A BLATENT FLAW in Cantor's diag proof
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From: William Hughes on 8 Jun 2010 23:09 On Jun 8, 11:21 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > "William Hughes" <wpihug...(a)hotmail.com> wrote > > > > > On Jun 8, 6:17 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > >> "William Hughes" <wpihug...(a)hotmail.com> wrote > > >> > On Jun 8, 5:52 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > >> >> "William Hughes" <wpihug...(a)hotmail.com> wrote > > >> >> > On Jun 8, 5:39 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > >> >> >> "William Hughes" <wpihug...(a)hotmail.com> wrote > > >> >> >> > On Jun 8, 5:29 pm, "|-|ercules" <radgray...(a)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? > Yes. However, it is not in the list. The list contains 3 31 314 3145 .... It does not contain pi. - William Hughes
From: George Greene on 8 Jun 2010 23:09 On Jun 8, 4:29 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > YOU CAN'T FIND A NEW DIGIT SEQUENCE AT ANY POSITION ON THE COMPUTABLE REALS. OF COURSE you can't find it "at any position". It is INFINITELY long and the differences occur at INFINITELY MANY DIFFERENT positions! >. - N O . N E W . D I G I T . S E Q U E N C E The anti-diagonal IS ALWAYS a new digit sequence, DUMBASS.
From: |-|ercules on 8 Jun 2010 23:10 "George Greene" <greeneg(a)email.unc.edu> wrote > On Jun 8, 4:19 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> All possible digit sequences are computable to all, as in an infinite amount of, finite lengths >> >> TRUE or NOT? > > Nonsensically stupid. Barely even grammatical. > All finite amounts IS NOT "an infinite amount", not EVEN ONE infinite > amount. > Infinitely many finite amounts, taken together as a totality, still > "contain" only ZERO > infinite amounts. Having infinitely many red things does not mean you > have a blue thing, > dumbass. Your reluctance to admit a simple truth is your signature admission of error. Try this simpler question. How many natural numbers are there in "all natural numbers"? >> All possible digit sequences are computable to all, as in an infinite amount of, finite lengths. Herc
From: George Greene on 8 Jun 2010 23:12 On Jun 8, 8:22 pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > Herc's ignorance can defeat any number of teachers, no matter > how knowledgeable and patient. Herc is not the only one. I think this is giving Herc too much credit. This topic is sort of in the FAQ, or at least it should be. The concept of a limit ordinal (which is what is really going on here) is just a pons asinorum for some people. The notion that you can collect "all of" some class of ordinal and wind up with A BIGGER KIND of ordinal is just too much for some people. They just don't see how, if every x in the collection has property p, the whole collection X could still fail to also have property p. The fact that in moving from x to X, you would have to move from property p to property P, is just lost on them.
From: |-|ercules on 8 Jun 2010 23:13
"William Hughes" <wpihughes(a)hotmail.com> wrote > On Jun 8, 11:21 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> "William Hughes" <wpihug...(a)hotmail.com> wrote >> >> >> >> > On Jun 8, 6:17 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> >> "William Hughes" <wpihug...(a)hotmail.com> wrote >> >> >> > On Jun 8, 5:52 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> >> >> "William Hughes" <wpihug...(a)hotmail.com> wrote >> >> >> >> > On Jun 8, 5:39 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> >> >> >> "William Hughes" <wpihug...(a)hotmail.com> wrote >> >> >> >> >> > On Jun 8, 5:29 pm, "|-|ercules" <radgray...(a)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? >> > > Yes. However, it is not in the list. The list contains > > 3 > 31 > 314 > 3145 > ... > > It does not contain pi. > > - William Hughes Your argument seems to be there is uncountable infinity because computers can only calculate a finite number of digits. Herc |