
Re: ZFC and God
Posted:
Jan 27, 2013 12:44 PM


WM <mueckenh@rz.fhaugsburg.de> writes:
> On 27 Jan., 18:18, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: >> WM <mueck...@rz.fhaugsburg.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 nonterminating 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

