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