
Re: ZFC and God
Posted:
Jan 27, 2013 1:21 PM


WM <mueckenh@rz.fhaugsburg.de> writes:
> On 27 Jan., 18:44, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: > >> 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. > > You are wrong. Can't you understand? All natural numbers are finite. > Why the heck should I define a single k?
Because, of course, you accepted 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.
Thus, if you claim that 0.777... has a terminating representation, then you must show that there is a natural number k and a function f as above such that
0.777... = sum_i=1^k f(i) * 10^i.
Else, you have no cause to claim that 0.777... has a terminating decimal representation.
>> > 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. > > Perhaps in order to show that matheology is not complete nonsense? > >> 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. > > I did this in my last posting. Please look it up there. Well as I have > it just at hand, here it is again: > 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 is written as > 0.777... > > This is the definition that I agreed to.
Er, no.
The definition that you agreed to is reproduced above. You have to show that the definition above is actually satisfied, i.e., that there is a natural number k and a function f satisfying the appropriate conditions.
You've done no such thing.
Frankly, I'm a bit stunned that you're arguing that 7/9 has a terminating decimal representation, but as long as you're claiming so, then you need to stick to the definition we've agreed on.
 "Being who I am, I know that's a solution that will run in polynomial time, but for the rest of you, it will take a while to figure that out and know why [...But] it's the same principle that makes n! such a rapidly growing number." James S. Harris solves Traveling Salesman

