Date: Jan 27, 2013 11:54 AM
Subject: Re: ZFC and God

On 27 Jan., 15:49, "Jesse F. Hughes" <> wrote:

> >> That is, for each i in N, the i'th digit of 0.777... is defined and is
> >> 7.

> > And do you have problems to find this confirmed as possible in the
> > complete set of terminating decimals? Any digit or index missing?

> I've no idea what you mean when you ask whether I can "find this
> confirmed as possible".  But, for each i in N, the i'th digit of
> 0.777... is defined and equals 7.  Is there anything more I need to
> know in order to claim that it is a non-terminating decimal?

You need to know whether this n is an element of a finite initial
segment of {1, 2, 3, ..., n, n+1, n+2, ..., n^n}.

> I didn't ask whether it was a finite definition.  I asked whether it
> was terminating.  And it is not sufficient to note that each t_i is a
> terminating decimal to conclude that d defined by
>   d(j) = 7 if j > k or t_j(j) != 7
>   d(j) = 6 if j <= k and t_j(j) = 7.
> is also terminating.  This simply does not follow.

Here you are simply wrong. Prove by induction: If every d(j) is
element of a finite initial segment, then all d(i) with i < j are
elements of one and the same finite initial segment. If you are unable
to prove this, try to find a counter example and fail. (I would be
very surprised if you would succeed. But I calmly await your answer.)
> This is analogous to the fact that limits of sequences of rational
> numbers may be irrational.

Please refrain from handwaving analogies.

> We begin with a list of terminating
> decimals, but it doesn't follow that the anti-diagonal is also
> terminating.  YOU HAVE TO PROVE THAT.

On the contrary, you have to prove that it is possible to express the
anti-diagonal or any irrational number by a sequence of digits. You
are so accustomed to that nonsense that you think the contrary must be

No, you have to prove that the 2500 years old proof of Hippasos is
wrong, saying that it is *not* possible to express sqrt(2) by a
fraction. You have to prove that not and "never" can be egalized by
"actually infinite".

Regards, WM