Date: Jan 27, 2013 12:44 PM Author: Jesse F. Hughes Subject: Re: ZFC and God WM <mueckenh@rz.fh-augsburg.de> writes:

> 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.

>

> 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 "...".

Are you suggesting that 0.777... is *both* an infinite and terminating

expansion?

Anyway, you haven't proved that there is a function

f:{1,...,k} -> {0,...,9}

as required by *your* definition of terminating decimal, so you have

not shown that 0.777... is a terminating decimal.

> 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.

Well, I don't understand why anyone would wish to show that. But,

regardless, this is beside the point. I'm asking for a proof that

0.777... is terminating according to the definition of terminating

that you agreed to.

Can you please show that proof? Much thanks.

--

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