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 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?).

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.

Thanks much.

-- 
Jesse F. Hughes

"Part of the problem here, Peter, is that you are an idiot."
                       -- Daryl McCullough gives a diagnosis