Why can no one in sci.math understand my simple point?
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From: Jesse F. Hughes on 20 Jun 2010 00:26 Tim Little <tim(a)little-possums.net> writes: > On 2010-06-20, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: >> Of course it is computable. >> >> Cantor provides an explicit algorithm for computing it. > > Cantor provides an explicit algorithm for computing it with the list L > as input. Look again at the definition of "computable real". Is "the > algorithm may take as input an auxiliary list of infinite many real > numbers" part of that definition? > > >> I make no mention of finite algorithms. I just do *exactly* the same >> construction, but applied to a purported list of all computable >> Reals instead of a purported list of all Reals. > > The definition of your term "computable real" includes references to > "finite algorithm". So you do make reference to a finite algorithm, > and Cantor does not. > > >> Cantor: Lets form an anti-diagonal using this explicit construction >> based upon the nth digit of the nth Real on the list. It is >> obviously a Real, and it is obviously missing. > > Cantor did not make any assumption that it was "obviously" a real. In > fact, he went to some trouble to *prove* that it specified a real > number. > > >> Peter: Lets form an anti-diagonal using this explicit construction >> based upon the nth digit of the nth computable Real on the list. It >> is obviously a computable Real, and it is obviously missing. > > You, on the other hand, make no effort whatsoever to prove that the > antidiagonal number is computable. If you actually did, you would run > into insuperable difficulties. > > The devil is in the details in any mathematical proof. You are > actively *avoiding* the details and just assuming they work out. That > renders your so-called proof invalid, and is exactly why I accept > Cantor's and not yours. > > > At this point, and in this respect, it is clear that you are just as > much of a crank as JSH and WM. Indeed. -- Jesse F. Hughes "Well, I guess that's what a teacher from Oklahoma State University considers proper as Ullrich has said it, and he is, in fact, a teacher at Oklahoma State University." -- James S. Harris presents a syllogism
From: Tim Little on 20 Jun 2010 00:36 On 2010-06-20, Peter Webb <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > "Daryl McCullough(" <stevendaryl3016(a)yahoo.com> wrote in message >> Cantor's proof assumes the *existence* of such a list. It doesn't >> assume that you know how to compute it. > > If that is what you believe, then I am happy for the same rules to > be applied to the purported list of all computable Reals. Chaitin's Omega exists. Its binary expansion exists, in which there exists an n'th digit for all n, which we can denote Omega_n. So a list L:N->R exists for which L(n) = (1 - Omega_n) / 2^n. For all k, the k'th entry in L is either 0 or 1/2^n. Both of these are computable, so L is a list of computable numbers. Do you agree so far? - Tim
From: Sylvia Else on 20 Jun 2010 01:42 On 20/06/2010 6:10 AM, |-|ercules wrote: > If all digits of a single infinite expansion can be contained with > increasing finite prefixes, > and the computable set of reals has EVERY finite prefix, then all digits > of EVERY infinite > expansion are contained. It's far from clear what that actually means, but in any case you certainly haven't proved it. Sylvia.
From: |-|ercules on 20 Jun 2010 01:48 "Sylvia Else" <sylvia(a)not.here.invalid> wrote ... > On 20/06/2010 6:10 AM, |-|ercules wrote: > >> If all digits of a single infinite expansion can be contained with >> increasing finite prefixes, >> and the computable set of reals has EVERY finite prefix, then all digits >> of EVERY infinite >> expansion are contained. > > It's far from clear what that actually means, but in any case you > certainly haven't proved it. > > Sylvia. You agreed that all digits of PI (in order) are contained in this list, right? 3 31 314 .... So you should get the first part, >> If all digits of a single infinite expansion can be contained with >> increasing finite prefixes, Herc
From: |-|ercules on 20 Jun 2010 01:57
"Sylvia Else" <sylvia(a)not.here.invalid> wrote > On 20/06/2010 6:10 AM, |-|ercules wrote: > >> If all digits of a single infinite expansion can be contained with >> increasing finite prefixes, >> and the computable set of reals has EVERY finite prefix, then all digits >> of EVERY infinite >> expansion are contained. > > It's far from clear what that actually means, but in any case you > certainly haven't proved it. > > Sylvia. Hypothesis: an real number contains a finite sequence that is not computable. Contradiction Therefore: all digits of every infinite expansion are contained in the list of computable digit sequences. QED Herc |