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