A BLATENT FLAW in Cantor's diag proof
in [Math]
|
Prev: I'm gonna try this one more time CANTOR DISPROOF
Next: Reliability: Treating the Building & the Fault Line As One & the Same
From: William Hughes on 8 Jun 2010 10:26 On Jun 8, 10:57 am, jbriggs444 <jbriggs...(a)gmail.com> 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. Sorry. I see. I had assumed that "It's not finished yet" meant there were more lines to add, not that there were more digits to add to each line. I would say your interpretation is correct. It is possible to construct a list which contains every terminating decimal and a bunch of non-terminating decimals (e.g. the algebraic numbers). - William Hughes
From: jbriggs444 on 8 Jun 2010 12:48 On Jun 8, 10:03 am, Marshall <marshall.spi...(a)gmail.com> wrote: > On Jun 7, 8:40 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > > > > > > > "Marshall" <marshall.spi...(a)gmail.com> wrote: > > > 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. > > > Here you go: > > > 1 000000 > > 2 31415 > > 3 2818 > > 4 141 > > 5 22 > > 6 7 > > > It's not finished yet! Next digit is the 7th 0 on the first number. > > At no point will this process ever produce even a single > infinite string of digits. > > Also I see no reason to think it's going to be anything vaguely > comprehensive in the vertical direction either. With a serpentine traversal, for every position on the grid there will be (for an algorithm that doesn't end up looping quietly or halting) a time by which the algorithm will have provided a digit for that position. In this sense, the algorithm specifies every digit on every line.
From: George Greene on 8 Jun 2010 13:00 On Jun 7, 10:27 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! It works on EVERY [square] list. PERIOD. > > 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. THERE IS a new digit sequence that can be formed, DUMBASS. THIS digit sequence, the one formed by THIS process, IS NEW, dumbass! THIS sequence IS NOT ON the list! This sequence differs from EVERY sequence ON the list AT A FINITE position (it's just a DIFFERENT, LATER finite position for every DIFFERENT, LATER sequence on the list). > You're all DIM! How can you form a new digit sequence when they're all > computed up to infinite length? They're NOT ALL computed to infinte length, DUMBASS! The ones ON THE LIST are computed to infinite length, but the far greater number of them NOT on the list are NOT computed AT ALL! Even though EVERY FINITE sublist of them is computed. EVERY FINITE ANYthing is computed! Computer programs can be ANY FINITE length! There is no upper limit! > > Or as George Greene puts it, they're all computed up to ALL (infinite) FINITE lengths. Right. > And as George Greene puts it there's a new digit sequence at some FINITE point. Oh, bullshit! I did NOT say THAT! The new sequence being computed IS NOT "new" at ANY "finite point"; EVERY FINITE sequence has the property that a great many elements on the list begin with that sequence!
From: George Greene on 8 Jun 2010 13:02 On Jun 7, 10:39 pm, Sylvia Else <syl...(a)not.here.invalid> wrote: > However, the computable reals are countable, so one could hardly expect > a diagonalisation argument to show that they're not, if that's where > you're coming from. He's coming from pure idiocy. He's coming from believing that the number constructed by the diagonalization argument somehow IS on the list after all (even though he can't say where), or believing that the diagonalization argument itself must be somehow flawed (since, if the thing it constructs existed, that would be a contradiction, so this construction algorithm must be failing somehow). What the diagonalization argument is intended to show is simply that the computable reals Are Not ALL the reals that there Are.
From: George Greene on 8 Jun 2010 13:08
On Jun 7, 10:41 pm, "|-|ercules" <radgray...(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? > "William Hughes" <wpihug...(a)hotmail.com> wrote > > You can't. So you have a contradiction. The assumption > > that there is a list of all real numbers is wrong. > You can't find a new sequence using diagonalization? > > Herc Of course you can find a new sequence using diagonalization (well, not you personally, since it's infinite and you're not). William Hughes doesn't know every damn thing. Your actual problem is that they're NOT "all computed up to infinite length". 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 list does NOT HAVE a LAST element! This list does not have any infinite elements! Or if you like, its elements are ALL infinite, but they just have all 0's after the initial finite segment of 3's. My point is, IF YOU HAVE A BRAIN, it is EASY to see how something could be computed to all finite lengths, yet NOT computed to ANY INfinite length. Clue: 1/3 IS NOT ON this list. Never |