Why can no one in sci.math understand my simple point?
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From: Virgil on 20 Jun 2010 16:08 In article <d55211a9-4dbf-48b3-8d8d-88d1fd859ab3(a)c33g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 20 Jun., 16:33, Newberry <newberr...(a)gmail.com> wrote: > > On Jun 19, 6:06 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > > > > > Newberry <newberr...(a)gmail.com> writes: > > > >> Because every infinite sequence of digits represents a real number? > > > >> And > > > >> the antidiagonal is one such sequence? > > > > > > If it does not exist then it does not represent anything let alone a > > > > number. > > > > > > Now it is clear that it does not exist. Since all the reals are on the > > > > list and the anti-diagonal would differ from any of them. This > > > > violates the assumption. Hence the anti-diagonal does not exist. > > > > > Wow. Are you saying that the *sequence of digits* specified by the > > > anti-diagonal does not exist? > > > > That's right. There is no formula or algorithm to construct the list. > > It means that you would have to flip each and every digit one by one. > > And that is impossible. > > Well, there are some lists defined by finite definitions like > 0.1 > 0.11 > 0.111 > ... > > But the number of these lists and the number of their diagonals is > countable. > > > > > Anyway, in a sense, you're right. If we assume that every real is > > > represented by a sequence on the list, then we can prove that every > > > sequence occurs on the list (ignoring the issue of multiple > > > representations). And yet, we can also show that a particular sequence > > > is not on the list. > > > > Which sequence is that? > > O, don't be unfair. Matheolgicians "prove" their claims. They are not > used to give examples. Remember Zermelo who "proved" that every set > can be well ordered. Although every sober mind today knows that > hisproof is wrong, set theory is built upon this lie. Given the axiom of choice, every set is in principle well orderable. But without the axiom of choice, it is not always possible. So WM must be rejecting the axiom of choice for himself. But he does not have either the right, or, more importantly, the power to impose that rejection on everyone else.
From: Virgil on 20 Jun 2010 16:18 In article <f92c169d-ee85-40c2-aa82-c8bdf06f7b55(a)j4g2000yqh.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 20 Jun., 17:51, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > > > > Cantor was this utterly insane freak who chose not to accept Newberry's > > word for it, and instead *prove* that there was no list of all real > > numbers. �Obviously, his proof is nonsense, because, after all, Newberry > > said there was no list. > > His proof is nonsense because it proves that a countable set, namely > the set of all reals of a Cantor-list and all diagonal numbers to be > constructed from it by a given rule an to be added to this list, > cannot be listed, hence, that this indisputably countable set is > uncountable. That is a deliberate misrepresentation of the so called "diagonal proof". What that proof (not actually by by Cantor himself) ays is that any listing of reals is necessarily incomplete because given such a listing one can always construct a real not listed in it. What Cantor himself proved was that for any list of infinite binary sequences there are binary sequences not listed in that list. Cantor had previously proved by quite a different method, which WM seems unable or unwilling to criticize, that the set of reals was not countable. > > Even completely blinded matheologicians should be able, perhaps after > some contemplation, to recognize that. What completely blinded matheologicians are able, perhaps after some contemplation,to recognize is that WM is not working in the same world as mathematicians do.
From: Jesse F. Hughes on 20 Jun 2010 18:13 "Mike Terry" <news.dead.person.stones(a)darjeeling.plus.com> writes: > "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote in message > news:4c1e3743$0$1028$afc38c87(a)news.optusnet.com.au... >> No, to calculate the first n terms of the anti-diagonal you require only > te >> first n items on the list to n places of accuracy. > > So you require as input: > - the 1st digit to 1 place of accuracy > - the 2nd digit to 2 places of accuracy > - the 3rd digit to 3 places of accuracy > - ... > > The ... above is the give away! For your algorithm to work its going to > require an INFINITE amount of input data. (Remember, you are trying to find > a SINGLE stand-alone FINITE algorithm that produces ALL the required digits > of the antidiagonal.) > > So Tim was quite right, and you haven't answered his point correctly > yet... No, you fail to understand the brilliance that Peter brings to this conversation. Every real is computable. You just have to produce a finite amount of data to compute the real to n digits, namely, the first n digits of the real you want to compute. All of you naysayers are just holding back inevitable progress. -- "And yes, I will be darkening the doors of some of you, sooner than you think, even if it is going to be a couple of years, and when you look in my eyes on that last day of work at your school, then maybe you'll understand mathematics." -- James S. Harris on Judgment Day
From: Newberry on 20 Jun 2010 20:18 On Jun 20, 1:18 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <f92c169d-ee85-40c2-aa82-c8bdf06f7...(a)j4g2000yqh.googlegroups.com>, > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > On 20 Jun., 17:51, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > > > > Cantor was this utterly insane freak who chose not to accept Newberry's > > > word for it, and instead *prove* that there was no list of all real > > > numbers. Obviously, his proof is nonsense, because, after all, Newberry > > > said there was no list. > > > His proof is nonsense because it proves that a countable set, namely > > the set of all reals of a Cantor-list and all diagonal numbers to be > > constructed from it by a given rule an to be added to this list, > > cannot be listed, hence, that this indisputably countable set is > > uncountable. > > That is a deliberate misrepresentation of the so called "diagonal proof". > > What that proof (not actually by by Cantor himself) ays is that any > listing of reals is necessarily incomplete because given such a listing > one can always construct a real not listed in it. Please CONSTRUCT the anti-diagonal in the space provided below. Virgil's anti-diagonal construction BEGIN END > What Cantor himself proved was that for any list of infinite binary > sequences there are binary sequences not listed in that list. > > Cantor had previously proved by quite a different method, which WM seems > unable or unwilling to criticize, that the set of reals was not > countable. > > > > > Even completely blinded matheologicians should be able, perhaps after > > some contemplation, to recognize that. > > What completely blinded matheologicians are able, perhaps after > some contemplation,to recognize is that WM is not working in the same > world as mathematicians do.
From: Peter Webb on 20 Jun 2010 20:16
"Mike Terry" <news.dead.person.stones(a)darjeeling.plus.com> wrote in message news:B8WdnX3itoZ7zYPRnZ2dnUVZ7qednZ2d(a)brightview.co.uk... > "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote in message > news:4c1e2b53$0$316$afc38c87(a)news.optusnet.com.au... >> >> "Tim Little" <tim(a)little-possums.net> wrote in message >> news:slrni1r0ra.jrj.tim(a)soprano.little-possums.net... >> > On 2010-06-20, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: >> >> Explain to me how my requirements for the input list of all >> >> computable Reals is different to Cantor's requirements for the input >> >> list of all Reals, other than the requirement that every item on the >> >> list is computable? >> > >> > It is not. Your error comes later. >> > >> > Cantor's proof includes a construction taking a list L and defining an >> > antidiagonal real antidiag(L) from it. Your error is supposing that >> > this construction is a finite algorithm fitting the definition of >> > "computable real". It is not. By stretching the definition of >> > "algorithm" somewhat, it can be supposed to be an algorithm accepting >> > finite input n and infinite input L, and producing the n'th >> > antidiagonal digit. This is a stretch since algorithms are normally >> > not supposed to have infinite inputs. >> >> The calculation of the nth digit only requires the first n digits to n >> decimal places, and hence it is clearly a finite algorithm. >> >> Its simply ludicrous to suggest that the antidiagonal is not computable >> given that a very simple algorithm will explicitly churn it out. > > Not so - the "very simple algorithm" you are suggesting requires "the > input > list" in its raw (infinite) form as input. No. Computing the the anti-diagonall to n places only requires the first n items on the list to n digits of accuracy. And it clearly is computable. I can compute it. I take the nth digit of the nth term, and ... blah blah. How is this number any less computable than the square root of 2? In both cases there is a finite algorithm to calculate the number to n decimal places for all n. > As such, this is not an > algorithm that can be implemented on a Turing machine. I.e. the algorithm > is not the sort of algorithm that counts as "computable". Maybe you will > dispute this, but it's been explained to you several times (including one > of > my posts), and you can always go away and look up the definition of > computable function / Turing Machine etc., but it seems you've not done > this. I don't have to look it up. I know it. Obviously the anti-diagonal the computable. I can compute it to any required degree of accuracy using a very simple algorithm which explicitly computes it; you can't get much more computable than that. > >> >> > >> > However, there is no way that you can then prove the existence of a >> > finite algorithm accepting only the *finite* input n and producing the >> > n'th antidiagonal digit. >> > >> >> Of course I can prove the existence of a finite algorithm. I can produce > it. >> >> To produce the nth decimal of the anti-diagonal, look at the first n >> items >> on the purported list of all computable Reals. > > There! Was I right, or was I right? :-) How is looking at the first n digits of the first n terms and using Cantor's digit substitution to calculate the decimal place not computable? > > There is an infinite amount of data on this "purported list", and you are > not allowed to base a computable function on an infinite amount of input > data. To do so shows you simply do not understand what computable means. > I don't use the whole list. To calculate the nth digit of the anti-diagonal I only need the first n terms to n decimal places accuracy. That is a finite amount of data. And, as a practical matter, if you want to give me a purported list of all computable Reals, I can actually calculate it to any required degree of accuracy in a finite number of steps. > Now IF you could codify all this data somehow into a finite program, so > you > had a computable function D(m,n) which produced the n'th digit of the m'th > number in the list, you would be OK. But you DON'T have such a computable > function D. > Well, Cantor asks for a list of Reals. And D(m,n) is computable in my case by definition; which means we can always evaluate this function (as a computable Real must be able to be specified to any arbitrary degree of accuracy). > What you DO have, is for each m, a computable function D_m (n) which > produces the n'th digit of the m'th number in the list. But how can you > use > these individual D_m in the definition of the antidiagonal? (You > can't...) I do. If D(n,n)=1 then antidiagonal(n) = 2, else antidiagonal(n) = 1. > > Do you not see this distinction? > > Knowing the existence of all the individual D_m is equivalent to knowing > that the list (sequence of numbers) consists of computable reals. > > Knowing the existence of D (which is a single computable function) is > equivalent to having a *computable* list of computable reals. > Cantor's proof asks to see the list and then produces a number not on it. Same as my proof. Yes, that requires that te list be specified ... but Cantor's proof and mine are no different in that respect. > Cantor's proofs do not require computable lists, they work with just any > old > lists... > Indeed. And my proof requires computable Reals in a list. That is the difference. >> Then substitute "1" for "2" >> blah blah. How is that not a finite algorithm? > > As explained it is not a finite algorithm because it uses as input an > infinite source of data. > No, to specify the anti-diagonal to n places requires only a finite amount of data for all n, specifically the first n decimal places of the first n terms. > By your reckoning every single real number is computable, because we can > have a TM that just accepts as input an infinite tape with the entire > digit > sequence encoded on it, and outputs that digit sequence. No. I can explicitly calculate the nth decimal places with only n terms to n places of accuracy. No infinite input is required. In any event, it is a farcical argument. Of course I can calculate the antidiagonal to any number of places of accuracy using only finite processes; Cantor's proof is constructive. > > Surely you must be aware that this is NOT what computability means? > Being able to specify the nth digit of a number in a finite number of steps for all n is a sufficient condition for compuatbility, and that is exactly what Cantor's diagonal proof provides - a means of calculating the nth digit of the anti-diagonal in a finite number of steps, using a finite input (the first n digits of the first n terms). > Regards, > Mike. > > > |