A BLATENT FLAW in Cantor's diag proof
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From: William Hughes on 8 Jun 2010 17:10 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
From: |-|ercules on 8 Jun 2010 17:17 "William Hughes" <wpihughes(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. All possible digit sequences are computable to all, as in an infinite amount of, finite lengths. Herc
From: William Hughes on 8 Jun 2010 17:29 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
From: James Burns on 8 Jun 2010 17:39 Marshall wrote: > At no point, ever, in the serpentine traversal, will > we have even a single infinitely long line of digits. > We never ever produce even one infinite digit > sequence this way; we certainly cannot say that > we are producing all of them. The way you phrased things created an impression of incompleteness. As a technical matter, I do not disagree with you in the slightest, but it might be worth remembering that there are different ways of phrasing things that create impressions of completeness, and, technically, those ways are just as accurate. For example, we could ask which digits remain uncalculated. There are none, you say? That sounds to me as though we /can/ say that we are producing all of the digits. The moral of the story seems to be, "Be wary of any translations from Math to English, even easy ones." Jim Burns
From: herbzet on 8 Jun 2010 18:59
herbzet wrote: > jbriggs444 wrote: > > William Hughes wrote: > > > jbriggs444 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. Each and every digit sequence produced ends in an infinite sequence of zeroes. But I'll bet Herk can jigger it further to make it work right! -- hz |