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Re: ZFC and God
Posted:
Jan 27, 2013 12:44 PM
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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.
--
"A signature block is meant for the author of a post. If the name of
someone other than the author of a post appears 'lastly in the
post'.[sic] It is taken to mean that such a person is the author."
-- Archimedes Plutonium doesn't get .sigs
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