*NEW DIGIT SEQUENCE* is an OXYMORON!
in [Math]
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From: |-|ercules on 13 Jun 2010 17:09 "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote > |-|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. OK, George's statement can 'hold' if you consider a complete antidiagonal. But for noncomputable reals with a single digit that matches the diagonal anywhere it still makes no sense. And you can shuffle the list to change the diagonal, so even for an antidiagonal I doubt it holds any water. It implies something like this... 0.xxx3xxx8xxx6xxxx8... that 3xxx8 doesn't occur anywhere on the computable reals list. Herc
From: |-|ercules on 13 Jun 2010 17:19 "Mike Terry" <news.dead.person.stones(a)darjeeling.plus.com> wrote > "|-|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 GREAT! > >> 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). GREAT GREAT! George is caught out lying yet again. I add in "initial finite substring" and you can magically parse English again! > > 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! :-) Yes. all finite substrings Vs single infinite sequence. > > 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.) It indirectly follows. Here is a QUANTITATIVE analysis that shows it is NOT implied. --------------------------------------------------------- For example, consider the list 3 3.1 3.14 3.141 3.1415 3.14159 etc. The real number pi is not on this list, because pi is not a terminating decimal, and each real on the list is a terminating decimal. On the other hand, pi doesn't have any *finite* digit sequence that isn't somewhere on the list. -- Daryl McCullough Ithaca, NY ------------------------------------------------------------------------------- Here is ANOTHER QUANTITATIVE analysis that shows it is NOT implied. Consider the sequence of reals. 0.0 0.1 0.2 0.3 ... 0.9 0.11 0.12 .... 0.999 0.101 .... IT TOO contains every finite prefix of all sequences, but 3.14159... is NOT on that list. ---------------------------------------------------------------------- These are 2 QUANTITATIVE proofs that 1 does not imply 2. However a simple QUALITATIVE analysis of EVERY FINITE PREFIX clearly contradicts any new digit sequence being formed. Herc
From: Daryl McCullough on 13 Jun 2010 20:13 |-|ercules says... > >"Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote >> |-|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. > > >OK, George's statement can 'hold' if you consider a complete antidiagonal. > >But for noncomputable reals with a single digit that matches the diagonal >anywhere it still makes no sense. That statement makes no sense. >And you can shuffle the list to change the diagonal, >so even for an antidiagonal I doubt it holds any water. Exactly right. The antidiagonal depends on the ordering of the reals on the list. If you reorder them, you get a different antidiagonal, which is *ALSO* missing from the list. That shows that there are actually infinitely many reals that are not computable, one for each way that you might rearrange the list of computable reals. >It implies something like this... > >0.xxx3xxx8xxx6xxxx8... > >that 3xxx8 doesn't occur anywhere on the computable reals list. No, no, no. That's not true. *Every* finite sequence of digits appears on the list of computable reals. Let r_1, ... be the list of all computable reals. Let d be its anti-diagonal. Then: 1. d is unequal to r_1 in the first decimal place. 2. d is unequal to r_2 in the second decimal place. Now, d *might* be equal to r_2 in the first decimal place, but it is different in the *second* decimal place. 3. d is unequal to r_3 in the third decimal place. It might be equal to r_3 in the first and second decimal places. 4. d is unequal to r_4 in the fourth decimal place. It might be equal to r_4 in the first, second, and third decimal places. etc. So nobody is claiming that the first decimal place of d is missing from the list. Nobody is claiming that the first two decimal places are missing. Nobody is claiming that the first three decimal places are missing. What they are claiming is that none of the reals on the list r_1, r_2, etc. is equal to d, in all of its decimal places. Once again: To say that d is not on the list is to say: 1. An Em d is unequal to r_n in decimal place number m. To say that d can be approximated, to any finite length, by reals on the list is to say: 2. Am En d is agrees with r_n in the first m decimal places. If the list contains every computable real, then 1&2 are both true. -- Daryl McCullough Ithaca, NY
From: |-|ercules on 13 Jun 2010 20:49 "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote ... > |-|ercules says... >> >>"Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote >>> |-|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. >> >> >>OK, George's statement can 'hold' if you consider a complete antidiagonal. >> >>But for noncomputable reals with a single digit that matches the diagonal >>anywhere it still makes no sense. > > That statement makes no sense. > >>And you can shuffle the list to change the diagonal, >>so even for an antidiagonal I doubt it holds any water. > > Exactly right. The antidiagonal depends on the ordering > of the reals on the list. If you reorder them, you get > a different antidiagonal, which is *ALSO* missing from > the list. > > That shows that there are actually infinitely many > reals that are not computable, one for each way that > you might rearrange the list of computable reals. > >>It implies something like this... >> >>0.xxx3xxx8xxx6xxxx8... >> >>that 3xxx8 doesn't occur anywhere on the computable reals list. > > No, no, no. That's not true. *Every* finite sequence of digits > appears on the list of computable reals. > > Let r_1, ... be the list of all computable reals. > Let d be its anti-diagonal. > Then: > > 1. d is unequal to r_1 in the first decimal place. > 2. d is unequal to r_2 in the second decimal place. Now, > d *might* be equal to r_2 in the first decimal place, but > it is different in the *second* decimal place. > 3. d is unequal to r_3 in the third decimal place. > It might be equal to r_3 in the first and second decimal > places. > 4. d is unequal to r_4 in the fourth decimal place. > It might be equal to r_4 in the first, second, and third > decimal places. > etc. > > So nobody is claiming that the first decimal place of d > is missing from the list. Nobody is claiming that the first > two decimal places are missing. Nobody is claiming that the > first three decimal places are missing. What they are claiming > is that none of the reals on the list r_1, r_2, etc. is > equal to d, in all of its decimal places. > > Once again: > > To say that d is not on the list is to say: > > 1. An Em d is unequal to r_n in decimal place number m. > > To say that d can be approximated, to any finite length, by reals > on the list is to say: > > 2. Am En d is agrees with r_n in the first m decimal places. > > If the list contains every computable real, then 1&2 are both true. > > -- > Daryl McCullough > Ithaca, NY > You completely obfuscated my point. You just posted that George Greene's comment about numerous occurrences of different digits along the expansion had the interpretation that, the antidiagonal's digits are different at ALL digits. Is that correct? Herc
From: Daryl McCullough on 13 Jun 2010 21:25
|-|ercules says... >> Once again: >> >> To say that d is not on the list is to say: >> >> 1. An Em d is unequal to r_n in decimal place number m. >> >> To say that d can be approximated, to any finite length, by reals >> on the list is to say: >> >> 2. Am En d is agrees with r_n in the first m decimal places. >> >> If the list contains every computable real, then 1&2 are both true. >You completely obfuscated my point. Points 1&2 above are perfectly clear to someone who understands mathematics. Which may let you out... >You just posted that George Greene's comment about numerous occurrences >of different digits along the expansion had the interpretation that, >the antidiagonal's digits are different at ALL digits. No, George didn't say that. George said "It [the antidiagonal] is INFINITELY long and the differences occur at INFINITELY MANY DIFFERENT positions!" He did *NOT* say "the antidiagonal's digits are different at ALL digits". -- Daryl McCullough Ithaca, NY |