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Re: Matheology § 233
Posted:
Mar 30, 2013 9:38 AM
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On 29 Mrz., 19:34, Virgil <vir...@ligriv.com> wrote:
>
> > So we have established the fact that an irrational number has no node
> > of its own.
>
> No number in any infinite binary tree has any node "of its own", as
> every node has two child nodes belonging to necessarily different
> numbers.
That is correct, but only establishes the fact that no actually
infinite path can be distinguished from all rational paths as should
be possible in a Cantor-list - but is not.
> > The first bit causes that the anti-
> > diagonal differs from 2^-1 of all entries and agrees with 2^-1, the
> > second bit leaves 2^2 entries and the n-th bits leaves 2-^n entries of
> > the list agreeing with the anti-diagonal. There is no bit that would
> > leave zero entries agreeing with the diagonal, since there is no bit
> > at position infinity.
>
> Thus there is always at least one bit of any listed entry disagreeing
> with the antidiagonanl, just as the Cantor proof requires.
In a list containing every rational: Is there always, i.e., up to
every digit, an infinite set of paths identical with the anti-
diagonal? Yes or no?
Regards, WM
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