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From: |-|ercules on 20 Jun 2010 03:14 Hypothesis: a real number contains a finite sequence that is not computable. Contradiction Therefore: all digits of every real are contained in the list of computable reals. _________________________________________________________________ This may not IMPLY that all infinite digit sequences are computable, but it trivially defeats this argument: 123 456 789 Diag = 159 AntiDiag = 260 A new digit sequence can be found on all real lists. Herc -- If you ever rob someone, even to get your own stuff back, don't use the phrase "Nobody leave the room!" ~ OJ Simpson
From: Math-a-nator on 20 Jun 2010 11:33 POST IT *AGAIN*!!!! *POST IT* TO SCI.MATH!!!! POST IT *EVERY SINGLE* DAY!!!! POST IT *BEFORE* YOU POST IT!!!! POST IT *AFTER* YOU POST IT!!!! POST IT *AGAIN*!!!! "|-|ercules" <radgray123(a)yahoo.com> wrote in message news:885tecF9rsU1(a)mid.individual.net... > Hypothesis: Ah have nothin to day and ah am sayin it.
From: |-|ercules on 20 Jun 2010 12:45 "Math-a-nator" <MorePornLips(a)example.com> wrote... >> Hypothesis: Ah have nothin to day and ah am sayin it. > > it's worth thinking what "anti-diagonals" entail. You're not just constructing 0.444454445544444445444.. a 4 for every non 4 digit and a 5 for a 4. You're constructing ALL 9 OTHER DIGITS to the diagonal digits. And it's not just the diagonal, it's the diagonal of ALL PERMUTATIONS OF THE LIST. So, the first digit of the list can be... well anything, so the antidiagonal starts with anything.. then the second digit of the second real can be anything, so the antidiagonals next digit is anything.. No wonder you think there are infinitely more irrationals for every rational. So here is your formula for finding a new real. 0.xxxxxx 0.yyyyy 0.zzzzzz 0.aaaaa 0.bbbbb Pick ANY row and pick ANY digit other than the first digit. So instead of 'a' we'll choose 7. Pick ANY row other than aaaa and pick ANY digit other than the second digit. So instead of 'y' we'll choose 4. Is that a *new sequence* or is it anything at all? Considering all possible digit sequences occur INFINITELY WIDE I don't think you can do it! Herc
From: Colin on 20 Jun 2010 13:19 On Jun 20, 11:45 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: > [snip] (Yawn.) So, you're spewing your usual jism that there's nothing more to the real numbers than the computable reals, i.e., that "real number" and "computable real" are synonyms? Well, Master Debater, riddle me this: everyone knows the real numbers are closed under the basic operation of taking the supremum of a bounded sequence. Can you prove this is true of the computable reals?
From: |-|ercules on 20 Jun 2010 13:39
"Colin" <colinpoakes(a)hotmail.com> wrote > On Jun 20, 11:45 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> [snip] > > (Yawn.) So, you're spewing your usual jism that there's nothing more > to the real numbers than the computable reals, i.e., that "real > number" and "computable real" are synonyms? Well, Master Debater, > riddle me this: everyone knows the real numbers are closed under the > basic operation of taking the supremum of a bounded sequence. Can you > prove this is true of the computable reals? You have 1000 proofs on real numbers all entrenched together, based on the all new "260". 123 456 789 Diag = 159 Anti-Diag = 260 I'm not writing your axioms for you, and I'm not claiming there are no uncomputable reals at this point, I'm merely stating what fools 99% of mathematicians are. Herc |