Another AC anomaly?
in [Logic]
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From: WM on 29 Nov 2009 07:05 On 28 Nov., 14:11, William Hughes <wpihug...(a)hotmail.com> wrote: > On Nov 28, 8:35 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 27 Nov., 22:42, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > Every diagonal number is in the list. > > > > Only in Wolkenmuekenheim. Outside of Wolkenmuekenheim > > > there is only one diagonal number and it is not > > > in the list. > > > How do you know, unless you have seen the last? > > You use induction to show that every entry > in the list has a final 1. So you don't have > to see an entry to know that it has a final 1 > (there is no last entry in any case). > There is a constructive proof that the > diagonal number does not have a final 1. Use induction to show that the diagonal number cannot have more digits than every entry of the list. There is a constructive proof that the list does not have a final entry. Reagrds, WM
From: WM on 29 Nov 2009 11:04 On 29 Nov., 05:38, Virgil <Vir...(a)home.esc> wrote: > In article > <e70292e5-e110-48dc-afda-8ff08c448...(a)g1g2000pra.googlegroups.com>, > "Ross A. Finlayson" <ross.finlay...(a)gmail.com> wrote: > > > > > > > On Nov 28, 12:43 pm, Virgil <Vir...(a)home.esc> wrote: > > > In article > > > <f4e15df0-a3c0-48e4-959f-e341a9adf...(a)j4g2000yqe.googlegroups.com>, > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > On 27 Nov., 22:42, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > On Nov 27, 5:24 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > On 27 Nov., 21:17, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > > > On Nov 27, 3:33 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > > > With only potential, i.e., not finished infinity, i.e., > > > > > > > > reasonable > > > > > > > > infinity, the diagonal number (exchanging 0 by 1) of the > > > > > > > > following > > > > > > > > list can be found in the list as an entry: > > > > > > > > > 0.0 > > > > > > > > 0.1 > > > > > > > > 0.11 > > > > > > > > 0.111 > > > > > > > > ... > > > > > > > > Only in Wolkenmuekenheim where the argument goes > > > > > > > > Every entry in the list has a fixed last 1 > > > > > > > The diagonal number does not have a fixed last 1 > > > > > > > There is not a fixed last entry > > > > > > So, every entry in the list has a fixed last 1. > > > > > (We don't need a fixed last entry to say this) > > > > > We still have > > > > > > Every entry in the list has a fixed last 1 > > > > > The diagonal number does not have a fixed last 1 > > > > > > > Every diagonal number is in the list. > > > > > > Only in Wolkenmuekenheim. Outside of Wolkenmuekenheim > > > > > there is only one diagonal number > > > > > How do you know, unless you have seen the last? > > > > > Regards, WM > > > > Given a specific list of endless binary sequences, the so called Cantor > > > diagonal is the result of a specific and unambiguous algorithm applied > > > to that list, so it is, for any given list, unique, and not a member of > > > the list from which it is constructed. > > > > Which WM would have known if he had any sense. > > > Binary sequences aren't unique representations of real numbers. > > The original Cantor diagonal argument did not deal with real numbers > either, so what is your point? > > > (Binary and ternary (trinary) anti-diagonal cases require refinement.) > > But as neither I nor Cantor were not dealing with numbers in any base, > your objections are, as usual, irrelevant. > > > > > For example, the list contains .1 then all zeros, the anti-diagonal > > is .0111... = .100..., anti-diagonal is on the list. > > But, in the Cantor argument, the lists in question are not of functions > from N to range {0,1} but of functions from N to range {m,w} with no > assumption that such a function corresponds to any sort of number. They correspond to sequences of w's m's. And WM has shown, that every initial sequence of the diagonal is in a Cantor's list. Now, you can believe that Cantor's argument (the diagonal is not in the list) is as powerful as my argument (every initial sequence of the diagonal is in the list - and the diagonal has no more symbols than every initial sequence). Then you are a matheologian. Or you can even believe that Cantor's argument outperforms my argument and that there are more real numbers than can be identified. Then you are what I call a Fool Of Matheology. Regards, WM
From: Jesse F. Hughes on 29 Nov 2009 11:47 WM <mueckenh(a)rz.fh-augsburg.de> writes: > Use induction to show that the diagonal number cannot have more > digits than every entry of the list. Great! Er, but how does induction *do* that? -- Jesse F. Hughes "Usenet is demonstrably dangerous. It needs to be regulated." --James S. Harris, voice of reason and moderation
From: William Hughes on 29 Nov 2009 12:35 On Nov 29, 8:05 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: WM has conceded that you can use induction to show that every element of the list has a final 1, and that there is a constructive proof that the diagonal number does not have a final 1. WM has a new argument. > Use induction to show that the diagonal number cannot have more digits > than every entry of the list. This cannot be done. All you do is show that every one of an infinite number of different numbers, none of which is the diagonal number, cannot have more digits than every entry of the list. [Outside of Wolkenmuekenheim where the diagonal number does not change. Inside of Wolkenmuekenheim the diagonal number changes, and everything it changes to has fewer digits than some entry in the list] - William Hughes
From: William Hughes on 29 Nov 2009 12:44
On Nov 29, 12:47 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > WM <mueck...(a)rz.fh-augsburg.de> writes: > > Use induction to show that the diagonal number cannot have more > > digits than every entry of the list. > > Great! Er, but how does induction *do* that? It is simple, Just move to Wolkenmuekenheim where the diagonal number changes. Use induction to show that whatever it changes to has fewer digits than some entry in the list. - William Hughes |