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Topic: ZFC and God
Replies: 15   Last Post: Jan 29, 2013 9:19 PM

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mueckenh@rz.fh-augsburg.de

Posts: 15,089
Registered: 1/29/05
Re: ZFC and God
Posted: Jan 29, 2013 3:42 AM
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On 29 Jan., 03:11, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> WM <mueck...@rz.fh-augsburg.de> writes:
> > On 28 Jan., 21:36, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
> [...]
>

> >> You still have to show that the diagonal itself has a terminating
> >> decimal representation.  You can't just assume that it does.

>
> > I can prove that it does. The list does not contain infinite decimals.
> > Every digit of the anti-diagonal is a (changed) digit of a terminating
> > decimal. If you do not agree that the anti-diagonal is terminating,
> > then you should find the border where its indices leave the domain of
> > terminating decimals. Unless you can, we remain inside.

>
> Oops!  More mysterious disappearing text, WM!  Here, I repeat my
> little hint regarding your inability to apply a definition:
>
>   > Now find out which of the digits of 0.777... that is constructed
>   > of numbers of F has not a finite index k.
>
>   Not relevant to the definition at hand.


You are too stupid to understand it. No question.
But even if you consider an actually infinite diagonal, what would it
help you. The diagonal of a list of terminating decimals can only
differ at a finite position from an entry, simply because there are no
further digits. So what does your "actually infinite diagonal" help.


> Is it possible that you
>   can't understand the definition itself?


Is it possible that you are not accustomed to work in the domain of
terminating decimals?
>
>   It's really very simple, isn't it?


It is always simpler to believe in nonsense of matheology than to
adhere to the given limits. And these limits restrict everything in
the present discussion to terminating decimals.

If you don't believe that this is possible, then find a first digit
that surpasses this domain.

Regards, WM



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