Date: Jan 27, 2013 5:27 PM
Author: Jesse F. Hughes
Subject: Re: ZFC and God

WM <mueckenh@rz.fh-augsburg.de> writes:

> On 27 Jan., 19:56, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> WM <mueck...@rz.fh-augsburg.de> writes:
>> > On 27 Jan., 19:21, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> >> WM <mueck...@rz.fh-augsburg.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 non-terminating 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 anti-diagonal 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