaticus should not pretend that he can understand how to determine invariance or otherwise of Maxwell's Equations under the Galilean Transformation, or under the Lorentz Transformation, until he understands the multivariable calculus which underlies such considerations. Eleaticus is a loud idiot. The homogeneous Maxwell equations are invariant under the Galilean Transformation, with transformation laws: E_x' = E_x, E_y' = E_y - v B_z, E_z' = E_z + v B_y, B_x' = B_x, B_y' = B_y, B_z' = B_z. The derivation of these transformation laws was determined using the transformation laws for the differential operators given above. These transformation laws have the additional advantage that they determine the correct transformation for the force law, thus providing further evidence in favour of the transformation law for the differential operators, as above. The inhomogeneous Maxwell equations are also invariant under the Galilean transformation, with transformation laws: E_x' = E_x, E_y' = E_y, E_z' = E_z, B_x' = B_x, B_y' = B_y + v/c^2 E_z, B_z' = B_z - v/c^2 E_y, rho' = rho, J_x' = J_x - v rho, J_y' = J_y, J_z' = J_z. Note the the transformation laws for the charge density and current density are as they should be under the Galilean transformation. Homogeneous equations are invariant under the Galilean Transformation, and inhomogeneous equations are invariant under the Galilean Transformation, but Maxwell's Equations as a whole are NOT invariant under the Galilean Transformation, since the transformation laws required for the EM field for the two cases are inconsistent with each other. The transformation law for the EM field which makes the homogeneous equations invariant will not also make the inhomogeneous equations invariant. The transformation law for the EM field which makes the inhomogeneous equations invariant will not also make the homogeneous equations invariant. On the other hand, all of Maxwell's equations are invariant under the Lorentz Transformation, with transformation laws: E_x' = E_x, E_y' = gamma (E_y - v B_z), E_z' = gamma (E_z + v B_y), B_x' = B_x, B_y' = gamma (B_y + v/c^2 E_z), B_z' = gamma (B_z - v/c^2 E_y), rho' = gamma (rho - v/c^2 J_x), J_x' = gamma (J_x - v rho), J_y' = J_y, J_z' = J_z, where gamma = 1/sqrt(1 - v^2/c^2). Idiot Oren Webster sees himself this way, http://www.mazepath.com/uncleal/effete6.jpg The entire remainder of the planet sees him this way, http://www.mazepath.com/uncleal/effete3.png http://www.mazepath.com/uncleal/sunshine.jpg http://www.apa.org/journals/psp/psp7761121.html http://insti.physics.sunysb.edu/~siegel/quack.html Hey, stooopid troll Eleaticus - Do you want EVIDENCE? Each of the 24 GPS satellites carries either four cesium atomic clocks or three rubidum atomic clocks in orbit, with full relativistic corrections being applied. Internal inconsistencies in SR (meaning inconsistencies of a purely mathematical logical nature) automatically lead to contradictions in number theory, itself, and arithmetic, since the mathematics of Minkowski geometry is equiconsistent with the theory of real numbers and with arithmetic. Mathematics of gravitation Equivalence Principle testing http://arXiv.org/abs/hep-th/0111236 Geometric structure of reality http://arxiv.org/abs/gr-qc/0103044 http://arXiv.org/abs/hep-th/0307140 GR structure, especially Part 4/p. 7 http://arXiv.org/abs/gr-qc/0311039 Experimental constraints on General Relativity http://www.eftaylor.com/pub/projecta.pdf Relativity in the GPS system http://arXiv.org/abs/gr-qc/9909014 falling light http://metrologyforum.tm.agilent.com/cesium.shtml http://arxiv.org/abs/physics/0008012 Hafele-Keating Experiment http://www.hawaii.edu/suremath/SRtwinParadox.html Twin Paradox http://arXiv.org/abs/astro-ph/0401086 http://arxiv.org/abs/astro-ph/0312071 Deeply relativistic neutron star binaries http://arxiv.org/abs/hep-th/0405160 Black hole evaporation http://physicstoday.org/vol-57/iss-7/p40.shtml No aether http://fsweb.berry.edu/academic/mans/clane/ No Lorentz violation http://arXiv.org/abs/gr-qc/0409089 Spin-2 gravitons have problems (so does the proposal) http://arXiv.org/abs/gr-qc/0411113 http://arXiv.org/abs/gr-qc/0301024 Nordtvedt Effect http://map.gsfc.nasa.gov/ http://arxiv.org/abs/astro-ph/0403292 http://arXiv.org/abs/astro-ph/0310723 WMAP + Sloane Digital Sky Survey http://arxiv.org/abs/hep-ph/0404175 Dark matter candidates Carroll on what it all means. Special Relativity is physics on a topologically trivial Lorentzian manifold with a metric whose curvature tensor is zero. This is a perfectly diffeomorphism-invariant condition and does not require any particular coordinate choice. It is invariant under the full group of diffeomorphisms. The Poincare group is the group of *isometries* of the metric in special relativity. The Special Relativity metric is *non-dynamical* (unlike GR). It defines the coupling *constants* of your theory. If you change the metric in any nontrivial way you are changing your theory. An operation can only be called a symmetry of a special-relativistic (non-gravitational) theory if it preserves the metric, and therefore the symmetry of special-relativistic theories is the Poincare group only. General Relativity (gravitation) has a dynamic metric. NIM A 355 537 (1995) Physics Letters B 328 103 (1994) Physical Review Letters 64 1697 (1990) Physical Review Letters 39 1051 (1977) Physical Review 135 B1071 (1964) Physics Letters 12 260 (1964) Europhysics Letters 56(2) 170-174 (2001) General Relativity and Gravitation 34(9) 1371 (2002) http://fourmilab.to/etexts/einstein/specrel/specrel.pdf Longitudinal and transverse mass Physics Today 58(3) 34 (2005) Time passage, equator vs. poles http://arxiv.org/abs/gr-qc/0306076.pdf http://www.navcen.uscg.gov/pubs/gps/gpsuser/gpsuser.pdf http://www.navcen.uscg.gov/pubs/gps/sigspec/default.htm http://www.navcen.uscg.gov/pubs/gps/icd200/default.htm http://www.trimble.com/gps/index.html http://sirius.chinalake.navy.mil/satpred/ http://www.phys.lsu.edu/mog/mog9/node9.html http://egtphysics.net/GPS/RelGPS.htm http://www.schriever.af.mil/gps/Current/current.oa1 http://edu-observatory.org/gps/gps_books.html -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: jensen for a derivative >Let f be a derivative of a function on interval I, and f satisfies >f( (x+y)/ 2 ) <= ( f(x) + f(y) ) / 2 (*) >Show that f is continuous. I'll assume I is open. Any derivative is the pointwise limit of continuous functions, and therefore is measurable. Let B_n = {x in I: f(x) < n}. Then I = union {B_n: n=1,2,...} so some B_n has positive measure. But then (B_n + B_n)/2 = {(x+y)/2: x, y in B_n} contains a nonempty open interval, and by (*) so does B_n. Now if (a,b) is contained in B_n and b < c < 2b-a with c in I, then [b,(b+c)/2) is contained in B_{(n+f(c))/2}. Repeating this, we find that f is bounded above on every compact subinterval of I. Now we can rewrite (*) as f(x-d) >= 2 f(x) - f(x+d), and then by induction f(x-kd) >= f(x) + k(f(x)-f(x+d)) for all positive integers k for which x - kd is in I. Similarly, f(x+kd) >= f(x) + k(f(x+d)-f(x)) if x + kd is in I. If x+kd and x-kd are in B_n, these inequalities imply (f(x)-n)/k <= f(x+d)-f(x) <= (n-f(x))/k. Given x in I and epsilon > 0, x is in some open interval J contained in some B_n. Take k large enough that 2n/k < epsilon, and delta small enough that x + k delta and x - k delta are in J, and then for |d| < delta we have |f(x+d)-f(x)| < epsilon. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: jensen for a derivative > >The assertion f((x + y)/2) <= (f(x) + f(y))/2 for all x and all y in I > >is equivalent to the assertion f is convex; this follows from the >fact that f is a derivative and therefore that it has the intermediate >value property (i. e. if x and y belong to I and if c is somewhere >betwenn f(x) and f(y), then there's some z between x and y such that >f(z) = c). >> >> >> I don't follow that why f is convex. >> What role does intermediate value property play here? > >Oops, I guess that I forgot something here! :-( Yeah, I was wondering about that as well - imagine my disappointment at your reply. (It seems clear to me that the result must be true, because a discontinuous midpoint-convex function must be _very_ discontinuous, and a derivative can't be all that discontinuous. But that's just a little too vague to count as a proof...) >[...]I was under the illusion that >the intermediate value property would be enough to establish the result, >but right now I am unable even to deduce it from the fact that f has a >primitive. > >I hope that this helps. Sorry for wasting your time. > > >Jose Carlos Santos ************************ David C. Ullrich === Subject: Re: (SR) Lorentz t', x' = Intervals [snip lies] > > > (SR) Lorentz t', x' = Intervals > (c) Eleaticus/Oren C. Webster > Thnktank@concentric.net [snip 700 lines of trolled garbage] eleaticus, Oren Webster, is a despised and stooopid troll, http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Crimes.html Several crimes against logic and science Ha ha ha! Originally trolled across sci.physics sci.physics.relativity alt.physics sci.math sci.answers alt.answers news.answers Psychotic ineducable boring troll Eleaticus, Were there to be internal inconsistencies in SR (meaning inconsistencies of a purely mathematical logical nature) that would automatically lead to contradictions in number theory, itself, and arithmetic, since the mathematics of Minkowski geometry is equiconsistent with the theory of real numbers and with arithmetic. Eleaticus explicitly demonstrates that he is completely ignorant of multivariable calculus. He has no concept of the Chain Rule in multivariable calculus. Consider his Galilean Transformation goo and dribble: t' = t, x' = x - vt, y' = y, z' = z. His refusal to accept that t' must be introduced as a separate variable springs from a massive emprical stupidity re space and time are described as a four-dimensional manifold, with four coordinates instead of a time evolution of a three-dimensional manifold, and that the change of coordinate system should be a change of four coordinates, and not a time-dependent change of three coordinates. This is particularly vital when it comes to fields over space and time (electric and magnetic fields for example). The transformation law for the differential operators under the Galilean transformation is given by: d/dt' = d/dt + v d/dx, d/dx' = d/dx, d/dy' = d/dy, d/dz' = d/dz. This shows the necessity of introducing a new variable t', since partial differentiation with respect to t' (constant x', y', z') is a different operation to partial differentiation with respect to t (constant x, y, z). The above transformation law is determined by the Chain Rule: d/dt' = dt/dt' d/dt + dx/dt' d/dx + dy/dt' d/dy + dz/dt' d/dz, d/dx' = dt/dx' d/dt + dx/dx' d/dx + dy/dx' d/dy + dz/dx' d/dz, d/dy' = dt/dy' d/dt + dx/dy' d/dx + dy/dy' d/dy + dz/dy' d/dz, d/dz' = dt/dz' d/dt + dx/dz' d/dx + dy/dz' d/dy + dz/dz' d/dz. The presence of the term involving d/dx in the expression for d/dt' is indicative of the fact that x depends on t' (x', y', z', being held constant), as can be seen from the fact that the coefficient of d/dx in the expression for d/dt' is dx/dt'. Because of the now demonstrated fact that Eleaticus has no formal education in multivariable calculus, he has managed, somehow, to get it into his head that the presence of the term involving d/dx in the expression for d/dt' is indicative of t' depending on x (t, y, z, being held constant). Because of his stupidty Eleaticus cannot get the correct transformation law for the differential operators under the Galilean Transformation, and he cannot determine the invariance or otherwise of Maxwell's Equations under the Galilean Transformation. The first advice to Eleaticus is to learn multivariable calculus. Eleaticus should not pretend that he can understand how to determine invariance or otherwise of Maxwell's Equations under the Galilean Transformation, or under the Lorentz Transformation, until he understands the multivariable calculus which underlies such considerations. Eleaticus is a loud idiot. The homogeneous Maxwell equations are invariant under the Galilean Transformation, with transformation laws: E_x' = E_x, E_y' = E_y - v B_z, E_z' = E_z + v B_y, B_x' = B_x, B_y' = B_y, B_z' = B_z. The derivation of these transformation laws was determined using the transformation laws for the differential operators given above. These transformation laws have the additional advantage that they determine the correct transformation for the force law, thus providing further evidence in favour of the transformation law for the differential operators, as above. The inhomogeneous Maxwell equations are also invariant under the Galilean transformation, with transformation laws: E_x' = E_x, E_y' = E_y, E_z' = E_z, B_x' = B_x, B_y' = B_y + v/c^2 E_z, B_z' = B_z - v/c^2 E_y, rho' = rho, J_x' = J_x - v rho, J_y' = J_y, J_z' = J_z. Note the the transformation laws for the charge density and current density are as they should be under the Galilean transformation. Homogeneous equations are invariant under the Galilean Transformation, and inhomogeneous equations are invariant under the Galilean Transformation, but Maxwell's Equations as a whole are NOT invariant under the Galilean Transformation, since the transformation laws required for the EM field for the two cases are inconsistent with each other. The transformation law for the EM field which makes the homogeneous equations invariant will not also make the inhomogeneous equations invariant. The transformation law for the EM field which makes the inhomogeneous equations invariant will not also make the homogeneous equations invariant. On the other hand, all of Maxwell's equations are invariant under the Lorentz Transformation, with transformation laws: E_x' = E_x, E_y' = gamma (E_y - v B_z), E_z' = gamma (E_z + v B_y), B_x' = B_x, B_y' = gamma (B_y + v/c^2 E_z), B_z' = gamma (B_z - v/c^2 E_y), rho' = gamma (rho - v/c^2 J_x), J_x' = gamma (J_x - v rho), J_y' = J_y, J_z' = J_z, where gamma = 1/sqrt(1 - v^2/c^2). Idiot Oren Webster sees himself this way, http://www.mazepath.com/uncleal/effete6.jpg The entire remainder of the planet sees him this way, http://www.mazepath.com/uncleal/effete3.png http://www.mazepath.com/uncleal/sunshine.jpg http://www.apa.org/journals/psp/psp7761121.html http://insti.physics.sunysb.edu/~siegel/quack.html Hey, stooopid troll Eleaticus - Do you want EVIDENCE? Each of the 24 GPS satellites carries either four cesium atomic clocks or three rubidum atomic clocks in orbit, with full relativistic corrections being applied. Internal inconsistencies in SR (meaning inconsistencies of a purely mathematical logical nature) automatically lead to contradictions in number theory, itself, and arithmetic, since the mathematics of Minkowski geometry is equiconsistent with the theory of real numbers and with arithmetic. Mathematics of gravitation Equivalence Principle testing http://arXiv.org/abs/hep-th/0111236 Geometric structure of reality http://arxiv.org/abs/gr-qc/0103044 http://arXiv.org/abs/hep-th/0307140 GR structure, especially Part 4/p. 7 http://arXiv.org/abs/gr-qc/0311039 Experimental constraints on General Relativity http://www.eftaylor.com/pub/projecta.pdf Relativity in the GPS system http://arXiv.org/abs/gr-qc/9909014 falling light http://metrologyforum.tm.agilent.com/cesium.shtml http://arxiv.org/abs/physics/0008012 Hafele-Keating Experiment http://www.hawaii.edu/suremath/SRtwinParadox.html Twin Paradox http://arXiv.org/abs/astro-ph/0401086 http://arxiv.org/abs/astro-ph/0312071 Deeply relativistic neutron star binaries http://arxiv.org/abs/hep-th/0405160 Black hole evaporation http://physicstoday.org/vol-57/iss-7/p40.shtml No aether http://fsweb.berry.edu/academic/mans/clane/ No Lorentz violation http://arXiv.org/abs/gr-qc/0409089 Spin-2 gravitons have problems (so does the proposal) http://arXiv.org/abs/gr-qc/0411113 http://arXiv.org/abs/gr-qc/0301024 Nordtvedt Effect http://map.gsfc.nasa.gov/ http://arxiv.org/abs/astro-ph/0403292 http://arXiv.org/abs/astro-ph/0310723 WMAP + Sloane Digital Sky Survey http://arxiv.org/abs/hep-ph/0404175 Dark matter candidates Carroll on what it all means. Special Relativity is physics on a topologically trivial Lorentzian manifold with a metric whose curvature tensor is zero. This is a perfectly diffeomorphism-invariant condition and does not require any particular coordinate choice. It is invariant under the full group of diffeomorphisms. The Poincare group is the group of *isometries* of the metric in special relativity. The Special Relativity metric is *non-dynamical* (unlike GR). It defines the coupling *constants* of your theory. If you change the metric in any nontrivial way you are changing your theory. An operation can only be called a symmetry of a special-relativistic (non-gravitational) theory if it preserves the metric, and therefore the symmetry of special-relativistic theories is the Poincare group only. General Relativity (gravitation) has a dynamic metric. NIM A 355 537 (1995) Physics Letters B 328 103 (1994) Physical Review Letters 64 1697 (1990) Physical Review Letters 39 1051 (1977) Physical Review 135 B1071 (1964) Physics Letters 12 260 (1964) Europhysics Letters 56(2) 170-174 (2001) General Relativity and Gravitation 34(9) 1371 (2002) http://fourmilab.to/etexts/einstein/specrel/specrel.pdf Longitudinal and transverse mass Physics Today 58(3) 34 (2005) Time passage, equator vs. poles http://arxiv.org/abs/gr-qc/0306076.pdf http://www.navcen.uscg.gov/pubs/gps/gpsuser/gpsuser.pdf http://www.navcen.uscg.gov/pubs/gps/sigspec/default.htm http://www.navcen.uscg.gov/pubs/gps/icd200/default.htm http://www.trimble.com/gps/index.html http://sirius.chinalake.navy.mil/satpred/ http://www.phys.lsu.edu/mog/mog9/node9.html http://egtphysics.net/GPS/RelGPS.htm http://www.schriever.af.mil/gps/Current/current.oa1 http://edu-observatory.org/gps/gps_books.html -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: Recovering from a computational mathematics education Mail-To-News-Contact: abuse@dizum.com >> I'd like to be able to upgrade myself from a human calculator to a thinker. >> Is there any way I can do this through self-study? > What you need is a book in rigourous advanced calculus or sometimes >called mathematical analysis .This is necessary to study theory of >anything. Really? For number theory? For algebra? For set theory? -- Michael F. Stemper #include This sentence no verb. === Subject: She wants a better sex? All you need's here! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with SMTP id j6PHpPJ15610

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 === Subject: Re: Uncle Al's ROAST --- was [Re: Reliving Chem 1120 aka GenChem2] > Hey stooopid Hanson, ya ain't never gonna get Cher, > Susan Sarandon, and Michelle Pfeiffer - not even in a morgue's > refrigerated section, not even if you bribe the Mexican janitor with > still warm bread-8. Hmm. Look what this guy fancies. I wouldn't any of them dead or alive with three condoms on. Al, stay off of jewish cunts for a while. They're better than nothing, that's for sure, but too much of that and then you'd want to next Susan Sarandon! Jeez.. . Talking about an ass that feels like noodle soup. Ah crap. -- irAni hanuz luleheng nemituneh besAzeh, chetor mituneh nafto 'pAlAyesh' koneh? - Razmara === Subject: Re: Uncle Al's ROAST --- was [Re: Reliving Chem 1120 aka GenChem2] > BTW, I was on the Westcoast for the last few weeks. Saw this week > a TV show about the above subject, on PBS, hosted by Jared > Diamond. The central question came from an abo in New Guinea > who wondered: Why does the US man have so much 'Cargo while > we people have almost none.... It was an impressive show, for > which Jared concluded that the start for said difference was pure > luck in climate and geography.... Jared is obviously a New Liberal, one who places the credit/blame for everything on life's lottery. The Cargo Cultists desperately want Cargo but have no idea how much nose-grindstoning is required to get it, OTOH their visitors see them as living in Paradise. Notice that just surviving a day in Paradise requires similar levels of effort and ingenuity to those required to survive a day on New York's streets, just different efforts and ingenuity. They aren't _stupid_, just primitive. Bottom line; decide what you want, determine how much work is required to get it, and decide if it's worth the effort. So, Hanson, getting your buck's worth of bang yet? Mark L. Fergerson === Subject: Re: Uncle Al's ROAST --- was [Re: Reliving Chem 1120 aka GenChem2] > >> BTW, I was on the Westcoast for the last few weeks. Saw this week >> a TV show about the above subject, on PBS, hosted by Jared >> Diamond. The central question came from an abo in New Guinea >> who wondered: Why does the US man have so much 'Cargo while >> we people have almost none.... It was an impressive show, for >> which Jared concluded that the start for said difference was pure >> luck in climate and geography.... > [Mark] > Jared is obviously a New Liberal, one who places the > credit/blame for everything on life's lottery. > [hanson] So, what? Even liberals have insights. My problem with them is that they want the implementation of their insights to be financed by others then themselves. I have been in biz long enough (1) to know such liberal behavior well. Liberals do have merit as the 'know-it-all consultants... ahahaha... So, I apply the occasional good suggestions of theirs and make a buck off them... ahaha.. [Mark] > The Cargo Cultists desperately want Cargo but have no > idea how much nose-grindstoning is required to get it, OTOH > their visitors see them as living in Paradise. > Notice that just surviving a day in Paradise requires similar > levels of effort and ingenuity to those required to survive a day > on New York's streets, just different efforts and ingenuity. > They aren't _stupid_, just primitive. > Bottom line; decide what you want, determine how much work > is required to get it, and decide if it's worth the effort. > [hanson] Dirk Bruere interjected here: > Well, if IQ measures anything relevant to economic > performance they are not going to do too well, with an > average IQ of around 70. which is a valid point that illustrates one of the reasons what your neo liberal friend was alluding to... and you inadvertantly agree with in your bottom line.... ahahaha.... BTW, have you really seen that show? If not, watch the encore. [Mark] > So, Hanson, getting your buck's worth of bang yet? > Mark L. Fergerson > [hanson] Of course, I always have, I do now and I still will. See above (1). Also, same of course, win a few, lose a few... the luck bit. But, I don't care where it comes from. Biz is biz.....ahahaha.... I do know that it takes few bucks only, to satisfy one's basic needs for food/clothing/housing... (I sleep fine&well whether it's in a barn on a layer of straw or in a castle on a gold plated Victorian bed in eiderdown pillows) ... so, in my experience anything above that is play money to engage/partake in the games of one's ambitions to influence/change one's environment. Therefore, to complete the circle of this conversation: == If you can never reach above the needed few bucks level, because of your less abundant starting environment, (climatic, geographic, social [mono- vs multi-culture], etc), then your and my bang for the buck scenario is not applicable... cuz you'll keep on scraping on the botton of the barrel for your very survival and existance. Hence, it is in great part simply having more luck then the poor slob has that allows you to rise above him and look at him as the primitive,.... and so, your neo Liberal Jared was right....still.... ahahahaha... AHAHAHA... ahaha... ahahahson > > === Subject: Re: Uncle Al's ROAST --- was [Re: Reliving Chem 1120 aka GenChem2] > > > >> BTW, I was on the Westcoast for the last few weeks. Saw this week >> a TV show about the above subject, on PBS, hosted by Jared >> Diamond. The central question came from an abo in New Guinea >> who wondered: Why does the US man have so much 'Cargo while >> we people have almost none.... It was an impressive show, for >> which Jared concluded that the start for said difference was pure >> luck in climate and geography.... > > > Jared is obviously a New Liberal, one who places the credit/blame for > everything on life's lottery. > > The Cargo Cultists desperately want Cargo but have no idea how much > nose-grindstoning is required to get it, OTOH their visitors see them as > living in Paradise. > > Notice that just surviving a day in Paradise requires similar levels > of effort and ingenuity to those required to survive a day on New York's > streets, just different efforts and ingenuity. They aren't _stupid_, > just primitive. Well, if IQ measures anything relevant to economic performance they are not going to do too well, with an average IQ of around 70. -- Dirk The Consensus:- The political party for the new millenium http://www.theconsensus.org === Subject: Help with Integral exp(-a*sqrt(1+x^2))*sin(b*x) ? The Integral with the cos instead of the sin is known to be a/sqrt(a^2+b^2)*besselk(1,sqrt(a^2+b^2)) Andreas === Subject: Hi hi my name is nikki i am 17 years old and i am in home schooling === Subject: Re: Hi >hi my name is nikki i am 17 years old and i am in home schooling Hello Nikki. Did your home schooling teach you that you use I instead of i for the personal pronoun, that you capitalize the first word in a sentence and end it with a period, and that you don't run sentences together? --Lynn === Subject: Re: Hi > > >hi my name is nikki i am 17 years old and i am in home schooling > > Hello Nikki. Did your home schooling teach you that you use I > instead of i for the personal pronoun, that you capitalize the first > word in a sentence and end it with a period, and that you don't run > sentences together? On 16/7/5 the same poster posted in alt.math.undergrad and I replied: > > hi i am 17 years old and i am in home schooling (Ecot) Is English among the subjects that you are being schooled in? -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === Subject: Re: Hi > > > >hi my name is nikki i am 17 years old and i am in home schooling > > Hello Nikki. Did your home schooling teach you that you use I > instead of i for the personal pronoun, that you capitalize the first > word in a sentence and end it with a period, and that you don't run > sentences together? > > On 16/7/5 the same poster posted in alt.math.undergrad and I replied: > > > hi i am 17 years old and i am in home schooling (Ecot) > > Is English among the subjects that you are being schooled in? On 22/2/5 Rachel Harris posted in alt.math.undergrad and I replied: > I need help on my pre-Algrebra problems that i have to do tonight for > home work because i have math 2 priod tomorrow i hope some one can > help me tonight please??? I am in 8th grade so next year i am going to > be a freshman next year so please can some one help me tinight > please?? How can we help with your homework unless we know what it is? Do you want help with your English as well? -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === Subject: lim(x to 0) (x.sin(1/x)) At planetmath.org/?op=getobj&from=objects&id=5597 I came across this limit: lim(x to 0) (x.sin(1/x)). Although I can see that this limit should indeed be =0 because of the x approaching 0 and the sinus part oscillating between -1 and +1, I do Frank === Subject: Re: lim(x to 0) (x.sin(1/x)) > > At planetmath.org/?op=getobj&from=objects&id=5597 I > came across this > limit: lim(x to 0) (x.sin(1/x)). > Although I can see that this limit should indeed be > =0 because of the x > approaching 0 and the sinus part oscillating between > -1 and +1, I do > not know how to calculate it formally. Can someone > > Frank By formally I assume you mean using the epsilon-delta definition of a limit. Let's set f(x) = x sin(1/x). Now recall the formal definition of a limit is lim[f(x)] = L as x --> k iff for every e > 0 there exists a d > 0 such that |f(x) - L| < e whenever |x - k|. So we want to prove that for every e > 0 there exists a d > 0 such that |x sin(1/x) - 0| < e whenever |x - 0| < d. First, we can write 0 <= |sin(1/x)| <= 1 for all real x =! 0 since the norm of sine is between 0 and 1. Therefore, given an e > 0 we can choose d < e to obtain |x sin(1/x)| = |x||sin(1/x)| <= |x| < d < e and hence lim[x sin(1/x)] = 0 as x --> 0. [] I know this limit can be proved via L'Hospital, and I think I recall it also being proved via the squeeze theorem in calculus 1. Kyle === Subject: Re: lim(x to 0) (x.sin(1/x)) I think a standard theorem called the squeezing theorem / sandwich theorem helps. === Subject: Grid Subdividing Solitaire Game Here is a solitaire game. Start with an (n-1)-by-(n-1) grid (of n-by-n lines) lightly drawn on paper. You also need 2 sets of cards, each deck of n cards labelled 1 through n. The player draws on each move one card from each deck. Don't replace the card. Let the value of the cards be A and B. either draws (along the grid's lines) a vertical line then a horizontal line, or draws a horizontal then vertical line, drawing to either vertex (A,B) or vertex (B,A). The line changes directions at most once (taking a right angle turn). The line may coincide with lines which were drawn earlier. (Vertex (1,1) is the upper left corner of the grid. Vertex (n,n) is the lower right corner of the grid.) After n moves (when all the cards have been drawn from both decks) the player gets a score equal to the *product* of the numbers of squares in each section the grid has been subdivided into by the player's lines. (Imagine the lines drawn during the game as a knife cutting the grid-cake into pieces. Multiply the sizes of each piece to get score.) An example will hopefully make this clear: (View with fixed-width font. Google will likely mess up my ascii art anyway, I bet.) Cards on each move (for n = 9): (A,B) = (4,2) (1,3) (6,5) (2,4) (8,6) (5,1) (9,8) (3,7) (7,9) 1. . *- - -+ . . . . ! ! 2. . . S ! . . . . ! 3. . . . ! . . . . ! 4. * -- - --+- + . . . ! ! ! 5*- +-- -- --+--+ . . . ! ! ! ! 6! +-- -- --* ! . . . ! ! 7! . *-- -- --+ -- -+ . ! ! ! ! 8! . ! . . * . ! . ! ! ! 9+-- --+-- -- -- --F -* . 1 2 3 4 5 6 7 8 9 Start at S. Finish at F. *'s are other vertexes that are either (A,B) or (B,A). Score: 13 * 23 * 3 * 1 * 3 * 11 * 10 = 296010. Leroy Quet === Subject: Re: Grid Subdividing Solitaire Game On 25 Jul 2005 12:13:29 -0700, Leroy Quet >Here is a solitaire game. > >Start with an (n-1)-by-(n-1) grid (of n-by-n lines) lightly drawn >on paper. >You also need 2 sets of cards, each deck of n cards labelled 1 through >n. > >The player draws on each move one card from each deck. >Don't replace the card. Let the value of the cards be A and B. > >either draws (along the grid's lines) a vertical line then a horizontal >line, or draws a horizontal then vertical line, drawing to either >vertex (A,B) or vertex (B,A). The line changes directions at most once >(taking a right angle turn). The line may coincide with lines which >were drawn earlier. >(Vertex (1,1) is the upper left corner of the grid. Vertex (n,n) is >the lower right corner of the grid.) > >After n moves (when all the cards have been drawn from both decks) >the player gets a score equal to the *product* of the numbers of >squares >in each section the grid has been subdivided into by the player's >lines. >(Imagine the lines drawn during the game as a knife cutting the >grid-cake >into pieces. Multiply the sizes of each piece to get score.) > >An example will hopefully make this clear: >(View with fixed-width font. Google will likely mess up my ascii art >anyway, I bet.) > >Cards on each move (for n = 9): (A,B) = >(4,2) (1,3) (6,5) (2,4) (8,6) (5,1) (9,8) (3,7) (7,9) > >1. . *- - -+ . . . . > ! ! >2. . . S ! . . . . > ! >3. . . . ! . . . . > ! >4. * -- - --+- + . . . > ! ! ! >5*- +-- -- --+--+ . . . > ! ! ! ! >6! +-- -- --* ! . . . > ! ! >7! . *-- -- --+ -- -+ . > ! ! ! ! >8! . ! . . * . ! . > ! ! ! >9+-- --+-- -- -- --F -* . > 1 2 3 4 5 6 7 8 9 > >Start at S. Finish at F. >*'s are other vertexes that are either (A,B) or (B,A). > >Score: 13 * 23 * 3 * 1 * 3 * 11 * 10 = 296010. > >Leroy Quet The game looks like it might be interesting, but besides presenting it, are you asking any question about it? quasi === Subject: Re: Grid Subdividing Solitaire Game <57qae19hj6jv44kpjc8iocg4p3vf20f9r0@4ax.com> > On 25 Jul 2005 12:13:29 -0700, Leroy Quet > >Here is a solitaire game. > >Start with an (n-1)-by-(n-1) grid (of n-by-n lines) lightly drawn >on paper. >You also need 2 sets of cards, each deck of n cards labelled 1 through >n. > >The player draws on each move one card from each deck. >Don't replace the card. Let the value of the cards be A and B. > >either draws (along the grid's lines) a vertical line then a horizontal >line, or draws a horizontal then vertical line, drawing to either >vertex (A,B) or vertex (B,A). The line changes directions at most once >(taking a right angle turn). The line may coincide with lines which >were drawn earlier. >(Vertex (1,1) is the upper left corner of the grid. Vertex (n,n) is >the lower right corner of the grid.) > >After n moves (when all the cards have been drawn from both decks) >the player gets a score equal to the *product* of the numbers of >squares >in each section the grid has been subdivided into by the player's >lines. >(Imagine the lines drawn during the game as a knife cutting the >grid-cake >into pieces. Multiply the sizes of each piece to get score.) > >An example will hopefully make this clear: >(View with fixed-width font. Google will likely mess up my ascii art >anyway, I bet.) > >Cards on each move (for n = 9): (A,B) = >(4,2) (1,3) (6,5) (2,4) (8,6) (5,1) (9,8) (3,7) (7,9) > >1. . *- - -+ . . . . > ! ! >2. . . S ! . . . . > ! >3. . . . ! . . . . > ! >4. * -- - --+- + . . . > ! ! ! >5*- +-- -- --+--+ . . . > ! ! ! ! >6! +-- -- --* ! . . . > ! ! >7! . *-- -- --+ -- -+ . > ! ! ! ! >8! . ! . . * . ! . > ! ! ! >9+-- --+-- -- -- --F -* . > 1 2 3 4 5 6 7 8 9 > >Start at S. Finish at F. >*'s are other vertexes that are either (A,B) or (B,A). > >Score: 13 * 23 * 3 * 1 * 3 * 11 * 10 = 296010. > >Leroy Quet > > The game looks like it might be interesting, but besides presenting > it, are you asking any question about it? > > quasi As with all my games I post to sci.math (and to rec.puzzles), I am mostly presenting this game; but I always wonder what would be good strategies for playing the games I post. Leroy Quet === Subject: + - * / Game Here is a game for 2 or more players. Start with a deck of cards with n cards labeled 1 to n. On each move the player whose move it is draws one card. Do not replace card. Write down first number drawn. On each move (after the first) a player draws a card to get the value k. He/she may either add k to the written down number, subtract k from the written down number (as long as the difference is not negative), multiply k with the written down number, or divide the written down number by k (as long as k divides evenly into the written down number). A new written down number, in this way, is obtained. A player gets a point every time the written down number they generate is a prime number. Play continues for a total of n moves (when the deck of cards runs out). Here is an example game (for 2 players). (n=10) First move: Player 1 draws 7. (Gets a point) 2nd move: Player 2 draws a 4. 7 + 4 = 11. (Gets a point) 3rd move:Player 1 draws a 1. 11/1 = 11. (Gets a point) 4th move: Player 2 draws a 9. 11-9 = 2. (Gets a point) 5th move: Player 1 draws a 2. 2/2=1. (No point) 6th move: Player 2 draws a 3. 1*3 = 3. (Gets a point) 7th move: Player 1 draws a 10. 3+10=13. (Gets a point) 8th move: Player 2 draws a 5. 13*5= 65. (No point) 9th move: Player 1 draws a 8. 65+8=73. (Gets a point) 10th move: Player 2 draws a 6. 73-6=67. (Gets a point) So player 1 gets 4 points, player 2 gets 4 points too, a tie. (What do you expect when I play myself?) In practice, it would probably be better to have a higher n. Also, if you prefer a pure-strategy game without luck, you can, instead of using cards, have the integers drawn be from a predetermined sequence (such as 1,2,3,4,....,n or such as the primes taken in order.) Leroy Quet === Subject: Darboux intermediate value theorem This is about an exercise in the textbook 'Introduction to Mathematical Analysis' by SA Douglass. I quote the text litterally: Prove Darboux's Intermediate Value Theorem for derivatives: If f is differentiable on [a,b] and if d is some number between f'(a) and f'(b), then there exists a c in (a,b) such that f'(c) = d. (Hint: For x in [a,b], let g(x) = d(x-a) - f(x). Show that g must have a critical point in (a,b).) Now my question is: can anybody see why g(x) is defined here as d(x-a) - f(x), and not as dx - f(x) (as it is for example in the proof found at planetmath.org/?op=getobj&from=objects&id=3056. Of course, the a does not matter further, because it vanishes in the derivative, but why bother then? I have a further problem with the proof of the Rolle theorem in the same book. That proof, in the book, starts like this: If f(x) = f(a) for all x in [a,b], then f'=0 at every point of [a,b] and ... Although this does not matter much for the proof, I would think that under the conditions of the Rolle theorem (f continuous on [a,b] and differentiable on (a,b) ), one is only allowed to say that If f(x) = f(a) for all x in [a,b], then f'=0 at every point of (a,b) and ... My reasoning is that the conditions of the Rolle theorem only imply that f is differentiable to the right in a, and to the left in b, not that f is differentiable in either points. Is this right or wrong? Frank === Subject: Re: Would you let your daughter marry a mathematician? > Wayne Brown, > You are a maggot in society. Why don't you produce something of > practical value > to improve society instead of pursuing your own selfish, hedonistic > study of useless esoterica? Well, I'm pretty busy, what with raising two sons (and currently putting the older one through college), working full-time as a system administrator/programmer for a successful manufacturing company, teaching a class at church, wasting all that time studying mathematics (and ancient languages, and archaeology, and whatever else strikes my interest), and so forth. But I'll try to find a spare moment or two to do something valuable. (In case you're wondering, I'm fully aware that I'm being trolled, but choose to answer anyway for my own amusement.) -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Would you let your daughter marry a mathematician? Uhh...guys: You're being felt out. This is one of those social experiments. === Subject: Re: Would you let your daughter marry a mathematician? > >> He is no different than any monk, drug addict, or hedonistic hippie >> voyeur who thought only of himself. Would you like to have been his >> wife? Do you think he made her happy? > > His wife? Do you even know who Erdos was? > Mark showed his ignorance again. I guess Paul Erdos has never married. If Mark considers math as his wife, he probably loved his wife more than anybody. Mr. Mark, here are something for your information to learn about Paul Erdos. Erd.9as won many prizes including the Wolf Prize of 50 000 dollars in 1983. However he had a lifestyle that needed little money and he gave away:- ... most of the money he earned from lecturing at mathematics conferences, donating it to help students or as prizes for solving problems he had posed. In 1976 Ulam gave this description of Erd.9as:- He had been a true child prodigy, publishing his first results at the age of eighteen in number theory and in combinatorial analysis. Being Jewish he had to leave Hungary, and as it turned out, this saved his live. In 1941 he was twenty-seven years old, homesick, unhappy, and constantly worried about the fate of his mother who remained in Hungary. ... Erd.9as is somewhat below medium height, an extremely nervous and agitated person. ... His eyes indicated he was always thinking about mathematics, a process interrupted only by his rather pessimistic statements on world affairs, politics, or human affairs in general, which he viewed darkly. ... His peculiarities are so numerous it is impossible to describe them all. ... Now over sixty, he has more than seven hundred papers to his credit. === Subject: Re: Would you let your daughter marry a mathematician? So, Aldar F. Chan's stated reasons for admiring Paul ERdos is that: a) He has no wife or family. b) He won the Wolf Prize woth 50,000 1983 dollars. c) He spent his life adicted to his work to the exclusion of everything else (including love or family) and built up an impressive ass-kicking resume of 700+ papers, maybe much more. d) He is unhappy, constantly worried, nervous, and extremely agitated, and very pessimistic. HOw is Erods any different from a random work-aholic investment banker who built up an impressive work resume, earned impressive salaries, and worked so hard he did not bother to have a LIFE outside of work? Betcha that like any workaholic banker or lawyer, Erdos worked on New Year's Eve too. Aldar, Paul Erdos exemplifies those Hong Kong investment banking lifestyles that you claim you want to escape. Or do you? > Erd.9as won many prizes including the Wolf Prize of 50 000 dollars in 1983. > However he had a lifestyle that needed little money and he gave away:- ... most of the money he earned from lecturing at mathematics conferences, > donating it to help students or as prizes for solving problems he had posed. > > In 1976 Ulam gave this description of Erd.9as:- He had been a true child prodigy, publishing his first results at the age > of eighteen in number theory and in combinatorial analysis. Being Jewish he > had to leave Hungary, and as it turned out, this saved his live. In 1941 he > was twenty-seven years old, homesick, unhappy, and constantly worried about > the fate of his mother who remained in Hungary. ... Erd.9as is somewhat below > medium height, an extremely nervous and agitated person. ... His eyes > indicated he was always thinking about mathematics, a process interrupted > only by his rather pessimistic statements on world affairs, politics, or > human affairs in general, which he viewed darkly. ... His peculiarities are > so numerous it is impossible to describe them all. ... Now over sixty, he > has more than seven hundred papers to his credit. === Subject: Re: Would you let your daughter marry a mathematician? Mr. Mark, I guess you have taken a wrong candidate to criticize on in this news group. You'd better stop. I guess everyone here respects Paul Erdos not merely for his prolific work of about 1500 in number, but also for his enthusiasm for math, his kindness to juniors and also kids. If you don't beleive, there is a book about his life -- try to look up from amazon. Unlike some mathematicians, he didn't lock himself in a room to work but collaborate with others, stimulated others' thought. Some problems he posed in number theory are still unsolved today. His work has real applications in computer science which in terms transform the way how we live, work and trade today. Stupid dude, an investment banker earns money for himself, not for the society. Many of them even don't pay a single dollar tax. You will probably say they help to allocate resources for the soceity. Yes they do for their own sake --- the resulting contribution is just a observation made by Adam Smith. Ethically, I wouldn't respect to those investment bankers. For Paul Erdos, he worked hard for advancing the knowledge of mankinds, not for himself despite the interests he had in math. So, Aldar F. Chan's stated reasons for admiring Paul ERdos is that: a) He has no wife or family. b) He won the Wolf Prize woth 50,000 1983 dollars. c) He spent his life adicted to his work to the exclusion of everything else (including love or family) and built up an impressive ass-kicking resume of 700+ papers, maybe much more. d) He is unhappy, constantly worried, nervous, and extremely agitated, and very pessimistic. HOw is Erods any different from a random work-aholic investment banker who built up an impressive work resume, earned impressive salaries, and worked so hard he did not bother to have a LIFE outside of work? Betcha that like any workaholic banker or lawyer, Erdos worked on New Year's Eve too. Aldar, Paul Erdos exemplifies those Hong Kong investment banking lifestyles that you claim you want to escape. Or do you? > Erd.9as won many prizes including the Wolf Prize of 50 000 dollars in 1983. > However he had a lifestyle that needed little money and he gave away:- ... most of the money he earned from lecturing at mathematics > conferences, > donating it to help students or as prizes for solving problems he had > posed. > > In 1976 Ulam gave this description of Erd.9as:- He had been a true child prodigy, publishing his first results at the > age > of eighteen in number theory and in combinatorial analysis. Being Jewish > he > had to leave Hungary, and as it turned out, this saved his live. In 1941 > he > was twenty-seven years old, homesick, unhappy, and constantly worried > about > the fate of his mother who remained in Hungary. ... Erd.9as is somewhat > below > medium height, an extremely nervous and agitated person. ... His eyes > indicated he was always thinking about mathematics, a process interrupted > only by his rather pessimistic statements on world affairs, politics, or > human affairs in general, which he viewed darkly. ... His peculiarities > are > so numerous it is impossible to describe them all. ... Now over sixty, he > has more than seven hundred papers to his credit. === Subject: Re: Would you let your daughter marry a mathematician? you are mesmerized by paul erdos just like a teeny bopper is mesmerized by a movie star or home run hitter, although the reality is that the movie star is, in real life, a despicable ill-formed self-ish prima donna. You are just mezmerized by the 1000 papers, just like some are mezmerized by babe ruths 714 home runs. You overlook erdos's freakish shortocomings just like kids overlook how Ruth was an alchoholic who was mean to his peers and possibly a racist. Paul Erdos was a selfish freak and prima donna, and no matter how many papers he has doesn't change that. You call yourself scientists? See the truth, buddy. === Subject: Re: Would you let your daughter marry a mathematician? !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > Stupid dude, an investment banker earns money for himself, not for > the society. Many of them even don't pay a single dollar tax. You > will probably say they help to allocate resources for the soceity. > Yes they do for their own sake --- the resulting contribution is > just a observation made by Adam Smith. Ethically, I wouldn't > respect to those investment bankers. For Paul Erdos, he worked hard > for advancing the knowledge of mankinds, not for himself despite the > interests he had in math. Oh, Erdës worked for himself certainly as well. Still, his work bears his name. Money is anomymous. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Would you let your daughter marry a mathematician? !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > So, Aldar F. Chan's stated reasons for admiring Paul ERdos is that: > a) He has no wife or family. > b) He won the Wolf Prize woth 50,000 1983 dollars. And distributed it mostly to beggars and scholars. > c) He spent his life adicted to his work to the exclusion of everything > else (including love or family) and built up an impressive ass-kicking > resume of 700+ papers, maybe much more. About double. > d) He is unhappy, constantly worried, nervous, and extremely agitated, > and very pessimistic. > > HOw is Erods any different from a random work-aholic investment banker > who built up an impressive work resume, earned impressive salaries, and > worked so hard he did not bother to have a LIFE outside of work? The money of the investment banker does not differ from anybody else's money after he is gone. Erdës made a difference to the world. His works stand out after his death, and will continue to do so, like the works of a Shakespeare and a Bach, other people that worked in their professions and left their mark on humanity. That's what makes the difference. > Aldar, Paul Erdos exemplifies those Hong Kong investment banking > lifestyles that you claim you want to escape. Or do you? But Erdës did not work as a means to an end. His work _was_ his purpose. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Would you let your daughter marry a mathematician? <85mzoaso3v.fsf@lola.goethe.zz> > And distributed it mostly to beggars and scholars. What do you think the richest people, like bankers, do with their multi-millions?! The Rockefeller Foundation, the Ford Foundation, the Sloan Kettering fellowships, Bill Gate's Africa foundation, the Wolf Prize (who donated that money?? a mathematician?), the Nobel Prize, Leland Stanford Jr University. Dude, every named chair at a university is there due to the generosity of some rich fella who donated the money. Every scholarship or fellowship you or Paul Erdos ever held was donated by a rich fellow (or a hardworking taxpayer). Why do you fail to appreciate all the donations of these hardworking people and fixate on (overhype) the self-aggrndizing Erdos and his relatively piddling donations? > > Erdos made a difference to the world. His works stand out after his > death, and will continue to do so, like the works of a Shakespeare and > a Bach, other people that worked in their professions and left their > mark on humanity. which are put together and funded and organized by the hard work of investment When you are deathly sick, you will go crying to doctor, will he give you pills made by Paul ERdos? Those pills are made by a company that invented the drug using money invested with the help of Marky and investment bankers, my friend. Think about that the next time you get cancer or angina. Where do you work every day? If you work at a company, you bet that bankers help to put together the initial capital that founded your company. And that people like Marky provided the start up funds. Think about that before you cash your paycheck, buddy. > But Erdos did not work as a means to an end. His work _was_ his > purpose. Yes, bankers provde an essential service to socity. Erdos licked his own ass and behaved like a prima donna baby, just like the worst of the selfish ball players or opera singers. He said F YOU, I am too good for you, and yet you still admire him. === Subject: Re: Would you let your daughter marry a mathematician? <853bq33ml8.fsf@lola.goethe.zz> <8564uzz64q.fsf@lola.goethe.zz> Mark read those obits years ago. Erdos had a wife who he suspected of trying to poison him. This goes to show he was not exactly a loving husband who thought of his wife or her needs. Mark bets Erdos never lifted a finger around the house too. Poor Mrs. Erdos -- imagine being married to an inconsiderate, self-centered freak like that! Erdos ignored her so he could travel and pursue math. Yeah, yeah, honey, shut up, Erdos is busy. As for his friends, Mark suspects Erdos' friends hosted him only because they wanted co-author with him or boost their careers by being associated with a famous mathematician. They are name droppers of the most blatant kind. Yeah, Erdos stayed at my house. In other words, Erdos took advantage of junior colleagues to live at their house for free. How's that for a boss? In short, Erdos is the consummate work-aholic who had no life and gave it all up for WORK. IBM would've been glad to hire this sucker for its staff. Erdos is more corporate than any dedicated corporate drone. Was Erdos a good advisor? Did he ever take any junior mathematician under his wing and personally taught him some math? No, Mark bets not; that's because Erdos was way too self-centered. Erdos would donate a few scholarships, but when it comes to actually teaching, forget it. Don't bother erdos. === Subject: Re: Would you let your daughter marry a mathematician? !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > Mark read those obits years ago. Erdos had a wife who he suspected > of trying to poison him. You are confusing G.9adel with Erdës. > This goes to show he was not exactly a loving husband who thought of > his wife or her needs. Mark bets Erdos never lifted a finger around > the house too. Poor Mrs. Erdos -- imagine being married to an > inconsiderate, self-centered freak like that! Erdos ignored her so > he could travel and pursue math. Yeah, yeah, honey, shut up, Erdos > is busy. Try getting your facts straight before you start fabricating fantasies from them. > As for his friends, Mark suspects Erdos' friends hosted him only > because they wanted co-author with him or boost their careers by > being associated with a famous mathematician. They are name > droppers of the most blatant kind. Yeah, Erdos stayed at my > house. In other words, Erdos took advantage of junior colleagues > to live at their house for free. How's that for a boss? Well, I suggest that you try acquiring a friend or two so that you have a better idea what you are talking about. > Was Erdos a good advisor? Did he ever take any junior mathematician > under his wing and personally taught him some math? No, Mark bets > not; that's because Erdos was way too self-centered. Erdos would > donate a few scholarships, but when it comes to actually teaching, > forget it. Don't bother erdos. Wow. You probably don't have an inkling of a clue just how _utterly_ absurd your remarks are in _particular_ with reference to Erdës. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Would you let your daughter marry a mathematician? On 2005-07-25 07:36:39 -0700, double d said: > Mark read those obits years ago. Erdos had a wife who he suspected of > trying to poison him. > > You're thinking of an apocryphal story about Kurt G.9adel, I think. Also, Erd.9as was famously good with kids. Was Erdos a good advisor? Did he ever take any junior mathematician > under his wing and personally taught him some math? > > He worked with hundreds of mathematicians, many of them more junior than he was. He did not teach many classes, but he gave many seminars. You are attempting to upset people by pissing on a great man. It is not flattering behavior. Best wishes, Christian === Subject: Re: Would you let your daughter marry a mathematician? <853bq33ml8.fsf@lola.goethe.zz> <8564uzz64q.fsf@lola.goethe.zz> > Mark read those obits years ago. Not very well, I'm afraid. > Erdos had a wife who he suspected of > trying to poison him. He never married and left no immediate survivors. (http://www.cs.elte.hu/erdos/NY-Times.html) > This goes to show he was not exactly a loving > husband who thought of his wife or her needs. He wasn't a husband at all. > Mark bets Erdos never > lifted a finger around the house too. What house? - Randy === Subject: Re: Would you let your daughter marry a mathematician? ó.8d.98.87.8b.8c .97.99.95 .92.86.94.9d.92.87 > He never married and left no immediate survivors. > (http://www.cs.elte.hu/erdos/NY-Times.html) > > This goes to show he was not exactly a loving > husband who thought of his wife or her needs. > > He wasn't a husband at all. > > Mark bets Erdos never > lifted a finger around the house too. > > What house? - Randy I just realized exactly what my Erdos number is. It's (population of Earth)(t)+1. Goes to show, one doesn't need an Erdos Number of infinity to be a non-famous mathematician :-) -- I. N. Galidakis http://users.forthnet.gr/ath/jgal/ Eventually, _everything_ is understandable === Subject: Re: Would you let your daughter marry a mathematician? > I just realized exactly what my Erdos number is. It's (population of > Earth)(t)+1. > > Goes to show, one doesn't need an Erdos Number of infinity to be a > non-famous mathematician :-) > -- > I. N. Galidakis > http://users.forthnet.gr/ath/jgal/ > Eventually, _everything_ is understandable > I'm sorry, where does the +1 come in? === Subject: Re: Would you let your daughter marry a mathematician? Ì Aldar C-F. Chan ó.8d.98.87.8b.8c .97.99.95 .92.86.94.9d.92.87 > > > I just realized exactly what my Erdos number is. It's (population of > Earth)(t)+1. > > Goes to show, one doesn't need an Erdos Number of infinity to be a > non-famous mathematician :-) > I'm sorry, where does the +1 come in? Oops, you are right. That should be (population of Earth)(t). What's also interesting is that this number is dynamic, contrary to small Erdos numbers. Can anyone improve on the bound for non-famous mathematicians? :-) -- I. N. Galidakis http://users.forthnet.gr/ath/jgal/ Eventually, _everything_ is understandable === Subject: Re: Would you let your daughter marry a mathematician? >.93 Aldar C-F. Chan [NonBreakingSpace].8b.96.87À.8c .97.99.95 .93[CapitalThorn].92.9b.93.87 >> >> >> I just realized exactly what my Erdos number is. It's (population of >> Earth)(t)+1. >> >> Goes to show, one doesn't need an Erdos Number of infinity to be a >> non-famous mathematician :-) > >> I'm sorry, where does the +1 come in? > > Oops, you are right. That should be (population of Earth)(t). > > What's also interesting is that this number is dynamic, contrary to small > Erdos numbers. > > Can anyone improve on the bound for non-famous mathematicians? :-) > -- Based on your assumption to ignore the definition of infinity in distance in a graph, the bound could be pushed down a little bit by some 400 since Paul Erdos should have about 400+ collaborators who are still alive. === Subject: System Solvability Can anyone check if the following system is analytically solvable for a,b,c,d in R, (with p,q,r,s given), with Maple? un:=sqrt(b^2+c^2+d^2) exp(a)*(a*cos(un)-sin(un)*un)=p b*exp(a)*(cos(un)+a*sin(un)/un)=q c*exp(a)*(cos(un)+a*sin(un)/un)=r d*exp(a)*(cos(un)+a*sin(un)/un)=s I tried various tricks with Maple to get a closed form solution (the exp(a)*a suggests Lambert's W function), but it always turns out very complicated. I know it has an infinite number of solutions a_n,b_n,c_n,d_n, so I also need a couple of hints on how to investigate this. Maple solves it with fsolve, but picks n seemingly randomly depending on internal convergence criteria. -- I. N. Galidakis http://users.forthnet.gr/ath/jgal/ Eventually, _everything_ is understandable === Subject: Cosmology Conference at SLAC: Sean Carroll General Relativity in a Nutshell in 3 hours http://www-conf.slac.stanford.edu/ssi/2005/default.htm I am sitting in Panofsky auditorium on line. Quotes by Sean Carroll GR is easy. Metric guc(t,x), geodesics x^u(s), geodesic eq d^2x^u/ds^2 + {^uvw}(dx^v/ds)(dx^w/ds) = 0 {^uvw} = (1/2)g^u^l(gvl,w + glw,v - gvw,l) R^luvw = {^luw},v - {^luv},w + {^l vr}{^ruw} - {^lwr}{^ruv} (Check order of uv indices I may have made a mistake) This is ALL of 1915 GR that the cosmologists use to interpret their data! GR is a field theory. There are two ways to get it. 1. Einstein's way - the curvature of spacetime. Newtonian gravity is inconsistent with special relativity. Scalar gravity in SR is NO GOOD. James Woodward of course uses this no-good theory in his propellantless propulsion machine. Rocket equation is for CM = Center of Mass Force(CM) = dP(CM)/dt = M(total)dV(CM)/dt + (dM(total)/dt)V(CM) Woodward claims a new way to get dMtotal/dt =/= 0 without actually having to eject mass. Remember that any unconventional flying machine that depends on dM/dt is still a rocket pushing the ship OFF a time-like geodesic even if no propellant, i.e. F(CM) =/= 0. The issue of propellant is not the essential signature of a true zero g-force WEIGHTLESS timelike geodesic warp drive. A true geodesic warp drive must locally modify the local geometry in which the ship is immersed with small amounts of on-board power expended. Puthoff et-al has never properly defined what warp drive is in metric engineering. Neither did anyone at the short-lived Nick Cook on History Channel) on the subject. The only place it is done quickly is in the Paramount Pictures Star Trek IV DVS Special Collector's Edition Time Travel: The Art of the Possible Disk 2 Introduce metric tensor guv(t,x) - how is curvature related to gravity? 2. Massless spin 2 quanta in flat background, the quantum field is huv = hvu The nonlinear couplings to energy-momentum tensor implies guv = nuv + huv + ... The total INFINITE sum acts like curved spacetime! Globally flat background disappears. Use perturbation theory for gravity waves. Back to Einstein's way the principle of equivalence thought experiment. Put Alice inside a closed steel box no windows. Do experiments, what is the gravity field inside the box? Drop balls made of different materials - everything drops the same like universal gravity. But you could be in a box inside an accelerating rocket ship. No experimental difference in sitting on surface of a planet or accelerating in space. Tidal forces? No, because you are in a small region below limit of detectability of tidal curvature says Sean. Remember Wigner's limits on measuring curvature. Hence the universal gravity field is not a force in the usual sense. Spherical space sum of angles of triangle > Pi e.g., can be (3/2)Pi for example. Curvature need not be uniform like a sphere is. Variable curvature, so what to do? Look at infinitesimal pieces. Do calculus on the curved space. Draw a tiny triangle. Assume smooth non-fractal space. dx & dy are sides of tiny triangle. Can't assume a right triangle with ds^2 = dx^2 + dy^2, but with curvature we need ds^2 = f(x,y)dx^2 + g(x,y)dxdy + h(x,y)dy^2 do this in 11 dim? UGH! Use summation convention ds^2 = gij(x)dx^idx^j = line element generalizes Pythagorean theorem. Tiny little interval in space-time is what we are after here. We want the gij(x) field. Ex 1. Do surface of a unit sphere, don't worry about dimensions for now, make it all dimensionless r = 1 ds^2 = dtheta^2 + sin^2thetadphi^2 gphiphi = sin^2theta gthetatheta = 1 everything else is 0 Ex 2. In contrast, do FLAT plane in polar coordinates gthetatheta = r^2 grr = 1 ds^2 = dr^2 + r^2dtheta^2 How do you get off-diagonal terms? Simply choose local basis arbitrarily. Suppose you have a curved geometry. What are the extremal geodesic paths? In Riemannian space of positive curvature at least the geodesics are minimal length paths (not so in Lorentzian space-time where they are maximal length proper time for timelike paths - twin effect). Use calculus of variations, many paths connect same two initial and final points. dx^u are Cartan basis 1-forms I = |ds = |(gijdx^idx^j)^1/2 velocity = v^i = dx^i/dl (ds/dl)^2 = gijv^iv^j = vi^v^i = v.v I = |(v.v)^1/2dl independent of scaling l Vary the path x^i(l) -> x^i(l) + &x^i(l) I -> I + &I dx^i/dl -> dxYi/dl + d(&x^i)/dl Do Taylor series expansion on the metric tensor gij(x^k) -> gij(x^k + &x^k) ~ gij(x^k) + gij,k&x^k + ... Do the algebra mess &I = 0 vary the independent &x therefore &gij comes along for the ride to get the geodesic eq. (not invariant for l -> f(l), which will give non-zero RHS, affine coordinates by definition have a zero RHS in the geodesic eq - i.e. affinely-parametrized geodesic eq} g^k^l is the inverse metric g^k^lgli = I^ki Kronecker delta 4x4 Identity Matrix { } are Christoffel symbols AKA connection coefficients In flat 2D space, the { } are all zero (for Cartesian observers - left out by Sean). Next to globally flat Minkowski spacetime of 1905 special relativity, no gravity, no curvature. x^i -> x^u, u = 0,1,2,3, x^0 = ct nuv for GLOBAL FLAT space-time, use -1, +1, +1, +1 Light is impossible without that ONE MINUS SIGN! Space-time diagrams t vertical x horizontal in 1 + 1 toy model ds^2 = 0 is light-like (null geodesic) -> light cones time-like ds^2 < 0 inside the light cone space-like ds^2 > 0 outside the light cone proper time is dtau^2 = - ds^2 tau is the time measured by clocks on the time-like world lines (need not be geodesic). dtau = (-guvdx^udx^v) These MAXIMIZE the proper time on the time-like geodesic compared to all neighboring non-geodesics with same start and finish. Sean Carroll's prize question When will there be evidence that Einstein's 1915 general theory of relativity need any revision and what will that evidence be? Non-flat metrics Ex 3 Expanding universe - this simple metric works for the data in our universe! ds^2 = - (cdt)^2 + a^2(t)[dx^2 + dy^2 + dz^2] for co-moving Hubble flow geodesic observers at LARGE SCALE only of course. Ex 4 Schwarzschild ds^2 = -(1 - 2GM/c^2)(cdt)^2 + (1 - 2GM/c^2r)^-1dr^2 + r^2dO^2 dO^2 = dtheta^2 + sin^2thetadphi^2 Birkhoff's theorem: Schwarzschild is UNIQUE SSS vacuum solution to Ruv = 0 Ex 5 Newtonian weak field limit of SSS GR ds^2 = - (1 + 2V/c^2)(cdt)^2 + (1 - 2V/c^2)[dx^2 + dy^2 + dz^2] Grad^2V = 4piGrho the timelike geodesic. Solve geodesic eq in this Newtonian metric v/c << 1 tau ~ t space components u = i = 1,2,3 d^2x^i/dt^2 + {^iuv}(dx^u/dt)(dx^v/dt) = 0 ---> d^2x^i/dt^2 + {^i00} = 0 Get only for {^i00} ~ to first order (1/2)g^ii(-g00,i) ~ -V^,i Therefore GR limits to Newton's gravity theory in the appropriate limit. i.e. we get d^2x^i/dt^2 = - V^,i Einstein's field eq limits to Poisson's equation. Next 4-vectors V^u = (V^0, V^1, V^2. V^3) There is no such thing as the velocity of a distant galaxy. v = Hr is not really a velocity in GR Tangent vector to a curve x^u(l), the tangent vectors are v^u = dx^u/dl Norm vuv^u of tangent vector < 0 is a time-like vector 4-velocity in space-time is dx^u/dtau = U^u UuU^u = -1 4-Momentum p^u = mU^u raising and lowering indices with the curved metric in curved spacetime AuB^u = guvA^uB^v is INVARIANT SCALAR GCTs, Xu^u' is the INVERSE of Xu'^u GCTs are ORTHONORMAL U^u' = X^u'uU^u All tensors transform HOMOGENEOUSLY this way multi-linearly note coordinate x^u is NOT A TENSOR ds^2 is the physical infinitesimal distance. It does not care about changes of coordinates whether mere passive relabelings for Alice alone or Bob alone, or active COINCIDENT observers Alice & Bob in COINCIDENCE (WIGNER'S COLLISION MATRIX). All passive relabelings of Alice alone, Bob alone et-al are the normal subgroup H. The complete GCT group is everything. Hence the jumps between Alice & Bob are in GCT/H Note this is Jack not Sean talking here. Note the general tensor transformation law is NOT to be confused with the non-commutative multiplication of 4x4 matrices on a vector. Different slots are not created equal. Tensors are HOMOGENEOUS multi-linear maps to the real numbers. I note that the quantum tensors in possibly nonlocal EPR entanglements are multi-linear maps to complex numbers in QUBIT Hilbert space, but the quantum tensors are faithful representations of various groups (not Diff(4) GCT in QM) See H. Weyl 'Theory of Groups in Quantum Theory (Dover). (k,l)x(m,n) = (k+m, l+n) right upper, left lower You can add tensors of the same type. You can contract products of tensors (even not of the same type) over repeated upper and lower indices one on each factor. Nonsymmetric metrics give WRONG PHYSICS. Things that are not tensors. Partial derivatives are not tensors, with exception of gradient on a scalar function, that is a good first rank GCT tensor (0,1) = dual vector But V^v,u is NOT a second-rank GCT tensor! V^v',u' = X^uu'(X^v'vV^v),u gives inhomogeneous term - work out the algebra Another non-tensor are the Christoffel symbols Still another non-tensor is the determinant of the metric tensor. The determinant is a tensor density, with Jacobian outside, i.e. g' = |X|^2g Another non-tensor is the bare volume element Make it into a scalar invariant under GCT (-g)^1/2d^nx = (-g')^1/2d^nx' GCT tensor equations work in all local coordinate frames! Covariant derivatives D produce GCT tensors of next higher rank. Use the transformation of the partial derivative (same idea as in EM gauge invariance where A under U(1) is also not a U(1) tensor). DuV^v = V^v,u + {^vuw}V^w Du'V^v' = Xu'^uX^v'vDuV^v Dual vectors DuWv = Wv,u - {^wuv}Ww METRICITY is Dwguv = 0 with respect to the Christoffel symbols only of course. But NO NEED to introduce any OTHER CONNECTIONS so far - at least in cosmology data seen so far - possible exception of dark energy? My question not Sean's. EM Gauge Theory Analogy psi = e^ichipsi Dupsi = psi,u - ieAupsi A -> Au + chi,u Dupsi -> e^ichiDupsi i.e. U(1) tensor transformation Fuv = Av,u - Au,v + non-abelian terms if Yang-Mills SU(N) instead of U(1) EM [Du,Dv]psi = Fuvpsi this is analogous to the curvature tensor in GR [Du,Dv]V^w = R^wuvl V^l R is the covariant curl of the connection with itself. Note anti-symmetry in uv analog to Fuv in EM, there are other conditions as well limiting the independent components of the curvature to 10 in vacuum and 20 if Tuv(Matter) =/= 0 (my note) In tetrad notation the half-way house is the spin connection Cartan 1-form whose covariant exterior derivative with itself is the curvature 2-form. The spin connection is completely determined by the Einstein-Cartan tetrad field from locally gauging T4 and not locally gauging SO(1,3), i.e. no Shipov torsion here. My note not Sean's. Bianchi identities on curvature tensor are analogous to dF = 0 F = dA in flat EM (my note) trace of antisymmetric object is zero Ruv = R^lulv R = g^uvRuv Bianchi identity -> D^u(Ruv - (1/2)Rguv) = 0 You need a large region of space-time to SEE curvature! That is, to see the tidal stretch-squeeze geodesic deviation. Really of course this is contingent (my note). My note, in terms of g-force you need to fire a rocket to stand still in a curved vacuum spacetime. deviation? covariant derivative of this small separation D^2s^w/dl^2 = R^wluvU^lU^vs^v Gravity wave detector looks at this geodesic deviation. 2 mirrors hang and laser beam between them, how do the mirrors wiggle when a gravity wave passes by? Best way to get local GR field theory equations is to vary the global action S with respect to guv S = |d^nx(-detguv))L What are possibilities? Not g^u^vguv = 4 Choose L = R which is first order in the Riemann tensor - start with lowest order. This is classical effective low-energy field theory. In quantum field theory we put in higher order terms. Do calculus of variations guv -> guv + &guv &S = |d^4x(-g)^1/2(Ruv - (1/2)Rguv)&guv = 0 Ruv - (1/2)Rguv = 0 for classical vacuum This is a messy nonlinear equation to try to solve by brute force. The gravitons self-interact if you try to do this as quantum field theory. Newton's Poisson eq Grad^2V = 0 in contrast is LINEAR in this context. Note R = 0 here easily shown So Ruv = 0 (still very complicated nonlinear function of guv). Problem, show that SSS metric for HOVERING LNIFs obeys Ruv = 0, r > 2rs. Sean does not do any measurement theory in his 3hr fast course. Add in matter Lagrangian, do calculus of variations to get (1/16piG)[Ruv - (1/2)Rguv] + (-g)^-1/2&Smatter/&guv = 0 Tuv(Matter) = - 2(-g)^-1/2&Dm/&guv Obviously &S(GRAVITY)/&guv = 0 in classical vacuum and that would be the stress-energy tensor of the pure classical vacuum gravity field (no dark energy). Isotropic fluids in the local rest frame Tuv(isotropic fluid) = (rho + p)UuUv + pguv Choose geodesic coordinates at a single point only at that point - Minkowski metric i.e. equivalence principle, and choose rest frame of fluid U^u = (1,0,0,0) Result is T00 = rho, no off-diagonal Tij = p&ij in the REST LIF of the isotropic fluid * This is all you need for observational precision cosmology! COSMOLOGY Flat Robinson-Walker metric - flat & isotropic in 3D space - INFLATION gives this ds^2 = - dt^2 + a^2(t)[dx^2 + dy^2 + dz^2] guv = (-1 a^2 a^2 a^2) = g^u^v = (-1,a^-2,a^-2,a^-2) {^0ij} = (1/2)g^0^0(-gij,0) = (da/dt)aIij {^ij0} = (1/a)(da/dt)Iij all other connection symbols are zero This is all in the LIF REST Hubble Flow! in this preferred Hubble flow 3+1 slice T^u^v(Isotropic Fluid) U^u = (1,0,0,0) T^u^v = (rho, a^-2pI^i^j) DuT^u^v = 0 v = 0 ---> T^u^0,u + {^uul}T^l0 + {^0ul}T^ul = 0 T^0^0,0 + {^ii0}T^0^0 + {^0ii}T^i^i = 0 (rho),t + 3(da/dt)(1/a)rho + 3(da/dt)(1/a)p = 0 rho,t = - 3(da/dt)(1/a)(rho + p) p = wrho rho(t) ~ a(t)^-3(1 + w) = time evolution of energy density of isotropic stuff with w w = 0 for cold matter w = +1/3 for EM radiation w = -1 for zero point vacuum fluctuations Of course it's not so simple on small scales where we do not have isotropy & homogeneity. Compute Ricci tensor for above simple cosmological flat inflation metric that fits the data! R00 = -3(1/a)(d^2a/dt^2) Rij = ((d^2a/dt^2)(da/dt) + 2(da/dt)^2)Iij Einstein equation is here H^2 = (8pi/3)Grho H(t) = a^-1da/dt = Hubble parameter === Subject: Re: Bizarritudes > > 1/9801 = 1.02030405... No it doesn't. You've got the decimal point in the wrong place. -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === Subject: Re: Bizarritudes > > 1/9801 = 1.02030405... > > No it doesn't. You've got the decimal point in the > wrong place. > > -- > I don't know who you are Sir, or where you come from, > > but you've done me a power of good. Yes thank you for the correction; had to be read : 1000/9801. === Subject: Re: Bizarritudes Did someone try to attribute 10 different colours to the differents digits and see the beautiful rainbowy result ? === Subject: Re: Bizarritudes In (pi+1)/pi there are lots of 8 it seems. === Subject: Re: Bizarritudes > <30391176.1122268862898.JavaMail.jakarta@nitrogen.math > forum.org>, >In pi * 10/9 there is one 34594444444 and one > 999991253999999. > > And probably infinitely many, both in this and in pi > * r for any nonzero > rational r. But yes, it's somewhat surprising that > these occur so > early (starting at the 758th and 19437th digits, > respectively). > > Robert Israel > israel@math.ubc.ca > Department of Mathematics > http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, > BC, Canada Yes always early because the future belongs to the sequences which appear early in pi variants. But there is no 102030405060708091011121314151617181920212223 2425262728293031323334353637383940414243444546474849505152 5354555657585960616263646566676869707172737475767778798081 828384858687888990919293949596979900 and no 010203050813213 0578846348115971310233356904737852308313971108192746742095 1611273866047075462167895747045156076371350641478937266390 5444994443883220527325992524497423982220426305687443176078 3917567431053641781998181634508536215779371653702394181230 4273158904939892918476613799373674108495807657339125164157 9957571471865845034852005253055864228709970704111526416809 778765531871906253156884533791292049701990100. Because the pi variants don't like the regular sequences, it likes chaos and disorder. What do you use to count the places of sequences ? Is there a 123456789 and where ? === Subject: Re: Bizarritudes >><30391176.1122268862898.JavaMail.jakarta@nitrogen.math >>forum.org>, >> > > But there is no 102030405060708091011121314151617181920212223 > 2425262728293031323334353637383940414243444546474849505152 > 5354555657585960616263646566676869707172737475767778798081 > 828384858687888990919293949596979900 and no 010203050813213 > 0578846348115971310233356904737852308313971108192746742095 > 1611273866047075462167895747045156076371350641478937266390 > 5444994443883220527325992524497423982220426305687443176078 > 3917567431053641781998181634508536215779371653702394181230 > 4273158904939892918476613799373674108495807657339125164157 > 9957571471865845034852005253055864228709970704111526416809 > 778765531871906253156884533791292049701990100. > A proof of either assertion would make you famous. > Because the pi variants don't like the regular sequences, it likes chaos and disorder. Nonsense. Setting aside the fact that pi doesn't like or dislike anything, there's still no reason to assume, for example, that 121212 is more or less likely to appear in the sequence of decimal digits of pi than, say 164710. (They're both there, BTW. The first sequence can be found starting at position 241987 in the digits of pi and the second starts at position 155956.) Many people believe that *any* finite sequence of digits can be found in pi (though no proof exists yet). > > What do you use to count the places of sequences ? Try http://www.angio.net/pi/piquery and have fun exploring. > > Is there a 123456789 and where ? Not among the first 200000000 decimal digits. Rick === Subject: Re: Bizarritudes Let's say that the chances aren't very elevated :) There's a 314151413 at position at position 88,008. According to the fact that pi decimals correspond to something real and not to the deliring imagination of our scientists to make up to a lack of knowledge. In this case I would have put my birth date or my phone number :) === Subject: Re: Bizarritudes >Yes always early because the future belongs to the sequences which >appear early in pi variants. Are you being intentionally obscure, or is it the result of a language barrier? >But there is no 102030405060708091011121314151617181920212223 >2425262728293031323334353637383940414243444546474849505152 >5354555657585960616263646566676869707172737475767778798081 >828384858687888990919293949596979900 and no 010203050813213 >0578846348115971310233356904737852308313971108192746742095 >1611273866047075462167895747045156076371350641478937266390 >5444994443883220527325992524497423982220426305687443176078 >3917567431053641781998181634508536215779371653702394181230 >4273158904939892918476613799373674108495807657339125164157 >9957571471865845034852005253055864228709970704111526416809 >778765531871906253156884533791292049701990100. How do you know? >What do you use to count the places of sequences ? Maple. For example: S:= convert(evalf(Pi*10/9, 100000),string): StringTools[Search](123456,S); 92411 So (taking into account one place for the decimal point) the first 123456 starts at the 92410th digit >Is there a 123456789 and where ? Probably, but I don't think my computer has enough memory to tell. There is no 1234567 in the first million digits. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Bizarritudes A language barrier ? No just the difficulty with which I wake up (I have nazi neighbours who experiment gases on me) :) Cool, I don't know Maple, better said, I downloaded it but this didn't seem to me so easy to use so I stopped the use of this one. Derive is cool and the interface is less based on the programmation. === Subject: Re: Bizarritudes Is there a program to search sequences in the decimals of any irrational number, or to count the number of any sequences ? === Subject: System of even and odd integers I'm going through some problems in a book, and since there's no answers to these exercises in the back of the book, I would appreciate if you could verify the proofs. (E is the even integers, ie. 2n, where n is an integer, and I is the odd integers, ie. 2k + 1, where k is a natural number). I must show that E behaves as a zero element for addition. I have: E + E = 2n + 2n = 2(n + n) = 2m (let m = 2n). And I must determine which of E and I behave like 1 (one): EI = 2n(2k + 1) = 2(2kn + n) = 2m (let m = 2kn + n). Does this prove that I behaves like a zero element for multiplication with E and I? Another exercise asks which of E and I behaves like the zero element for multiplication: EI = 2n(2k + 1) = 2^2nk + 2n = 0 + 2m (let m = 2kn + n). Any help is appreciated -- Dvorak2 === Subject: Re: System of even and odd integers > (E is the even integers, ie. 2n, where n is an integer, and I is the > odd integers, ie. 2k + 1, where k is a natural number). I must show > that E behaves as a zero element for addition. I have: > > E + E = 2n + 2n = 2(n + n) = 2m (let m = 2n). > > And I must determine which of E and I behave like 1 (one): > > EI = 2n(2k + 1) = 2(2kn + n) = 2m (let m = 2kn + n). > > Does this prove that I behaves like a zero element for multiplication > with E and I? > > Another exercise asks which of E and I behaves like the zero element > for multiplication: > > EI = 2n(2k + 1) = 2^2nk + 2n = 0 + 2m (let m = 2kn + n). Right from the start, I don't understand E is the even integers, ie. 2n, where n is an integer. Do you mean E is the set of even integers, i.e. {2n : n is an integer}? Do sets X and Y get added and multiplied as follows: X + Y = (x + y: x in X, y in Y} XY = {xy: x in X, y in Y}? When you write E behaves as a zero element for addition what addition are you referring to? Addition of sets of integers? If so, then you need prove that E + X = X + E = X for all sets of integers X. -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === Subject: Re: System of even and odd integers > >Right from the start, I don't understand E is the even integers, ie. 2n, where n is an integer. > >Do you mean E is the set of even integers, i.e. {2n : n is an integer}? > >Do sets X and Y get added and multiplied as follows: X + Y = (x + y: x in X, y in Y} XY = {xy: x in X, y in Y}? > >When you write E behaves as a zero element for addition > >what addition are you referring to? Addition of sets of integers? If >so, then you need prove that E + X = X + E = X > >for all sets of integers X. Sorry, I wasn't making myself clear. E and I are just abbrevations for the even and odd integers. The text has stated the basic laws for addition and multiplication (associativity, commutativity, distributivity, 0a = 0, and 1a = a), so I assume that I don't need to prove them. The text has the following list: E + E = E E + I = I + E = I I + I = E EE = E II = I IE = EI = E. As I understand it, this list just says that 2n + 2n = 2(2n + n), and so on. -- Dvorak2 === Subject: Cahn-Hilliard a priori estimates I am trying to get some a priori estimates to show a global solution to u_t + bi-laplacian u = laplacian f(u) where f(u) = u^3 - u Take u and laplacian u to have zero normal derivative on the boundary. Take the initial condition as smooth as you like . I need to show that for fixed T>0 that u(t) is bounded in the H^2(Omega) norm for 0 le t le T. Then i could get a global solution by using the blow-up alternative. craig === Subject: Re: a non-godelian system > The Halting proof is amenible to a similar attack. > Define a Partial Halting Function where PH(f, a) = TRUE if f(a) halts, > or PH(f, a) = FALSE if f(a) halts, and f(a) doesn't halt -> PH(f,a) = > FALSE. > > This is like a witness function, if you are lucky and find a PH function > that determines a function halts then you prove it halts. > conversely, if you test many PH functions and they all return FALSE, > this gives a good indication the function won't halt. > > All that remains is to define a ratio of an infinite sequence, similar > to the prime witnesses where 1/2 of the numbers should work, similary > half of the PHalt functions should work for the halt test to give both > values, halt and not-halt. Well, if you know what happen in research, then you know that such partial halting functions do exist. Among others, you can find primitive recursion (*), termination orderings, polynomial interpretations, Size Change Principle, Dependency pairs and much more. [Dershowitz, Lankford, Manna & Ness, Jouannaud, Jones, Giesl, ...] Actually, there are even such partial complexity functions that answer either yes or no if the analysed algorithm as a given complexity (typically Ptime) but always no if it has higher complexity. Among other, bounded recursion, safe recursion, tiering, non-size increasingness, quasi-interpretations. [Cobhal, Bellantoni & Cook, Leivant, Marion, Hofmann, Jones, Moyen, Amadio, ...] Yet, we're still searching more and more of these partial functions in order to capture more and more algorithms. The main problem being that all these caracterisations are usually extentionnaly complete but far from intentionnal complexity. Typically, there is no primitive recursive algorithm to compute min(x,y) (in unary) with complexity O(min(x,y)) [Colson]. Among other things, we handle very poorly divide and conquer algorithms and similar nested recursions. There are regularly PhD. funding in the field... (*) PR is a syntax but you can define a function PH_{PR}(x) = true iff x is a primitive recursive program and this will give you a partial halting function like you want. So, yes, the halting problem is amendible to a similar attack but it does not change the fact that Turing's theroem is true. My dream would be to have a powerful enough termination tool to be able to plug it into gcc and prove termination of, say, 50% of the daily programs (that would already be a big step forward). -- Hypocoristiquement, Jym. Adresse mail plus valide .88 partir de septembre 2005. Utiliser l'adresse de redirection permanente : Jean-Yves.Moyen `at` ens-lyon.org === Subject: Re: a non-godelian system >>So you are proposing some kind of probability method to decide whether >>it is likely whether a program halts or not. An interesting idea, >>although I can see many many difficulties in working out the probabilities. >> >>But even if you can get this to work, how does this contradict Turing's >>Theorem? > > > The halting proof shows no SINGLE ALGORITHM can be consistently correct outputting halt values, but using a > set of algorithms can achieve the desired result. Your CONCLUSION there is no s y s t e m to determine > halt values is not proven if not wrong. > > There is no algorithm to do 'xyz > =/= > There is no computerised system to do xyz. > > Proving that no single man can play football might be a valid proof, but the conclusion is null. > > Your proposed algorithm is not deterministic. Turing's Theorem says > that no deterministic algorithm exists. > > You are postulating the existence of probabilistic algorithm. In a > sense, there is one that already exists - run the program for a > googolplex years. If it doesn't stop in that time, you can be very > certain that it will never stop. In fact for all practical purposes the > algorithm is going to be 100% effective, because there is no human built > Turing machine that can have enough states so that effectively you prove > it doesn't halt by an exhaustive search. The only problem with this > algorithm is that it is totally impractical - aside from physical > problems like finding material that will last that long out of which to > build the machine - I don't know how to build a machine using only all > the material in the universe that can count to googolplex. > > If you can find a mostly effective algorithm that is practical, I guess > that would be a great boon for everyone. I'm guessing that this does > not violate Turing's Theorem. which is? Herc === Subject: Re: a non-godelian system >>If you can find a mostly effective algorithm that is practical, I guess >>that would be a great boon for everyone. I'm guessing that this does >>not violate Turing's Theorem. > > > which is? > > Herc which is that there is no deterministic algorithm for deciding whether a computer program halts. Algorithm means anything (within reason) that makes use of the same set of tools that you allow for your proposed algorithm - typically Church's thesis is that this will be whatever a Turing machine can do. Determininist means that it gets the answer right every single time, with zero margin of error. Virgil said: > > > And yet, one cannot apply an increment an infinite number of time without > adding infinity. > > > That TO says it cannot be done does not prove that it can't be done. > > Anyone but TO, and a few others of equally limited capabilities, can do > it quite easily. > You can add an infinite number of 1's and get a finite result? Wow, you're good! -- Smiles, Tony Virgil said: > > > What it really shows is that digital systems with a given number of > digits have more strings than digits. it is not necessary to have > aleph_0 digits, if you allow for smaller infinities. > > Smaller than what? The set of naturals is as small as infinite sets get. I disagree, obviously. > TO seems to be saying that because the set of all naturals has no limit > on thet number of digits needed that there must be some one neatural > with no limit on the number of digits needed, but it does not follow. I never said that. > > It is just another instance of TO's quantifier dyslexia. You are full of crap. You simply don't understand the meaning of iff. > -- Smiles, Tony Dik T. Winter said: > ... > > Now this case is quite dissimilar. Cantor's proof (about the cardinality > > of powersets vs. the cardinality of the base sets) is indeed simple and > > can be written in very few steps. > ... > > You know, if the only conclusion drawn from Cantor's proofs was that a power > > set is necessarily larger than its base set, I would have absolutely no > > problem. > > Actually that is the only conclusion. But that conclusion can be shown to > be equivalent to other conclusions. > > > That fact is always the case and I don't dispute it. Infinite sets of > > finite natural numbers, on the other hand, are self contradictory, > > Yes, you have told that already numerous times without actually showing > that is true. But let us go from the binary to the dyadic notation. > In that case the digits used are 1 and 2. 0 can not be represented, but > each natural number can be represented in only *one* way by a string of > those digits. A short table to show the idea: > 1 = 1 > 2 = 2 > 3 = 11 > 4 = 12 > 5 = 21 > 6 = 22 > 7 = 111 > 8 = 112 > 9 = 121 > 10 = 122 > etc. > In this representation each finite natural number is represented by a single > finite string. Now how many of such finite strings are there, given that the > stringlength is unbounded? This is precisely the kind of question which cannot be answered. There is no bound to the lengths of the strings, but you claim they are finite. What number am I supposed to use to calculate this set size? Still, the question is interesting if infinite digits are allowed. If you allow infinite strings and values, then I say there are N numbers, by definition. Now, this is a little different from normal digital number systems. At first I said two symbols, S=2, but the problem here is that the strings have different lengths and cannot be simply extended using leading zeroes and using S=3, because the zeroes are ONLY allowed to the the left of the 1's and 2's, so we cannot have some constant infinite L. So, we can say there are 2^L strings for each L, and therefore sum(x=1->n: 2^x) strings less than or equal to length x, or 2^(n+1)-2. Therefore, if 2^(n+1)-2 =N, then 2^(n+1)=N+2, n+1=log2(N+2), and n=log2(N+2)-1 digits required for N numbers. In other words, it has one fewer digits than would be required to make a binary number that is 2 greater in value. > How would your number N be represented in > that representation? The largest number would be 2222...222 obviously. Since the digit places represent powers of two, as in binary, this would appear to be twice as large as 111...111 in binary, but as noted above, with one less digit we can make a number that is two greater. When we have one less binary digit, we are essentially dividing by 2, which we can do by shifting each digit to the right, or by dividing each digit by 2, since each digit is a 2 in this case. Then we get 111...111 in normal binary, which is what we expect. This is the normal number of digits, which is one less than would be required to represent a number that is 2 greater, since that would require the additon of another digit. That was a hard one, but not impossible. > > > and the > > conflation of larger and infinite with uncountable is arbitrary. > > You are missing the point. Each set larger in some sense than the naturals > as mathematicians think about them is (by definition) uncountable. Sure, unless you consider sets like the halves or square roots to be larger than N, which I do. besides, the reals are only uncountable, without bijection with the naturals, only because the naturals are not allowed to be infinite. > > > The > > diagonal proof has hidden assumptions that mathematicians seem to > > acknowledge and not question, > > Wrong, the diagonal proof proves the same thing as the proof that the > powerset of a set is larger than the base set. The reals can be put in > a 1-1 relation with the powerset of the natural numbers. (But only when > you consider natural numbers in the sense of the mathematicians.) > And if you allow infinite naturals, then the naturals can also be put in 1-1 correspondence with the power set of the naturals via binary strings. -- Smiles, Tony !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > This is precisely the kind of question which cannot be > answered. There is no bound to the lengths of the strings, but you > claim they are finite. Can you tell us the difference between arbitrarily large and infinite? -- David Kastrup, Kriemhildstr. 15, 44793 Bochum <85sly2r44d.fsf@lola.goethe.zz> > > This is precisely the kind of question which cannot be > answered. There is no bound to the lengths of the strings, but you > claim they are finite. > > Can you tell us the difference between arbitrarily large and > infinite? Leading the discussion into a circle won't help. karl m > > The mathematics or the physicists, whoever does it. The empirical content of any mathematically formulated theory of physics lies in the mapping of the mathematatical theory into the space of measurables and physical operations. Mathematics, as such, has no empirical content whatsoever. Bob Kolker > Mathematicians were solving real world problems long before Cantor's > Theory came along. Set theory made it possible to get much more profound > mathematical results that were then applied to real systems where such > predictions agreed with observation. I doubt you could come up with an example that survives close scrutiny. (i.e. where it was undeniably set theory that made the result possible) > Applied mathematicians never need to think about > completeness, and I suspect many don't know what it is. > > That's the whole point of completeness, that you can just take limits > willy nilly and assume you get a real number back as long as the > sequence is cauchy So it helps applied mathematicians sleep better at night? Funny how mathematicians before Cantor would never have had any doubts about taking limits. > My argument is that the mathematics that is relevant to real > world problems necessarily has a property of observability - that is, > mathematical statements must have observable implications, where > we think of the computer as the mathematician's microscope which > lets us make observations. > What is this obsession with computer as a microscope? Computer is a > very limitted device capable of making some limited formal > manipulations on strings of meaningless characters. I'm talking about computers in the abstract, as a conceptual aid to clarify what mathematics is all about. Virgil said: > > David Kastrup said: > > What is supposed to be a constant equality? > > > An equality that holds true for n=1, and for n=n+1 given true for n. > > In mathematics, n=n+1 is always false. > excuse the typo. again -- Smiles, Tony Martin Shobe said: > >Barb Knox said: >> >>Infinite whole numbers are required for an infinite set of whole numbers. >> >> Good grief -- shake the anti-Cantorian tree a little and out drops a >> Phillite. Here's a clue: ALL whole numbers are finite. Here's a >> (2nd-order) proof outline, using mathematical induction (which I >> assume/hope you accept): >> 0 is finite. >> If k is finite then k+1 is finite. >> Therefore all natural numbers are finite. >> >> >That's the standard inductive proof that is always used, and in fact, the ONLY >proof I have ever seen of this fact. Is there any other? I have three proofs >that contradict this one. Do you have any others that support it? > >Inductive proof proves properties true for the entire set of naturals, right? > > Yep. > >That entire set is infinite right? > > Yep. > >Therfore, the number of times you are adding >1 and saying, yep, still finite, is infinite, right? > > Yep. But be careful here, at *every* stage of this process, we have > still only done it a finite number of times. Uhhh.... Look at what you just agreed to. The number of times you are adding 1 is infinite. But, now you say it is always a finite number of times? make up your mind. > > So, you have some way of >adding an infinite number of 1's and getting a finite result? > > Nope. You weren't careful. You contradicted yourself. Do you apply successor and increment the value a finite number of times, or an infinite number of times? Be careful. > > You might want to >discuss this with your colleagues specializing in infinite series. There is a >very simple rules that says no infinite series can converge to a finite value >unless the terms of the series have a limit of zero as n approaches infinity. >Does this constant term, 1, have a limit of zero? > > Nope. > > No it doesn't, and the >infinite series of constant 1's cannot converge, but diverges to infinity. > > Yep. > > Can >you actually deny this? If so, then Poincare was right. > > BTW, there is a caveat on convergence. You have to assume the > standard topology. In other topologies, that sequence can converge. You mean with a ring? That's really not what we're talking about, unless you agree that the number line is a circle, and even then it's not relevant. In pure quantitative terms, a sum of infinite 1's is infinite. So, please make up your mind. Do we increment to get a successor an infinite number of times, or only a finite number of times, to get N? > > Martin > > -- Smiles, Tony Virgil said: > > Virgil occasionally needs his ass kicked. Maybe I do too, from time > to time. Such is life. > > TO seems to take such offense at my proofs of his manifold errors and > delusions as to want to kick my ass. > > If TO could do it by showing my logic flawed, he would have done it long > since. Been there, done that, without response from you. You tried to prove that the paths in an infinite tree were uncountable but the branches were countable, even though there are precisely half as many paths as branches, by using one tree where the branches were bits in binary numbers, and another where the paths represented subsets of N. In other words, you changed trees midproof, which was a wonderful example of the garbage possible within set theory. All that nonsense rests on your insistence that integers be finite, which I have shown to be incompatible with the infinite set. > > My occasional errors have been pointed out by those competent to find > them, which TO is not. I have tried to thank them always correction > those errors. > > What TO has claimed as my errors are, in fact, his own. I didn't claim it was an error. I said it was dishonest. > -- Smiles, Tony W. Dale Hall said: > > > The World Wide Wade said: > >>, >> >> >The idea of uncountability as being equivalent to larger than the set of >naturals is unfounded. There is no reason to believe that larger sets cannot >be enumerated. the power set of the naturals can be enumerated and bijected >with the naturals, as I described in another post, as long as infinite >natural >numbers are allowed. >> >>Sort of like saying if 0 = 1 is allowed. >> > > No, more like saying if infinite digits are allowed, which is what is > required to have an infinite set of digital numbers using a finite base. That > was a useless comment. I shouldn't even be responding. > > This is untrue, with the possible exception of whether it's worthwhile > infinitely long numbers to accommodate an infinite set of numbers. It > is necessarily to have numbers of arbitrarily finite length, but no > integer requires infinitely many digits. Actually, whether you are talking about reals in [0,1) or n in N, or ANY infinite set of elements represented as strings of symbols, in order to have an infinite set, you either need an infinite set of symbols, or infinitely long strings. To the right of the point, these infinitely long digital strings represent finite values, whereas to the left, they represent infinite values (mostly). This is a fact of any symbolic systems. > > I'm sure I'm not the only one to suggest the following, but here goes: > > 1. Note that the successor function is a 1:1 function from the set of > natural numbers to a proper subset. This establishes N as an infinite > set, since it is in 1:1 correspondence with a proper subset. > > 2. Note that the induction axiom (one of the Peano axioms) states: > > If A is a subset of N that is closed under the successor > function, and which contains 0, then A = N. > > The set of all finite natural numbers satisfies the requirements of the > induction axiom. Therefore, the set of finite natural numbers is equal > to the set of all natural numbers. In other words, every natural number > is finite. Take a look at the rewrite of Peano's axioms I posted earlier today. > > 3. Note that the set of finite natural numbers is (without the > induction axiom) already an infinite set by virtue of (1) above: > the successor of every finite natural number is another finite natural > number, and the set of all successor numbers is a proper subset. > > Whatever set you take to be the set of natural numbers, if it contains > infinite natural numbers, is not the same set that mathematicians are > referring to when they use the term natural numbers. Okay. Let's just call them integers. > > Perhaps you mean to be claiming that there is no model of Peano > arithmetic? I am claiming that after infinite succession, we have changed values by an infinite amount. > > I'll also note that your infinite series argument is not germane. The > issue is not whether one *can* formulate infinite series, but whether > one *must* formulate these. Arithmetic by itself does not mandate such, > nor does algebra. In fact, you nod to that point by raising the notion > of convergence, a notion that is decidedly non-algebraic, but instead > topological, in nature. The very act of defining convergence and of > assigning numerical values to the formal sums that can be written > involves an expansion of the notion of summation; if it were not so, > then there would be no choice in the matter of assigning the value > of sums. Instead, there is a great deal of choice in the matter, > viz. p-adic numbers, being completions of the rationals wrt norms > other than the usual one. In addition, note that the phenomenon of > conditional convergence carries with it the ability to assign the > value of a sum to any real value whatsoever by merely rearranging terms. That may all be true, but none of it makes it possible to increment a value an infinite number of times and get a finite value. You are adding 1 an infinite number of times to produce the set. You have infinite numbers in it. > > Dale. > -- Smiles, Tony Virgil said: > > Virgil said: > > David Kastrup said: > > David Kastrup said: >> >> Now, I am not familiar, I think, with the proof concerning >> subsets of the natural numbers. Certainly a power set is a larger >> set than the set it's derived from, but that is no proof that it >> cannot be enumerated. >> >> Uh, not? > > Yes, not. Larger is not a synonym for uncountable except in > Cantorland, and that is a leap and an assumption. > > Larger is a synonymon for can't be surjected onto from in set > theory. And uncountable is a synonymon for larger than the set of > naturals. It is not a leap or an assumption, but simply a > definition. > >> Is this the same as the proof concerning the uncountability of >> the reals? >> >> It's pretty similar. > Figures. >> >> Assume a set X can be put into complete bijection with its powerset >> P(X) such that we have a mapping x->f(x) where x is an element from >> X >> and f(x) is an element from P(X). Now consider >> Q = {x in X|x not in f(x)}. Clearly, for all x in X we have >> Q unequal to f(x), since x is a member of exactly one of f(x) and Q. >> So Q is missing from the bijection. >> >> > Again with the Clearly. You might want to refrain from using the > word, and just try to be clear, without hand-waving. > > There is no requirement that subset number x include x as a member, > > Quite so. But there is a requirement that subset number x _either_ > include x as a member _or_ not include x as a member. Only one of > those two statements can be true. And then Q _either_ not includes x > as a member _or_ does include it, respectively. > > You are free to choose your mapping as you want to. But once you have > chosen your mapping, each subset number x _either_ includes x as a > member _or_ it doesn't. Whether it does, can be chosen independently > for every x. But once you are through, for every particular x, x will > be in f(x), or it won't. And depending on that, x won't be in Q, or > it will. > Actually, as I think about it, given this natural mapping of the naturals > to > the subsets of naturals, subset number x will ONLY include x as a member > for > subsets 0, 1 and 2. Beyond that, subset x will NEVER contain x. So, you > have > non-empty Q for the null set, and the singletons {1} and {2}. So, what > does > that prove? > > The mapping from X to P(X) is not natural, it is 'arbitrary', meaning > that it can be anything and cannot be assumed to have any special > properties. > > In particular, f(0) = f(1) = f(2) = X is possible, whenever {0,1,2} is > a subset of X, . > > I defined the mapping and it's as natural as it gets. Each succesive bit > represents membership of each successive element. It doesn't get any more > natural than that.n Every unique infinite string of bits represents a unique > subset of the naturals, so I don't know WHAT that last sentence means. > > There is a 1-1 correspondence between infinite bit strings and > subsets. > > But there is no 1-1 correspondence between naturals and INFINITE bit > strings (only with finite bit strings). This is just another instance of > that same delusion that TO has that there exist naturals with more than > finitely many naturals as predecessors. > If they have all 1's in finite positions, then there is indeed a bijection between infinite bit strings and finite naturals. If they have any 1's in positions infinitely to the left of the point, then they represent sets that include infinite integers, and also have infinite values as binary numbers. The bijection works perfectly, IF one allows infinite integers. This restriction on the whole numbers is the only thing standing in the way of bijection betweem [0,1) and N. -- Smiles, Tony Chris Menzel said: > Chris Menzel said: said: >> ... >>> To say that a list of the reals is aleph_1 in length assumes the >>> continuum hypothesis; is that what you intend? (And if so, why?) >>> Moreover, granted, aleph_1 is omega_1 in pure set theory, but one should >>> use ordinal numbers rather than cardinals when talking about such things >>> as the length of a list, as there are many lists of different ordinal >>> lengths that are the same size. >> >> I don't think it assumes any such thing. >> >> If the assumption you are referring to is that there is a list of >> reals of length omega_1 (= aleph_1 in pure set theory), then all you >> are doing by denying that this assumption requires the continuum >> hypothesis is showing your abject ignorance of the subject. > The continuum hypothesis states, if I'm not mistaken, that there are > no infinities between aleph_0 and aleph_1. > > Sorry, dude, no, you are quite mistaken. aleph_1 is *by definition* the > next cardinal after aleph_0. The continuum hypothesis is that aleph_1 > is the size of the set of real numbers. Interesting. MathWorld says: The proposal originally made by Georg Cantor that there is no infinite set with a cardinal number between that of the small infinite set of integers aleph_0 and the large infinite set of real numbers c (the continuum). Symbolically, the continuum hypothesis is that aleph_1=c. Apparently it says both, dude. > > So, what does this have to do with assuming that a complete list of > the reals has aleph_1 members? > > Get it now? > > The continuum hypothesis is mularkey as far as I can see. > > Yes, and we've seen how far you can see. No, you really haven't looked. > > There is a whole spectrum of infinities between those alephs. > > Really, you are making such a fool of yourself by talking through your > hat instead of just learning some simple, basic set theory. Do yourself > a favor. I don't need a favor. > >> In fact, it is my position that there is an infinite spectrum of >> infinities between that of the naturals and that of the reals. > > Guess what? That means you reject the continuum hypothesis! Isn't that > exciting? Don't you just want to go out and learn all about it rather > than just spouting vague, uninformed, and often silly nonsense? Why do I want to spend my time studying something that I disagree with? Why don't you go study Mormonism? Wouldn't that be exciting? > >> Well, then your assumption that there is a list of reals of length >> omega_1 is curious indeed. > > If you are curious, why don't you try asking a question, instead of > making declarations? What, exactly, do you find curious? > > Hope that's been explained now that I've explained a bit of the > mathematics you didn't understand to you. Not really, and that doesn't answer the question. > >> I have not immersed myself deeply in set theory, no. But, I can see >> clearly that some of the conclusions are wrong, and in arguing >> this, I have stumbled upon a few of the obvious flaws in the logic. >> >> David Kastrup has noted the problems with this reasoning. Many >> thinkers through the years have thought they clearly knew one thing >> or another, only to be shown wrong by deeper advances in the relevant >> discipline. As I noted earlier, actually learning the material you >> criticize as an intelligent but uneducated amateur will help you to >> see where your conceptual errors lie. > > Well, it might help if you identified some conceptual errors on my > part, instead of sending me off to read books. Obviously, you haven't > indentified any major flaws in my logic, or you would be pointing them > out specifically, wouldn't you? > > Oh but I have -- just pointed out one conceptual error above re CH. Not really. besides, that has nothing to do with the arguments I put forth. Try again. > Another recently identified is your belief that if there is no upper > bound on the length of the strings in a set, then the set must contain a > string of infinite length. That was not my statement, but someone else's paraphrase of their understanding of my position. Pay attention. I have repeatedly stated that N is infinite IFF N contains infinite n. > Now, unfortunately, you are confusing your > inability to see your errors when they are clearly pointed out to you > with the idea that no one has found any. That is why I am, quite > sincerely, pointing you to texts containing real mathematics, in the > hope that some sincere study will enable you to see the errors you are > making. If I read as well as you do, I won't learn anything anyway. Try responding to MY statements, if you want to claim that you are doing so. > >> Hopefully I will find time soon to decipher the specific axioms of >> ZF and see whether the roots of the problem lie there, or with >> subsequent assumptions. >> >> Or with your own thinking -- an important part of being a scholar is >> always to bear in mind that one *might* be wrong. > > Yes, I have been corrected in the past, and that's okay. I haven't > been corrected correctly on this, and refuse to concede simply on the > basis of peer pressure. The flaws I see are glaring in my eyes, and > the refutations of them all have holes I can stick my hand through. > > Well, look. There are a lot of people who have a lot of years of > advanced study behind them who are claiming that you are making some > mistakes, and who, moreover, are going to some lengths to try to explain > them to you. Actually, not. Generally my claims are refuted by being mischaracterized and made to look stupid. That is not a valid mathematical refutation, or an honest communication. > Moreover, it is more than evident, as you yourself admit, > that you in fact are quite ignorant of the fields in which the things > you criticize are studied; you don't understand some of their most > elementary concepts. Now, just probabilistically, what are the odds > that *all* of those folks (not to mention several generations of great > mathematicians) are rather hopelessly confused about the foundations of > their disciplines, and that you -- uneducated in those disciplines -- > are right? Now, I grant you, it is not logically impossible -- perhaps > God is playing a trick on everyone but you. But ponder the odds. Oh, I have. What are the odds of being alive to begin with. What are the odds of miracles? Perhaps they happen every day, and perhaps every sun has some life around it. What were the chances of every scholar in Europe being wrong about the Sun revolving around the Earth? Truth is not decided democratically. > >> Now, if I came up with a theory that proved that 1=2, would you >> believe it, at first glance? >> >> Of course not at first glance. But I would not dismiss it out of >> hand, either, given what we know from G.9adel about the unprovability >> of consistency. I would be greatly skeptical, of course, if the >> proof came from someone who obviously doesn't know much about >> arithmetic or set theory. > > Basically what Godel proved was that no system can prove itself > correct from within, so that proving any system correct becomes an > infinite regress into the systems from which it is derived or the > context in which it exists. Godel did some good work there. It's too > bad he got involved with the continuum hypothesis and ended up with > similar mental issues as Cantor. > > Ah, yes, working on CH causes mental illness. Perhaps. It's certainly not a fruitful endeavor, obviously. > > Metacomment: What could *possibly* drive someone who probably *knows* > he doesn't understand G.9adel's theorem, and probably *knows* he doesn't > have any idea what G.9adel proved about CH, to make pronouncements like > this on a public forum? Conviction? > > Chris Menzel > > -- Smiles, Tony > Affirmative. After I discovered (in 1973) that Set Theory was just a > burden on analysis (Lie groups), I've managed to do analysis without > set theory for another 30 years. Your ignorance and incompetence has no empirical or philosophical import. And I still cannot imagine what its > additional value is, Your problem. Why don't you learn something? Bob Kolker > > > That's true. But mathematical axioms start to behave _as if_ they were > physical laws, as soon as they become being _applied_ to i.e. physics. The mathematical axioms remain so. It is the mapping of an abstract mathematical system into a system of physical measurements and operations that produces empirical content. The mathematics, per se, has no empirical content whatsoever. Here is the bottom line. Mathematics has no empirical content whatsoever. None. Zip. Nada. Zero. Bupkis. K'ducchis. Got it? > >> The axioms of set theory (ZFC) make a really, really interesting game. > > > Set theory doesn't deserve such a predominant place in mathematics. > After the discovery of Russell's paradox et all, everybody should have > become most reluctant. Without set theory there would be no rigorous theory of real or complex variables. Without set theory there would be no point set topology, a general theory of spaces based on the abstract notion of nearness. Set theory has made modern mathematics what it is today. Bob Kolker Chris Menzel said: > said: >> That's not true. If S is an infinite set of strings, then there is a >> difference between (1) There is no finite bound on the lengths of >> strings in S. (2) There is a string in S that is infinite. > > Except that TO claims that (1) implies (2), though I can't > even get far enough into his head to see why he thinks it, > never mind finding his 'argument' convincing. > > This seems to be a fairly common element in crankitude. I've seen > several folks argue here and elsewhere that there can be infinitely many > natural numbers only if there is an infinite natural number. (Indeed, I > think TO believes this, as I believe I saw reference to an infinite > natural in one of his posts.) The origin of this idea sometimes seems > to reside in imagination -- the poor afflicted fellows picture the > number sequence as something like an endless string of beads that > eventually disappears to nothing. There is thus no perceptual > difference between *really really long* proper initial segments of the > string and the entire string itself. So the entire string is the same > sort of thing as its really really long proper initial segments. Other > times, there seems to be some sort of a priori cardinality principle at > work: for every set of natural numbers there is a natural number that > numbers them. No finite natural number numbers all the finite natural > numbers, so (obviously) there is an infinite natural number. > > Whatever. Kinda sad. > > What is sad is that you say you have heard the same arguments repeatedly, and still don't seem to understand them. If there is a constant finite difference of one between any two consecutive natural numbers, and there are an infinite number of them, then indeed the overall range of values becomes infinite. You might want to try that imagination thing you were just making fun of. Maybe you can rediscover the chemical structure of benzene. -- Smiles, Tony <42DC3F5F.30904@netscape.net> <85ll43dvil.fsf@lola.goethe.zz> <3k7oh5Ft39kpU1@individual.net> > What is sad is that there are a number of informed people here who are willing to help you, but your stubborn unwillingness to even inform yourself about the subject prevents you from benefiting from the knowledge of these people. What is sad is that people who have a tremendous amount of knowledge and wisdom to share with you, if you'd only let them, spend so much energy, fruitlessly trying to raise you from your self-imposed ignorance. If you would just read a book on the subject, then you'd appreciate your fortune at having someone like Chris Menzel to answer a question now and then and even to disagree with you when there is some rational basis for disagreement. MoeBlee !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > What is sad is that you say you have heard the same arguments > repeatedly, and still don't seem to understand them. If there is a > constant finite difference of one between any two consecutive > natural numbers, and there are an infinite number of them, then > indeed the overall range of values becomes infinite. Sure does. But the difference between any given two fixed numbers remains finite, even though there is no finite overall range. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum Daryl McCullough said: > >> Bigger in the sense of no surjection from the smaller set to the >> larger, is one thing, bigger in the sense of having the smaller >> set as a proper subset is different. While these two measures happen to >> coincide for finite sets, they do not coincide for infinite sets, as the >> definition of infinite for sets should hint to you. >> >gee, they coincide for finite sets AND infinite sets, under Bigulosity >but I don't suppose you consider that extra consistency any sort of >progress. > > No, that's no progress at all. You can prove that for finite sets > if there is a bijection between set A and set B, then they have > the same Bigulosity. But that fails for infinite sets. So Bigulosity > is an inconsistent notion of size. > > -- > Daryl McCullough > Ithaca, NY > > Huh? Bigulosity doesn't rely on bijections. That why it DOES work. -- Smiles, Tony Daryl McCullough said: > >> 1. Phi(0). >> 2. for all natural numbers x, Phi(x) implies Phi(x+1). > >The proof has a finite form, much like a recursive algorithm. A recursive >algorithm will run forever if it doesn't have some stop condition, like running >out of nodes in a tree path, which is bad for a computer program. > > But unlike an algorithm, there is no implied infinite number of > steps. Yes there is. You show the proprty true for n=0, then for succ(n), then succ (succ(n)), etc. This is the justification for the axiom. That is why inductive proof is said to work. > >Your #2 above is the recursive part of the proof; it proves something >true based on the truth of its predecessor. > > No, the only thing that you prove is the implication > Phi(x) implies Phi(x+1). Yeah that's what i said. > >When you actually run the proof, > > You don't run proofs. That's why I put it in quotes, but it is really a recursive proof with an implied loop. > >#2 acts like a loop, iterating its way through all the members >of the set, with no stop condition. > > No, it's not like that, because you don't run proofs. Why do you think inductive proof is agreed to work? Think. > > -- > Daryl McCullough > Ithaca, NY > > -- Smiles, Tony <42DC3F5F.30904@netscape.net> <85pste8vvp.fsf@lola.goethe.zz> <854qaq8u5i.fsf@lola.goethe.zz> <85mzoi5wgm.fsf@lola.goethe.zz> <85fyu9tu92.fsf@lola.goethe.zz> > Why do you think inductive proof is agreed to work? It doesn't have to be just agreed to work. It is proven to work. It is proven by the fact that inductive proofs are applied to inductive sets. The very nature of a set being inductive is what ensures inductive provability. > Think. You think. And read a book about the subject, would you please? !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > Daryl McCullough said: >> >>> 1. Phi(0). >>> 2. for all natural numbers x, Phi(x) implies Phi(x+1). >> >>The proof has a finite form, much like a recursive algorithm. A >>recursive algorithm will run forever if it doesn't have some stop >>condition, like running out of nodes in a tree path, which is bad >>for a computer program. >> >> But unlike an algorithm, there is no implied infinite number of >> steps. > Yes there is. You show the proprty true for n=0, then for succ(n), > then succ (succ(n)), etc. This is the justification for the > axiom. It may be the _motivation_ for the axiom. But the axiom's consequences are independent from its motivation. > That is why inductive proof is said to work. That is why the axiom was chosen: to have a better mechanism than needing to check each case individually. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum hale@tulane.edu said: > > > malbrain@yahoo.com said: > [cut] > That's not what our axiom says. It says that induction covers all the > natural numbers in a single step, a single leap-of-faith. > That's really not the case. It is a recursive proof where the property is > proven true for each element depending on its truth for the preceding element. > f(n)->f(n+1), for n=1 to oo. Otherwise, how do you think it proves things for > each and every n in N? > > Do you agree that one can prove the statement If n is even, then n+1 > is odd > without using induction? > > I would say that one could. > > In such a proof, one would start off with something like: Suppose n > is even. Then, there is an m such that n = 2 * n. etc > > In such a proof, you would not be running through elements n = 1 to oo. > Or, do you disagree with this? If you disagree, then we can reduce the > problem to this case and eliminate the discussion about how induction > works. Are you offering a different type of proof that all whole numbers are finite? I have seen only the inductive form, so that might be interesting. > > Then, to answer your last question, I prove that f(n) is true for > each and every n in N by invoking axiom of induction after I have > proved f(1) is true and f(n)->f(n+1) is true (this without invoking > induction). Okay, so you ARE using induction, invoking it after not invoking it. Right? > > Induction is an axiom, not a recursive proof. The idea of recursing > from n = 1 to oo is an intuitive justification of the axiom of > induction. But, an axiom doesn't need a justification (in a certain > sense). > > -- Bill Hale > > Well, Bill, I have come to realize that this is one of my most central complaints about mathematics as it is today. There is great emphasis put on axiomatic systems, and axioms are given this status as unquestionable atomic fact without justification, when really facts NEED to be justified somehow, from outside of the axiomatic system. Axioms are taken as fact within an axiomatic system and used for proofs and arguments, but that doesn't mean every axiom is universally true as stated, or true at all outside of the system where they reside. In this case, there is justification for the axiom of induction. Peano didn't make it up randomly. It is a statement about how a recursive proof can cover an infinite set without requiring an infinite statement. As such, it really is an axiom about an infinite recursive process, and if we keep this in mind, and note that our inductive step (f(n)->f(n+1)) involves an increment (n+ 1 is finite), then we can see that the infinite loop that this statement represents ends up incrementing the value an infinite number of times, to produce an infinite value. By keeping in mind the justifications for our axioms, we can more carefully see where they do and do not apply, and refine them as needed. In my opinion, mathematics needs to be more fully integrated, and not further decomposed into independent axiomatic systems without justification. -- Smiles, Tony <42DC3F5F.30904@netscape.net> <85pste8vvp.fsf@lola.goethe.zz> <854qaq8u5i.fsf@lola.goethe.zz> <85mzoi5wgm.fsf@lola.goethe.zz> > Well, Bill, I have come to realize that this is one of my most central > complaints about mathematics as it is today. There is great emphasis put on > axiomatic systems, and axioms are given this status as unquestionable atomic > fact without justification Such unquestioned status giving is not required for axiomatics. > Axioms are taken as fact within an > axiomatic system and used for proofs and arguments, but that doesn't mean every > axiom is universally true as stated, or true at all outside of the system where > they reside. Non-logical axioms, by definition, are not universally true. Nor does axiomatics depend on claiming truth in any model that is not indeed a model of the theory. > our inductive step (f(n)->f(n+1)) involves an increment (n+ > 1 is finite), then we can see that the infinite loop There is not an infinite loop. > to > produce an infinite value. No infinite value is produced. > In my opinion, mathematics needs to be more fully integrated, and not further > decomposed into independent axiomatic systems without justification. Enter set theory, which integrates axiomatic systems. MoeBlee Daryl McCullough said: > >Daryl McCullough said: > >> Once again, if your claims had any merit whatsoever, then you >> would be able to rephrase them in a way that does not rely on >> unorthodox meanings of terms. Rephrase your claim without using >> the word finite or infinite. Is that possible? > >Are you talking to me or Dik? > > You. You're the one taking about infinite naturals. Dik made a statement, which you have now snipped, to which you were responding, I believe. > >No one else seems to have a definition for infinite as a >stand-alone word. > > Mathematicians have a definition for infinite set, > namely a set such that there exists a bijection between > that set and a proper subset of itself. yeah, I know. > >Finite means with an end > > I don't want more words. I want a *mathematical* definition. > Give a definition in terms of the standard mathematical concepts > such as subset, element of, etc. > >If the leftmost possible significant digit is at position n, >then the largest number possible in base b is b^n. > > Who says that there is a leftmost possible significant digit? GIVEN n as finite, b^n is finite. Iff n can be infinite, then b^n can be infinite. > > You are confusing two different things: > > (1) For every string, there is a finite number n such that > the length of the string is less than n. > > (2) There is a finite number n such that for every string > the length of the string is less than n. > > (1) is true, but (2) is false. They are both false if you allow infinite strings. What makes you think i am confusing those two statements? I am not. > >If b and n are known, then b^n is known. If b >and n are finite, then b^n is finite. This is just using the >standard definition of finite, independent of cardinality. > > There is no standard mathematical definition, other than the > one you've rejected. Your definition of finite vs. infinite sets is okay. I don't have an issue with it. It's your treatment of infinite sets which doesn't work for me. The bijection being used distinguishes the infinity. > > -- > Daryl McCullough > Ithaca, NY > > -- Smiles, Tony <85hdeq8vck.fsf@lola.goethe.zz> > Daryl McCullough said: > You are confusing two different things: (1) For every string, there is a finite number n such that > the length of the string is less than n. (2) There is a finite number n such that for every string > the length of the string is less than n. > > (1) is true, but (2) is false. > They are both false if you allow infinite strings. What makes you think i am > confusing those two statements? I am not. Well, yes, if you consider a set of strings which may be infinite strings (having only one end), then we all agree that both (1) and (2) are false. But the reason I'm sure I'm not alone in thinking you are confusing the two statements is that in the case in which one is true and the other is not, you appear to be unable to distinguish them, and use the truth of one permuted into the truth of the other. When we consider the set of finite strings, by definition of 'finite' (the ditty stops) (1) is true, but if there is no limit on the length of the finite strings, (2) is false. Oh dear, I'm repeating someone else. Well, it's hopeless, isn't it? I suppose I felt prompted to respond because you claim you are not confusing (1) and (2). If you are not, you should be able to give us a different example where one is true and the other is false, no? (Otherwise you are not confusing them, because they are equivalent statements.) Brian Chandler http://imaginatorium.org David Kastrup said: > > >> Instead of using the term size to refer to sets, we could >> refer to the bloppitude. >> >> Instead of using the words infinite, we could use the term >> mega-bloppity. >> >> Nothing of any importance about mathematics would change >> if we substituted different words for the basic concepts. >> > > Shouldn't there be one p in blopptitude? > > Anyway, nothing in mathematics would change, but surely the interest > of the resulting propositions would diminish, should the word size > disappear. People think (and IMHO in error) that the definition > captures correctly the concept size. > > Increase your set size! This all-natural new formula will give you > the true superset you always dreamt of! No injections required! Now > you can fill all pigeonholes and still have your member available for > more! > > Hear the testimonials: > > WM: It actually has infinite potential. Certainly the largest finite > possible. > > TO: It goes beyond finite. I did things on the Peano that others > claimed impossible. Gives a whole new meaning to the word. > > You are probably right. The interest of the resulting propositions > would diminish. > > Pretty Funny! At least you have a sense of humor. :) -- Smiles, Tony stephen@nomail.com said: >> Inductive proof proves properties true for the entire set of naturals, right? > > Wrong! It proves things only for the MEMBERS of that set, not the set > itself! > > Definitions (Cantor): > (1) a set is finite if and only if there do not exist any > injective mappings from the set to any proper subset > (2) a set is infinite if and only if there exists any > injection from the set to any proper subset. > Clearly then, a set is finite if and only if it is not infinite. > Definitions (Auxiliary): > (3) a natural number, n, is finite if and only if the set > of naturals up to it, {m in N: m <= n}, is finite > (4) a natural number, n, is infinite if and only if the set > of naturals up to it, {m in N: m <= n}, is infinite > > If these definitions are valid, then it is easy to prove buy induction > that there are no such things as infinite naturals: > > (a) The first natural is finite, since there is clearly no > injection from a one member set the empty set. > > (b) If any n in N is finite then n+1 is also finite. > This is also while quite clear, though a comprehensive proof > would involvev a lot of details. > > I can flesh it out a bit. > > Inductive step. Show that if n is finite, then n+1 is finite. > Proof by contradiction. > > Suppose that n is finite, but that n+1 is infinite. This > means there exists a bijection f from { 1, 2, 3, ... n+1} > to some proper subset S of { 1, 2, 3, ... n+1}. Without > loss of generality we can assume that S does not contain n+1. > > If we apply the function f to { 1, 2, 3, ... n} we > get the set S-f(n+1). Because S does not contain n+1, > S-f(n+1) is a proper subset of {1, 2, 3, ... n}. This > means there exists a bijection from {1, 2, 3, .. n} > to a proper subset of {1, 2, 3, ... n}, which means > that n is infinite which contadicts the assumption that > n was finite. > > Stephen > Yes, it relies, as usual, on the lack of a largest finite integer, but that doesn't mean that there are no infinite integers. There is no smallest infinite integer, but that doesn't mean there are no finite integers either. -- Smiles, Tony > stephen@nomail.com said: > ... >> I can flesh it out a bit. >> >> Inductive step. Show that if n is finite, then n+1 is finite. >> Proof by contradiction. >> >> Suppose that n is finite, but that n+1 is infinite. This >> means there exists a bijection f from { 1, 2, 3, ... n+1} >> to some proper subset S of { 1, 2, 3, ... n+1}. Without >> loss of generality we can assume that S does not contain n+1. >> >> If we apply the function f to { 1, 2, 3, ... n} we >> get the set S-f(n+1). Because S does not contain n+1, >> S-f(n+1) is a proper subset of {1, 2, 3, ... n}. This >> means there exists a bijection from {1, 2, 3, .. n} >> to a proper subset of {1, 2, 3, ... n}, which means >> that n is infinite which contadicts the assumption that >> n was finite. >> >> Stephen >> > Yes, it relies, as usual, on the lack of a largest finite integer, but > that doesn't mean that there are no infinite integers. It means EXACTLY that; he just proved that the successor of a finite number is finite. Since 0 is finite, it follows by induction -- you say you understand induction, right? -- that ALL natural numbers are finite. > There is no smallest infinite integer, ... If you actually understood induction, as you claim, you'd know that it is equivalent to the principle that, for any property P, if any natural number has P, there is a smallest natural number that has P. So if there are any infinite natural numbers, there has to be a smallest one. What can we conclude (from other well known axioms of arithmetic) about infinite natural numbers? Again, your ignorance of even basic arithmetic, let alone the more advanced topics you presume to address, is evident for all to see. Haven't you suffered enough shame? Daryl McCullough said: > >Daryl McCullough said: > >> Not quite. Uncountable set means set with a larger cardinality >> than the set of naturals. >Yes, that seems to be the Cantorian definition. Does it actually follow >that an infinite set larger than the naturals can't be enumerated? > > Yes, it certainly does follow. The definition of enumerated means > put in one-to-one correspondence with the set of naturals (or a subset > of the naturals), which implies having the same (or smaller) cardinality > as the set of naturals. Okay, well that is only because you view all countable sets as equivalent. In my book, the set of multiples of 1/2 is twice the size of the whole numbers. It seems perfectly countable. And, the powerset of a countable set, as far as I can see, is also countable, as demonstrated by denoting each subset by a unique binary string, which also represents a whole number. The power set is larger than the root set, but that doesn't mean onecan't enumerate all the subsets in a set linearly. > >Personally, I don't see the connection at all, and view it as a >conflation. > > I don't see how you could fail to see the connection. Maybe the above will help you. > >That's interesting, since the mathematiicians here seem to >like to haggle over words that they themselves can't define, >like infinite. > > That's false. Nobody likes to haggle over the definition > of infinite. It has a perfectly good definition. No, infinite set seems to have an established definition. No one seems able to define infinite. > >I use Bigulosity, to distinguish my measures from cardinality. > > Fine. Give a definition of Bigulosity. What *I* mean by > a definition is a rewrite rule so that any sentence involving > the word Bigulosity can be rephrased into an equivalent > sentence involving standard mathematical and logical concepts: > > existential quantification, > universal quantification, > addition, > multiplication, > equality, > set membership, the sizes of infinite sets. > >Try unending. Or, as I requested, give ANY synonym, or >definition of the word itself. > > For mathematical purposes, a word is defined if you are > able to rewrite any sentence involving that word into > an equivalent sentence involving only standard concepts. Gee, if I stick to standard concepts, then don't I just come up with standard analysis? There's already a million of you hard at work toeing that line. You don't need me there. > >> Nothing of any importance about mathematics would change >> if we substituted different words for the basic concepts. > >Then you shouldn't be having a word problem with me, right? > > I have no idea what you are talking about when you use > the words infinite, finite, larger, smaller, > size, etc. You don't know what an infinite number, set, string, tree, or process are? They go on forever, without end, sometimes in more than one direction. Finite ones have an end, which is where the word comes from. Smaller means closer to zero, and larger means further from zero. For set sizes, smaller means closer to the null set and larger means further from it, with more elements. Strictly speaking, a set is larger if one or more elements is added, and smaller if one or more elements is removed. > >My arguments have NOTHING to do with terminology. > > Then rephrase them without using the terminology infinite, > finite, size, larger, smaller, without end etc. > Rephrase it using only *mathematical* concepts. > > When I say There are infinitely many natural numbers I > mean exactly There exists a function f from naturals > to a subset of the naturals. I mean that there is no end to the set. > >If I say that the set of evens is smaller than the set of naturals, > > Don't say that. Say what you mean mathematically. For example, maybe > you mean > > The set of evens is a subset of the set of naturals. > > I agree with that. But don't use the word smaller because > that word doesn't mean anything definite. Yes it does. It is a proper subset. Elements have been removed. Therefore it is smaller. This is not the only criterion for size, but in the case of a proper subset, the relationship is obvious. We could say most generally, that if we can tranform one set into another by adding and removing elements, then the first is larger if we remove more elements than we add, and smaller if we add more than we remove. > >I think you all know what I mean > > No, I don't. > >Cardinality is supposed to be a measure of set size, > >> As a challenge, see if you can express your claims about >> infinite sets, or infinite naturals, or set size, or whatever, >> *without* using the words infinite, larger, size, etc. >Yeah sure, and you describe your fluffy pink flying elephant >without using the words fluffy, pink, elephant or flying. > > If I'm talking about fluffy pink flying elephants, then I'm > not talking about mathematics. When you talk about infinite objects > without defining them, then you aren't talking about mathematics. No, but you want me to talk about them without using the words that denote them. > >How do you expect me to talk about infinity or infinite >sets without using the word infinite? > > If you can't, then you aren't talking about mathematics. The word has been defined. If you don't understand the definition as unending, then i recommend contemplating Turing machines, trees, number systems and symbolic systems, as well as spece and time. I can't define end for you if you don't already know what that means. I am not getting into infinite regress as a distraction. Hopefully I clarified things a little for you above. > > -- > Daryl McCullough > Ithaca, NY > > -- Smiles, Tony <42DC3F5F.30904@netscape.net> <87hderrxf7.fsf@phiwumbda.org> <3k51ijFsefj4U2@individual.net> > You don't know what an infinite number, set, string, tree, or process are? They > go on forever, without end, sometimes in more than one direction. Oh, that's odd. I thought you said that the reason the diagonal proof was wrong, was that the diagonal hit the right hand side of the rectangle of infinite digit strings. But now you say that if a digit string is infinite it doesn't have an end. So how does the diagonal manage to stop at this end that isn't there? Brian Chandler http://imaginatorium.org Dave Rusin said: > Bored today, I peek in again at this year's Tiresome Poster of The Year > winner and find, to my surprise, that there has been a tiny bit of > progress. Something like actual definitions have been offered. Imagine! > > >> Tony has any number* of proofs that an infinite set of natural >> numbers must include infinite naturals, but these are generally >> circular. The one from information theory says that since there can >> only be a finite number of strings of finite length (even if the length >> has no limit), then to get an infinite set of numbers, you must include >> some that are infinitely long. The bit after since is a restatement >> of what he purports to prove, but he ignores people pointing this out. > >Ahem! That is another misrepresentation. The bit after since is a statement >about symbolic systems, and is a fact outside of the natural numbers. > > True! (The last bit, I mean) -- maybe even truer than you think. Whatever that means.....probably meant to be insulting. It's a fact in the area of symbolic systems, not quantitative systems. > >Given a set of symbols of size S, > > (To be precise, it's the SET which has size S, not the symbols themselves. > Here set ... of size S should mean that S is a cardinal; I don't know > whether Master Orlow is going to use a _finite_ set of symbols or not, but > in fact what he's about to say is TRUE even for non-finite sets S.) symbols that all have some certain size..... But I should have known you would find some way to misinterpret things. > >one can construct a set of all strings of length L, > > True! (But this time set ... of length L does NOT mean the _set_ has > length L; this time it's the strings IN the set that have length L. > Isn't natural language a bear? Everyone says things ambiguously > this way, but only T.O. is actually tripped up by the ambiguity.) Yeah, natural language is a bear when you deliberately make it that way. Who ever talked about sets having lengths? People don't talk about the length of a set. You are going out of your way to try to make me sound stupid, but all it proves is that you can't read. Please try to be constructive. > > We'd better be clear what strings are. It's not enough to have a > bunch of symbols; the symbols have to come in a certain order, e.g., > 1A$ is not the same string as A1$. So unless you want a very > generous notion of what strings are, you'll have to interpret > length L as meaning that L is an _ordinal_, or at the very > least that L is a linearly ordered set (not necessarily > well-ordered, I suppose). Yes, strings are ordered sets of symbols which need not be unique. L is a number, a quantity, representing the number of symbols in the string. It is not ordinal. > For strings of finite length, this is > unnecessarily fussy since the finite ordinals and finite cardinals are > identical, and we can just say ...of length 3 without much fuss and > bother. On the other hand, there is a very clear difference between > what one might call strings of length 1+omega and strings of length > omega+1 -- one of them has a zeroth term, then a 1st, 2nd, 3rd, etc; > the other instead has a 1st term, then a 2nd, 3rd, etc. AND a last term. > Failure to notice the distinction has led our hero to spout all kinds > of gobbledygook. Actually, that distinction is artificial. Numbers are numbers, really, whether finite or infinite. Ordinals are simply numbers that denote position in a discrete coordinate system, whereas cardinals are number representing size or number. What major difference do you see between starting with an index of 0 vs. 1? The origin of the system is arbitrary. > > But again, yes, for any ordered set L one can indeed construct the > set of all L-strings of elements of S. (It's isomorphic to, or by > some definitions equal to, the set of functions L --> S .) > >and the set of strings has size S^L. > > True again! Even when S and L are not finite (assuming ... has size ... > means ... has cardinality ...). Indeed, this is the usual _definition_ > of what cardinal exponentiation is. As a bonus, we get to wink at the > subtleties of the last section, since if L and M are two ordinals > of the same cardinality, then S^L = S^M. > >This is a fact, > > True! > >which > > [mal-formed sentence alert: we've got a subject to a sentence for which > no main verb follows. I think what was intended was that the verb proves > was to follow ...(S is finite), below. Or something like that.] > >when combined with >the fact that digital strings are strings on a finite alphabet (S is finite), > > True! -- sort of. I'm not sure who uses the term digital strings. > Do they have in mind strings of a _particular_ finite length? Of arbitrary > but finite lengths? Or what? Hmm, we'll let this slide for a moment. > >S^L can only be infinite if L is infinite. > > True again! Indeed, ( S^L is finite ) iff > ( ( S is finite AND L is finite ) or ( L is empty ) or ( S is empty) ) . > >Therefore, an infinite set of >digital numbers MUST contain numbers with infinite numbers of digits. > > Aha! Tony wins! Yes indeed, it is true -- IF set of digital numbers means > set of strings in S^L, where S = {0, 1, ..., 9} (I'm guessing here), > then indeed, such a set cannot be infinite if L is finite. That's > absolutely correct! If we specify _A_ finite set L and consider the > set of digit strings of length L, then the set of all of those is finite. > Sure enough! Someone actually understood what I was saying???? Miracle of miracles. Dave wins! > > BUT --- what has that got to do with the set of natural numbers? > No one has ever said that each natural number is a digit string of some > _single_ finite length L. (Indeed, no one but hacks ever says that natural > numbers are digit strings in the first place, but never mind that now.) Are you saying that not all natural numbers can be represented as strings of binary or decimal digits, or that not all unique infinite strings of digits represent unique whole number values? > > It is true that EACH natural number n (individually) can be represented by a > digit string of SOME finite length L(n). But (duh!) there is no single > finite ordinal L which is at least as large as every one of these, > in other words, bigger numbers need more digits. Duh. > > If you want to consider all the natural numbers at once as being members > of a single set S^L so that you can apply the previous line of reasoning, > you'll have to use an infinite L . You're welcome to do so, and you > may choose for example to use the ordinal omega , so that today's date > would be > ...0002005 - ...0007 - ...00020 > > That's pretty cumbersome but not wrong. It's damning for your line of > reasoning because now the L in your arguments is not a finite set, so > you can't conclude N is finite. Huh? Firstly, L is not a set, but a string length, and we are considering how the size of the set of strings varies with the length of strings we include, given a certain set of symbols of size S. I don't see anything damning to my argument, especially because you have just drawn the conclusion I drew a long time ago. If L is infinite then so is the set of strings, and vice versa. The finiteness (not the size) of the set of strings is the same as the finiteness of the string length. > It's also a potently misleading notation > because S^L will include other things besides natural numbers -- things > like ...01010101 which are most emphatically not natural numbers. But they are whole numbers, which is really the point, and are required in the infinite set. > And yes, you can use other, larger ordinals too, embedding the natural > numbers into S^(omega+1), for example --- a set which includes not only > the previous non-numbers like ...010101 (which I would now assume is > shorthand for 0 ...010101) but also things like 1...000000 . > This last bit of freedom allows you to make an even bigger fool of > yourself by continually changing your choice of ordered set L, > thus changing the set of things about which you make your wild claims. Again, L is not a set. And you were doing so well.... > (For example, even using L = omega+1 does not allow an interpretation > of what 101...0101 means. In which S^L does this string lie, grandmaster?) In which S^L? Are you asking what number that is? It's 5N/7+1. Treat the infinite whole numbers as fractions of N and you'll suddenly see how well this all works. > > Summary: To his credit, Tony has actually said some correct things about sets > of the form S^L . But he mistakenly assumes that the set of (finite) > natural numbers is a subset of S^L for some finite L, and based on > some of his other screed seems to believe that the natural numbers > coincides with S^L for some infinite-but-never-quite-specified-and- > indeed-of-time-shifting-value ordered set L. So his correct > statements about S^L have no bearing on questions about the set > of natural numbers. You mean there is no bijection between finite digital strings and natural numbers? You really think that is a valid dismissal? It's not even remotely true. The set of finite naturals is equivalent to the set of finite digital strings. Claiming you have an infinite number of finite whole numbers is equivalent to claiming you have an infinite number of finite strings, which as I understand it, standard analysis also claims. However, you have just agreed that one CANNOT have an infinite number of ANY kind of finite-length strings, if they are constructed from a finite set of symbols. So, despite your infinite-but-never-quite-specified-and-indeed-of-time-shifting-value ordered set, whatever that means (I think it was meant to make me feel stupid), you have not pointed out how you can represent all infinity of these finite naturals in a digital system without using infinite strings, or how you can require infinite strings without representing infinite values. You have simply erroneously attributed some senseless claim to me and dismissed it using an equally senseless response. So, please clarify your position. Is there a 1-1 correspondence between digital strings and whole numbers? Can you have an infinite set of strings without infinite lengths? How can you have an infinite set of digital whole numbers, without having infinitely long strings representing infinite values? > > HTH! HAND! Yeah mecca lecca hey mecca hiney ho! > > dave > -- Smiles, Tony said: > You do come across as sincere in your differing opinions. I can only > suppose that in some strange manner that your brain is wired > differently than ours are. What seems completely logical and sensible > to us, seems to be nonsense to you, and conversely, what seems to be a > proper argument to you, is so weird and strange to us that we seem > unable to even know where to start it trying to disuade you from your > point of view. Oh please. There isn't just a difference in points of view here. TO is making demonstrable, elementary mistakes of both logic and mathematics. It would be one thing if he were to point out clearly some particular principle (Axiom of infinity? Power set? Excluded middle?) that he disagreed with -- we might then be able just to acknowledge a difference in intuitions, in which case the idea of different wiring might have some purchase. But TO has at least implicitly acknowledged such things as the existence infinite sets and cardinals, and the power set axiom, and he reasons using excluded middle, so his rejection of the theorems these principles entail shows that his views are at least implicitly contradictory. In addition, however, he has explicitly made many elementary mathematical errors and committed numerous logical howlers, all of which have been pointed out to him very clearly. That may indeed be a matter of wiring, but that would be a shame; I would rather hope it is a curable combination of ignorance and rather appallingly shameless hubris. Chris Menzel Robert Low said: > Daryl McCullough said: >>Not quite. Uncountable set means set with a larger cardinality >>than the set of naturals. > Yes, that seems to be the Cantorian definition. Does it actually follow that an > infinite set larger than the naturals can't be enumerated? Personally, I don't > see the connection at all, > > The connection is fundamental. Set A is bigger than set B if there > is no surjection from B to A. If set B is the integers, then > saying that set A is bigger than set B is saying that there is > no surjection from the integers to A, which implies that there > is no bijection between them, and hence that there is no enumeration > of A (by definition of enumeration). > > That's how the game works. We agree on the basic definitions > of words like 'bigger', 'enumeration' and so on, and then > work out logical consequences of them. > > Now, until you give a definition of 'bigger', and explain > why my first paragraph above is rendered irrelevant, we'll > all be very grateful. Or at the very least, very > surprised. > Your definition of bigger works fine for finite sets. When it comes to infinite sets, however, it gives a lot of false positives in the form of equivalences. Sizes of such sets need to be built upon the properties of their members, so the method differs a little for whole numbers vs strings, although all of these sets are related in ways. An infinite set of numbers, for instance, can be compared to another infinite set of numbers by looking at the functions which describe them; if one is defined by a function that is always larger than the function describing the other, then it is a smaller set. Sets of strings are measured using N=S^L, so we could say, for instance, that all decimal numbers which include only 1's and 0's would be 2^(log10(N)) in size. A more precise notion of size for infinite sets requires a slightly more complex method than bijection, as far as I can see. -- Smiles, Tony > Robert Low said: >>The connection is fundamental. Set A is bigger than set B if there >>is no surjection from B to A. If set B is the integers, then >>saying that set A is bigger than set B is saying that there is >>no surjection from the integers to A, which implies that there >>is no bijection between them, and hence that there is no enumeration >>of A (by definition of enumeration). >>Now, until you give a definition of 'bigger', and explain >>why my first paragraph above is rendered irrelevant, we'll >>all be very grateful. Or at the very least, very >>surprised. > Your definition of bigger works fine for finite sets. And so far, you have given no evidence of even having any hint of a definition of what it means for one infinite set to be bigger than, or the same size as, another. All you have is a vague intuition which you insist we all agree with. (In spite of the fact that it leads you to patently wrong conclusions.) When you can actually come up with a better definition of what it means for two sets to be the same size (heck, even one that's no worse) then there'll be some point in listening to you. At the moment, I see no evidence whatever that you're capable of rational discourse. > Robert Low said: > Now, until you give a definition of 'bigger', and explain why my > first paragraph above is rendered irrelevant, we'll all be very > grateful. Or at the very least, very surprised. > > Your definition of bigger works fine for finite sets. When it comes > to infinite sets, however, it gives a lot of false positives in the > form of equivalences. Sizes of such sets need to be built upon the > properties of their members, so the method differs a little for whole > numbers vs strings, although all of these sets are related in ways. > An infinite set of numbers, for instance, can be compared to another > infinite set of numbers by looking at the functions which describe > them; if one is defined by a function that is always larger than the > function describing the other, then it is a smaller set. Sets of > strings are measured using N=S^L, so we could say, for instance, that > all decimal numbers which include only 1's and 0's would be > 2^(log10(N)) in size. A more precise notion of size for infinite sets > requires a slightly more complex method than bijection, as far as I > can see. And the silliness of his argument shows that TO cannot see very far at all. said: > >> Every bit of Cantorianism has been well enough defined for the >> understanding of thousands upon thousands of people. That TO fails where >> so many have succeeded says more about TO than about the adequacy of >> Cantorianism's explanations. > > The fact that a faith has millions of adherants doesn't say anything > about its validity. It says something about the society wherein it is > accepted, though. Faith?? Cantorianism is embodied in a completely rigorous axiomatic theory. The propositions you find unacceptable are demonstrably valid in that theory. There is not a lick of faith involved. Instead of tossing off idiotic comparisons to religious belief -- an inevitable rhetorical haven for cranks and crackpots -- you might consider a genuinely mathematical response: point out the axiom(s) of set theory you consider unacceptable and defend your rejection of them with arguments; or simply embark straightaway on the development of an equally rigorous alternative. Responses like yours only show you haven't the least clue what mathematics is. Chris Menzel <42DC3F5F.30904@netscape.net> <85ll43dvil.fsf@lola.goethe.zz> <84c62$42df5560$82a1e3ad$10805@news1.tudelft.nl> >Faith?? Cantorianism is embodied in a completely rigorous axiomatic >theory. How can axioms be rigorous? Aren't they supposed to be self-evident? Doesn't that sort of imply that you believe them to be true? Isn't that faith? On 25 Jul 2005 11:41:05 -0700, georgie said: >>Faith?? Cantorianism is embodied in a completely rigorous axiomatic >>theory. > > How can axioms be rigorous? Aren't they supposed to be self-evident? > Doesn't that sort of imply that you believe them to be true? Isn't > that faith? I am using axiom in a purely formal sense: A sentence in a formal language. Completely rigorous. The question of their self-evidence is a completely separate issue. <42DC3F5F.30904@netscape.net> <85ll43dvil.fsf@lola.goethe.zz> <84c62$42df5560$82a1e3ad$10805@news1.tudelft.nl> >Faith?? Cantorianism is embodied in a completely rigorous axiomatic >theory. > > How can axioms be rigorous? Aren't they supposed to be self-evident? > Doesn't that sort of imply that you believe them to be true? Isn't > that faith? Axioms are rigorous in that there is an effective method by which to determine whether a formula is an axiom. Non-logical axioms are, by definition, true in some models and not true in others. Just to study a theory, one does not have to commit to a belief that a particular model is one of the real world, whatever one takes 'the real world' to mean. <42DC3F5F.30904@netscape.net> <85ll43dvil.fsf@lola.goethe.zz> <84c62$42df5560$82a1e3ad$10805@news1.tudelft.nl> >Axioms are rigorous in that there is an effective method by which to >determine whether a formula is an axiom. Please provide an effective method by which we may determine whether a formula is an axiom. <42DC3F5F.30904@netscape.net> <85ll43dvil.fsf@lola.goethe.zz> <84c62$42df5560$82a1e3ad$10805@news1.tudelft.nl> > Please provide an effective method by which we may determine > whether a formula is an axiom. Each step in the definition of 'phi is an axiom of T' is given recursively, therefore, there is an effective procedure to determine whether phi is an axiom of T. At the very least, to determine whether a formula is an axiom, one needs only to consult the recursive definition. To see proofs that the defintions are recursive, consult any book on the subject, such as Herbert B. Enderton's [i]A Mathematical Introduction To Logic[/i], or, for example, consult whatever online postings there are of Godel's incompleteness paper. Since it is well known, and not controversial, that formal theories have effective methods for determining whether a formula is an axiom, I don't know the point of your question. Do you have difficulties following recursive definitions to determine whether formulas are axioms? Robert Low said: > That's not true. If S is an infinite set of strings, then there > is a difference between (1) There is no finite bound on > the lengths of strings in S. (2) There is a string in S that is > infinite. > > Except that TO claims that (1) implies (2), though I can't > even get far enough into his head to see why he thinks it, > never mind finding his 'argument' convincing. > > Funnily enough, there is a similar sounding statement that > is true in non-standard analysis: any set containing arbitrarily > large finite integers must also contain an infinite integer. > But in that game, the class of all finite integers isn't > a set :-) > > I only mentioned this because I thought it might muddy the > waters in an entertaining way... > Sure, that sounds like good mud. The set of all finite integers is a poorly defined set, an indeetrminate set, with no clear boundary. I can see, intuitively, that a set that contains arbitrarily large finite values must include an infinite value, although i am not sure what their proof relies on. Now, what do you not understand about N=S^L. The number of binary strings of length L is 2^L, so you cannot have an infinite set of binary strings unless L is allowed to be infinite, in which case the binary value is infinite. There can only be a finite number of finite strings with a finite alphabet. -- Smiles, Tony > Now, what do you not understand about N=S^L. The number of binary > strings of length L is 2^L, so you cannot have an infinite set of > binary strings unless L is allowed to be infinite, in which case the > binary value is infinite. There can only be a finite number of finite > strings with a finite alphabet. What is the largest number of different characters that one can use in any one position in a string? What is the largest number of characters that one can string together into a string? TO assumes some finite bound on both the number of different characters in the character set and a limit on the number of characters in a character string, neither of which bounds exist. herefore there is no finite bound on the number of strings, nor on the number of naturals which they can represent. <42DC3F5F.30904@netscape.net> <85ll43dvil.fsf@lola.goethe.zz> <3k7oh5Ft39kpU1@individual.net> > Robert Low said: > Sure, that sounds like good mud. The set of all finite integers is a poorly > defined set, an indeetrminate set, with no clear boundary. I can see, > intuitively, that a set that contains arbitrarily large finite values must > include an infinite value, although i am not sure what their proof relies on. Ah. So do I understand then, that despite ranting endlessly at people who have studied normal mathematics, and have proofs of things, you don't actually have a proof of this bit of Orlovian math, merely a hunch that somehow it has to be like this. How is work on the axioms progressing? Will there be an axiom of indefiniteness, that sort of blurs things when there would otherwise be a contradiction? How about your axiom of indeterminateness of set membership? After all, you can't let (e.g.) the (infinite) set of naturals have a clearcut subset of members which are infinite, since if you did, we would ask you about its complement. And group theory? There _is_ going to be an infinite cyclic O-group, is there? > Now, what do you not understand about N=S^L. The number of binary strings of > length L is 2^L, so you cannot have an infinite set of binary strings unless L > is allowed to be infinite, in which case the binary value is infinite. There > can only be a finite number of finite strings with a finite alphabet. For any particular value of L, yes, of course. I wish you'd stop repeating this trivial and obvious fact, and concentrate on understanding why no-one but you thinks it helps your (circular) proof. <42DC3F5F.30904@netscape.net> <85ll43dvil.fsf@lola.goethe.zz> <3k7oh5Ft39kpU1@individual.net> Tony Orlow: What is your logistic system, your primitive terms, and your axioms? MoeBlee said: >> Guess what? That means you reject the continuum hypothesis! Isn't >> that exciting? Don't you just want to go out and learn all about it >> rather than just spouting vague, uninformed, and often silly >> nonsense? > > Really, do you mean that _accepting_ the Continuum Hypothesis somehow > represents no-nonsense behaviour ? Really, do you mean that you think my comment above has any such implication? TO, in the comment to which I was responding, essentially asserted that he believed CH, though with typical cluelessness he was unable to recognize the fact. I was simply it pointing out to him, and suggesting he might want actually to study the mathematics instead of continuing to waste his time making an utter fool of himself. Chris Menzel > ... > Really, do you mean that you think my comment above has any such > implication? TO, in the comment to which I was responding, essentially > asserted that he believed CH, though with typical cluelessness he was > unable to recognize the fact. I was simply it pointing out to him... it pointing -> pointing it Stephen Montgomery-Smith said: > Stephen Montgomery-Smith said: > > I love the way mathematikers love to spew insults instead of replies. The > conclusion of the proof is that there isn't such a list, which is what I > disagree with, so your statement is either totally confused, or deliberate > obfuscation, which is more likely. > > Whatever you got, you didn't catch it from me. > > I admit that I was poking fun at your argument. But I see so many other > people trying to discuss things with you at a more logical level, and > simply getting nowhere. You are definitely not getting what they have. > > Now let me, for a moment, put aside ideas like correct or incorrect. > There is something in my very depths of my being, that when I read one > of the mathematikers proofs, that it just makes complete sense to me. I > can try to justify why I think our approach is the correct approach > using the collected thoughts of philosophers and logicians through the > ages, but in the final analysis, I just *know* that we are correct. > > You do come across as sincere in your differing opinions. I can only > suppose that in some strange manner that your brain is wired differently > than ours are. What seems completely logical and sensible to us, seems > to be nonsense to you, and conversely, what seems to be a proper > argument to you, is so weird and strange to us that we seem unable to > even know where to start it trying to disuade you from your point of view. > > Stephen > Well, it's nice to see someone who allows for different modes of thought among participants in these arenas. I do indeed think I am wired a little differently, working more with visual and constructive methods than with symbols and equations. As you say you simply know when what you are reading is correct, I feel the same in my position. However, that really doesn't fly in math or science, and one needs to test one's ideas to see if they pan out or fall flat. In science one can do an experiment or study. In math, one has to compare results from different methods to see if they agree, which is precisely what I am doing. Some of the conclusions of set theory directly contradict results from other areas of math, and are at the root of the problems in cardinality. When you speak about proofs seeming correct, you are talking about intuition, and I agree it is important and useful. This area of math is admittedly about the most counterintuitive around, and people tell me it takes a couple years to fully wrap one's head around it. That, to me, is an indication of problems. It's counterintuitive because it's incorrect. I know that rubs some the wrong way, especially if they have invested a lot into understanding it, and I'm sorry, but it's better to lance the boil thatn to let it fester any longer than necessary. As far as I can see, it has far-reaching harmful effects. But, I don't expect anyone to see what I see until after the fact. Correcting this problem, and the more general problem within the study which makes this disconnect possible, will lead eventually to the basic mathematical truths which translate directly into laws of physical and psychic reality, in my opinion. Of course, there's no point in trying to argue about that. I don't expect the world to see through my eyes, but I do expect mathematicians to be able to follow a couple of simple equations and accept minor corrections without getting nasty. Well, I did, anyway...... -- Smiles, Tony > Tony has no clue what mathematics is, nor how it is done, so he doesn't > normally bother with definitions. The closest we got from him for a > definition of finite was that a finite number is less than an > infinite one. And you can guess the definition of infinite. > Well, that's about as close to a lie as one can get, eh? I asked for a > definition of infinite, and no one could give me a definition of that word. The > best I could get was that an infinite set can have a bijection with a proper > subset, which is hardly a definition of the word infinite. > > Here's the real problem: you do not understand what definition means > in mathematics. It does not mean something that gives the reader an > etymologically warm feeling; it is not seeking to convey an intuitive > grasp of some concept. Rather it seeks to provide some sort of > mechanical test that can be used to divide objects into those that fall > in the scope of the definition from those that do not. > > Of course, in a natural language dictionary, the definition of > infinite will be vague (to a mathematician), because it is seeking to > transfer to the reader the whole set of ideas for which the word may be > used. But in a mathematical treatise, the definition above is extremely > effective: given a set, it is a straightforward matter to determine > whether there is or is not a 1-1 mapping from the set to a proper > subset of itself. All of these terms, set, mapping, proper > subset, and so on, are clearly defined without relying on anything > about being infinite, so the definition 'works' - we determine > whether a particular set is an infinite set or a finite set by > investigating mappings. > > I have said this many times, but I do not think there is any hope you > will ever understand the mathematical concept of infinity, unless you > first go through a stage of studying it under a different name, because > you simply have too many preconceived (wrong) notions about it. > > ... In fact I went to > the etymology, which literally means without end. Finite means with a known > end or bound, and infinite means without end. Of course, I got all sorts of > flack for my definition, from those that couldn't even suggest one outside of > the set theory they were regurgitating. let's try to be straight here, and no > more insulting than necessary, so it doesn't come back to bite us, why don't > we? > > As best I can grasp it, the central principle of Orlovian pseudo-maths > is that infinite numbers, never clearly being defined, but reached by > continuing from finite numbers through a twilight zone (whose > existence is anecdotally stated), have essentially the same sort of > properties as ordinary numbers. The Orlow-refutation of the diagonal > proof rests on the fact that a list of infinite sequences of digits > is not a quarter-plane, as one might imagine, but a rectangle, of width > P and height Q (P and Q being some Orlovian infinite numbers), so of > course the diagonal hits the (infinite) side at some point. The > mathematical concept of a sequence being endless means (to us) that > there is no end, but that doesn't stop Tony using the end to prove > something. You'll notice he gets a bit irritable when people point out > that one of his proofs doesn't work because there *isn't* a largest > integer (or whatever). > > (sigh) Yes you are on a bull roll here, Brian. At least I offered a > definition for infinite, which none of you did. The twilight zone we > discussed is that impassable zone between finite and infinite, where your > impossible largest finite and you fictitious omega meet, which I repeatedly > agreed could not be transcended through finite addition or incrementation. > > fictitious? I think all of maths is fictitious. If it were not > fictitious, we could measure it, or do experiments to find out (for > example) the precise value of 3. > > meet?? There is no meet. If you read page 1* of Conway's ONAG > again, you'll see that he defines 0, 1, 2, and so on, using set > notation. Then he defines w (omega, the first ordinal), by writing > the whole set of {0, 1, 2, ...} on one side of the number notation. > Later on, he creates (the surreals form a field, remember, even if you > don't know what that means yet) w-1. But w-1 does not 'arrive' until > *AFTER* w. There is no twilight zone; there is no sense [UIMM] in which > one could meaningfully start up the pofnats (0, 1, 2, ...) and somehow > later find oneself arriving at w having come (in reverse order) from > w-1, w-2, w-3, w-4, ... . > > * Value of 1 may be increased somewhat as required. > > I > asserted that many of the same properties hold on either side of this zone, in > mirror image, which I still contend is true. You assessment of my objection to > the diagonal proof is essentially correct, and I still stand by it. > > Yes, but you are asserting and contending. There is no mathematical > basis to these assertions and contentions - you are just reciting your > intuitions. > > Yes I get sick of the largest finite mantra, ... > > It's only a mantra because you keep making false claims, which rely > on the existence of a last pofnat. These are refuted by the mantra. > If you kept relying on dividing by zero, you might get annoyed by being > reminded that you can't. Tough. > > > ... especially in the context of > your equally impossible omega, your smallest infinite, which you only maintain > through the use of non-standard arithmetic where omega-1=omega. I offered > three proofs regarding the naturals, only one of which had anything to do with > a largest member. One of the others was refuted by saying that induction > doesn't prove things for an infinite set (bull), and that I was trying to > prove things about sets, not numbers, which is also bull, since I was > proving a property regarding a set DEFINED by a natural number, which is > ultimately a property of that number. The one using digital representations has > not been refuted at all, but largely ignored, since you CANNOT have an infinite > number of digital numbers, or strings on any finite alphabet, without allowing > infinitely long strings. > > That's (if I've remembered correctly) exactly the one I went through > and pointed out the precise error. Consider the set of finite > string-lengths over this finite alphabet: you claim that there are only > a finite number of these lengths, right? But is this the result you are > trying to prove, or is it something you rely on in your proof? (It's > both, of course.) Is that what you call pointing out a precise error? If I made such a statement, you would accuse me of not doing math, and the accusation would be correct. Suggestions that there are errors are not precise statements of flaw, but in this case, baseless innuendo and smoke. Point out the precise flaw, and don't claim to have done so when you haven't. The proof regarding strings is so simple, you really can't complain. N=S^L precisely describes the relationship between the number of strings N, the symbol set size S, and the length of strings L. I am not assuming anything except for this fact, and for the fact that S^L can only be infinite with either infinite S or L. Do you argue with either of these facts? If not, then can you see any way to have your infinite set of strings without strings of infinite length, besides having an infinite alphabet? No? Well, then, infinite N -> infinite L, and finite L -> finite N. Where is the flaw in this reasoning? > > > That's how it went, for the record. What really > irritates me is deliberate bull, and lies regarding what I have said. I > do not need people summarizing my position, thank you. I can do that very well > myself. > > Tony has any number* of proofs that an infinite set of natural > numbers must include infinite naturals, but these are generally > circular. The one from information theory says that since there can > only be a finite number of strings of finite length (even if the length > has no limit), then to get an infinite set of numbers, you must include > some that are infinitely long. The bit after since is a restatement > of what he purports to prove, but he ignores people pointing this out. > > Ahem! That is another misrepresentation. The bit after since is a statement > about symbolic systems, and is a fact outside of the natural numbers. Given a > set of symbols of size S, one can construct a set of all strings of length L, > and the set of strings has size S^L. This is a fact, which when combined with > the fact that digital strings are strings on a finite alphabet (S is finite), > S^L can only be infinite if L is infinite. Therefore, an infinite set of > digital numbers MUST contain numbers with infinite numbers of digits. If there > are infinite numbers of significant digits to the left of the digital point, as > would be the case with infinitely long whole numbers, then by the definition of > digital systems, such strings represent infinite values. > > Refute that, specifically. > > It's obvious nonsense. You have quantifier dyslexia, which is generally > incurable: for a particular (finite) value of L, there are S^L strings, > and S^L has a finite value. But above you said: > > [There can] only be a finite number of strings of finite length (even > if the length > has no limit) > > If the length has no limit, then there is not a single value L, but any > number of values L, and if these values are without limit, by the > (etymological!) definition even, the number of values of S^L (well, > there isn't a single value L at all, is there) is without limit. This > is what we mean by infinite. If the strings can be any finite length, > then no finite limit can be put on them. The absence of a finite limit > (and a finite limit means a real, normal, finite pofnat, like 57, (or > Mueck's latest 10^100^100 whatever it is) means that the value is > infinite. That's how it works. > > Well, you see, typing is tiring, particularly as one has to keep > inserting totally superfluous words like finite everywhere. Good > night! > > > Brian Chandler > http://imaginatorium.org > > I don't see where you pointed out any specific flaw, except to rant about your largest finite number again. If no string in the set can be infinitely long, then how, without infinite L, can S^L EVER be infinite? It can't. You say it is obvious nonsense, and that I have quantifier dyslexia, but that is bull and you know it. Those are typical Cantorian defense mechanisms, trying to discredit any other math that contradicts your already self- contradictory mess. If those values can increase without bound, then they are infinite like the set that increases without bound. But, you simply refuse to allow this, by definition. That's pretty piss poor. Talk about obvious nonsense. -- Smiles, Tony > > The proof regarding strings is so simple, you really can't complain. N=S^L > precisely describes the relationship between the number of strings N, the > symbol set size S, and the length of strings L. I am not assuming anything > except for this fact, and for the fact that S^L can only be infinite with > either infinite S or L. That would indicate that one character can only produce one number, which is one error. But also, if either the S or L may be unboundedly large, as they both are, so may S^L be unboundedly large. Every non-empty set of naturals has a smallest member. TO claims that the set of string lengths has an upper bound, so the set of naturals exceeding that bound must be non-empty. Can TO prove that there is a natural number greater that any possbile string length? If not, TO fails again. > Do you argue with either of these facts? If not, then > can you see any way to have your infinite set of strings without strings of > infinite length, besides having an infinite alphabet? By having a set of strings of unbounded lengths. > I don't see where you pointed out any specific flaw, except to rant about > your > largest finite number again. If no string in the set can be infinitely long, > then how, without infinite L, can S^L EVER be infinite? It can't. Willful ignorance, such as TO exhibits, does not cinvince anyone. > > You say it is obvious nonsense, and that I have quantifier dyslexia, but that > is bull and you know it. If the naturals satisfy the Peano properties, then any non-empty subset of the naturals will have a smallest (first) member. If the set of infinite naturals is not empty, it must have a smallest member. Subtract one and that must be the largest finite natural. Unless TO can produce evidence either of a largest finite natural or a smallest infinite natural, he argues in violation of the Peano properties. So that whatever TO's numbers are, they are not the Peano naturals. <42DC3F5F.30904@netscape.net> <85pste8vvp.fsf@lola.goethe.zz> <854qaq8u5i.fsf@lola.goethe.zz> <85mzoi5wgm.fsf@lola.goethe.zz> > I don't see where you pointed out any specific flaw, except to rant about your > largest finite number again. No, well, I give up. Just for my curiosity, though, I still cannot understand your point when you complain about ranting about my[sic] largest finite number. It has been pointed out to you so many times - with absolutely no effect - that the Peano axioms (or any similar more informal notion of pofnats) imply that there cannot be a largest pofnat. Just tell me: do you claim... (1) There _is_ a largest pofnat. (2) There is no largest pofnat (but the contradictions with your ideas escape you) (3) The answer to Is there a largest pofnat? is somehow neither 'Yes' nor 'No'. Brian Chandler http://imaginatorium.org >The proof regarding strings is so simple, you really can't complain. N=S^L >precisely describes the relationship between the number of strings N, the >symbol set size S, and the length of strings L. But for the set we are talking about, there *is* no L. We're talking about the set of *all* finite strings. That's an infinite union: If A_n = the set of all strings of length n, then the set of all possible finite strings is the set A = union of all A_n = { s | for some natural number n, s is in A_n } This set has strings of all possible lengths. So there is no L such that size(A) = S^L. >I am not assuming anything except for this fact. You are assuming that every set of strings has a natural number L such that every string has length L or less. That's false. -- Daryl McCullough Ithaca, NY David Kastrup said: > > David Kastrup said: >> >> Barb Knox said: >>> >>>Infinite whole numbers are required for an infinite set of whole numbers. >>> >>> Good grief -- shake the anti-Cantorian tree a little and out drops a >>> Phillite. Here's a clue: ALL whole numbers are finite. Here's a >>> (2nd-order) proof outline, using mathematical induction (which I >>> assume/hope you accept): >>> 0 is finite. >>> If k is finite then k+1 is finite. >>> Therefore all natural numbers are finite. >>> >>> >> That's the standard inductive proof that is always used, and in >> fact, the ONLY proof I have ever seen of this fact. Is there any >> other? I have three proofs that contradict this one. Do you have any >> others that support it? >> >> Inductive proof proves properties true for the entire set of >> naturals, right? >> >> For each of its members. >> >> That entire set is infinite right? Therfore, the number of times you >> are adding 1 and saying, yep, still finite, is infinite, right? >> >> No. He is not adding 1 more than a single time, just to check that >> for n finite (which means that the set 0..n obeys the pigeon-hole >> principle) n+1 is still finite (the case of one additional pigeon-hole >> can be reduced to the case n if you check for the hole in position >> n+1 and in 0..n both before and after permutation). > > How do you know 6598367 is finite? Because you got it by adding 1 to > 6598366, the 6598366th number. > > Certainly not. Don't tell me that you have spent the effort of > counting to 6000000 manually just to be able to talk about that > number. > > I know it is finite because it has a form (a finite number of digits) > that can be shown to always refer to a finite number. I see that the > number has 7 digits, and I know it is finite. > > That is, I know it is finite because it obeys certain laws, and those > laws can be shown to hold for all natural numbers, by induction. I > don't need to carry out any individual steps to make use of that > knowledge. Then why do you ignore those laws of representation, which are based on N=S^L, and pretend you can have an infinite set of such finite strings? Why do you feel entitled to violate established laws of math? You cannot form an infinite number of strings with a finite alphabet, without strings of infinite length. This should be obvious. > > So how to you get the aleph_0th number? What IS that number? it's > aleph_0. > > But that number does not obey a form I can recognize as belonging to a > finite number. And indeed, taking a look at its properties, it > becomes clear that it can't be a finite number. Well, no, it's infinite. Aleph_0 is basically 111...111+1. This is a clear contradiction within set theory, this assumption of all finite values. Try my new axioms. You can count from both sides at once. > >> So, you have some way of adding an infinite number of 1's and >> getting a finite result? >> >> No, he is just checking that the conditions for the fifth Peano axiom >> hold. The whole point of the axioms is not to have to check an >> infinite number of steps in order to get a statement about the members >> of a particular infinite set, the naturals. > > And yet, one cannot apply an increment an infinite number of time > without adding infinity. > > It is neither necessary nor feasible to increment an infinite number > of time or add infinity. The Peano axioms do just that. they start at 0 (or 1) and define successors, which are interpreted to be the previous value plus 1. That's called an increment, and if you do it x times, your value increases by x, wether x is finite or infinite. > > -- Smiles, Tony > David Kastrup said: > > And yet, one cannot apply an increment an infinite number of time > without adding infinity. > > It is neither necessary nor feasible to increment an infinite number > of time or add infinity. > The Peano axioms do just that. Those axioms only do it once per number. One of the trivial consequences of those axiom is the theorem that any non-empty set of naturals has a first (smallest) member. If TO were right that the set of infinite naturals is non-empty, there would have to be a smallest one, anddit would have to have a predecessor which is finite, unless TO wants to say that all naturals, including 1, are infinite. So that either there are no infintie naturals or all naturals are infinite. !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > David Kastrup said: >> >> How do you know 6598367 is finite? Because you got it by adding 1 >> to 6598366, the 6598366th number. >> >> Certainly not. Don't tell me that you have spent the effort of >> counting to 6000000 manually just to be able to talk about that >> number. >> >> I know it is finite because it has a form (a finite number of >> digits) that can be shown to always refer to a finite number. I >> see that the number has 7 digits, and I know it is finite. >> >> That is, I know it is finite because it obeys certain laws, and >> those laws can be shown to hold for all natural numbers, by >> induction. I don't need to carry out any individual steps to make >> use of that knowledge. > Then why do you ignore those laws of representation, which are based > on N=S^L, and pretend you can have an infinite set of such finite > strings? Because L can become arbitrarily large, there is no fixed S^L. > Why do you feel entitled to violate established laws of math? You > cannot form an infinite number of strings with a finite alphabet, > without strings of infinite length. This should be obvious. Without strings of _arbitrary_ length. If you don't get the difference between infinite length and arbitrary length, you will never understand this. >> >> So how to you get the aleph_0th number? What IS that number? it's >> aleph_0. >> >> But that number does not obey a form I can recognize as belonging to a >> finite number. And indeed, taking a look at its properties, it >> becomes clear that it can't be a finite number. > Well, no, it's infinite. Aleph_0 is basically 111...111+1. This is a > clear contradiction within set theory, this assumption of all finite > values. Try my new axioms. You can count from both sides at once. No, I can't. oo is not the successor of any number according to your axioms, so I can't go backwards from there. > The Peano axioms do just that. they start at 0 (or 1) and define > successors, which are interpreted The axioms don't interpret. > to be the previous value plus 1. That's called an increment, and if > you do it x times, It is your problem if you want to do anything in a particular order or a number of times. The axioms are not bothered about that. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum > ...You cannot form an infinite number of strings with a finite > alphabet, without strings of infinite length. Oh jeez, you're as clueless about formal languages as you are about set theory, not that this is a surprise. Do you know that what you are saying here is contradicted in the early chapters of any book on formal languages? Really! Go have a look! Here, let me save you some trouble: On page 1 of their famous standard text *Introduction to Automata Theory, Languages and Computation*, Hopcroft and Ullman define a string to be *finite* sequence of symbols. (Note: they don't think the notion of string can't be generalized to the infinite, it's just that in their text they are only interested in the finite ones.) Then, on the very next page, they point out that The set of palindromes (strings that read the same forward and backward) over the alphabet {0,1} is an infinite language. See? Do you realize what a fool you are making of yourself by making assertions about set theory, transfinite arithmetic, formal languages, computability, etc. that show you don't understand even the most elementary concepts, or grasp the most elementary theorems, of set theory, transfinite arithmetic, formal languages, computability, etc? C'mon, stop embarrassing yourself. Go *learn* some mathematics before you start spouting off about it. It's painful to watch. Chris Menzel David Kastrup said: > > David Kastrup said: >> >> Don't be a jerk. >> >> That's your job description. > > I didn't accuse anyone of sulking, especially when they were doing > anything but that. Typical Cantorian ad hominem. > Nice clipping of your original obnoxiousness. Also very typical. > >> No, the proof is a misapplication of the axiom. The inductive proof >> method works well for constant equalities, >> >> What is supposed to be a constant equality? > > An equality that holds true for n=1, and for n=n+1 given true for n. > > For n=n+1. Do you even read the nonsense you write? oops. sue me in typo court. > > An equality, as opposed to an inequality, or a vague property such > as finite, which is really an inequality that is constantly > decreasing. > > This is hand-waving hogwash. Please come up with a proper definition > of constant equality if you want to argue using it. A formula using one of these =. Where I come from, we call it an equal sign, and it means is the same as. I am sure you have come across these from time to time. Do you honestly not know the difference between an equality and an inequality? If you do, can you tell which one is finite might be most like? > >> but in the case of proving that all naturals are finite, you are >> incrementing the value at each of an infinite number of steps >> >> Nonsense. Complete bull. I do nothing at all like that. >> There are no steps whatsoever involved here, and certainly not an >> infinite number of them. n is finite for n=0, and it is shown that >> for any finite n, n+1 is finite. > > So, inductive proof does not rely on proving for n+1 based on n? > > It relies on exactly that. But you need not check this explicitly for > any n except 0. It is sufficient to show the implication from n to > n+1, and the case for 0. Those two steps are all that is required. Yes, I am well aware of the construction of inductive proof. Are you aware of WHY it is understood to work? Do you understand the concept of a recursive stepwise process with a starting point, but no ending point? The lack of endpoint is what we call infinite. If it ends, it's finite. Inductive proof does not end, or have a stopping point where it has covered all cases. Conceptually, it defines an infinite linked list of implication. > > The infinite number of successive naturals for which you prove your > property do not constitute an infinite number of implied steps in > inductive proof? > > No. The inductive proof consists of two steps. Once you have covered > them, the proof is established for _all_ n. That's what induction is > all about. > > Because you only require three lines for the construction of the > proof, you think there are only three steps implied? > > Yes, exactly that. Exactly that is the purpose of defining the > naturals by a small set of axioms: being able to deduce general > properties without having to check each number individually. > > Talk about nonsense, and complete bull. > > It is called generalization. It is called bull. It's like you really don't understand what inductive proof is doing. You memorized all three steps and now you think you're smart. But, if you can't see that it proves the property true for each n in N, in turn, recursively, then I can't imagine what you think is going on. This is simple denial in the face of an obvious contradiction between counting an infinite number of times and still having only finite values. Now you want to say induction has no steps, when the step that gets repeated an infinite number of times is the meat of the proof, and in the proof of finiteness, that step involves adding 1. What is in the water at these institutions, anyway? > >> That is not an infinite number of steps, but at best 2 steps. >> >> and maintaining that the value remains finite, because you are >> looking only at individual steps. >> >> That's what induction is about. Showing f(0), and proving >> f(n)->f(n+1). Two steps, and this covers the naturals. If you >> want to crosscheck more than that, feel free to do so, but the >> axioms don't require doing that. That's their whole point: not >> having to check things piece by piece. > > The truth for n+1 depends on the truth for n, for all n in N, from 1 > to oo, an infinite number of times, proving the property true for an > infinite number of numbers. > > It does not depend an infinite number of times. It is just one > general dependency. And yes, as a general dependency it obviously > holds for an infinite amount of numbers. But that does not mean that > I have to check each number individually. No, of course not. Otherwise it wouldn't be a proof, but fact-checking. Still, you need to be careful about your property in this case, and keep in mind that inductive proof IS an infinite loop, so that incrementing in the loop creates infinite values, and the quality of finiteness is not maintained over those infinite iterations of the loop. > > If the inductive portion of the proof, where truth is derived for > n+1 from truth for n, increments a value, > > f(n)->f(n+1) is a static relation between two different statements > depending on n. It does not increment anything. > >> It is irrelevant. The axioms don't talk about steps, and certainly >> not an infinite number of them. You are babbling what you consider >> intuitive, but it bears no relation to the math. > > the first step is 1, and subsequent steps are denoted by n, and > there are an infinite number, as many as you have n in N. > > The Peano axioms don't involve steps or any progress or whatever. > It is utterly irrelevant in what order you determine f(6)->f(7), > f(3)->f(4). If you have ascertained f(0), and the validity of > f(n)->f(n+1) by whatever means and in whatever order, the validity of > f(n) for all n is established. No need for an infinite number of > steps. The Peano axioms define an infinite set as a self-referential recursive loop. Naturals are defined as the successor of the previously-generated natural, and properties are proven for a given natural based on their truth for the predecessor. You say we can go in any order, but that's not what axioms say. They explicitly talk about the successor to each natural. You're just spouting defensive nonsense here. > > The whole idea of proof by induction is to prove with two steps a > statement valid for all elements of N (which happens to be an > infinite set). > >> Basically, adding X repeatedly Y times is the same as adding Y >> repeatedly X times, so adding 1 and inifnite number of times is the >> same as adding infinity once. >> >> Derive that from the axioms. If you can't, it is irrelevant. > > So you contend that X*Y<>Y*X? Interesting. > > Wrong. I contend that the Peano axioms neither define nor use > multiplication, and so multiplication is irrelevant in contexts where > just the Peano axioms are used. So, you can add 1 an infinite number of times and get a finite value. Also very interesting. Whatever math doesn't suit your case is simply irrelvant. Time to grow up, Dave. > > I am sure you must have an axiom in your bag for the commutative > property of multiplication? Check the side pocket.... > > Indeed, for _arithmetic_ on natural and whole numbers, there is > another bag of axioms. You can't expect to be talking about > multiplication before you have even defined the term. We both know what multiplication is. Don't you ever get tired of playing dumb? > >> There is a strictly limited number of steps involved in an >> induction proof, and it provides proof for all natural numbers, >> which happen to form an infinite set. > > And I guess only three of them are involved in the proof? How do you > prove it for number x? Does that not depend on x-1? The number of > inductive steps implied in proving a property true for all n in N is > infinite. > > Here is an example for a proof by induction: > > Proposition: > The number of arrangements of k items is k!, > where 0!=1, and (n+1)!=(n+1)n!. > > Proof: > There is exactly one way to arrange 0 items, and 0! = 1 by > definition. > > Now let us arrange n+1 items. This can be done by arranging n items > (which offers us n! possibilities by the induction premise). After > arranging the n items, there are n+1 possibilities to place the last > item: for n=0, there is 1 possibility, and for larger n we can > place the remaining item to the left of the n items, or to the right > of the n items, or in any of the n-1 positions in between. So we > have (n+1)n! possibilities for the combined placement. > > Finished. I don't need to check for n=7. This is already covered by > the proof. Very good. You used an equality as a property. That works every time. In fact, I don't think I can think of any other case where this type of proof doesn't work, except for the proof of finiteness of naturals. Then again, finite not be a very good mathematical term to use in such a proof. > >> At each step in the proof, the set has a largest element which is >> precisely the same as the size of the set. >> >> Fine, and so you prove something for a set that has a largest >> element. The set of natural numbers is no such set, and so your proof >> does not hold for it. > > What the proof shows is that, no matter how large the set gets, it > never has more members than the number represented by its largest > member. > > For every set that is characterized by a largest member. The set of > natural numbers is no such set, and so your proof does not hold for > it. No largest finite. Shake rattle The proof holds for all n in N. Your refusal to even think about the proof is silly. There is no way the size of the set can be larger than every element within it. Just think about it a minute and stop playing lawyer-math. However large your set, you cannot fill it with only values that are strictly smaller than that size, or you run out. > > Therefore, if the elements are all finite, then the set size cannot > be infinite, > > If the set has a largest member. The set of natural numbers has no > largest number, and so your reasoning does not hold for it. Your reasoning is simply not happening at all. > > since that would be larger than the value of every member in the > set. I am sure I can work out a proof without any reference to > largest element. > > Do so and then come back. I'll work on it. > >> This constant equality holds for all such sets defined by ANY >> natural number in the set. >> >> Certainly. Unfortunately, the set of natural numbers is no such set >> (since there is always a larger number than any given number in it), >> and so your proof does not apply to it. > > It applies to ALL n in N. Sorry. Take your complaint to Peano. > > Sure does. And that means that it applies to all sets defined by some > n in N as its last element. The set of natural numbers is no such > set, since it has no last element. Tough. At every step in the proof, the new element is the largest element and equal to the sizeof the set. At no time can you ever have the set be bigger than all elements, but it is always equal to the last element added. Whatever the size of your set, that IS the upper bound on element values. This is too obvious. Stop playing dumb, with your largest element excuse. > >> Most generally, it is impossible to have ANY set of natural >> numbers, finite or infinite, which has a number of elements that >> is greater than its largest element value. >> >> If it has a largest element value. The set of natural numbers has >> no largest element, and so your proof does not apply to it. > > Yes, it does. No set of naturals can contain a greater number of > elements than all element numbers in the set. > > What is this? Proof by whining? The set of positive unit fractions > has 0 as its lower bound, yet 0 is not a member of that set. What is this? Proof by non-sequitur? This has nothing to do with what we're talking about. > >> This should be clear to anyone who thinks about it. It is >> impossible to have an infinite set of strings, of which digital >> numbers are a type, without having either an infinite base, or an >> infinite number of digits. >> >> Oh, there is an infinite number of digits, but every single string >> only occupies a finite number of them. > > You mean it only has non-zero values in a finite number of digits? > > Every single number has non-zero values only in a finite number of > digits. But there is no finite number of digits that would contain > _all_ numbers. Prove it. > > That's the same as only having a finite number of digits, in which > case one can only have a finite number of unique digital numbers. > > Each number has a finite number of digits, but there is no fixed > finite number of digits that would contain all numbers. Then what makes you think they are all finite? How do you have an infinite set of finite length strings on a finite alphabet, when the formula is N=S^L. No one seems to want to address this issue. How do you get infinite N with finite S and L? You don't. > >>> So, which inductive proof do you believe? You cannot add 1 an >>> infinite number of times to your maximal element, >>> >>> There is no maximal element, and the Peano axioms don't define >>> infinite number of times or similar processes. >> >> If Peano's fifth is correct, and inductive proof applies to the >> entire infinite set in a stepwise manner, >> >> There is no stepwise manner in the axioms. > true for n+1, given true for n, for each n in N. steps, an infinite number of > them. > > Where is there a step in the fifth axiom? You are fantasizing your > own rules. The recursive statement, the meat of the proof, where it is demonstrated that f (n)->f(n+1). > >> then there are indeed an infinite number of steps implied. It's an >> immediate consequence, as you said above. >> >> It is an immediate consequence of the axioms that the described set is >> infinite, since the successor relation gives a bijection to a proper >> subset. But there are no steps involved at all. You just make them >> up for the sake of your personal intuition. They are a personal >> crutch of your own, and cause you to make mistakes that are not >> inherent in the axioms. > > Uh huh. And the inductive construction of the naturals using the > successor operator at each step also involves no steps. > > There are no steps, so it does not make sense to talk about at each > step. The fifth axiom relies on for every n. It is irrelevant > whether you establish that in any order, forwards, backwards, even > numbers first, then odds, or simultaneously, so it is nonsensical to > talk of steps. There is no prescribed order in the axioms. Gee, what was that n doing in there again? What IS that thing anyway? Maybe we should go ask Gumby and Pokey.... > > Maybe you could use a stepwise crutch so you don't have to crawl on > your belly. It is amazing that you cannot see that incrementing an > infinite number of times to generate an infinite set of naturals > does not result in infinite values. Poincare was right. > > But there is no incrementing in the axioms. If there is, point it > out. And certainly not a number of times. What did n+1 mean again? I fell down and banged my head, and now I don't remember..... > >> and are proven using assumptions that are unfounded, >> >> Axioms are not unfounded, and not assumptions. > > Excuse me, but axioms are just that, for the most part. They are > statements assumed to be true for the sake of argument and > proof. > > But they are not assumed anew for every argument and proof. They are > cornerstones of the mathematics built upon them. You can choose them > arbitrarily, but there is a cost of throwing them away afterwards, and > this cost gets higher as mathematics progresses. Sometimes it is > still worth to pay the price for abandoning an established axiom. But > some bumbling bozo not understanding their implications is not a > worthwhile reason. Don't be so hard on yourself. You can probably be made to understand, if you try hard. > >> and told that the Banach-Tarski result is a paradox and not a >> proof by contradiction. There really seems to be a bad influence >> going on that allows people to think they have an infinite language, >> when they only allow finite strings, or that the number of paths in >> a binary tree, which is always half of the number of branches, is >> suddenly infinitely larger than the number of branches, when those >> numbers become infinite. >> >> Then look up and understand the definition of an infinite set. It is >> _the_ decisive mark of an infinite set that it does no longer obey the >> pigeon hole principle. > > That's one conception, and one I reject. 1 is 1, whether it's in N, > or in {1,2,3}. > > It is not a conception, it is the bloody _definition_. The word is > no longer free for the taking. The very least you have to do is to > _define_ it if you are going to use it with a different than the > established meaning. Everything else is nonsensical. What's nonsensical is pretending you can talk about infinity solely in terms of counting and get anywhere, and then pretending you got somewhere, but using axiomatic sleight of hand. It is certainly possible for some to think about infinite sets without resorting to bijections. That is not the only way to conceive of them, and your insistence that it is, is what needs correction here. Cantor does not have a monopoly on infinity. > >> I simply can't understand how mathematics has come to accept such >> illogical results as counterintuitive rather than incorrect. >> >> Then get yourself books and learn. > > Yes, spend years trying to qwrap my head around concepts that are > clearly wrong. What a grand use of my time. > > Then just shut up. There is no sense in spewing off about something > which you do not even plan to understand. Apparently I understand it better than you experts, since I see clearly where is makes foolish assumptions, and you can't even follow connections to infinite series or symbolic systems, with simple formulas provided. DO you go and read lots of books on Hitler so you can be sure nazism is wrong? The only reason I am slogging tis out with you all is that it's become clear to me just how entrenched and nonsensical cardinality is, and what its effects are on the mind. It's not good. > > I am afraid I WILL have to tear it apart bit by bit to convince > anyone, but even the most obvious objections to your set of naturals > are rejected with insults, and no real refutation except your own > established definitions. > > Uh, that's what the definitions are for. Establishing what one is > talking about. If you want to talk about something different, you > need to start with defining things differently. If you don't, and > then people prove you wrong _by_ _the_ _definitions_, then you are > just wrong, period. You can't prove things by definition. If I am talking about something different, and you throw definitions I am not using as proof that I'm wrong, then I tell you to get your head out of your axioms and listen to the point I am making. You can't prove to me that you have an infinite set of finite naturals, based on a proof with a flaw that is identified, and your declarations of immediately follows. > >> You got it all confused again. There are no infinite natural >> numbers, so the Cantorians don't insist on anything for those >> non-existing entities. > > You are doing exactly that when I suggest the inclusion of infinite > naturals. You are saying they don't behave the same way, so they > are different, and not in the set. > > Quite so, since the set is defined by the axioms. Then why did you say what you said above? You're not making sense. This is getting boring. > > I might as well say that infinite sets are not sets at all, because > my concept of a set is that every set is finite. > > Yes, and this would be a perfectly valid argument. Once you define a > set as something finite, it is clear from the Peano axioms that the > natural numbers can't form such a set, since the successor relation > clearly violates the pigeon hole principle that a finite set obeys. > > But as long as we are talking about natural numbers and sets and > infinite sets without redefining them, we are talking about > established concepts. And apparently well established misconceptions. > > Don't throw your definitions at me as if they are proofs. They're > not. > > But a proof that can't be traced back to the definitions is not > sound. You can't prove things if you don't even know what you are > talking about. You mean like infinite recursion. Read up a little. > >> Does this seem a little unfair to you? Why do you insist that >> eveything one can do with a finite number should work exactly the >> same for an infinite number? >> >> There is no infinite natural number. The finite natural numbers >> are defined by the Peano axioms, and yes, everything that works >> with finite naturals works with all of them. That is the whole >> point of defining them with those axioms: to have a body of numbers >> governed by the same rules. > > And why don't you want all sets to have the same rules? if infinite > sets can have slightly different rules, why can't infinite numbers? > > Oh, they can. But they are not part of the naturals, because the > definition of the naturals happens to rule them out. So you need to > find a different context to talk about infinite numbers rather than > them being natural numbers. Fine. Call them whole numbers. If you insist that all naturals are finite, that they each differ by at least 1 from all others, and that there is an infinite set, then i don't believe in natural numbers, and will henceforth not use the term. Can you live with infinite WHOLE numbers? > >> If you don't like it, you can invent your own axioms and numbers, >> but then you can't cry foul that your laws don't hold with the >> numbers the others are talking about. > > Sure, except my numbers are a proper superset of your, so if > anything holds for all of them, then it holds for yours too. > > Feel free to come up with axioms and definitions that meet that > criterion. Until you do, you'll look less stupid if you shut up. Unfortunately for you, I don't mind looking stupid, expecially when that's in the eye of the beholder, so I don't think I'll shut up. Why don't you try it, and maybe you'll look like less of a crybaby. > >> The determination of divisibility by a number which is prime >> relative to the number base depends on a termination to the >> process of division. >> >> Good. And since all natural numbers are finite, every division >> among those naturals is guaranteed to finish. > > For finite naturals, yes. > > Which are the only ones there are, since the Peano axioms defining > them leave no room for other naturals. > >> Whether the number that is represented by a binary string which >> represents a subset is a member of that subset is irrelevant. >> >> Not if you are establishing a bijection. > > What? Now it's my turn to say bull. There is no requirement > that the subset of N associated with the number x must contain x as > an element. > > No, but it must either contain or not contain x as an element, and > that is all that is required to establish that you can't biject > between set and powerset. Given any subset denoted by a binary string, and any element in the set of all whole numbers, one can easily determine whether the subset contains the whole number denoted by the binary string. Simply look at the bit at the position corresponding to that element and see if it's a 1. If this is your criterion, it is met. > >>> If you divide this number by 3 (11) you find it is divisible >>> or not, depending on whether you have an odd or even number of >>> 100's in your infinite string. Of course, this question is not >>> really answerable, so I don't have an answer for you. What do >>> you think? What is aleph_0 mod 3? >>> >>> Oh, I never claimed that aleph_0 was a member of the natural >>> numbers, so I don't need to make claims about aleph_0 mod 3. >> >> But the answer to this question depends on the number of digits >> in the infinite number. >> >> Since there are no infinite numbers, I need not answer the >> question. You claim that there are, so it is up to you to provide >> an analysis of your claims. > > My claim is not that all arithmetic holds for infinite whole > numbers, but that infinite whole numbers are required in any > infinite set of whole numbers. > > They aren't. You require an unlimited amount of finite numbers, but > each one of them is finite, even though there is no maximum to them. Which is entirely impossible if they each differ from each other by at least some finite amount. Then, you have an infinite sum of finite differences, for an overall infinite range of values, which means some values muct be infinite. How can you not see this? > > The arithmetic is a sidebar, and if you want to claim that I have to > work all that out to assert their existence, you're full of > ptooey. That is an entirely diferent question. > > The divisability is a sideeffect of the Peano axioms. They leave no > room for numbers suddenly stopping to be either divisable or not > divisable by 3. Huh? Whatever.... > >> Which is why infinite numbers have not been admitted into the >> naturals. It would be impractical to have a set of numbers that >> are not governed by the same laws. > > They certainly need to be handled differently, > > Yes, and thus they are not natural numbers. Since they can't be > handled with the Peano axioms. They are whole numbers that are required for the infinite set. Peano's axioms can easily be inverted to produce them. > > but that's what Cantorians say, unapologetically, about infinite > sets. > > Sure. There is nothing wrong with claiming infinite numbers or > whatsoever, as long as you don't make the mistake of assuming that the > Peano axioms or anything else derived from them would extend to those > numbers. It is perfectly fine to talk about such entities as long as > one finds axioms that fit them. The Peano axioms don't, and so you > can't call them natural numbers. They do if inverted. One can start at infinity and count down quite easily. > > When asked why a certain property is not true of infinite sets, the > typical answer is, because not everything that holds for finite > sets must hold for infinite sets, which doesn't answer the question > at all. So, you get no apologies from me for including infinite > values in the set of whole numbers (okay not naturals, see?), in > order for that set to be infinite. > > Unfortunately, the moniker whole numbers is already taken as well. > Call them TO numbers, and devise a coherent set of axioms for them, > and nobody will complain. Can I quote you on that? Let's see how your prediction pans out: 1. oo and 0 are numbers 2. If x is a number, the successor of x is a number. 3. oo and 0 are not successors of any number. 4. Two numbers of which the successors are equal are themselves equal. 5. (induction axiom.) If a set T of numbers contains oo and 0, and also the successor of every number in T, then every number is in T. Any complaints? > > By your definition of naturals as all finite, and the basic math > governing quantities and strings, I can only conclude that you have > a finite set. > > Wrong. But since you have been hit with the proof for that about five > dozen times, it appears useless to do it again. I have pointed out the flaw in that proof repeatedly, and shown it to be wrong using other means, so I am not even sure what I am doing on this planet, talking in circles with people who refuse to understand, and simply sling around. Try thinking about it a minute. > > -- Smiles, Tony > By your definition of naturals as all finite, and the basic math > governing quantities and strings, I can only conclude that you > have a finite set. > > Wrong. But since you have been hit with the proof for that about > five dozen times, it appears useless to do it again. > I have pointed out the flaw in that proof repeatedly TO's proof is based on his wrongheaded idea that a set of finite naturals, even though unbounded, cannot be infinite in the Cantor sense. > I am not even sure what I am doing on > this planet, You presence here puzzles us, too. > talking in circles with people who refuse to understand, > and simply sling around. TO's description fits himself better thatn anyone else. Except possibly WM. Try thinking about it a minute, TO. !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > David Kastrup said: >> >>> No, the proof is a misapplication of the axiom. The inductive proof >>> method works well for constant equalities, >>> >>> What is supposed to be a constant equality? >> >> An equality that holds true for n=1, and for n=n+1 given true for n. >> >> For n=n+1. Do you even read the nonsense you write? > oops. sue me in typo court. >> >> An equality, as opposed to an inequality, or a vague property such >> as finite, which is really an inequality that is constantly >> decreasing. >> >> This is hand-waving hogwash. Please come up with a proper definition >> of constant equality if you want to argue using it. > A formula using one of these =. Where I come from, we call it an > equal sign, and it means is the same as. So that is a constant equality. And what is a non-constant equality? And a constantly decreasing inequality? This is all a bunch of nonsense. > I am sure you have come across these from time to time. Do you > honestly not know the difference between an equality and an > inequality? If you do, can you tell which one is finite might be > most like? is finite is neither an equality, nor an inequality, but a property. And there are no decreasing inequalities and no non-constant equalities. >>> but in the case of proving that all naturals are finite, you are >>> incrementing the value at each of an infinite number of steps >>> >>> Nonsense. Complete bull. I do nothing at all like that. >>> There are no steps whatsoever involved here, and certainly not an >>> infinite number of them. n is finite for n=0, and it is shown that >>> for any finite n, n+1 is finite. >> >> So, inductive proof does not rely on proving for n+1 based on n? >> >> It relies on exactly that. But you need not check this explicitly for >> any n except 0. It is sufficient to show the implication from n to >> n+1, and the case for 0. Those two steps are all that is required. > Yes, I am well aware of the construction of inductive proof. Are you > aware of WHY it is understood to work? Because it is the most basic application of the fifth Peano axiom. > Do you understand the concept of a recursive stepwise process with a > starting point, but no ending point? The lack of endpoint is what we > call infinite. Anyway, you are putting the cart before the horse. The _idea_ of a process that one can repeat indefinitely certainly underlies the selection of the fifth Peano axiom. But that's it. The axiom stands on its own. The idea might well be still around, but it is not relevant. > If it ends, it's finite. Inductive proof does not end, Inductive proof consists of two steps, and that is that. And this is _exactly_ the reason why the set of natural numbers is defined in terms of the induction property instead of some indefinite process: because mathematical laws that would require you to check an indefinite number of conditions would be mostly useless as long as you can't systematize the process. The induction axiom provides the necessary shortcut. It _replaces_ a non-terminating process, but does neither require nor perform it. >> The infinite number of successive naturals for which you prove >> your property do not constitute an infinite number of implied >> steps in inductive proof? >> >> No. The inductive proof consists of two steps. Once you have >> covered them, the proof is established for _all_ n. That's what >> induction is all about. >> >> Because you only require three lines for the construction of the >> proof, you think there are only three steps implied? >> >> Yes, exactly that. Exactly that is the purpose of defining the >> naturals by a small set of axioms: being able to deduce general >> properties without having to check each number individually. >> >> Talk about nonsense, and complete bull. >> >> It is called generalization. > > It is called bull. It's like you really don't understand what > inductive proof is doing. It is a complete and exhaustive replacement for the much worse idea of checking every value. And since it is so much more useful, it is _induction_ that has become the _axiom_, not some fuzzy unfinished process. > You memorized all three steps and now you think you're smart. But, > if you can't see that it proves the property true for each n in N, It proves the property true for each n in N, period. > in turn, There is no order implied. > recursively, There is no stepwise process implied. > then I can't imagine what you think is going on. One checks the preconditions for applying the fifth axiom to a property of integers, and if they are met, one has shown the property to hold for all integers. That is what is going on. If you feel like it, you are, of course, free to check a few consecutive cases manually. But that is not a part of the proof. > This is simple denial in the face of an obvious contradiction > between counting an infinite number of times and still having only > finite values. There is nothing like counting an infinite number of times. > Now you want to say induction has no steps, when the step that > gets repeated an infinite number of times is the meat of the proof, No step gets repeated an infinite number of times in induction. Induction checks two conditions, and if they are met, the property holds for all integers. It is a bloody axiom. Any process you want to invent is a construct of your own, and is not relevant to the proof technique. > and in the proof of finiteness, that step involves adding 1. What is > in the water at these institutions, anyway? Ask your guard if the taste offends you. >> The truth for n+1 depends on the truth for n, for all n in N, from 1 >> to oo, an infinite number of times, proving the property true for an >> infinite number of numbers. >> >> It does not depend an infinite number of times. It is just one >> general dependency. And yes, as a general dependency it obviously >> holds for an infinite amount of numbers. But that does not mean >> that I have to check each number individually. > No, of course not. Otherwise it wouldn't be a proof, but > fact-checking. See, you get it if you really try. > Still, you need to be careful about your property in this case, and > keep in mind that inductive proof IS an infinite loop, Too bad. You lost it again. > so that incrementing in the loop creates infinite values, and the > quality of finiteness is not maintained over those infinite > iterations of the loop. Please point out where a loop or process is in the following axiom: A set containing 0, and for each of its elements k also containing its successor S(k), contains all natural numbers. I can switch the order around without changing the logic: A set containing for each of its elements k also the successor S(k), contains all natural numbers as long as it contains 0. See? I can start the statement with the successor stuff without even having specified a starting point or whatever. >> The Peano axioms don't involve steps or any progress or whatever. >> It is utterly irrelevant in what order you determine f(6)->f(7), >> f(3)->f(4). If you have ascertained f(0), and the validity of >> f(n)->f(n+1) by whatever means and in whatever order, the validity of >> f(n) for all n is established. No need for an infinite number of >> steps. > The Peano axioms define an infinite set as a self-referential > recursive loop. No. They might have been inspired by such an idea, but they most certainly and definitely don't define any such thing. See above. > Naturals are defined as the successor of the previously-generated > natural, Wrong. There is one Peano axiom that says if n is a natural, S(n) is also a natural. But this single axiom alone does not define the natural numbers. It takes all five axioms together to do that. > You say we can go in any order, but that's not what axioms say. The axioms don't prescribe an order at all, and so there is no order. > They explicitly talk about the successor to each natural. The successor relation has nothing to do with a succession in time. > You're just spouting defensive nonsense here. That is a nice characterization of what _you_ do. Please come up with any temporal order or process defined in the Peano axiom. And no, that one call a defining relation successor is just picturesque language for the sake of children. It is a name, nothing else. And unless I am mistaken, it was never written out in the original formulation of the axioms, anyway, but was just S(n). >> The whole idea of proof by induction is to prove with two steps a >> statement valid for all elements of N (which happens to be an >> infinite set). >> >>> Basically, adding X repeatedly Y times is the same as adding Y >>> repeatedly X times, so adding 1 and inifnite number of times is the >>> same as adding infinity once. >>> >>> Derive that from the axioms. If you can't, it is irrelevant. >> >> So you contend that X*Y<>Y*X? Interesting. >> >> Wrong. I contend that the Peano axioms neither define nor use >> multiplication, and so multiplication is irrelevant in contexts where >> just the Peano axioms are used. > > So, you can add 1 an infinite number of times and get a finite > value. Also very interesting. You are babbling incoherent nonsense without any apparent relation to what you pretend to be answering. Smokescreen. > Whatever math doesn't suit your case is simply irrelvant. Whatever math is not present in the axioms and definitions is irrelevant. Yes. > Time to grow up, Dave. A fine resolution. Go to, go to. >> Indeed, for _arithmetic_ on natural and whole numbers, there is >> another bag of axioms. You can't expect to be talking about >> multiplication before you have even defined the term. > We both know what multiplication is. Apparently you don't know it. It is not defined in the context of the basic five Peano axioms. You need more axioms (usually associated with the whole numbers) to get it into the picture. Since we are not talking about anything but natural numbers, multiplication does not apply. > Don't you ever get tired of playing dumb? Don't you ever get tired of blaming your deficiencies on others? >>> There is a strictly limited number of steps involved in an >>> induction proof, and it provides proof for all natural numbers, >>> which happen to form an infinite set. >> >> And I guess only three of them are involved in the proof? How do >> you prove it for number x? Does that not depend on x-1? The >> number of inductive steps implied in proving a property true for >> all n in N is infinite. >> >> Here is an example for a proof by induction: >> >> Proposition: >> The number of arrangements of k items is k!, >> where 0!=1, and (n+1)!=(n+1)n!. >> >> Proof: >> There is exactly one way to arrange 0 items, and 0! = 1 by >> definition. >> >> Now let us arrange n+1 items. This can be done by arranging n items >> (which offers us n! possibilities by the induction premise). After >> arranging the n items, there are n+1 possibilities to place the last >> item: for n=0, there is 1 possibility, and for larger n we can >> place the remaining item to the left of the n items, or to the right >> of the n items, or in any of the n-1 positions in between. So we >> have (n+1)n! possibilities for the combined placement. >> >> Finished. I don't need to check for n=7. This is already covered by >> the proof. > Very good. You used an equality as a property. That works every > time. In fact, I don't think I can think of any other case where > this type of proof doesn't work, except for the proof of finiteness > of naturals. It works there as well, since it is based on an axiom. > Then again, finite not be a very good mathematical term to use in > such a proof. Go via finite sets. Theorem: If f(0) = {}, the empty set, and f(n+1) = f(n) U {n+1} for all n, then f(n) for all n can't be put into 1:1 correspondence with a proper subset of it. Proof 1: we prove that f(n) has n members by induction. f(0) has 0 members. f(n+1) has f(n)+1 = n+1 members (this is a shortcut since we have to show that n+1 is not already in the set, but that also can be done by induction). So we need to put n+1 members into correspondent with n members. This is impossible for all n. So the set f(n) is a finite set for all n, and we call all n by assocation also finite numbers, since they indicate the cardinality of a finite set. >>> At each step in the proof, the set has a largest element which >>> is precisely the same as the size of the set. >>> >>> Fine, and so you prove something for a set that has a largest >>> element. The set of natural numbers is no such set, and so your >>> proof does not hold for it. >> >> What the proof shows is that, no matter how large the set gets, >> it never has more members than the number represented by its >> largest member. >> >> For every set that is characterized by a largest member. The set >> of natural numbers is no such set, and so your proof does not hold >> for it. > No largest finite. Shake rattle The proof holds for all n in N. For all sets that end with an n in N. N itself is no such set. > Your refusal to even think about the proof is silly. There is no way > the size of the set can be larger than every element within it. Just sulking does not make it so. > Just think about it a minute and stop playing lawyer-math. Math is about laws, much more rigidly than the legal profession could ever make it. > However large your set, you cannot fill it with only values that are > strictly smaller than that size, or you run out. Just sulking does not make it so. Math requires proof, not conviction. >> Therefore, if the elements are all finite, then the set size >> cannot be infinite, >> >> If the set has a largest member. The set of natural numbers has no >> largest number, and so your reasoning does not hold for it. > > Your reasoning is simply not happening at all. What is this supposed to be? A paraphrase of LA LA LA CAN'T HEAR YOU? >> since that would be larger than the value of every member in the >> set. I am sure I can work out a proof without any reference to >> largest element. >> >> Do so and then come back. > I'll work on it. Fine. Until then, shut up. >>> This constant equality holds for all such sets defined by ANY >>> natural number in the set. >>> >>> Certainly. Unfortunately, the set of natural numbers is no such set >>> (since there is always a larger number than any given number in it), >>> and so your proof does not apply to it. >> >> It applies to ALL n in N. Sorry. Take your complaint to Peano. >> >> Sure does. And that means that it applies to all sets defined by some >> n in N as its last element. The set of natural numbers is no such >> set, since it has no last element. Tough. > > At every step in the proof, the new element is the largest element > and equal to the sizeof the set. You are confusing steps in the proof with some weird iteration scheme again. Anyway, you are always proving something for a finite set. N is no finite set. > At no time can you ever have the set be bigger than all elements, Because you are always talking about finite sets here. N is no finite set. Your proof does not apply to it. > but it is always equal to the last element added. N is no set with a last element. So your proof does not apply to it. > Whatever the size of your set, that IS the upper bound on element > values. This is too obvious. For sets that end at a given n. N is no such set. > Stop playing dumb, with your largest element excuse. It is not an excuse. It is what is relevant. Even though you pretend not to get it. >>> Most generally, it is impossible to have ANY set of natural >>> numbers, finite or infinite, which has a number of elements >>> that is greater than its largest element value. >>> >>> If it has a largest element value. The set of natural numbers >>> has no largest element, and so your proof does not apply to it. >> >> Yes, it does. No set of naturals can contain a greater number of >> elements than all element numbers in the set. >> >> What is this? Proof by whining? The set of positive unit >> fractions has 0 as its lower bound, yet 0 is not a member of that >> set. > What is this? Proof by non-sequitur? This has nothing to do with > what we're talking about. Correct. But since you don't get what we're talking about, maybe you would have understood a similar setting. >>> This should be clear to anyone who thinks about it. It is >>> impossible to have an infinite set of strings, of which >>> digital numbers are a type, without having either an infinite >>> base, or an infinite number of digits. >>> >>> Oh, there is an infinite number of digits, but every single >>> string only occupies a finite number of them. >> >> You mean it only has non-zero values in a finite number of >> digits? >> >> Every single number has non-zero values only in a finite number of >> digits. But there is no finite number of digits that would contain >> _all_ numbers. > Prove it. Assume that a finite number of digits is sufficient to contain all natural numbers. A finite number is characterized by not being able to be put into bijection with a proper subset of itself. A finite number k of digits can assume a number of states 10^k. Now we use the successor relation S(n) to map all possible numbers to a different number. Different numbers have different successors, so we again get 10^k different numbers. However, 0 is not the successor of any number, so our 10^k states have been mapped to just 10^k-1 states if the finite number k would have been sufficient to represent all numbers. >> That's the same as only having a finite number of digits, in which >> case one can only have a finite number of unique digital numbers. >> >> Each number has a finite number of digits, but there is no fixed >> finite number of digits that would contain all numbers. > Then what makes you think they are all finite? How do you have an > infinite set of finite length strings on a finite alphabet, when the > formula is N=S^L. No one seems to want to address this issue. How do > you get infinite N with finite S and L? You don't. Probably nobody wants to address this issue because you are not bothering to define your variables. Guessing their meaning, I'd say your mistake is to assume L is limited. L can take on only finite values, but arbitrarily large ones. In consequence, N takes on arbitrarily large finite values, depending on the value of L. >>> If Peano's fifth is correct, and inductive proof applies to the >>> entire infinite set in a stepwise manner, >>> >>> There is no stepwise manner in the axioms. >> true for n+1, given true for n, for each n in N. steps, an >> infinite number of them. >> >> Where is there a step in the fifth axiom? You are fantasizing >> your own rules. > The recursive statement, the meat of the proof, where it is > demonstrated that f (n)->f(n+1). That is a single step you need to show for all n (with no implied order). Not an infinite number of them. >>> then there are indeed an infinite number of steps >>> implied. It's an immediate consequence, as you said above. >>> >>> It is an immediate consequence of the axioms that the described >>> set is infinite, since the successor relation gives a bijection >>> to a proper subset. But there are no steps involved at all. >>> You just make them up for the sake of your personal intuition. >>> They are a personal crutch of your own, and cause you to make >>> mistakes that are not inherent in the axioms. >> >> Uh huh. And the inductive construction of the naturals using the >> successor operator at each step also involves no steps. >> >> There are no steps, so it does not make sense to talk about at >> each step. The fifth axiom relies on for every n. It is >> irrelevant whether you establish that in any order, forwards, >> backwards, even numbers first, then odds, or simultaneously, so it >> is nonsensical to talk of steps. There is no prescribed order in >> the axioms. > Gee, what was that n doing in there again? What IS that thing > anyway? Maybe we should go ask Gumby and Pokey.... For all n. No order implied, no steps implied. >> Maybe you could use a stepwise crutch so you don't have to crawl >> on your belly. It is amazing that you cannot see that >> incrementing an infinite number of times to generate an infinite >> set of naturals does not result in infinite values. Poincare was >> right. >> >> But there is no incrementing in the axioms. If there is, point it >> out. And certainly not a number of times. > > What did n+1 mean again? I fell down and banged my head, and now I > don't remember..... n->n+1. That is the only step involved here. There is no infinite number of steps or anything like that. >>> and are proven using assumptions that are unfounded, >>> >>> Axioms are not unfounded, and not assumptions. >> >> Excuse me, but axioms are just that, for the most part. They are >> statements assumed to be true for the sake of argument and proof. >> >> But they are not assumed anew for every argument and proof. They >> are cornerstones of the mathematics built upon them. You can >> choose them arbitrarily, but there is a cost of throwing them away >> afterwards, and this cost gets higher as mathematics progresses. >> Sometimes it is still worth to pay the price for abandoning an >> established axiom. But some bumbling bozo not understanding their >> implications is not a worthwhile reason. > > Don't be so hard on yourself. You can probably be made to > understand, if you try hard. Well, I am not the one fantasizing about infinite steps and recursion and all other manner of junk that is not in the axioms. >>> and told that the Banach-Tarski result is a paradox and not >>> a proof by contradiction. There really seems to be a bad >>> influence going on that allows people to think they have an >>> infinite language, when they only allow finite strings, or >>> that the number of paths in a binary tree, which is always >>> half of the number of branches, is suddenly infinitely larger >>> than the number of branches, when those numbers become >>> infinite. >>> >>> Then look up and understand the definition of an infinite set. >>> It is _the_ decisive mark of an infinite set that it does no >>> longer obey the pigeon hole principle. >> >> That's one conception, and one I reject. 1 is 1, whether it's in >> N, or in {1,2,3}. >> >> It is not a conception, it is the bloody _definition_. The word >> is no longer free for the taking. The very least you have to do is >> to _define_ it if you are going to use it with a different than the >> established meaning. Everything else is nonsensical. > > What's nonsensical is pretending you can talk about infinity solely > in terms of counting and get anywhere, and then pretending you got > somewhere, but using axiomatic sleight of hand. You are again babbling incoherent nonsense. > It is certainly possible for some to think about infinite sets > without resorting to bijections. But then they are not thinking about what is called infinite set in mathematics. > That is not the only way to conceive of them, and your insistence > that it is, is what needs correction here. Cantor does not have a > monopoly on infinity. Oh, he does in a manner: his definition was adopted as the most useful one. You are, of course, free to disagree, but then you bloody have to properly define your different terms instead of using established words to mean something different and complaining that people don't follow your usage. >> Yes, spend years trying to qwrap my head around concepts that are >> clearly wrong. What a grand use of my time. >> >> Then just shut up. There is no sense in spewing off about >> something which you do not even plan to understand. > > Apparently I understand it better than you experts, since I see > clearly where is makes foolish assumptions, Look, you can't even bloody see where you make foolish assumptions when one rubs your nose into it twenty times in a row. So you are probably the least qualified person on Earth to spot fallacious assumptions in anybody else's reasoning. > and you can't even follow connections to infinite series or symbolic > systems, with simple formulas provided. You are putting the cart before the horse. The Peano axioms are not bothered about infinite series or symbolic systems. If infinite series or symbolic systems had a problem with the Peano axioms (which they don't), it would be entirely the problem of the infinite series or symbolic systems, and they would have to sort that out. > DO you go and read lots of books on Hitler so you can be sure nazism > is wrong? The only reason I am slogging tis out with you all is that > it's become clear to me just how entrenched and nonsensical > cardinality is, and what its effects are on the mind. It's not > good. Since it has not yet registered on your mind, I'd call that sour grapes. >> I am afraid I WILL have to tear it apart bit by bit to convince >> anyone, but even the most obvious objections to your set of >> naturals are rejected with insults, and no real refutation except >> your own established definitions. >> >> Uh, that's what the definitions are for. Establishing what one is >> talking about. If you want to talk about something different, you >> need to start with defining things differently. If you don't, and >> then people prove you wrong _by_ _the_ _definitions_, then you are >> just wrong, period. > You can't prove things by definition. I did above, a few times. > If I am talking about something different, and you throw definitions > I am not using If you are not using them, then you have no business complaining about their consequences. That's what others are talking about when they are using those words. If you want to talk about something different, invent your own bloody terminology. > as proof that I'm wrong, then I tell you to get your head out of > your axioms and listen to the point I am making. I am afraid that the axioms are more relevant to the matter than any point you try making. >> Don't throw your definitions at me as if they are proofs. They're >> not. >> >> But a proof that can't be traced back to the definitions is not >> sound. You can't prove things if you don't even know what you are >> talking about. > You mean like infinite recursion. Read up a little. Not an axiom. And if you want to make yourself ridiculous in yet another area: I have been working as a programmer since the 70s. >>> Does this seem a little unfair to you? Why do you insist that >>> eveything one can do with a finite number should work exactly >>> the same for an infinite number? >>> >>> There is no infinite natural number. The finite natural numbers >>> are defined by the Peano axioms, and yes, everything that works >>> with finite naturals works with all of them. That is the whole >>> point of defining them with those axioms: to have a body of >>> numbers governed by the same rules. >> >> And why don't you want all sets to have the same rules? if >> infinite sets can have slightly different rules, why can't >> infinite numbers? >> >> Oh, they can. But they are not part of the naturals, because the >> definition of the naturals happens to rule them out. So you need >> to find a different context to talk about infinite numbers rather >> than them being natural numbers. > > Fine. Call them whole numbers. If you insist that all naturals are > finite, that they each differ by at least 1 from all others, and > that there is an infinite set, then i don't believe in natural > numbers, and will henceforth not use the term. Can you live with > infinite WHOLE numbers? You'd have to define them first. >>> If you don't like it, you can invent your own axioms and >>> numbers, but then you can't cry foul that your laws don't hold >>> with the numbers the others are talking about. >> >> Sure, except my numbers are a proper superset of your, so if >> anything holds for all of them, then it holds for yours too. >> >> Feel free to come up with axioms and definitions that meet that >> criterion. Until you do, you'll look less stupid if you shut up. > > Unfortunately for you, I don't mind looking stupid, expecially when > that's in the eye of the beholder, so I don't think I'll shut > up. Why don't you try it, and maybe you'll look like less of a > crybaby. So, you want to talk about some numbers that are not supposed to be the natural numbers and which you are not going to define. And that is going to prove exaxtly what to whom? >>> The determination of divisibility by a number which is prime >>> relative to the number base depends on a termination to the >>> process of division. >>> >>> Good. And since all natural numbers are finite, every division >>> among those naturals is guaranteed to finish. >> >> For finite naturals, yes. >> >> Which are the only ones there are, since the Peano axioms defining >> them leave no room for other naturals. >> >>> Whether the number that is represented by a binary string >>> which represents a subset is a member of that subset is >>> irrelevant. >>> >>> Not if you are establishing a bijection. >> >> What? Now it's my turn to say bull. There is no requirement >> that the subset of N associated with the number x must contain x as >> an element. >> >> No, but it must either contain or not contain x as an element, and >> that is all that is required to establish that you can't biject >> between set and powerset. > > Given any subset denoted by a binary string, and any element in the > set of all whole numbers, one can easily determine whether the > subset contains the whole number denoted by the binary > string. Simply look at the bit at the position corresponding to that > element and see if it's a 1. If this is your criterion, it is met. Uh, no. There is no number mapping to the set of numbers dividable by 3. You seem to have a bad memory. >>>> If you divide this number by 3 (11) you find it is divisible >>>> or not, depending on whether you have an odd or even number of >>>> 100's in your infinite string. Of course, this question is not >>>> really answerable, so I don't have an answer for you. What do >>>> you think? What is aleph_0 mod 3? >>>> >>>> Oh, I never claimed that aleph_0 was a member of the natural >>>> numbers, so I don't need to make claims about aleph_0 mod 3. >>> >>> But the answer to this question depends on the number of digits >>> in the infinite number. >>> >>> Since there are no infinite numbers, I need not answer the >>> question. You claim that there are, so it is up to you to provide >>> an analysis of your claims. >> >> My claim is not that all arithmetic holds for infinite whole >> numbers, but that infinite whole numbers are required in any >> infinite set of whole numbers. >> >> They aren't. You require an unlimited amount of finite numbers, >> but each one of them is finite, even though there is no maximum to >> them. > > Which is entirely impossible if they each differ from each other by > at least some finite amount. Sulking won't make it so. > Then, you have an infinite sum of finite differences, for an overall > infinite range of values, which means some values muct be infinite. You are babbling again. Try to form coherent words. >>> Which is why infinite numbers have not been admitted into the >>> naturals. It would be impractical to have a set of numbers that >>> are not governed by the same laws. >> >> They certainly need to be handled differently, >> >> Yes, and thus they are not natural numbers. Since they can't be >> handled with the Peano axioms. > > They are whole numbers that are required for the infinite > set. Peano's axioms can easily be inverted to produce them. You are babbling again. >> but that's what Cantorians say, unapologetically, about infinite >> sets. >> >> Sure. There is nothing wrong with claiming infinite numbers or >> whatsoever, as long as you don't make the mistake of assuming that >> the Peano axioms or anything else derived from them would extend to >> those numbers. It is perfectly fine to talk about such entities as >> long as one finds axioms that fit them. The Peano axioms don't, >> and so you can't call them natural numbers. > > They do if inverted. One can start at infinity and count down quite > easily. You never get to finite values that way. Besides, you'd have to prove that the inverted axioms are equivalent, and I don't see how you will invert both if n is a natural number, S(n) is a natural number as well as 0 is not equal to S(n) for any natural number n. >> When asked why a certain property is not true of infinite sets, the >> typical answer is, because not everything that holds for finite >> sets must hold for infinite sets, which doesn't answer the question >> at all. So, you get no apologies from me for including infinite >> values in the set of whole numbers (okay not naturals, see?), in >> order for that set to be infinite. >> >> Unfortunately, the moniker whole numbers is already taken as well. >> Call them TO numbers, and devise a coherent set of axioms for them, >> and nobody will complain. > > Can I quote you on that? Let's see how your prediction pans out: > > > 1. oo and 0 are numbers > > 2. If x is a number, the successor of x is a number. > > 3. oo and 0 are not successors of any number. > > 4. Two numbers of which the successors are equal are themselves equal. > > 5. (induction axiom.) If a set T of numbers contains oo and 0, and also the > successor of every number in T, then every number is in T. > > > Any complaints? Well, then your numbers basically consist of the union of the disjoint sets {0, S(0), S(S(0))...} and {oo, S(oo), S(S(oo))...} with both of the sets being isomorphic to N. It is a pretty pointless structure, but not yet invalid. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum stephen@nomail.com said: >> > Tony has no clue what mathematics is, nor how it is done, so he doesn't > normally bother with definitions. The closest we got from him for a > definition of finite was that a finite number is less than an > infinite one. And you can guess the definition of infinite. >>Well, that's about as close to a lie as one can get, eh? > >>I asked for a definition of infinite, and no one could give me a >>definition of that word. The best I could get was that an infinite >>set can have a bijection with a proper subset, which is hardly a >>definition of the word infinite. > > On the contrary, that's a perfectly good definition of the concept > infinite set. > > These seems to be another common misconception among > the anti-Cantorians that words cannot have specific > meanings in specific contexts. Somehow they > think an all-encompassing definition of 'infinite' must > be provided before someone can say what an infinite set is. > I am not sure what they mental hangup is. I wonder > how any of them would ever learn a foreign language. > > What is so hard about > A set is infinite if there exists a bijection between > the set and a proper subset of the set. > ? There is no need to get all mystical and metaphysical > just because the word 'infinite' shows up. The word 'infinite' > may have other definitions in other contexts, but that is > true of all words, and is irrelevant in a discussion of > infinite sets. > > Stephen > The problem comes whenever someone wants to use the word infinite in any other context. There ARE no infinite numbers, only infinite sets, in your lexicon. Talk about needing to learn about language. What is so hard about infinite numbers, or trying to define the word itself, independent of bijection genuflection? -- Smiles, Tony > The problem comes whenever someone wants to use the word infinite > in any other context. There ARE no infinite numbers, only infinite > sets, in your lexicon. There are no infinite natural numbers, but there are plenty of infinite 'cardinal numbers', infinitely many of them, at least in ZFC. > Talk about needing to learn about language. What is so hard about > infinite numbers, or trying to define the word itself, independent of > bijection genuflection? We already have cardinalities, which are sufficient to our needs. Let's see TO's definition of an infinite number. > The problem comes whenever someone wants to use the word infinite in any > other context. The problem comes when people use a word in one context as if its meaning were specified by a different context. That's what an amphiboly is. > There ARE no infinite numbers, only infinite sets, in your > lexicon. Not only do we know about infinite numbers, but there are at least two different kinds of them: cardinal and ordinal ones. (Still other notions of infinite number occur in other contexts.) Again, clear definition of what is meant in a particular context is required. > Talk about needing to learn about language. What is so hard about > infinite numbers, or trying to define the word itself, independent of bijection > genuflection? What is so hard about the idea of context sensitivity, and accepting that the word 'infinite' can mean different things in different contexts? Does it bother you than 'bread' can refer to products made from different grains, or can even refer to 'money', depending on the context? Virgil said: > > Virgil said: > > If TO's assumprtions were actually the case, there would have to be a > finite natural so large that adding 1 to it would produce an infinite > natural. But TO cannot produce either a largest finite nor a smallest > infinite, so the set of all finite naturals is already big enough. > > We have been through all this before. You lay these requirement on me, but > when > I say you cannot have a smallest infinite omega, or omega-1 would be finite, > you say omega-1=omega. > > I do not say anything at all about omega. I merely go by the Peano > properties, which prohibit TO's vision of infinite naturals. > > > So, I can play your stupid game, and say that alpha is > the largest finite, but alpha+1=alpha. Tada! I am as senseless as you! Isn't > that great? Let me know when you want to get off the largest finite kick. I > won't be responding to it any more. > > It is not me who keeps requiring a largest finite, but TO, by his > delusional insistence that there must be non-finite naturals within the > Peano system. > While it may seem that counting ala Peano can never produce any infinite numbers, the insistence on the set being infinite requires the existence of infinite values in the set. Perhaps you should call your set boundless or bigfinite or something, but it's not infinite if you only count a finite number of times and only have finite values. You know, the Peano axioms can easily be inverted to produce infinite whole numbers that count down, with exact symmetry compared to the finite end of the number circle. If one uses such a set of axioms, does that mean that finite whole numbers cannot exist, because that set of axioms doesn't seem to allow us to count down that far? This insistence on determining the dividing point between finite and infinite is clearly a waste of time, and your repetitions of this non-point don't make it any more important. You can't count through the divide in any finite number of steps. So what? That doesn't mean that one side exists and the other doesn't. Infinite set sizes ARE infinite whole numbers. I can't understand why this isn't clear to everyone. -- Smiles, Tony > So, I can play your stupid game, and say that alpha is > the largest finite, but alpha+1=alpha. Tada! I am as senseless as you! > Isn't > that great? Let me know when you want to get off the largest finite > kick. I > won't be responding to it any more. > > It is not me who keeps requiring a largest finite, but TO, by his > delusional insistence that there must be non-finite naturals within the > Peano system. > > While it may seem that counting ala Peano can never produce any infinite > numbers, the insistence on the set being infinite requires the existence of > infinite values in the set. Perhaps you should call your set boundless or > bigfinite or something, but it's not infinite if you only count a finite > number of times and only have finite values. TO must be having a different definition of an infinite set than the rest of us. The mapping n -> n+1 on N clearly injects N into a proper subset of itself without ever mapping any finite member of N onto any infinite member of N. So the finite members of N are enough. > You know, the Peano axioms can > easily be inverted to produce infinite whole numbers that count down, with > exact symmetry compared to the finite end of the number circle. If one uses > such a set of axioms, does that mean that finite whole numbers cannot exist, > because that set of axioms doesn't seem to allow us to count down that far? > This insistence on determining the dividing point between finite and infinite > is clearly a waste of time, and your repetitions of this non-point don't make > it any more important. You can't count through the divide in any finite > number > of steps. So what? Since there is no step from finite naturals to infinite naturals, and all stepping is done one step at a time, from some n to n+1, there is no getting from finite naturals to anything else by one-stepping. That doesn't mean that one side exists and the other > doesn't. Infinite set sizes ARE infinite whole numbers. I can't > understand why this isn't clear to everyone. Because everyone else reserves the phrase whole numbers for finite numbers. Whatever TO wants to call them, it should not be some word or phrase that already has another meaning. The common term is 'cardinals'. !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > Virgil said: >> >> It is not me who keeps requiring a largest finite, but TO, by his >> delusional insistence that there must be non-finite naturals within >> the Peano system. >> > While it may seem that counting ala Peano can never produce any > infinite numbers, It does not seem so. It is provable from the axioms quite easily. > the insistence on the set being infinite requires the existence of > infinite values in the set. No. Sulking won't make it so. > Perhaps you should call your set boundless or bigfinite or > something, but it's not infinite It is infinite if it can be bijected to a proper subset of it, and the successor relation does that. > if you only count a finite number of times You can only count finite numbers, but there is no finite limit to how large they may be, even though every single one of them is finite. > and only have finite values. You know, the Peano axioms can easily > be inverted Different axioms, different game. Irrelevant. If you invert the Peano axioms, you get a rule every number has a predecessor, and of course this means that the set we get can't have 0 as a member, since it then would also need -1. > to produce infinite whole numbers that count down, with exact > symmetry compared to the finite end of the number circle. Problem is that then you'll never reach 0, since there is no predecessor operation that will magically cross from the infinite to the finite. > If one uses such a set of axioms, does that mean that finite whole > numbers cannot exist, because that set of axioms doesn't seem to > allow us to count down that far? Of course not. But the axioms for natural numbers don't count down, so your silliness does not apply. > This insistence on determining the dividing point between finite and > infinite is clearly a waste of time, and your repetitions of this > non-point don't make it any more important. You can't count through > the divide in any finite number of steps. So what? That doesn't mean > that one side exists and the other doesn't. It means that the other side is not part of natural numbers. > Infinite set sizes ARE infinite whole numbers. Whatever you want to call them: they are not natural numbers as determined by the Peano axioms. > I can't understand why this isn't clear to everyone. Because the natural numbers are defined in a way that does not admit the size of the set of natural numbers into the set itself. That does not mean that the set size does not exist or is not a number or whatever else. But it means that it is not a natural number. Wherever you want to go looking for it, it is useless looking _in_ the natural numbers for it. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum Virgil said: > > Virgil said: > > > There is a simple, demonstrably valid proof of Cantor's Theorem > in ZF set theory. So you must think the proof is unsound. > Which axiom of ZF do you believe to be false? > > Chris Menzel > > > I was asked that before, and never got around to fully analyzing > the axioms for lack of time, but the diagonal proof suffers from > the fatal flaw of assuming that the diaginal traversal actually > covers all the numbers in the list. > > If it misses any of them, it must miss a first one. Which is the > first one misssed? > > If there is not a first one missed, then none are missed. > > The first one missed is the first one directly below the diagonal > traversal. > > > If any are missed then there must be an n in N corresponding to that > first one missed. Why? > > If TO cannot produce such a finite n from N, then no such n exists, and > that anti-diagonal is complete. That is non-logic. You might as well say fsirnlskrnlsnglk QED. > -- Smiles, Tony > Virgil said: > > Virgil said: > > > There is a simple, demonstrably valid proof of Cantor's Theorem > in ZF set theory. So you must think the proof is unsound. > Which axiom of ZF do you believe to be false? > > Chris Menzel > > > I was asked that before, and never got around to fully analyzing > the axioms for lack of time, but the diagonal proof suffers from > the fatal flaw of assuming that the diaginal traversal actually > covers all the numbers in the list. > > If it misses any of them, it must miss a first one. Which is the > first one misssed? > > If there is not a first one missed, then none are missed. > > The first one missed is the first one directly below the diagonal > traversal. > > > If any are missed then there must be an n in N corresponding to that > first one missed. > Why? Because in ZF and ZFC and other similar axiom systems, the set of naturals is well-ordered, meaning that every non-empty subset contains a smallest member, so the set of n in N for which f(n) is 'missed' is either empty or has a smallest (first) member. > > If TO cannot produce such a finite n from N, then no such n exists, and > that anti-diagonal is complete. > That is non-logic. You might as well say fsirnlskrnlsnglk QED. TO must at least prove that such a value exists in order to successfully claim one exists, even if he cannot give a value for it. So far TO has only claimed, and not proved, that any such number exists. Considering TO's track record on claims, that's just not good enough. > Daryl McCullough said: > > I offered, and you saw, a deductive proof that proves that the > largest natural in a set must be at least as large as the set size. > > Yes. So we have: If S is a set of natural numbers, and S has a largest member N, > then N >= the cardinality of S. > > How do you prove that every set of natural numbers has a largest > member? > > -- > Daryl McCullough > Ithaca, NY > > > [Annotating with * and # for clarity] > > Essentially, the proof shows that no set can have a larger number[*] of naturals > in it than the values[#] of all the naturals in it. > > What does it mean to compare a number [*] with multiple values[#]? > Let's consider the set {1, 2, 3}. You appear to be saying that the > number [*] of naturals in this set (which I hope we agree is three) > cannot be larger than the values[#] of all the naturals in it. I > don't quite understand this comparison - I don't think you intend to > refer to the sum of the values[?], so perhaps you mean one at a time. > Well, there are only three cases: let's look at them. > > First case: the element 1. Oh dear, 3>1, so plainly the set can have > more naturals in it than this first value. > > Second case: the element 2. Oh dear, 3>2, so plainly the set can have > more naturals in it than this second value. > > Third case: the element 3. Ah! 3 is not more than 3, so in this one > case only, your claim is indeed true. > > One out of three. Not enough to pass, I'm afraid. But what about that > one out of three: which one was it? Aha! It was 3, and 3 is the largest > element in the set. So here's a revised claim: > > Essentially, the proof ought to show that a set _can_ have a larger > number[*] of naturals in it than any value of a natural in it that is > not the largest member. Uh, no. Incorrect. The fact is that any set of whole numbers greater than or equal to 1 MUST have at least one element with a value at least equal to the size of the set. That is a fact and has been proven. > > It seems to me this gets three out of three. (Infinity out of infinity, > even ^_^) > > I will devise a proof without > a largest element, if need be, but that largest element argument is a waste > of time. I still don't see how you can say that for all finite sets this is > true, but that one can get an infinite set, and still have all finite numbers. > If each finite n in N is the size of the set including all m<=n, then each of > them corresponds to a finite set. How do we get an infinite set, then, if m<=n > is finite for any finite n in N? > > Because this set goes on and on, from one finite number to the next, > each one totally 100% finite, because it is one more than the previous > one, and this set goes on and on and on, and this going on and on never > stops, ever, never gets to an end because there isn't one, as Wolf > Kirchmeir's grandchild could tell you. I bet if Wolf asked his grandchild what number you would get if you added 1 forever, he or she would say infinity. You, on the other hand, despite the knowledge of infinite series, seem to believe you can add an infinite number of 1's and NOT get infinity. Really quite fascinating, this theory of yours.... > > > Brian Chandler > http://imaginatorium.org > > -- Smiles, Tony > The fact is that any set of whole numbers greater than or equal to 1 > MUST have at least one element with a value at least equal to the > size of the set. That is a fact and has been proven. That is not a fact, and cannot be proven, without some additional assumption, such as that the set is finite or, equivalently, that it contains a maximal member. If everything that is true about finite sets had to carry over to infinite sets, infinite sets would be finite, too. !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > The fact is that any set of whole numbers greater than or equal to 1 > MUST have at least one element with a value at least equal to the > size of the set. That is a fact and has been proven. Sulking won't make it so. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum >The fact is that any set of whole numbers greater than or equal to 1 MUST have >at least one element with a value at least equal to the size of the set. That >is a fact and has been proven. No, it's not a fact, and no, it hasn't been proven. You are assuming that the size of every set of naturals must be a natural. What is the basis for believing that? What definition of size are you using? -- Daryl McCullough Ithaca, NY <854qaq8u5i.fsf@lola.goethe.zz> > The fact is that any set of whole numbers Correction: any finite set of whole numbers > greater than or equal to 1 MUST have > at least one element with a value at least equal to the size of the set. That > is a fact for finite sets > and has been proven. by induction on finite sets. - Randy Daryl McCullough said: > >> Tony has no clue what mathematics is, nor how it is done, so he doesn't >> normally bother with definitions. The closest we got from him for a >> definition of finite was that a finite number is less than an >> infinite one. And you can guess the definition of infinite. >Well, that's about as close to a lie as one can get, eh? > >I asked for a definition of infinite, and no one could give me a >definition of that word. The best I could get was that an infinite >set can have a bijection with a proper subset, which is hardly a >definition of the word infinite. > > On the contrary, that's a perfectly good definition of the concept > infinite set. > > -- > Daryl McCullough > Ithaca, NY > > But not the word infinite on its own. Do you think the dictionary has the word set in the definition of infinite? Are sets the only way to think about infinity? Not hardly. -- Smiles, Tony > Daryl McCullough said: > >> Tony has no clue what mathematics is, nor how it is done, so he doesn't >> normally bother with definitions. The closest we got from him for a >> definition of finite was that a finite number is less than an >> infinite one. And you can guess the definition of infinite. >Well, that's about as close to a lie as one can get, eh? > >I asked for a definition of infinite, and no one could give me a >definition of that word. The best I could get was that an infinite >set can have a bijection with a proper subset, which is hardly a >definition of the word infinite. > > On the contrary, that's a perfectly good definition of the concept > infinite set. > > -- > Daryl McCullough > Ithaca, NY > > > But not the word infinite on its own. Do you think the dictionary has the > word set in the definition of infinite? Are sets the only way to think > about infinity? Not hardly. We have also given a meaning to finite natural as an n in N for which the set {m in N: m <= n} is a finite set. This, of course, means that there are no infinite naturals, which TO does not like, but it is our definition, not his. For functions f:N -> R, we say that they converge to the finite limit L in R if and only if for every positive real epsilon (however small) the set {n in N: | f(n) - L | > epsilon } is finite. For functions f:N -> R, we say that they diverge to positive infinity if and only if for every positive real epsilon (however large) the set {n in N: f(n) < epsilon } is finite. For functions f:N -> R, we say that they diverge to negative infinity if and only if for every positive real epsilon (however large) the set {n in N: f(n) > -epsilon } is finite. Dictionaries, unless they are mathematical dictionaries, are not commonly precise about the mathemaical usage of words. The meaning of infinite, in mathematics, is context sensitive. There is no meaning that is universal. Give us a precise context, and we can give a precise definition. >Daryl McCullough said: >>I asked for a definition of infinite, and no one could give me a >>definition of that word. The best I could get was that an infinite >>set can have a bijection with a proper subset, which is hardly a >>definition of the word infinite. >> >> On the contrary, that's a perfectly good definition of the concept >> infinite set. >But not the word infinite on its own. There is no need for such a definition. Words are never used on their own, they are used in *sentences*. The role of a definition is to understand the meaning of *sentences* involving a word. Saying a set S is infinite if there is a bijection f between S and a proper subset of S is all that you need to know about the meanings of sentences about infinite sets. It doesn't tell you what infinite natural means, but that's actually because it is a meaningless phrase. -- Daryl McCullough Ithaca, NY Daryl McCullough said: > Tony Orlow said: > >Daryl McCullough said: > >> Yes, that's exactly what the diagonal argument proves. There is >> no list of length aleph_0 that contains all real numbers. > >Okay, that I can agree with, at least in digital terms, which is what the proof >relies on. > > That's all that Cantor's proof shows, and it's all then anyone > has ever claimed that it shows. > >What it really shows is that digital systems with a given number of digits have >more strings than digits. it is not necessary to have aleph_0 digits, if you >allow for smaller infinities. You only need SOME infinite number of digits. > > That's not true. If S is an infinite set of strings, then there > is a difference between (1) There is no finite bound on > the lengths of strings in S. (2) There is a string in S that is > infinite. Yes, I understand the difference between those two statements, and in this case the two are equivalent. If the length of strings is L and the symbol set has a finite size of S, then you have S^L strings, which is infinite IF AND ONLY IF L is infinite. Infinite Set <-> Infinite Element. This is not the case for the real numbers, because they do not have any minimum finite difference between them. That is why you can have [0,1), an infinite set with all finite numbers. You cannot have such a set with integers or strings. > > that you are mixing up the order of quantifiers: > > (1) forall b, exists s in S, > (if b is a finite bound, then length(s) > b) > > (2) exists s in S, forall b > (if b is a finite bound, then length(s) > b) > > Statement (1) says that the *set* S has no finite bound. > Statement (2) says that S contains an *element* that has > no finite bound. Those are two different statements. of quantifier dyslexia. It's because folks here refuse to see what is abundantly clear and obvious, no matter how carefully it is explained, or how many times. It's Cantorian dyslexia. This point could not be more clear, and yet you seem almost to insist that S^L can be infinite with finite S and L, and that you can increment a value an infinite number of times and still have a finite value. Can this really be the position of the professional mathematical community? Do you really not understand what if and only if means? > > -- > Daryl McCullough > Ithaca, NY > > -- Smiles, Tony > Daryl McCullough said: > Tony Orlow said: > That's not true. If S is an infinite set of strings, then there is > a difference between (1) There is no finite bound on the lengths of > strings in S. (2) There is a string in S that is infinite. > > Yes, I understand the difference between those two statements, and in > this case the two are equivalent. TO is off on his insane hobbyhorse of infinite in the sense ofendless meaning that one must have an end to the endless. That is like saying that because time is endless, at least as far as we can tell, that time will eventually reach infinity. > If the length of strings is L and > the symbol set has a finite size of S, then you have S^L strings, > which is infinite IF AND ONLY IF L is infinite. Same false assumption that an infinite sequence must contain its limit. If a non-empty set of digit strings is finite, it will contain a longest string. Just line them up by length and take the last one. TO's claim requires that at some point adding one more character to a finite string makes it actually infinite. > Infinite Set <-> Infinite Element. TO <-> WRONG! > you are mixing up the order of quantifiers: > > (1) forall b, exists s in S, > (if b is a finite bound, then length(s) > b) > > (2) exists s in S, forall b > (if b is a finite bound, then length(s) > b) > > Statement (1) says that the *set* S has no finite bound. Statement > (2) says that S contains an *element* that has no finite bound. > Those are two different statements. > You need something to break you out of your delusions, TO. > I see now why WM is > constantly accused of quantifier dyslexia. Because he cannot see the difference between for all x there is a y such that ... and there is a y such that for all x ... > It's because folks here > refuse to see what is abundantly clear and obvious, no matter how > carefully it is explained, or how many times. It's Cantorian > dyslexia. This point could not be more clear, and yet you seem almost > to insist that S^L can be infinite with finite S and L, No one is saying anything like that. What we ARE saying is that neither S nor L need be bounded by any finite value, so that S^L need not be bounded by any finite value either. Perhaps if TO ever bothered to read what we actually say, he would not misrepresent us so. > and that you > can increment a value an infinite number of times and still have a > finite value. Again a misrepresentation! What we actaully say is that there is no finite number of iterations at which we must stop, and so there is no finite limit on the size of naturals. Given any finite limit, there are naturals larger than that finite limit. > Can this really be the position of the professional > mathematical community? My statement is, TO's misrepresentation is not. Do you really not understand what if and > only if means? A good deal better than TO does, at least on the evidence he presents. Tony Orlow said: Yes, Tony. That is exactly what you need. You seem to be trying to learn some elementary facts about logic and mathematics by arguing on the internet. That isn't a very efficient way for a beginner to learn. If you're at Cornell, then go take a course in logic. It's an excellent institution. -- Daryl McCullough Ithaca, NY Tony Orlow said: >Daryl McCullough said: >> That's not true. If S is an infinite set of strings, then there >> is a difference between (1) There is no finite bound on >> the lengths of strings in S. (2) There is a string in S that is >> infinite. > >Yes, I understand the difference between those two statements, and >in this case the two are equivalent. Good. >If the length of strings is L and the symbol set has a >finite size of S, then you have S^L strings, which is infinite >IF AND ONLY IF L is infinite. Infinite Set <-> Infinite Element. You've just contradicted yourself. I thought you agreed that there is a difference between the two sentences (1) There is no finite bound on the lengths of strings in S. (2) There is a string in S that is infinite in length. In case (1), the set is infinite, but there is no infinite element. You seem to be thinking that there is some L characterizing the lengths of strings. There isn't. There are strings of length 1, there are strings of length 2. There are strings of length 3. Each string has a length, but there is no length associated with the set of finite strings as a *whole*. >> that you are mixing up the order of quantifiers: >> >> (1) forall b, exists s in S, >> (if b is a finite bound, then length(s) > b) >> >> (2) exists s in S, forall b >> (if b is a finite bound, then length(s) > b) >> >> Statement (1) says that the *set* S has no finite bound. >> Statement (2) says that S contains an *element* that has >> no finite bound. Those are two different statements. > Yes, Tony, you certainly do. Since you are at Cornell, there are a number of courses you could take to fill the gaps in your knowledge. To learn about logic, I suggest starting with Math 281: Deductive Logic. To actually get up to speed on the theory of natural numbers, you need to take MATH 481 Mathematical Logic. Or you could just take advantage of the excellent faculty. Go to Dexter Kozen or Anil Nerode or Richard Shore and ask them about your claim that an infinite set must contain an infinite object. (But please don't tell them that I put you up to it.) -- Daryl McCullough Ithaca, NY Sorry, I've been away..... Daryl McCullough said: > >Essentially, the proof shows that no set can have a larger number >of naturals in it than the values of all the naturals in it. > > What does the number of naturals in a set mean? The size of the set, the quantity of elements included in the set, each of which is a unique natural number. > >If each finite n in N is the size of the set including all m<=n, then each of >them corresponds to a finite set. > > Right. Each natural number n corresponds to a finite set: the set > of all natural numbers less than n. > >How do we get an infinite set, then, if m<=n is finite for any finite >n in N? > > You get an infinite set by (1) Pick some starting number a. > (2) Pick an operation f(x) that, given a number n, returns a new > number that is greater than n. (3) Then form the set > > { a, f(a), f(f(a)), ... } > > That's guaranteed to be infinite. Meaning it goes on forever? If it goes on forever, for an infinite number of iterations, each time incrementing the value of the next element (assuming your f() is successor/increment), then the value of the next element will become infinite. Therefore, if the set contains an infinite number of elements, it will contain elements of infinite value. If it doesn't include any elements of infinite value, then the set cannot be infinite. > > -- > Daryl McCullough > Ithaca, NY > > -- Smiles, Tony > Sorry, I've been away..... Daryl McCullough said: > > >How do we get an infinite set, then, if m<=n is finite for any >finite n in N? > > You get an infinite set by (1) Pick some starting number a. (2) > Pick an operation f(x) that, given a number n, returns a new number > that is greater than n. (3) Then form the set > > { a, f(a), f(f(a)), ... } > > That's guaranteed to be infinite. > Meaning it goes on forever? If it goes on forever, for an infinite > number of iterations, each time incrementing the value of the next > element (assuming your f() is successor/increment), then the value of > the next element will become infinite. If it becomes infinite by operating one step at a time, there must be a step at which it becomes infinite. Which step is that, TO? On the other hand, if infinite here only means unending, there is no problem, as the sequence merely does not ever end. >Daryl McCullough said: >>How do we get an infinite set, then, if m<=n is finite for any finite >>n in N? >> >> You get an infinite set by (1) Pick some starting number a. >> (2) Pick an operation f(x) that, given a number n, returns a new >> number that is greater than n. (3) Then form the set >> >> { a, f(a), f(f(a)), ... } >> >> That's guaranteed to be infinite. >Meaning it goes on forever? Meaning that if you start enumerating the elements of the set (say one per second), you will never finish. >If it goes on forever, for an infinite number of >iterations, each time incrementing the value of the >next element (assuming your f() is successor/increment), >then the value of the next element will become >infinite. I'm telling you that a certain process never completes. You are saying that *after* it completes, the next element will be infinite. That's completely nonsensical. You seem to be thinking that *first* you generate all the finite elements, and *then* you start on the infinite elements. But the first step (generate the finite elements) *never* completes. -- Daryl McCullough Ithaca, NY > There is no room for a bit of error in the tenth place if you are > factoring primes. I can factor primes without looking at any of their places. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > >> There is no room for a bit of error in the tenth place if you are >> factoring primes. > > I can factor primes without looking at any of their places. Give me half the bucks of Bill Gates, and I won't mind repeating his mathematical blunders. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum > >> Clearly, it has never crossed their minds that such a nice relationship >> between topology and calculus could possibly esists. > > Apart from trivialities like de Rham cohomology and > the Atiyah-Singer Index theorem, anyway. Allright, I've been fishing. Now give me some fish: references please. Han de Bruijn > Clearly, it has never crossed their minds that such a nice relationship > between topology and calculus could possibly esists. >> Apart from trivialities like de Rham cohomology and >> the Atiyah-Singer Index theorem, anyway. > Allright, I've been fishing. Now give me some fish: references please. http://en.wikipedia.org/wiki/De_Rham_cohomology http://en.wikipedia.org/wiki/Atiyah-Singer_index_theorem and references therein. > >> > >> Clearly, it has never crossed their minds that such a nice relationship >> between topology and calculus could possibly esists. > > Apart from trivialities like de Rham cohomology and > the Atiyah-Singer Index theorem, anyway. >> >> Allright, I've been fishing. Now give me some fish: references please. > > http://en.wikipedia.org/wiki/De_Rham_cohomology > http://en.wikipedia.org/wiki/Atiyah-Singer_index_theorem > > and references therein. > But, as I suspected, these don't compare with my remarkably _simple_ result. Which nevertheless went unnoticed by mainstream mathematics. Han de Bruijn > Clearly, it has never crossed their minds that such a nice > relationship > between topology and calculus could possibly esists. >> Apart from trivialities like de Rham cohomology and >> the Atiyah-Singer Index theorem, anyway. > But, as I suspected, these don't compare with my remarkably _simple_ > result. Which nevertheless went unnoticed by mainstream mathematics. I was referring you to large bodies of mathematics which relate topology and calculus. If you want a simple one, then the *usual* integral to find the winding number is already well-known to compute the crossing number. But the complaint that something is a large body of well-developed mathematics is hardly strengthening the claim that mainstream mathematics is ignorant of nice relationships between calculus and topology. > >> > >> > >> Clearly, it has never crossed their minds that such a nice >> relationship >> between topology and calculus could possibly esists. > > Apart from trivialities like de Rham cohomology and > the Atiyah-Singer Index theorem, anyway. >> >> But, as I suspected, these don't compare with my remarkably _simple_ >> result. Which nevertheless went unnoticed by mainstream mathematics. > > I was referring you to large bodies of mathematics which > relate topology and calculus. If you want a simple one, > then the *usual* integral to find the winding number > is already well-known to compute the crossing number. But my awkward integral to find the crossing number is NOT well known. I find this very strange, if not unbelievable. > But the complaint that something is a large body of > well-developed mathematics is hardly strengthening the > claim that mainstream mathematics is ignorant of > nice relationships between calculus and topology. Somewhat overdone, I admit. But nevertheless ... Han de Bruijn === Subject: Re: Update: Objections to Cantor's Theory <3kcbnlFtkksnU1@individual.net> <22610$42e102a1$82a1e3ad$8186@news1.tudelft.nl> <42E2EE05.D37761E0@osu.edu> > > > Now, you can say 'I don't like the axiom of infinity', > and work in what's left. That's your choice. But including > the axiom of infinity and then using the existence of > a bijection between two sets to say that they have the > same number of elements works well: for those of us > who are happy with the axiom of infinity and wish to > be able to use infinite sets, there is no competitor > that I know of. At least, none that makes as much sense. > > Oh yeah ? Let's see. > > Theorem > ------- > The number of even naturals is half the number of all naturals. > > [...] > > I've written out a rather long-winded counter-example to your > result above. My point is not just that your system is > inconsistent, though. I am trying to illustrate why more complex > notions of set size are susceptible to developing inconsistencies > to which the simpler Cantorian notion is provably immune. Let's just look and see. > I'm going to make some assertions that I think you > would agree with. Please believe that they are a good faith > effort to see things from your point of view. I suspect > you will not like the conclusion, though. Uhm, not _like_ the conclusion .. > a) The number of naturals is equal to the number of decimal > string representations of the naturals: > Card({ 0, 1, 2, 3, ... }) = > Card({ '0', '1', '2', '3', ... }) I'll go along with you, but on the condition that you don't throw away a crucial step in my argument, namely that the above set is actually _finite_: assume the size of this set is finite and N . > b) The number of decimal string representations of the naturals > is equal to the number of binary string representations: > Card({ '0', '1', '2', '3', ... }) = > Card({ '0', '1', '10', '11', ... }) > > c) The number of binary string representations of the naturals > is equal to the number of binary string representations of the > naturals with a '0' appended on the right of each string: > Card({ '0', '1', '10', '11', ... }) = > Card({ '00', '10', '100', '110', ... }) > > **** D is probably the step you're most likely to object to. No. > d) The set of binary string representations of the naturals > with a '0' appended on the right of each string actually /is/ > the set of binary string representations of the even naturals, > and so their number is the same. > Card({ '00', '10', '100', '110', ... }) = > Card({ '00', '10', '100', '110', ... }) > > e) The number of binary string representations of the even > naturals is equal to the number of decimal representations of > the even naturals: > Card({ '00', '10', '100', '110', ... }) = > Card({ '0', '2', '4', '6', ... }) > > f) The number of decimal representations of the even naturals > is equal to the number of even naturals: > Card({ '0', '2', '4', '6', ... }) = > Card({ 0, 2, 4, 6, ... }) > > g) By steps a, b, c, d, e and f, the number of naturals is > equal to the number of even naturals. Apart from a minor refinement. See below. [ ... rest deleted ... ] The number of (all) naturals in the set { 0, 1, 2, 3, ... , N-1 } is equal to the number of (even) naturals in the set { 0, 2, 4, 6, ... , 2(N-1) } The cardinality of both these sets is N. So far so good. But Cantorians say more. They say that the second set is a _subset_ of the first. But this is not true, if you go along with me in my finitary reasoning. Only part of the second set, namely { 0, 2, 4, 6, ... , N-1 } or { 0, 2, 4, 6, ... , N-2 } is a subset of { 0, 1, 2, 3, ... , N-1 } . Now take the limit for N -> oo and everything is just fine. You will not like the conclusion, though. :-( Han de Bruijn === Subject: Re: Update: Objections to Cantor's Theory > >> >> Abstractions are useful fictions. When we create fictions, we are >> free to assign to them any laws of logic we may wish, but if we >> want them to be useful, we are restricted. Maybe I'm not explaining >> things especially well, but you are not even trying to understand. > > > There is no iron-clad logical law reqiring the fictions to be useful. > The requirement of usefulness is not a mathematical requirement. Only > one thing counts --- logical consistency. G.H. Hardy would have loved you for saying that. I really think that you have been brainwashed by mid-20th century formalism. Even if mathematicians say they agree with you, their behavior reveals that they think otherwise. === Subject: Re: Update: Objections to Cantor's Theory > Well, a lot of people are reasonably happy with the > cumulative hierarchy view of what kind of sets the > axioms of ZF(C) are supposed to be. ZFC hasn't even been around 100 years. I don't think that we can say that it has stood the test of time. > > |For example, in my opinion, the replacement > |axiom is completely unnecessary for almost all useful mathematics - > |personally I would ditch it. > > It's true that it seldom is needed, but it is sometimes > used. I believe Borel determinacy uses it. It allows > you to prove the consistency of Z. It's equivalent to > saying that the levels of the cumulative hierarchy > include all ordinals, and not just ordinals up to > omega+omega. If you ordinarily don't care about sets > up there, that's fine, but it's different from saying > there's any actual problem with their existence. But you probably would agree that if replacement were found to be inconsistent, this would effect less than 1% of mathematics in an average math university math library. > Perhaps more important, it's another one of those odd > situations where mathematicians are able to use the axiom > correctly better than they're able to distinguish when > they're using it and when they're not. This also occurs > with the axiom of choice. If we took it out, we'd have > the slightly silly situation where a mathematician would > define a sequence of objects A1,A2,A3,..., and then do a > little double-take in order to ask whether they accidentally > used the axiom of replacement there. Usually, what they > did turns out not to need the axiom of replacment, but > why bother teaching them how to tell the difference? > For the time being, I don't see any tangible benefit to > avoiding the axiom of replacement. Yes, for the time being, I don't think that replacement does any harm. When it does hit the dust, it will be easy to remedy most of these issues. But by then we will have completely reformulated the foundations math, so it won't matter anyway. And of course this whole matter will seem completely irrelevent to most working mathematicians. Also, I really think that AC is far more reasonable than replacement. (And yes, I do know that I am making many many unproven assertions here.) > Moreover, ZFC wasn't derived on some kind of a priori basis. > The original version-- I forget which of the axioms it had-- > was produced in order to list the principles required for a > proof of the well-ordering theorem that had been troubling > some people. This approach seems to have succeeded better > than trying to intuit principles entirely a priori. > > So I think the idea that these axioms correspond to > *a* natural way to do set theory isn't so far from true. I wasn't aware of this. In my opinion this certainly does make ZF more natural. (For example, isn't replacement needed to obtain what I think is called Hartogg's Lemma? Isn't this needed to show that well-ordering is equivalant to AC?) > > Incidentally, it's the power set axiom that considered > problematic for the limitation of size philosophy. The > axiom of replacement says that some things that have a > one-to-one relationship with the members of a set (hence > not bigger) are members of a set. But the power set axiom > says that all of some more open-ended type of thing are > members of some set. My guess is that it is replacement that will prove more problematical than the powerset axiom. Or maybe the two in combination. For example, I am somewhat sure that you do need replacement to construct the set union_{n=0}^infty P^n(N) where P^n(N) denotes the powerset operation applied n times to the natural numbers. > | Now I > |have put forward my opinion elsewhere that I personally think that > |Cantor's diagonal argument really does mean something. But I do not > |think that anyone who disagrees with me is necessarily a crackpot. > > I agree, but there's a very long list of opinions that I > think are in much greater need of being taken seriously > first. That is a matter of personal values and priorities. There are some people I know (none I think on this newsgroup) who think that arguing about set theory on sci.math is a waste of time. Stephen === Subject: Re: Update: Objections to Cantor's Theory > >>See? Nothing else is needed than the classical, pre-Cantorian concept >>of a limit, which is accepted by all kind of mathematicians I know of. >> >> Sure. However, it leaves you completely unable to compare the cardinality >> of these two sets >> >> {1, 2, 3, 4, .... } and { one, two, three, four, ..... } >> >> or, for that matter >> >> {1, 2, 3, 4, .... } and { 1, 2, watermelon, 4, ..... } >> >> without adding more axioms. > >Sure. But that's quite another question. Doesn't it seem unfortunate that your definition of cardinality doesn't let you compare most sets (without invoking an as yet unspecified number of additional rules) and the definition that you reject has no such limitations? Alan -- Defendit numerus === Subject: Re: Update: Objections to Cantor's Theory >> For simplicity, assume the even natural numbers are in a list. > > OK, here's my list: > 2 > 4 > 6 > 8 > 10 > ... > I should have stated that I include 0 in the natural numbers. So the list of all even natural numbers would be: 0 2 4 6 8 ... >> I will generate an even number not on the list using Cantor's >> diagonal argument. >> Call this missing number CMN (Cantor's missing number). >> >> Take the first even number in the list. >> If the least significant (first) digit is equal to 0 then >> the first digit of the CMN is 2, else the first digit >> of the CMN is 0. > > OK, so the first digit of the CMN is 0. > > >> Take the second even number. >> If the second position is this number is 0 then >> the second position in the CMN is 2, else the >> CMN's second position is 0. > > OK, so the second digit is 2. > > >> Continue this process for the entire list of even >> natural numbers. > > OK, so the CMN is 0222222... I defined the first digit to be the least significant digit. So, the CMN for all even natural numbers in order would be: ...2222 > >> The CMN will be an even natural number not on the list. > > The CMN is not a natural number. Why do you say CMN is not a natural number? We started with a list of natural numbers. Each position in the list is indexed by a natural number. I only set digits of CMN in positions that are indexed by natural numbers. I never said CMN would be a small natural number. CMN was a natural number after I set the first digit. It is still a natural number after I set the second digit. Exactly when did CMN stop being a natural number? Are you suggesting there is a maximum number of digits that a natural number can have? While you decide how many digits a natural number can have, I will give another proof where CMN is easily proven to be a natural number. I will prove the set of natural numbers is uncountable. Assume we have a bijection from N to N. For simplicity, assume there is a list of natural numbers. Start by setting CMN to 0. Take the ith number in the list. If this ith number is equal to or larger than CMN then set CMN to the successor of the ith number. After we have examined every number in the list, CMN will be equal to a natural number not in the list. It is easy to show that CMN is a natural number. CMN is either 0 or the successor of some number in the list. 0 is a natural number and, by definition, any successor of a natural number is a natural number. Every number in the list is a natural number. Therefore, CMN is a natural number. Russell - 2 many 2 count === Subject: Re: Update: Objections to Cantor's Theory <8764v3a2jb.fsf@phiwumbda.org> >Only one thing counts --- logical consistency. But how do you *know* that? Did God tell you that? === Subject: Re: Update: Objections to Cantor's Theory <1aHEe.176365$x96.112796@attbi_s72> <25VEe.205424$nG6.192354@attbi_s22> >If you accept the constructive version of Cantor's diagonal argument, >then as far as I am can see you accept the meat of the argument. *** The anti-Cantorians claim that while infinite sets and power sets of those infinite sets are undeniably useful abstractions, what Cantor did was to take an argument (the diagonalization argument), which is perfectly valid in concrete mathematics, and wrecklessly apply it to the abstractions of infinity, ultimately producing garbage. *** === Subject: Re: Update: Objections to Cantor's Theory > > >>If you accept the constructive version of Cantor's diagonal argument, >>then as far as I am can see you accept the meat of the argument. > > > > *** > The anti-Cantorians claim that while infinite sets and > power sets of those infinite sets are undeniably useful > abstractions, what Cantor did was to take an argument > (the diagonalization argument), which is perfectly valid > in concrete mathematics, and wrecklessly apply it to > the abstractions of infinity, ultimately producing garbage. > *** Yes, and if you read my answers to your original emails, I basically made the point that this is a reasonable point of view. But honestly, when I have seen anti-Cantorian arguments on sci.math, they haven't been at this level. As far as I am concerned, your only mistakes were to call yourself anti-Cantorian, and to identify with the other anti-Cantorian crackpots on sci.math. And to be fair to me, I don't think that you responded in a manner that would clarify my misunderstanding (but then also to be fair to you, perhaps my replies were not totally professional.) Stephen === Subject: Re: Update: Objections to Cantor's Theory > > > What is absurd about there not existing a bijection between > a set and its powerset? > > The view is that one can be constructed. karl m > > But the view is not justified! The bijection as a meaningful tool to > measure set-sizes is no even laid down in the axioms. > > The C in ZFC specifically allows construction of injections/bijections > between sets. For any set A, there is a function f on A such that f(x) belongs to x for x belonging to A and being not empty. There is not the least faint glimmer of a definition, not even of function. Neither is bijection proved to be of any meaning with respect to measure the cardinality of infinite sets. You see: Without tangible mathematics, there is no chance to apply set theory. But if it does not fit into their dream world the use of tangible mathematics is refused by set theorists. Then only the axioms are allowed. These axioms alone would not even lead to any uncountable set. Anyhow, there is nothing in the axioms of standard set theory that justifies the use of a bijection. === Subject: Re: Update: Objections to Cantor's Theory > There is not the least faint glimmer of a definition, not even of > function. Neither is bijection proved to be of any meaning with > respect to measure the cardinality of infinite sets. > > You see: Without tangible mathematics, there is no chance to apply set > theory. But if it does not fit into their dream world the use of > tangible mathematics is refused by set theorists. Then only the > axioms are allowed. These axioms alone would not even lead to any > uncountable set. > > Anyhow, there is nothing in the axioms of standard set theory that > justifies the use of a bijection. 'function', 'bijection', and 'cardinality' are given precise definitions in set theory. What set theory book are you using? If you don't find definitions, then you need to check the book for pages that have been torn out. MoeBlee === Subject: Re: Update: Objections to Cantor's Theory > Set theory is a disease from which mathematics > will one day recover (Poincare) > > > > Poincare never said that. You will not be able to give a valid > reference for this. > > There is no actual infinity. The Catorians have forgotten that and have > fallen into contradictons. [H. Poincar.8e, Les math.8ematiques et la > logique III, Rev. m.8etaphys. morale 14, p. 316, (1906).] > > Poincare never said that. And now it is that in just this abominable mummy, as Kant is, Monsieur Poincar.8e felt quite enarmoured, if he is not bewitched by him. So I understand quite well the opposition of Mons. Poincar.8e, by which I felt myself honoured, so he never had in his mind to honour me, as I am sure. If he perhaps expect, that I will answer him for defending myself, he is certainly in great a mistake. Why then did Cantor (in a letter to Russell, Sept. 19. 1911) say that? And here for those who are able to read German: Wie bisher r.9fhre ich keinen Finge zu meiner Vertheidigung wider die seit Jahren fortgesetzten boshaften Angriffe Poincar.8e's. Die letzte Attaque desselben hat meine Augen nur noch sehender gemacht in Bezug auf die D.9frftigkeit, Oberfl.8achlichkeit, das Schaumschlagen, die zu Grunde liegende gemeine Gesinnung dieses Gelichters, das sich einbildet, an der Spitze der Wissenschaft zu stehen und ihr Vorschriften machen zu k.9annen, wie sie sich in Zukunft zu gestalten habe. Cantor to Hilbert, June. 24, 1908 === Subject: Re: Update: Objections to Cantor's Theory Obscurity, linux) > >> >> Set theory is a disease from which mathematics >> will one day recover (Poincare) >> >> >> >> Poincare never said that. You will not be able to give a valid >> reference for this. >> >> There is no actual infinity. The Catorians have forgotten that and have >> fallen into contradictons. [H. Poincar.8e, Les math.8ematiques et la >> logique III, Rev. m.8etaphys. morale 14, p. 316, (1906).] > > Poincare never said that. His original text is at www.ac-nancy-metz.fr/enseign/philo/textesph/default.htm *Il n'y a pas d'infini actuel*; les Cantoriens l'ont oubli.8e, et ils sont tomb.8es dans la contradiction. -- Alan Smaill === Subject: Re: Update: Objections to Cantor's Theory >|The axioms of ZF are, in my opinion, even worse. This is because we >are >|not quite sure what sets are really supposed to be, and so we are kind > >|of groping in the dark. > >Well, a lot of people are reasonably happy with the >cumulative hierarchy view of what kind of sets the >axioms of ZF(C) are supposed to be. > >|For example, in my opinion, the replacement >|axiom is completely unnecessary for almost all useful mathematics - >|personally I would ditch it. > >It's true that it seldom is needed, but it is sometimes >used. I believe Borel determinacy uses it. It allows >you to prove the consistency of Z. It's equivalent to >saying that the levels of the cumulative hierarchy >include all ordinals, and not just ordinals up to >omega+omega. If you ordinarily don't care about sets >up there, that's fine, but it's different from saying >there's any actual problem with their existence. > >Perhaps more important, it's another one of those odd >situations where mathematicians are able to use the axiom >correctly better than they're able to distinguish when >they're using it and when they're not. This also occurs >with the axiom of choice. If we took it out, we'd have >the slightly silly situation where a mathematician would >define a sequence of objects A1,A2,A3,..., and then do a >little double-take in order to ask whether they accidentally >used the axiom of replacement there. Usually, what they >did turns out not to need the axiom of replacment, but >why bother teaching them how to tell the difference? >For the time being, I don't see any tangible benefit to >avoiding the axiom of replacement. > >| The history of math also reveals this >|difficulty - at the turn of the century a very bright mathematician, >|axioms on the basis of what he intuitively knew a set should be. Then > >|Bertrand Russell told him of an elementary contradiction that >destroyed >|his whole scheme. So Russell and Whitehead (also extremely bright >|people) developed a very contrived system of set theory in their >efforts >|to get around this. Only decades later did we get ZF and NBG axiom >|systems, that today we just take for granted are completely natural. >(I >|have also read a book by Quine giving yet another axiom scheme, but I >|think that it is more enjoyed by logicians than mathematicians.) > >Someone pointed out that Aristotle had already indicated that >existing thing shouldn't be regarded as a genus. Frege's >system suffered from proceeding as if everything could be >regarded as one big domain. > >Frege has a comment in his preface where he says that he's >arranged his development in such a logical way that if the >reader has a qualm with some result in it (to paraphrase), >it can be traced back to the axioms. He also mentions the >axiom that turned out to be self-contradictory as being the >one with the greatest chance of failure (although he was >still pretty confident of it). > >Before Frege was confronted with his contradiction, however, >Cantor had already developed a point of view that didn't >suffer from the paradox. It's sometimes described as the >limitation of size principle. I'm not sure it's known how >he came to feel that it was important not to treat classes >that are too big (like everything) as sets, whether it >was by knowing of some paradox or what. In any case, this >fix hasn't been shown inadequate by any of the later >developments in the theory. > >Moreover, ZFC wasn't derived on some kind of a priori basis. >The original version-- I forget which of the axioms it had-- >was produced in order to list the principles required for a >proof of the well-ordering theorem that had been troubling >some people. This approach seems to have succeeded better >than trying to intuit principles entirely a priori. > >So I think the idea that these axioms correspond to >*a* natural way to do set theory isn't so far from true. > >Incidentally, it's the power set axiom that considered >problematic for the limitation of size philosophy. The >axiom of replacement says that some things that have a >one-to-one relationship with the members of a set (hence >not bigger) are members of a set. But the power set axiom >says that all of some more open-ended type of thing are >members of some set. > >|It is beyond dispute that Cantor's diagonal argument can be proved in >|ZF. But it is not beyond dispute whether this means anything. > >ZF is overkill. One just wants enough to do real analysis, >essentially axioms for the integers and the reals. Treating >the argument axiomatically is unlikely to be helpful to >anybody who doesn't find it convincing to begin with. > >| Now I >|have put forward my opinion elsewhere that I personally think that >|Cantor's diagonal argument really does mean something. But I do not >|think that anyone who disagrees with me is necessarily a crackpot. > >I agree, but there's a very long list of opinions that I >think are in much greater need of being taken seriously >first. Haha, very true. > >Keith Ramsay But this discussion I found very insightful. quasi === Subject: Re: Update: Objections to Cantor's Theory [ .. later .. ] OK, Jim. This is a very nice writeup. But I'll think about it first. Han de Bruijn === Subject: Re: Update: Objections to Cantor's Theory <1aHEe.176365$x96.112796@attbi_s72> > On 24 Jul 2005 03:56:12 -0700, Ross A. Finlayson > > ... >quasi? Uh... you're a moron. > ... > You're a frickin moron, ... > ... > and you're a frickin idiot. > ... >Oh, by the way, we have better theories already, ya little >know-nothing-yet. > ... > > reference? > > ... >Hey, you're a moron, idiot, etcetera, various appelations of low >intelligence and intellectual character, but hey, at least you're >pseudonymous, ya frickin moron. > ... > > haha > > quasi A lot of people study the foundations of mathematics, and for most the reasons are quite individual: the reflection into the reasons why something can be said to be true and provable can serve as a psychological underpinning for rationality, salve for curiousity, or competitive advantage in arguments about truth in the foundations of mathematics. Where you've been reading these newsgroups, unmoderated and widely distributed discussion fora open to basically all comers, it might seem obvious to you as it is to me that over the past years there is some change on focus of what the foundations of mathematical logic are and there has been reevaluation by many of some of the critical results upon which a variety of basically standard statements about the roots of mathematical logic rest. For some it is not a reevaluation but an initial exposure to various contentious issues that frame the large part of the contemporary arguments about the foundations of mathematical logic and what they mean and so on. Not so long ago, these various alternatives to basically set theory were not available nor thus profferred. It was quite personally entertaining to lambast you as mathematical cranks, morons, and idiots, I confess, and I'm glad you were able to see humor in that. The point is that in mathematical logic, in are a variety of reasons to consider what are currently alternatives to the etablished schools of foundations. So anyways, the universe is infinite and infinite sets are equivalent. There are true, logically sound reasons to say that. That is thus an objection to Cantor's powerset result and the transfinite cardinals and their arithmetic and so on, but it would be correspondingly false or unacceptably ignorant to deny what may be true within those arguments. Thus, the reevaluation leads to the necessity of the consideration of why those things are each true, in particular to the solution of what otherwise would be logical paradoxes. There plainly are reasons, and reason, to reject some wrong statements of foundations of mathematical logic. Hey, the infinite: it's always one more. That's pretty funny. Ross === Subject: Re: Update: Objections to Cantor's Theory > I just mentioned a set that one wants to have that doesn't seem > avaiable without the axiom of replacement. It seems that in order to > prove that, even for a countable set x, the set of all tuples of > members of x is a set, one needs the axiom schema of replacement. Put > another way, it seems one needs the axiom schema of replacement to > prove the existence of this set: > > {y | En(n is a natural number & y e A^n)} where 'A^n' stands for 'the > Cartesian n-product of A', i.e., the set of all n-tuples of members of > A. Well the reason I think that I can do this without replacement is that before I read a real book on set theory and logic, I perused through Halmos's Naive Set Theory book, and for some reason never noticed the axiom of replacement. So I used to do all my math without it, never even knowing it existed. Anyway, I think your set can be expressed as a subset of the power set of NxA (here N = natural numbers, x is cross product). > > If I'm not mistaken, it's proven that if Z is consistent, then ZF is > consistent. (Is that not right?) Since V_omega is a model for Z, you can prove Z consistent in ZF. If you can also prove Z consistent implies ZF consistent, then you can prove ZF consistent in ZF. By Godel's Theorem, ZF will then by inconsistent. === Subject: Re: Update: Objections to Cantor's Theory > [Ross Finlayson insults quasi] > > |I know that quasi is not a theist, but even he, I think, will > appreciate > |the words of Proverbs 9:7-9 > | > | Whoever corrects a mocker invites insult; > | whoever rebukes a wicked man incurs abuse. > ... > > Answer a fool as his folly deserves, > Lest he be wise in his own eyes. > > Proverbs 26:5 > > This wisdom stuff can be very tricky. :-) Especially in light of Proverbs 26:4 Do not answer a fool according to his folly, Lest you also be like him. === Subject: Re: Update: Objections to Cantor's Theory > mueckenh@rz.fh-augsburg.de says... > > >> You've got that completely backwards. The one thing that (classical) >> mathematicians insist on is that logic works the same regardless >> of whether the domain is naturals, reals, infinite sets, or whatever. > >Then they should agree that there cannot exist more irrationals than >rationals in the real continuum, because, in normal order <, there does >never exist a pair of irrational numbers without a rational number >between them. There is no logic available to circumvent this fact. > > What *logical* claim are you saying that mathematicians are violating? > Betweenness, order, rational, irrational, are *mathematical* > concepts, not logical concepts. Yes, of course, the *mathematics* > of infinite sets is different from the mathematics of finite sets. > But the *logic* is the same. > > The logical operators, to refresh your memory are: and > or > implies > not > exists > forall > > What *logical* statement do you think mathematicians are > saying is true for finite sets but not infinite sets? If, in a linear order, between any pair of irrationals at least one rational is existing, this fact *implies* *forall* cases that there *exist* at least as many rationals as irrationals. Here only basic mathematical and logical notions have been used. === Subject: Re: Update: Objections to Cantor's Theory > mueckenh@rz.fh-augsburg.de says... > > >> You've got that completely backwards. The one thing that (classical) >> mathematicians insist on is that logic works the same regardless >> of whether the domain is naturals, reals, infinite sets, or whatever. > >Then they should agree that there cannot exist more irrationals than >rationals in the real continuum, because, in normal order <, there does >never exist a pair of irrational numbers without a rational number >between them. There is no logic available to circumvent this fact. > > What *logical* claim are you saying that mathematicians are violating? > Betweenness, order, rational, irrational, are *mathematical* > concepts, not logical concepts. Yes, of course, the *mathematics* > of infinite sets is different from the mathematics of finite sets. > But the *logic* is the same. > > The logical operators, to refresh your memory are: and > or > implies > not > exists > forall > > What *logical* statement do you think mathematicians are > saying is true for finite sets but not infinite sets? > > If, in a linear order, between any pair of irrationals at least one > rational is existing, this fact *implies* *forall* cases that there > *exist* at least as many rationals as irrationals. Not in ZFC. === Subject: Re: Update: Objections to Cantor's Theory > If, in a linear order, between any pair of irrationals at least one > rational is existing, this fact *implies* *forall* cases that there > *exist* at least as many rationals as irrationals. Saying it doesn't make it so. === Subject: Re: Update: Objections to Cantor's Theory <3kcbnlFtkksnU1@individual.netThe bijection as a meaningful tool to > measure set-sizes is no even laid down in the axioms. It is nothing but > a thoughtless extrapolation from the finite into the infinite. > ^^^^^^^^^^^ > > You mis-spelled 'sensible'. > > Of course it's an extrapolation. We're trying to develop something > beyond what our intuition readily handles. The only strategy > we have is to take something that works where we know what > the answer ought to be, and try to generalise it. This > generalisation also has many properties that we'd like to > see: the Schroeder-Bernstein theorem, for example. > > Now, you can say 'I don't like the axiom of infinity', > and work in what's left. That's your choice. But including > the axiom of infinity and then using the existence of > a bijection between two sets to say that they have the > same number of elements works well: for those of us > who are happy with the axiom of infinity and wish to > be able to use infinite sets, there is no competitor > that I know of. The bijection rests on the foundation of induction: If I can map n on 2n, then I can map n+1 on 2(n+1) and so define that bijection for all elements of the sets. The same induction proves that every set of even numbers contains numbers which are larger than the number of elements of this set. Both conclusion are valid for *all* natural (even) numbers, including the whole sets (if they exist). Or both are not valid. But set theorists insist that the first case is correct whereas the second is not. That is inconsequent. === Subject: Re: Update: Objections to Cantor's Theory > The bijection rests on the foundation of induction: If I can map n on > 2n, then I can map n+1 on 2(n+1) and so define that bijection for all > elements of the sets. The same induction proves that every set of even > numbers contains numbers which are larger than the number of elements > of this set. WM's claim that every set of even numbers contains numbers which are larger than the number of elements of this set is only valid for sets for which the number of elements is a natural number. This is not true of sets which of naturals which do not have a largest member, as the number of members is not, in this case, a natural number. And the set of even naturals does not have a largest member so that WM's claim cannot be true for it. > > Both conclusion are valid for *all* natural (even) numbers, including > the whole sets (if they exist). WM's claim is only true for sets whose cardinality is a natural number. The set of all even naturals is not one of them. Or both are not valid. WRONG, AGAIN! > But set > theorists insist that the first case is correct whereas the second is > not. And, within the axiom system in which they are operating, they are right. === Subject: Re: Update: Objections to Cantor's Theory <7bb9e$42e0b7eb$82a1e3ad$6138@news2.tudelft.nl> > > anti-Cantorian > > Read: mathematical crackpot > > There are no other anti-Cantorians. > > I do not know. I would say that Kronecker was as anti-Cantorian as you > can get. On the other hand, he did not reject Cantorianism on the > grounds that it was invalid, inconsistent, or whatever, I don't know what predominates in Cantor's theory - philosophy or theology, but I am sure that there is no mathematics there (Kronecker, quoted from the web). > but on the > grounds that in his opinion mathematicians should only operate with > finite numbers and with a finite numbers of operations. So in his > opinion transcendental numbers did not exist. (Note the contrast with > Mueckenheim who is of the opinion that irrational numbers do not exist. > For Kronecker, algebraic numbers did exist, that is a field were his > major contributions are.) Und ich glaube auch, da¤ es dereinst gelingen wird ... (namentlich die Hinzunahme der irrationalen und kontinuierlichen Gr.9a¤en) wieder abzustreifen, welche zumeist durch die Anwendungen auf Geometrie und Mechanik veranla¤t worden sind. Kronecker, Werke III, p. 253. === Subject: Re: Update: Objections to Cantor's Theory > I don't know what predominates in Cantor's theory - philosophy or > theology, but I am sure that there is no mathematics there (Kronecker, > quoted from the web). The same is even more obviously true of WM. === Subject: Re: Update: Objections to Cantor's Theory Nntp-Posting-Host: hera.cwi.nl ... > Not a prediction, it is an axiom. > > I'll look through the Peano axioms again, and then I'll look through > the axioms of ZFC, and then ... Oh, well, sometimes it is an axiom, sometimes it is a theorem, depends on how you look at it; it is an axiom in ring theory. You think a theorem is a prediction? On the other hand as an axiom in ring theory, it is true (the distributive law) if you are working in a ring, in a non-ring it is not necessarily true. > > Here's another that hasn't actually been proved yet. Calculate > > one of the complex roots of the Reimann Zeta function. Then > > the prediction is that the real part of that root will be 1/2, to > > within the accuracy of your calculation. > > You thoroughly misstate the Riemann hypothesis. The term to within > the accuracy of your calculation is not part of the hypothesis. > > What I have been saying is that mathematical statements can be > interpreted as statements making predictions about the results of > computational experiments. Even though that is not the way the > Riemann Hypothesis is usually stated, it could be restated to make > predictions. But the Riemann hypothesis is not more than what the name states. It is an hypothesis. And until the truth has been established (and it has been shown to be actually a theorem), mathematics do not use the hypothesis, unless it is explicitly stated that under the Riemann hypothesis, such and so is valid). It is *not* a prediction. > But, is it testable? > > Yes, once we have ascertained that the roots of the Zeta function can > actually be computed to arbitrary accuracy. (i.e. that's not actually > part of the Riemann Hypothesis) Once we have ascertained that the roots of the Zeta function can actually be computed to arbitrary accuracy, you still have not a necessarily testable hypothesis. The hypothesis states that it is true for *all* roots, the only thing that you can establish with testing is that the hypothesis is false, not that it is true. On the other hand, you can test it for each individual zero. But for that you do not need the zero in arbitrary precision. The best way to go is that in the neighbourhood of the zero, and with some specific function, that function is positive at one side and negative on the other side. I will not go into details, but that was about the procedure when the first 1.5*10^9 zeros were found to be lying on the critical line. -- 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: Update: Objections to Cantor's Theory > You think a theorem is a prediction? A theorem can be interpreted as making predictions about the results of computations. I really didn't understand the rest of your post, but let me try to clarify things. A statement makes predictions. Those predictions may be wrong. When we have a proof for the statement, then we are entitled to say that mathematical theory makes the prediction. So the Riemann Hypothesis, together with the knowledge that the roots of the Zeta function really are computable, makes predictions about what will be observed when we perform the computations (the prediction being that the real part will equal one half, to within the accuracy of the computation). The predictions may be wrong; we don't have a proof. === Subject: Re: Update: Objections to Cantor's Theory >> Oh yeah ? Let's see. >> >> Theorem >> ------- >> The number of even naturals is half the number of all naturals. > > By much the same argument, we note that the sum of all even > naturals is strictly less than the sum of all naturals. > But unfortunately, is is also twice the sum of > all naturals. > > That's what I mean by not making as much sense. Please show us this much the same argument. Challenge us to find a hole in it. Han de Bruijn === Subject: Re: Update: Objections to Cantor's Theory > > > > Oh yeah ? Let's see. > > Theorem > ------- > The number of even naturals is half the number of all naturals. >> >> >> By much the same argument, we note that the sum of all even >> naturals is strictly less than the sum of all naturals. >> But unfortunately, is is also twice the sum of >> all naturals. >> >> That's what I mean by not making as much sense. > > > Please show us this much the same argument. Challenge us to find > a hole in it. Consider the sequence of naturals 1...N. Add up the whole sequence. Add up the even ones. Take the ratio as N tends to infinity. You get 1/2. So the sum of the even numbers is half the sum of the naturals. But 2+4+6+8+... = 2(1+2+3=4+5+...) = twice the sum of all natural numbers. So depending on how you calculate it, the sum of the even naturals is half, or twice the sume of all naturals. I don't see this as any worse than your argument for the number of evens vs the number of naturals: it does, however, expose the problem of reasoning about quantities (such as the number, or the sum of all even numbers) which haven't been defined. === Subject: Re: Update: Objections to Cantor's Theory > >> Please show us this much the same argument. Challenge us to find >> a hole in it. > > Consider the sequence of naturals 1...N. > > Add up the whole sequence. 1 + 2 + 3 + ... + N = N.(N+1)/2 > Add up the even ones. Distinguish the cases where N is even: 2 + 4 + 6 + ... N = 2.N/2.(N/2+1)/2 = N/2.(N/2+1) And where N is odd, thus (N-1) is even: 2 + 4 + 6 + ... (N-1) = 2.(N-1)/2.((N-1)/2 + 1) > Take the ratio as N tends to infinity. You get > 1/2. So the sum of the even numbers is half the > sum of the naturals. Take the even case for convenience: N/2.(N/2+1) 1/2 + 1/N ----------- = ---------- = 1/2 as N -> oo N/2.(N+1) 1 + 1/N Agreed. > But > > 2+4+6+8+... = 2(1+2+3+4+5+...) = twice the sum of all natural numbers. > > So depending on how you calculate it, the sum of the > even naturals is half, or twice the sume of all naturals. Hey, hey! That is not true, of course: 2 + 4 + 6 + ... + N = 2.(1 + 2 + 3 + ... N/2) ^^^ And not: 2 + 4 + 6 + ... + N = 2.(1 + 2 + 3 + ... N) So the tricky part comes with your all of the natural numbers. No further comment needed, I suppose? Han de Bruijn === Subject: Re: Update: Objections to Cantor's Theory <3kcbnlFtkksnU1@individual.net> <22610$42e102a1$82a1e3ad$8186@news1.tudelft.nl> <3ki71jFuggabU3@individual.net> <3kk3ufFu5ageU1@individual.net> <66a88$42e4f4b6$82a1e3ad$4079@news1.tudelft.nl> > > >> Please show us this much the same argument. Challenge us to find >> a hole in it. > > Consider the sequence of naturals 1...N. > > Add up the whole sequence. > > 1 + 2 + 3 + ... + N = N.(N+1)/2 What meaning does this have? What is N+1? Is it equal to N? What is N*(N+1)? What makes you think the finite formula is valid for a sum up to N? It is proved by induction, which only covers finite numbers. N is not finite and is not a natural number. What does a sum up to N even mean? The sequence 1, 2, 3, ... will never end at anything. It certainly won't ever arrive at N. - Randy === Subject: Re: Update: Objections to Cantor's Theory > >> > >>Please show us this much the same argument. Challenge us to find >>a hole in it. > >Consider the sequence of naturals 1...N. > >Add up the whole sequence. >> >>1 + 2 + 3 + ... + N = N.(N+1)/2 > > What meaning does this have? What is N+1? Is it equal > to N? What is N*(N+1)? > > What makes you think the finite formula is valid for > a sum up to N? It is proved by induction, which only > covers finite numbers. N is not finite and is not a > natural number. If you have followed the thread, then you know that N _is_ finite and _is_ a natural number here. > What does a sum up to N even mean? The sequence > 1, 2, 3, ... will never end at anything. It certainly > won't ever arrive at N. Han de Bruijn === Subject: Re: Update: Objections to Cantor's Theory <3kcbnlFtkksnU1@individual.net> <22610$42e102a1$82a1e3ad$8186@news1.tudelft.nl> <3ki71jFuggabU3@individual.net> <3kk3ufFu5ageU1@individual.net> <66a88$42e4f4b6$82a1e3ad$4079@news1.tudelft.nl> > If you have followed the thread, then you know that N _is_ finite > and _is_ a natural number here. OK, mea culpa. I thought you were doing Tony Orlow math. - Randy === Subject: Re: Update: Objections to Cantor's Theory >> 2+4+6+8+... = 2(1+2+3+4+5+...) = twice the sum of all natural numbers. > Hey, hey! That is not true, of course: > > 2 + 4 + 6 + ... + N = 2.(1 + 2 + 3 + ... N/2) > ^^^ > And not: > > 2 + 4 + 6 + ... + N = 2.(1 + 2 + 3 + ... N) > > So the tricky part comes with your all of the natural numbers. > > No further comment needed, I suppose? So adding up 1+2+3+...+[N/2] with N ranging over all naturals gives a different answer from adding up 1+2+3...+N with N ranging over all naturals. Quite a lot of further comment needed, I suppose. === Subject: Re: Update: Objections to Cantor's Theory > So adding up 1+2+3+...+[N/2] with N ranging over > all naturals gives a different answer from adding > up 1+2+3...+N with N ranging over all naturals. Yep. But don't forget the division part. > Quite a lot of further comment needed, I suppose. Can somebody send this guy to a course on calculus and teach him how to work with limits? Maybe all Cantorians need such a course? Han de Bruijn === Subject: Re: Update: Objections to Cantor's Theory > >> So adding up 1+2+3+...+[N/2] with N ranging over >> all naturals gives a different answer from adding >> up 1+2+3...+N with N ranging over all naturals. > > > Yep. But don't forget the division part. > >> Quite a lot of further comment needed, I suppose. > > > Can somebody send this guy to a course on calculus and teach him > how to work with limits? Ipse dixit. === Subject: Re: Update: Objections to Cantor's Theory > One objection I have about what he has done, is that I can reorder the > set, and get any rational limit I want strictly between 0 and 1. Sure. That's why I used the sequence, not the set. But, does the sequence 1,2,..,N not converge to the set of all naturals for N -> oo ? Han de Bruijn === Subject: Re: Update: Objections to Cantor's Theory >>Did I suggest somewhere that I would consider what I have proved >>here as neat stuff ? Have you got the message anyway ? > > What message? According to your logic, there are the same number > of primes as perfect squares, and the same number of > composites as natural numbers. However I suspect that > you do not agree with the conclusions of your own logic. I think that you are asking for something different. With my (not so) neat method, it's impossible to come up with results that you find worthwile. That's true. Actually what you want is something like the following (I think): > The density of the perfect squares within the natural integers is > NOT infinitesimal - as I understand it - but equal to 1/sqrt(x) . > And the density of the primes - as I understand it - is 1/ln(x) > approximately (: Prime Number Theorem). Here x is a sufficiently > large natural number. These densities become zero (infinitesimal) > for x -> oo . > > So, indeed, the primes are more numerous than the squares, because > ln(x) < sqrt(x) ==> 1/ln(x) > 1/sqrt(x) for x > 2 . > > O yeah, I find that the density of the cubes goes like x^(-2/3) . But the above implies densities which are defined _locally_, while the proof of my even = all div 2 theorem is only a spitting image of Cantor's cardinalities theory (: got it?). The latter doesn't need a more complicated proof mechanism than the one presented. Han de Bruijn === Subject: Re: Update: Objections to Cantor's Theory >Did I suggest somewhere that I would consider what I have proved >here as neat stuff ? Have you got the message anyway ? >> >> What message? According to your logic, there are the same number >> of primes as perfect squares, and the same number of >> composites as natural numbers. However I suspect that >> you do not agree with the conclusions of your own logic. > I think that you are asking for something different. With my (not so) > neat method, it's impossible to come up with results that you find > worthwile. That's true. Actually what you want is something like the > following (I think): You are not paying attention. With your method you can prove that the number of perfect squares equals 0 times the number of natural numers. You can also prove that the number of primes equals 0 times the natural numbers. Those are inescapable conclusions of your logic. Now does 0 times the number of natural numbers equal 0? Does 0 times the number of natural numbers equal 0 times the number of natural numbers? Who knows, as this is your logic, and despite your claim, you need to define some arithmetic for you infinite numbers. > But the above implies densities which are defined _locally_, while > the proof of my even = all div 2 theorem is only a spitting image of > Cantor's cardinalities theory (: got it?). The latter doesn't need a > more complicated proof mechanism than the one presented. Your proof has nothing to do with Cantor's cardinality. You are the one who does not get it. You presented a proof that used nothing but limits. I presented similar proofs using identical arguments to conclude that there are 0 primes and 0 perfect squares. You must accept my proof if you accept yours, as they are the same. Stephen === Subject: Re: Update: Objections to Cantor's Theory > >Can you name a testable prediction made by mathematics? >> >>A few. >> >>Take three numbers a,b,c, and perform the following two >>calculations: a*(b+c) and a*b + a*c. Mathematics predicts >>that the results will be the same. > > Well, do it on a computer, and you'll find that this result does not > transfer to the world of computation. That's true. Because an idealization must be materialized first. The computer does this by adding a bit of sloppyness. Then you will find that your result _does_ transfer to the world of computation, quite good enough, that is. Han de Bruijn === Subject: Re: Update: Objections to Cantor's Theory !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > >> >>Can you name a testable prediction made by mathematics? > >A few. > >Take three numbers a,b,c, and perform the following two >calculations: a*(b+c) and a*b + a*c. Mathematics predicts >that the results will be the same. >> Well, do it on a computer, and you'll find that this result does not >> transfer to the world of computation. > > That's true. Because an idealization must be materialized first. > The computer does this by adding a bit of sloppyness. Then you will > find that your result _does_ transfer to the world of computation, > quite good enough, that is. Well, I am quite glad that your sloppy hand-waving does not transfer back to the world of mathematics. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Update: Objections to Cantor's Theory > >> > >Can you name a testable prediction made by mathematics? >> >>A few. >> >>Take three numbers a,b,c, and perform the following two >>calculations: a*(b+c) and a*b + a*c. Mathematics predicts >>that the results will be the same. > >Well, do it on a computer, and you'll find that this result does not >transfer to the world of computation. >> >>That's true. Because an idealization must be materialized first. >>The computer does this by adding a bit of sloppyness. Then you will >>find that your result _does_ transfer to the world of computation, >>quite good enough, that is. > > Well, I am quite glad that your sloppy hand-waving does not transfer > back to the world of mathematics. The materialization process works only one-way: from mathematics towards the real world. So there doesn't exist a transfer back, indeed. Han de Bruijn === Subject: Re: Update: Objections to Cantor's Theory > >> >> >>See? Nothing else is needed than the classical, pre-Cantorian concept >>of a limit, which is accepted by all kind of mathematicians I know of. > >You need quite a bit more, since you need some sort of arithmetic with >infinite numbers, which, I believe, wasn't developed until Cantor. >> >>No, I don't need some sort of arithmetic with infinite numbers. >> >>Just the limit. Nothing else. > > Yes you do. Without it, there is no ratio of even numbers to natural > numbers. No I don't. Keep trying. Han de Bruijn === Subject: Re: Update: Objections to Cantor's Theory > >> > > >See? Nothing else is needed than the classical, pre-Cantorian concept >of a limit, which is accepted by all kind of mathematicians I know of. >> >>You need quite a bit more, since you need some sort of arithmetic with >>infinite numbers, which, I believe, wasn't developed until Cantor. > >No, I don't need some sort of arithmetic with infinite numbers. > >Just the limit. Nothing else. >> >> Yes you do. Without it, there is no ratio of even numbers to natural >> numbers. > >No I don't. Keep trying. Yes you do. Talking about the ratio of infinite set sizes requires that you have some notion of size for infinite sets, and some notion of division for those sizes. Otherwise, the ratio is just nonsense. Now, you seem to want to implicitly define these things as a limit of density. This has at least the following problems. 1) For infinite sets, the density depends on the order. 2) The results depend upon what set your measuring the densities in. 3) Once you've fixed an order and a base set, the limits don't always converge. Martin === Subject: Re: Update: Objections to Cantor's Theory > Now, you seem to want to implicitly define these things as a limit of > density. Heh, heh, seems that we are getting somewhere. > This has at least the following problems. > > 1) For infinite sets, the density depends on the order. Sure. That's why I took an (ordered) sequence, not a set. > 2) The results depend upon what set your measuring the densities in. Sure. Therefore I took the naturals and nothing but the naturals. > 3) Once you've fixed an order and a base set, the limits don't always > converge. Maybe not always. But in this very special case, they DO converge. But I think your real problem is that my approach seems to be lacking a flavour of generality. Well, maybe this is deliberately so. Han de Bruijn === Subject: Re: Mathematical physics for graduate school > Right now I am going into my senior year as a physics major. I am > interested in applying to graduate school, and I want to study > mathematical physics. In particular, I think I would enjoy studying > ways to place physical theories like quantum field theory or general > relativity on rigorous mathematical foundations. > > .... If anybody has other advice relating to > my graduate school search, please feel free to tell me. since you're at caltech, perhaps kip thorne would be a good person to start with. he may not personally know which math departments are best suited to your interests, but i'd be surprised if he doesn't know several people well qualified to make suggestions. vale, rip -- NB eddress is r i p 1 AT c o m c a s t DOT n e t === Subject: Re: collection of mathematically elegant tricks windows-nt) > [...] > Wow, this is a forum that supports math equations... very nice. What kind of > math equations does it support? Latex? > > I have been always looking for such a forum to write and communicate math > more efficiently... You might try LaTeX2HTML or TtH. The UK TeX FAQ, http://www.tex.ac.uk/cgi-bin/texfaq2html?label=whereFAQ has a couple more. -- % Randy Yates % Maybe one day I'll feel her cold embrace, %% Fuquay-Varina, NC % and kiss her interface, %%% 919-577-9882 % til then, I'll leave her alone. %%%% % 'Yours Truly, 2095', *Time*, ELO http://home.earthlink.net/~yatescr === Subject: Re: collection of mathematically elegant tricks > Hi all, > > I am an engineering student who is interested in math. I don't remember how > often I have been amazed by nice and elegant mathematical tricks that make > difficult problem suddenly very easy and trivial... The proof of sin(a + b) = (sin a) * (cos b) + (sin b) * (cos a) according to E. Schmidt runs as follows: f(x) := sin(a + b - x)*cos(x) + cos(a + b - x)*sin(x) f(0) = f(b) follows the desired theorem. Michael === Subject: Re: collection of mathematically elegant tricks > >> Hi all, >> >> I am an engineering student who is interested in math. I don't remember how >> often I have been amazed by nice and elegant mathematical tricks that make >> difficult problem suddenly very easy and trivial... > >The proof of sin(a + b) = (sin a) * (cos b) + (sin b) * (cos a) > >according to E. Schmidt runs as follows: f(x) := sin(a + b - x)*cos(x) + cos(a + b - x)*sin(x) > f(0) = f(b) > >follows the desired theorem. > >Michael The above proof looks great except for one thing -- the reasoning is circular, since the usual proof that the derivative of sin(x) is cos(x) makes use of the formula for sin(a+b). quasi === Subject: Re: collection of mathematically elegant tricks > >> > Hi all, > > I am an engineering student who is interested in math. I don't remember > how > often I have been amazed by nice and elegant mathematical tricks that > make > difficult problem suddenly very easy and trivial... >> >>The proof of >> >> sin(a + b) = (sin a) * (cos b) + (sin b) * (cos a) >> >>according to E. Schmidt runs as follows: >> >> f(x) := sin(a + b - x)*cos(x) + cos(a + b - x)*sin(x) >> >> >> f(0) = f(b) >> >>follows the desired theorem. >> >>Michael > > The above proof looks great except for one thing -- the reasoning is > circular, since the usual proof that the derivative of sin(x) is > cos(x) makes use of the formula for sin(a+b). > > quasi such a proof always seemed silly to me. how do you define sin and cos if not by the exponential function? if you define it that way, showing that the derivative of sin is cos is trivial. === Subject: Re: collection of mathematically elegant tricks > >> > >> Hi all, >> >> I am an engineering student who is interested in math. I don't remember >> how >> often I have been amazed by nice and elegant mathematical tricks that >> make >> difficult problem suddenly very easy and trivial... > >The proof of sin(a + b) = (sin a) * (cos b) + (sin b) * (cos a) > >according to E. Schmidt runs as follows: f(x) := sin(a + b - x)*cos(x) + cos(a + b - x)*sin(x) > f(0) = f(b) > >follows the desired theorem. > >Michael >> >> The above proof looks great except for one thing -- the reasoning is >> circular, since the usual proof that the derivative of sin(x) is >> cos(x) makes use of the formula for sin(a+b). >> >> quasi > >such a proof always seemed silly to me. >how do you define sin and cos if not by the exponential function? >if you define it that way, showing that the derivative of sin is cos is >trivial. Ok, yes, now I see what the proof intended, so I withdraw my objection. By mentioning that it's also true over C, that suggests that the author assumes a prerequisite of complex analysis, so the proof is redeemed, and it's definitely elegant. It certainly beats expanding the formal power series for sin(a+b) with a,b as unknowns and comparing it to the power series for sin(a)cos(b)+cos(a)sin(b). quasi === Subject: Re: collection of mathematically elegant tricks > Hi all, > > I am an engineering student who is interested in math. I don't remember how > often I have been amazed by nice and elegant mathematical tricks that make > difficult problem suddenly very easy and trivial... Many nice tricks > frequently appear in these newsgroups... I am wondering if anybody has seen > a collection of mathematical tricks ranging from high school math up to > graduate school math? Any website, Internet resources, books that have these > kind bags of tricks? If not, I may want to start collecting and compile one > such resources. > > Recently, one very striking trick is offered by Scott Hemphill in > computing the expected waiting time for a certain pattern to occur in coin > tossing... I also remember many of the other tricks that have been > contributed by many other authors in these newgroups... > You will probably enjoy: Proofs from the Book by Martin Aigner, Gunter M. Ziegler ISBN: 3540636986 It's a collection of the most elegant mathematical proofs* accesible to non professional mathematicians. You don't need lots of background knowledge, and the problems are easy to grasp intuitively. I certainly liked it. * in the authors opinion of course. === Subject: Re: collection of mathematically elegant tricks one of the excellent math book availabe in net. http://www.physics.miami.edu/nearing/mathmethods/ Ajith > Hi all, > > I am an engineering student who is interested in math. I don't remember how > often I have been amazed by nice and elegant mathematical tricks that make > difficult problem suddenly very easy and trivial... Many nice tricks > frequently appear in these newsgroups... I am wondering if anybody has seen > a collection of mathematical tricks ranging from high school math up to > graduate school math? Any website, Internet resources, books that have these > kind bags of tricks? If not, I may want to start collecting and compile one > such resources. > > Recently, one very striking trick is offered by Scott Hemphill in > computing the expected waiting time for a certain pattern to occur in coin > tossing... I also remember many of the other tricks that have been > contributed by many other authors in these newgroups... > > > You will probably enjoy: > > Proofs from the Book > by Martin Aigner, Gunter M. Ziegler > ISBN: 3540636986 > > It's a collection of the most elegant mathematical proofs* accesible to > non professional mathematicians. You don't need lots of background > knowledge, and the problems are easy to grasp intuitively. > > I certainly liked it. > > * in the authors opinion of course. === Subject: Re: collection of mathematically elegant tricks Two tricks that you see fairly often are adding zero or multiply by one. example: show why 0^0 is undefined... rewrite 0^(1-1) ** trick of adding zero you show 0^0 is equivalent to 0/0 ** my apologies to the group and anyone who teaches undergrad mathematics. simply writing 0/0 gives me a headache. Reading it, moreso. but suffice to say, there are a lot of proofs/shows that utilize one of these two. > > one of the excellent math book availabe in net. > http://www.physics.miami.edu/nearing/mathmethods/ > > Ajith > > > Hi all, > > I am an engineering student who is interested in math. I don't remember how > often I have been amazed by nice and elegant mathematical tricks that make > difficult problem suddenly very easy and trivial... Many nice tricks > frequently appear in these newsgroups... I am wondering if anybody has seen > a collection of mathematical tricks ranging from high school math up to > graduate school math? Any website, Internet resources, books that have these > kind bags of tricks? If not, I may want to start collecting and compile one > such resources. > > Recently, one very striking trick is offered by Scott Hemphill in > computing the expected waiting time for a certain pattern to occur in coin > tossing... I also remember many of the other tricks that have been > contributed by many other authors in these newgroups... > > > You will probably enjoy: > > Proofs from the Book > by Martin Aigner, Gunter M. Ziegler > ISBN: 3540636986 > > It's a collection of the most elegant mathematical proofs* accesible to > non professional mathematicians. You don't need lots of background > knowledge, and the problems are easy to grasp intuitively. > > I certainly liked it. > > * in the authors opinion of course. === Subject: Re: collection of mathematically elegant tricks !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > Two tricks that you see fairly often are adding zero or multiply by > one. > > example: > show why 0^0 is undefined... > rewrite 0^(1-1) ** trick of adding zero > you show 0^0 is equivalent to 0/0 Nonsense. According to your reasoning: Show why 0^1 is undefined... rewrite 0^(2-1) You show 0^1 is equivalent to 0/0 -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: collection of mathematically elegant tricks > Two tricks that you see fairly often are adding zero or multiply by > one. > example: > show why 0^0 is undefined... > rewrite 0^(1-1) ** trick of adding zero > you show 0^0 is equivalent to 0/0 Bad example, because 0^0 *is* defined. Rewriting 0^0 as 0^(1-1) doesn't demonstrate anything except that the trick doesn't work in this case. It doesn't show that 0^0 is undefined. A better trick is to write 0^0 = (1-1)^0 and expand using the binomial theorem, obtaining 0^0 = 1. See the sci.math FAQ for more about why 0^0 makes perfectly good sense. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: fastest convolution > > Does anybody have the fastest method (using paper and pencil) to calculate > the convolution: > > y[n]=x[n]*h[n] > > where x[n]=u[n-3] - u[n-14], h[n]=u[n-5] - u[n-8], where u[n] is the step > function, u[n]=1 for n>=0, and u[n]=0 for n<0. > > Can you do it in 2 minutes? Easily. I did it in just over 1 minute. Here's how: I drew a cartoon of both functions to set the specifications. Then I went and got a timer. Set the timer and started: I realized that one of them has to be reversed before sliding, so I drew another cartoon of it. Then I noted that 8 seconds had to pass before the first overlap and put a sample at 8 seconds. I continued figuring out where the next overlap would be in time and put a new sample at that point. It was easier because only one pair of samples coincided at the same time. It was easier because no samples had other than unity weight. All you have to do is figure out when the samples coincide - which is very simple to visualize. When I was done the clock had gone 1 minute and 5 seconds. There was no temporal scaling except to label the times of the samples. It was easy to do the simple arithmetic in my head and drawing scaled cartoons wasn't necessary. If it had been more complex then perhaps scaled drawings (in time) would have been helpful - assuming no computer. So, I think that visualization with a little help from the pencil and paper is the fastest method. Fred === Subject: Re: fastest convolution > >> > >> >>Does anybody have the fastest method (using paper and pencil) to >>calculate >>the convolution: >> >>y[n]=x[n]*h[n] >> >>where x[n]=u[n-3] - u[n-14], h[n]=u[n-5] - u[n-8], where u[n] is the >>step >>function, u[n]=1 for n>=0, and u[n]=0 for n<0. >> >>Can you do it in 2 minutes? > >No, maybe 30 seconds. Tops... You really have to understand how the >convolution works. Lets say you >write it as y[n] = sum_m x[m] h[n-m] > >a) Figure out how x[m] and h[m] look like, and draw them with horizontal >axis 'm'. >b) Mirror h[m] around m=0. This gives you h[n-m] for n=0. >c) By shifting h[n-m] to the right, you observe the convolution for n>0. >Now look at four special >points: >1) h[n-m] and x[m] start to overlap, this is where output y[n] becomes >non-zero >2) h[n-m] and x[m] start to overlap completely, this is where output y[n] >has max value >3) h[n-m] and x[m] end to overlap completely, this is where output y[n] >still has max value >4) h[n-m] and x[m] end to overlap, this is where output y[n] becomes zero >again > >That gives y[n]. Fill in the details of the convolution yourself. Jeroen >> >>Slow slow slow slow slow slow ... need so many steps... :=) > > > Looks a lot of work maybe, but like I said: if you have some experience you can make the sketch of > y[n] in seconds... > > Jeroen I did it in under 3 minutes, but I had to find the quadrule pad first. Jerry -- Engineering is the art of making what you want from things you can get. żżżżż żżż[OS lash]żżżż[DownQuest ion]żżż żżżżż żżż[OSl ash]żżżż[DownQuesti on]żżż żżżżż żżż[OSl ash]żżżż[DownQuesti on]żżż żżżżż żżż[OSl ash]żżżż[DownQuesti on]żżż żżż === Subject: Re: fastest convolution The method Jeroen mentioned is the standard way of doing convolution by hand. You can find in any introductory text. When you say slow, and too many steps, are you saying you have a new way of doing this? Can you share your idea? John > > Slow slow slow slow slow slow ... need so many steps... :=) > === Subject: Re: fastest convolution Noting that u[n]*u[n] = n u[n] =r[n] (which is a ramp with slope 1), y[n] = r[n-8] - r[n-11] - r[n-19] + r[n-22] which is a ramp starting at 8, becoming horizontal at 11, decreasing with slope 1 at 19 and zero starting from 22. J. > > Does anybody have the fastest method (using paper and pencil) to calculate > the convolution: > > y[n]=x[n]*h[n] > > where x[n]=u[n-3] - u[n-14], h[n]=u[n-5] - u[n-8], where u[n] is the step > function, u[n]=1 for n>=0, and u[n]=0 for n<0. > > Can you do it in 2 minutes? > > === Subject: Re: fastest convolution Jane skrev: > > Does anybody have the fastest method (using paper and pencil) to calculate > the convolution: > > y[n]=x[n]*h[n] > > where x[n]=u[n-3] - u[n-14], h[n]=u[n-5] - u[n-8], where u[n] is the step > function, u[n]=1 for n>=0, and u[n]=0 for n<0. > > Can you do it in 2 minutes? Yes, I can. Even if this seems to be a homework question, I'll show you how. Use an A4 paper (or your local equivalent) of the kind with 5mm x 5mm squares already in it. Start with laying out three coordinate systems with horizontal axis 0 <= x <= 25 and vertical axis 0 <= y <= 1, then a fourth coordinate system with 0 <= y <= 5. Make sure the horizontal axes are aligned vertically. This is a maths question, so I don't mention the term impulse response, but if you were an engineer, I would explain the next step in terms of x[n] being the impulse response and h[n] being the input signal to an LTI system. But I don't mention those terms. Now, in the uppermost axis, plot x[n] scaled by the first non-zero coefficient of h[n], and delayed by an amount corresponding to the same coefficient. Repeat for the non-zero coefficients of h[n], there are three non-zero coefficients (OK, maybe there are four, it's too hot to day to work out absolutely all details). Having done that, add vertically across the axes to produce y[n]. Your paper ought to look something like this after you are through (use a fized-width font to view): ^ 1 | o o o o o o o o o o o o | | | | | | | | | | | | | +-o-o-o-o-o-o-o-------------------------o-o-o-o-o-o-o------> : : ^ : 1 | o o o o o o o o o o o o | | | | | | | | | | | | | +-o-o-o-o-o-o-o-o-------------------------o-o-o-o-o-o------> : : ^ : 1 | o o o o o o o o o o o o | | | | | | | | | | | | | +-o-o-o-o-o-o-o-o-o-------------------------o-o-o-o-o------> : : : : ^ y[n] : | : 3 | o o o o o o o o o o 2 | o | | | | | | | | | | o 1 | o | | | | | | | | | | | | o +-o-o-o-o-o-o-o-----------------------------o-o-o-o-o-o-o--> 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 n 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 The vertical line of colon's shows what samples to add to produce sample y[11]. Once you know the technique, it ought not to take more than 2 minutes to do, for a reasonable number of rectangular functions like above. As a bonus, extend the technique to the case where h[n] = {3 2 1} for n={0 1 2} 0 otherwise. You got as much time for the computation as you needed for the original problem, plus another 10 s. Rune