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Re: ZFC and God
Posted:
Jan 27, 2013 12:37 PM
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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 "...".
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.
Regards, WM
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