```Date: Jan 27, 2013 12:18 PM
Author: Jesse F. Hughes
Subject: Re: ZFC and God

WM <mueckenh@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 responsediverts from the issue I want clarified.  (Once again, you'veinadvertently 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?).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 concludeyou have no proof that 0.777... is terminating.Thanks much.-- Jesse F. Hughes"Part of the problem here, Peter, is that you are an idiot."                       -- Daryl McCullough gives a diagnosis
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