|
|
Re: A BLATENT FLAW in Cantor's diag proof
Posted:
Jun 8, 2010 10:21 PM
|
|
"William Hughes" <wpihughes@hotmail.com> wrote
> On Jun 8, 6:17 pm, "|-|ercules" <radgray...@yahoo.com> wrote:
>> "William Hughes" <wpihug...@hotmail.com> wrote
>>
>>
>>
>> > On Jun 8, 5:52 pm, "|-|ercules" <radgray...@yahoo.com> wrote:
>> >> "William Hughes" <wpihug...@hotmail.com> wrote
>>
>> >> > On Jun 8, 5:39 pm, "|-|ercules" <radgray...@yahoo.com> wrote:
>> >> >> "William Hughes" <wpihug...@hotmail.com> wrote
>>
>> >> >> > On Jun 8, 5:29 pm, "|-|ercules" <radgray...@yahoo.com> wrote:
>>
>> >> >> >> The infinitely many long sequences of all possible digit sequences DOESN'T MISS A SEQUENCE OF DIGITS.
>>
>> >> >> > Since every sequence has a last digit, any sequence
>> >> >> > of digits that does not have a last digit is missed.
>>
>> >> >> You're as confused as when you said there is no algorithm to produce an infinite list.
>>
>> >> >> Herc
>>
>> >> > Is this digit sequence (which does not have a last 3)
>>
>> >> > 33333...
>>
>> >> > in this list
>>
>> >> > 1 3
>> >> > 2 33
>> >> > 3 333
>> >> > ...
>>
>> >> > of sequences (all of which have a last 3).
>>
>> >> > Yes or No.
>>
>> >> > - William Hughes
>>
>> >> No.
>>
>> > So a list of sequences with last digit will miss
>> > a sequence without last digit.
>>
>> > - William Hughes
>>
>> Yes, this is a fine QUANTITATIVE argument that SUPPORTS *missing sequences*, using an example of a converging sequence.
>>
>> Unfortunately a rudimentary QUALITATIVE analysis contradicts that modifying the diagonal results in a new sequence of digits.
>
> Nope, The list contains only sequences with last digit. A sequence
> without last digit is a sequence that
> is not contained in the list.
>
> - William Hughes
>
Is pi computable?
Herc
|
|
