A BLATENT FLAW in Cantor's diag proof
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From: herbzet on 8 Jun 2010 15:40 jbriggs444 wrote: > On Jun 8, 9:14 am, William Hughes <wpihug...(a)hotmail.com> wrote: > > On Jun 8, 10:06 am, jbriggs444 <jbriggs...(a)gmail.com> wrote: > > > > > Non-terminating decimals are not a problem for this output > > > convention. At a guess, lines 2 and 3 at least are just > > > such digit sequences (pi and e). > > > > You live on a very strange planet. On my planet line > > 2 has five digits and line 3 has four digits. > > I flatter myself that I understand roughly where Herc is > heading. His algorithm has not yet generated the remaining > digits on lines 2 or 3. As he wrote, it's just getting > ready to output the next digit on line 1. > > This overlooks the fact that he promised us an algorithm > and in its place gave us some partial output. Right. The algorism is to systematically produce every possible digit sequence. Limiting ourselves to two digits, we start with the idea of enumerating every finite digit sequence -- sort of like an infinite truth table: a b c ... ----- 1) 0 0 0 ... 2) 1 0 0 ... 3) 0 1 0 ... 4) 1 1 0 ... 5) 0 0 1 ... 6) 1 0 1 ... 7) 0 1 1 ... 8) 1 1 1 ... . . . . . . . . . (The pattern under each column a,b,c, is familiar.) Of course we have to do this in a serpentine fashion: a b c d e f g h i j etc. ------------------- 1) 0 0 0 0 0 0 0 0 0 0 2) 1 0 0 0 0 0 0 0 0 3) 0 1 0 0 0 0 0 0 4) 1 1 0 0 0 0 0 5) 0 0 1 0 0 0 6) 1 0 1 0 0 7) 0 1 1 0 8) 1 1 1 9) 0 0 10) 1 etc. Now if we let this process run "to infinity", then we get every possible infinite binary digit sequence! Not. -- hz
From: |-|ercules on 8 Jun 2010 16:19 "George Greene" <greeneg(a)email.unc.edu> wrote > On Jun 7, 11:17 pm, Tim Little <t...(a)little-possums.net> wrote: >> On 2010-06-08, |-|ercules <radgray...(a)yahoo.com> wrote: >> 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. > > It makes a LITTLE bit of sense. What he MEANT was that > there are infiniteLY MANY of these different finite lengths and > that there is NO finite upper bound on them. > Partial credit for George. All possible digit sequences are computable to all, as in an infinite amount of, finite lengths TRUE or NOT? Herc
From: |-|ercules on 8 Jun 2010 16:23 "George Greene" <greeneg(a)email.unc.edu> wrote >> I can compute the list of all computable reals. >> There's just some numbers that show up blank. > > Are these "numbers that show up blank" > computable reals or NON-computable reals? > Haven't you lost the argument if you admit that non-computable reals > even exist at all??? This is the SECOND reason you FAIL. The one or two uncomputable reals you defined FIT REALLY WELL WITH THE THEORY OF *UNCOUNTABLE* REALS! IT'S A WITNESS TO SUPERINFINITY! IT MUST BE TRUE dreams George! JUST ANSWER THE QUESTION YES OR NO. All possible digit sequences are computable to all, as in an infinite amount of, finite lengths. That's the FIRST reason you FAIL. You think a NEW DIGIT SEQUENCE is found just like 159 and 260. Herc
From: William Hughes on 8 Jun 2010 16:28 On Jun 8, 5:19 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > "George Greene" <gree...(a)email.unc.edu> wrote > > > On Jun 7, 11:17 pm, Tim Little <t...(a)little-possums.net> wrote: > >> On 2010-06-08, |-|ercules <radgray...(a)yahoo.com> wrote: > >> 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. > > > It makes a LITTLE bit of sense. What he MEANT was that > > there are infiniteLY MANY of these different finite lengths and > > that there is NO finite upper bound on them. > > Partial credit for George. > > All possible digit sequences are computable to all, as in an infinite amount of, finite lengths > > TRUE or NOT? > > Herc True. However, this does not mean what you think it means. The fact that for a number y, every subsequence of the digit sequence of y that has a last digit is computable, does not mean that the digit sequence of y (which does not have a last digit) is computable. There is no way to combine all the computable subsequences to get a computable sequence. - William Hughes
From: |-|ercules on 8 Jun 2010 16:29
"George Greene" <greeneg(a)email.unc.edu> wrote > They're all computed up to ALL FINITE lengths. > Hint: > Suppose you were trying to decimally approximate 1/3. > Suppose you had an infinite list OF FINITE sequences approximating it: > .3 > .33 > .333 > .3333 > .33333 > etc., i.e., the nth element on this list is a decimal point followed > by n 3's. > The list has denumerably (i.e. countably many) elements and EVERY > finite > approximation (or element or finite sequence) on the list occurs at a > KNOWN FINITE > position on the list; there is an element for every (positive) natural > n, and every natural > n has an element (namely, the string of a decimal point followed by n > 3's). > > Despite the fact that the decimal approximation for 1/3 is computed > UP TO EVERY FINITE length, on this list, THE INFINITE approximation > that is EXACTLY 1/3 IS *NOT* ON this list! This is the THIRD reason you fail. ironically it's the ole 0.3 0.33 0.333 ... analogy. SO WHAT DUMBASS. THAT SEQUENCE CONVERGES. It's easy to show a DIFFERENT number. The infinitely many long sequences of all possible digit sequences DOESN'T MISS A SEQUENCE OF DIGITS. It's 100% dense, covering every digit sequence vertically to 100%, of digit positions horizontally to 100%. YOU CAN'T FIND A NEW DIGIT SEQUENCE AT ANY POSITION ON THE COMPUTABLE REALS. FIND A BRAIN ONE OF YOU. - N O . N E W . D I G I T . S E Q U E N C E Herc |