Virgil
Posts:
6,967
Registered:
1/6/11


Re: ZFC and God
Posted:
Jan 27, 2013 4:14 PM


In article <6cfca275f73b481080c53b24ee884692@m12g2000yqp.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 27 Jan., 18:18, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: > > WM <mueck...@rz.fhaugsburg.de> writes: > > > On 27 Jan., 15:49, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: > > > > >> >> That is, for each i in N, the i'th digit of 0.777... is defined and is > > >> >> 7. > > > > >> > And do you have problems to find this confirmed as possible in the > > >> > complete set of terminating decimals? Any digit or index missing? > > > > >> I've no idea what you mean when you ask whether I can "find this > > >> confirmed as possible". But, for each i in N, the i'th digit of > > >> 0.777... is defined and equals 7. Is there anything more I need to > > >> know in order to claim that it is a nonterminating decimal? > > > > > You need to know whether this n is an element of a finite initial > > > segment of {1, 2, 3, ..., n, n+1, n+2, ..., n^n}. > > > > [SNIP] > > > > Sorry, let's focus on the question at hand. I fear that your response > > diverts from the issue I want clarified. (Once again, you've > > inadvertently snipped my primary question.) > > > > By definition, > > > > 0.777... = sum_i=1^oo 7*10^1. > > > > You claim that 0.777... has a terminating decimal representation > > (right?). > > We are working in the domain of terminating decimals. Unless you can > find an index of a digit of 0.777... that does not belong to a finite > initial segment {1, 2, ..., n} of the natural numbers, 0.777... > belongs to that domain. > > > > > You accept the following definition: > > > > Let x be a real number in [0,1]. We say that x has a terminating > > decimal representation iff there is a natural number k and a > > function f:{1,...,k} > {0,...,9} such that > > > > x = sum_i=1^k f(i) * 10^i. > > > > Therefore, I request a proof that there is a function > > > > f:{1,...,k} > {0,...,9} > > > > such that > > > > sum_i=1^k f(i)*10^i = sum_i=1^oo 7*10^i. > > > > Unless you can prove that there is such a function, we must conclude > > you have no proof that 0.777... is terminating. > > Unless you can prove that there is a digit 7_i with an i that does not > belong to a finite initial segment of the natural numbers, I see no > necessity to prove anything. We must conclude you have no proof that > 0.777... is longer than every terminating sequence, namely actually > infinity.
The union of any set of sets is itself a set so the union of the set of all fisons is a set and it contains for every fison a natural number not in it is, so that union cannot be itself a fison. > > But here is the proof that we can work in the domain of terminating > decimals including 0.777...: > > 0.7 is terminating. > if 0.777...777 with n digits is terminating, then also 0.777...7777 > with n+1 digits is terminating. Therefore there is no upper limit for > the number of digits in a terminating decimal. This fact is usually > denoted by "infinite" and abbreviated by "...".
But you still have no proof that we MUST work with only terminating decimals, and until you do we won't. > > Note, there is another meaning of infinite, namely "actually > infinite". Those who adhere to that notion *in mathematics* should > show that it differs from "potentially infinite" *in mathematics*, > i.e., expressible by digits.
We find that the union of all fisons is a set containing all and only natural numbers, which set we denote by N.
Our definition of a set being infinite is that there exist an injection from N to that set.
We do not recognize any distinction between what WM calls "potential infiniteness" and "Actual infiniteness". 

