*NEW DIGIT SEQUENCE* is an OXYMORON!
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From: |-|ercules on 13 Jun 2010 02:21 This is proof of higher infinities! Take any list of real numbers digit expansions 123 456 789 DIAG = 159 ANTI-DIAG = 260 EVERYONE in sci.math thinks it's a NEW (NON COMPUTABLE EVEN!) digit sequence! EVERY digit sequence is computable up to ALL finite initial substrings. This means EVERY digit sequence is computable to EVERY (INFINITE AMOUNT OF) initial substrings. NOBODY ON SCI.MATH GETS IT! They think, to quote George Greene... > OF COURSE you can't find it (a new digit sequence) "at any position". > It is INFINITELY long and the differences occur at INFINITELY MANY > DIFFERENT positions! Funny thing is, nobody can paraphrase what George is talking about here! 0.xxxxxxxxxxxxx3xxxxxxxxxx5xxxxxxxxxxxxx9xxxxxxxxxxx6xxxxxxxxx... Apparently 3xxxxxxxxxx5 is not on the computable reals list! Herc -- the nonexistence of a box that contains the numbers of all the boxes that don't contain their own box number implies higher infinities. - Cantor's Proof (the holy grail of paradise in mathematics)
From: Daryl McCullough on 13 Jun 2010 08:37 |-|ercules says... >> OF COURSE you can't find it (a new digit sequence) "at any position". >> It is INFINITELY long and the differences occur at INFINITELY MANY >> DIFFERENT positions! > > >Funny thing is, nobody can paraphrase what George is talking about here! Don't say "nobody can" when the truth is that everyone can, except you. What George said was perfectly clear to anyone who is at all competent at mathematics. It was not clear to you, however. If r_0, r_1, ..., is the list of all computable reals, and d is its antidiagonal, then what George is saying is this: For each natural number n, r_n differs from d in infinitely many decimal places. What does it mean to say that they differ in infinitely many decimal places? It means that there is no *last* decimal place in which they differ. So for example: 0.12343333.... and 0.33333333... differ only in decimal places 1, 2, and 4. So decimal place number 4 is the last decimal place in which they differ. In contrast, let's compare: 0.12121212.... and 0.13131313.... These two reals differ in decimal place 2, 4, 6, etc. There is no last decimal place in which they differ. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 13 Jun 2010 08:24 |-|ercules says... >EVERY digit sequence is computable up to ALL finite initial substrings. Let r_0, r_1, ... be the list of all computable reals. Let d be the antidiagonal. Then you are confusing two different statements: 1. Forall d' such that d' is a finite approximations to d, there exists a real r_n on the list such that d' agrees with r_n to the number of decimal places in d'. 2. There exists a real r_n on the list such that forall d' such that d' is a finite approximations to d, d' agrees with r_n to the number of decimal places in d' The first sentence is true. The second sentence is false. The first sentence has nothing to do with Cantor's theorem. The second sentence is provably false by Cantor's theorem. You keep bringing up the first sentence, which has *nothing* to do with Cantor's theorem. It has nothing to do with the question of whether there are uncountably many reals. -- Daryl McCullough Ithaca, NY
From: Mike Terry on 13 Jun 2010 09:59 "|-|ercules" <radgray123(a)yahoo.com> wrote in message news:87jbo4Fe5iU1(a)mid.individual.net... > This is proof of higher infinities! > > Take any list of real numbers digit expansions > > 123 > 456 > 789 > > DIAG = 159 > ANTI-DIAG = 260 > > EVERYONE in sci.math thinks it's a NEW (NON COMPUTABLE EVEN!) digit sequence! Not quite everyone - there's one guy who's confused... :-) Anyway, I've worked out where you're going wrong below: you don't understand quantifiers properly, in particular that "seemingly minor" changes to the quantifiers can completely alter the meaning of a statement! See comments below... > > EVERY digit sequence is computable up to ALL finite initial substrings. > Yes > This means EVERY digit sequence is computable to EVERY (INFINITE AMOUNT OF) initial substrings. Yes, yes. Initial finite substrings of course (of which there are infinitely many). OK, let me state this more formally with quantifiers. (If you used quantifiers like I'm about to in all your posts, you'd avoid confusing yourself, as you'd see instantly what you were doing wrong, so I recommend you do this from now on!) // What we all agree: For all infinite digit sequence D: For all n in N: // there is a TM can compute the digit sequence // as far as position n: There exists a TM, TM_n: For all k<=n: TM_n(k) = D(k) Note that while the above is correct, this is just saying that corresponding to any D there is a sequence of TMs (TM_n) where TM_n computes D to n positions. It could be the case that all the TM_n are different, no? We HAVE NOT shown that there is a *single* TM, TM_ALL_D that works for all n similtaneously. I.e. which satisfies the following: // (WRONG:) For all infinite digit sequence D: // (a TM can compute the digit sequence for all n:) // (which is to say, the sequence D is "computable") There exists a TM, TM_ALL_D: For all k in N: TM_ALL_D(k) = D(k) Can you see that the two quantified statements are not the same? (Look at them carefully! :-) The first is saying that any D is computable to any finite length, and is correct. The second is saying that any D is computable, and is incorrect. That it is incorrect requires proof, and is maybe not obvious. However, the point is that you are claiming that the second statement is the same as the first - but it is not, as is apparent when they are written out accurately. If you think the second statement *follows* from the first somehow, then present a proof of this. (Just repeating it over and over does not constitute a proof.) > > NOBODY ON SCI.MATH GETS IT! Everyone gets it but you... > > They think, to quote George Greene... > > > OF COURSE you can't find it (a new digit sequence) "at any position". > > It is INFINITELY long and the differences occur at INFINITELY MANY > > DIFFERENT positions! > > > Funny thing is, nobody can paraphrase what George is talking about here! > Eh? You mean *you* can't paraphrase it. See my paraphrasing above. > 0.xxxxxxxxxxxxx3xxxxxxxxxx5xxxxxxxxxxxxx9xxxxxxxxxxx6xxxxxxxxx... > > Apparently 3xxxxxxxxxx5 is not on the computable reals list! Apparently to whom? (Not to George or anyone other than you...) Just goes to show it's just you who doesn't get it. Regards, Mike.
From: Colin on 13 Jun 2010 13:13 On Jun 13, 1:21 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: > This is proof of higher infinities! > > Take any list of real numbers digit expansions > > 123 > 456 > 789 > > DIAG = 159 > ANTI-DIAG = 260 > > EVERYONE in sci.math thinks it's a NEW (NON COMPUTABLE EVEN!) digit sequence! > > EVERY digit sequence is computable up to ALL finite initial substrings. > > This means EVERY digit sequence is computable to EVERY (INFINITE AMOUNT OF) initial substrings. > No, this is where you've gone off the rails. If you have an infinite sequence of digits, then, yes, every finite subsequence is computable, i.e., there's a recursive function that will enumerate such a finite subsequence. It does NOT follow that the entire infinite sequence will be computable, i.e., there is NOT a recursive function that will enumerate the entire sequence. Just because every finite subset of a set is recursive doesn't mean the original set is recursive.
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