
Re: ZFC and God
Posted:
Jan 27, 2013 5:27 PM


WM <mueckenh@rz.fhaugsburg.de> writes:
> On 27 Jan., 19:56, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: >> WM <mueck...@rz.fhaugsburg.de> writes: >> > On 27 Jan., 19:21, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: >> >> WM <mueck...@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. >> >> > I did not fix k but only assumed that it is a natural number. >> >> Pardon me? >> >> Look, you said that a number has terminating decimal representation >> iff there is some k and f such that blah blah blah. Thus, to show >> that a number *does* have a terminating decimal representation is >> equivalent to showing that there is, in fact, a k and f satisfying the >> above. > > Of course. But why should we agree on a special k? Every natural > number will do. So we only have to know that k is one of those natural > numbers that belong to FISONs. As long as we work in FISONs we cannot > have a nonterminating decimal.
I honestly have no idea what you're trying to say here. Why not simply prove that there is such a k and f?
>> >> What could be more obvious than this? > > Remember the Binary Tree to answer your doubts. Do you believe that > the set of all FISONs is definable in ZF? Is every FISON finite? Is > there a fixed k limiting all FISONs? No.
As you said before, let's keep analogies out of this.
>> > You have no cause to claim the contrary, since there is no index >> > (natural number) infinitely many counts away from the decimal point. >> >> Are you asking me to prove that 0.777... does not have a terminating >> decimal representation? I would be happy to do so, if needed. > > I am doubting that when working in FISONs you can find an index of a > diagonal number that requires to leave the set of FISONs (with respect > to the indexes of the digits). > >> it follows *FROM THE AGREED DEFINITION* that 0.777... has no >> terminating decimal representation. > > Show that in your 0.777..., and in particular in the antidiagonal of > a list of terminating decimals, there is an index k that does not > belong to a FISON {1, 2, ..., n}.
Utterly irrelevant to the matter at hand. Evidently, you don't understand the definition you agreed to.
> As long as you refuse there is no reason to believe you, because > every index in a FISON belongs to a FISON (finite initial set or > sequence of natural numbers). Why do you try to deny that?
Who ever denied that?
My interest is waning here. We have a definition, but you refuse to apply it (or, perhaps, are too stupid to understand it) in order to prove what you claim.
Please, just show me that there is a function
f:{1,...,k} > {0,...,9}
such that
0.777... = sum_i=1^k f(k) * 10^i,
and I will concede that 0.777... has a terminating decimal representation.
And if you cannot do that, then I will not concede the point but rather conclude you cannot show what you claim.
 "This sucks," said a Pennsylvania State University student [...] " Why can't the college let me do what I want to do with my computer? The school computer security guys are being way more annoying than the spyware was."  A student pines for his disabled spyware

