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Re: A BLATENT FLAW in Cantor's diag proof
Posted:
Jun 7, 2010 11:04 PM
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On Jun 7, 11:51 pm, "|-|ercules" <radgray...@yahoo.com> wrote:
> "William Hughes" <wpihug...@hotmail.com> wrote
>
>
>
> > On Jun 7, 11:41 pm, "|-|ercules" <radgray...@yahoo.com> wrote:
> >> "William Hughes" <wpihug...@hotmail.com> wrote
>
> >> > On Jun 7, 11:27 pm, "|-|ercules" <radgray...@yahoo.com> wrote:
> >> >> Here is an example of diagonalization
>
> >> >> 123
> >> >> 456
> >> >> 789
>
> >> >> Diag = 159
>
> >> >> AntiDiag = 260 <<<<<<<NEW SEQUENCE NOT ON THE LIST!
>
> >> >> YOU ALL THINK THIS WORKS ON THE LIST OF COMPUTABLE REALS!
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> >> >> DON'T YOU!!!
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> >> >> Gee it works for 159, must work in the infinite case too, who cares if there's
> >> >> no new digit sequence that can be formed.
>
> >> >> You're all DIM! How can you form a new digit sequence when they're all
> >> >> computed up to infinite length?
>
> >> > You can't. So you have a contradiction. The assumption
> >> > that there is a list of all real numbers is wrong.
>
> >> > - William Hughes
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> >> You can't find a new sequence using diagonalization?
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> > Not if you start with a list that does not exist.
>
> > - William Hughes
>
> I can compute the list of all computable reals.
> There's just some numbers that show
> up blank.
>
> It's trivial to compute a list that covers every digit sequence to all (infinite) finite lengths.
>
Indeed, however, although such a list
contains an infinite number
of sequences with a last digit,
it does not contain a sequence that
does not have a last digit. Since there are
computable sequences that do not have a last
digit, this is not a list of all computable
sequences.
-William Hughes
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