```Date: Jan 27, 2013 4:14 PM
Author: Virgil
Subject: Re: ZFC and God

In article <6cfca275-f73b-4810-80c5-3b24ee884692@m12g2000yqp.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:> On 27 Jan., 18:18, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:> > WM <mueck...@rz.fh-augsburg.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 non-terminating 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".--
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