A BLATENT FLAW in Cantor's diag proof
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From: Marshall on 7 Jun 2010 22:58 On Jun 7, 7:51 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > > 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. How are you going to do that? Write a program that first prints out an infinite sequence of zeroes, and then... Oops! Already a problem! There is no "then" that comes after writing the zeroes, because the process of writing the zeroes will never finish. Please show us this trivial program that computes every infinite digit sequence. Marshall
From: Tim Little on 7 Jun 2010 23:00 On 2010-06-08, |-|ercules <radgray123(a)yahoo.com> wrote: > You're all DIM! How can you form a new digit sequence when they're all > computed up to infinite length? You're begging the question. > Well I can't see it. I'm not surprised. - Tim
From: William Hughes on 7 Jun 2010 23:04 On Jun 7, 11:51 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > "William Hughes" <wpihug...(a)hotmail.com> wrote > > > > > On Jun 7, 11:41 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > >> "William Hughes" <wpihug...(a)hotmail.com> wrote > > >> > On Jun 7, 11:27 pm, "|-|ercules" <radgray...(a)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! > > >> >> DON'T YOU!!! > > >> >> 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 > > >> You can't find a new sequence using diagonalization? > > > 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
From: Tim Little on 7 Jun 2010 23:17 On 2010-06-08, |-|ercules <radgray123(a)yahoo.com> wrote: > I can compute the list of all computable reals. No, you can't. > It's trivial to compute a list that covers every digit sequence to > all (infinite) finite lengths. Delete "(infinite)", and your statement is correct. There is no such thing as an infinite finite length though, so inserting the word "infinite" there makes no sense. - Tim
From: |-|ercules on 7 Jun 2010 23:38
"Tim Little" <tim(a)little-possums.net> wrote > On 2010-06-08, |-|ercules <radgray123(a)yahoo.com> wrote: >> I can compute the list of all computable reals. > > No, you can't. > > >> It's trivial to compute a list that covers every digit sequence to >> all (infinite) finite lengths. > > Delete "(infinite)", and your statement is correct. There is no such > thing as an infinite finite length though, so inserting the word > "infinite" there makes no sense. > How about, all possible digit sequences are computable to all, as in an infinite amount of, finite lengths. Herc |