*NEW DIGIT SEQUENCE* is an OXYMORON!
in [Math]
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From: |-|ercules on 13 Jun 2010 21:39 "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote ... > |-|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". What about this statement which another poster made, but no point in searching for it, hopefully. "Reals that are not on the computable reals list differ at one, numerous, infinite, or all digits." Of forget it, you won't address the question you'll just reiterate your digit(n,n) is different because we defined it that way" argument. You're like talking to an encyclopedia. Herc
From: Mike Terry on 13 Jun 2010 19:31 "|-|ercules" <radgray123(a)yahoo.com> wrote in message news:87l0amF4k2U1(a)mid.individual.net... > "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. > Correct - the digit sequence for Pi is clearly not in either of the lists. > However a simple QUALITATIVE analysis of EVERY FINITE PREFIX > clearly contradicts any new digit sequence being formed. You're just pointing out that maybe no new finite digit prefix is formed. However, a new infinite digit sequence is formed. Cantor's proof is discussing *Real* numbers, which correspond to *infinite* digit sequences, not finite ones. So the antidiagonal gives a new Real not in the original list. (Although maybe all finite prefixes are in the list, as you've shown is possible.) Regards, Mike. > > Herc >
From: |-|ercules on 13 Jun 2010 21:53 "George Greene" <greeneg(a)email.unc.edu> wrote > If you mean all of them collectively, then yes, that is an infinite number > of things, but that does NOT mean that > the infinite COLLECTION has any property (LIKE COMPUTABILITY) that the > finite individual parts (or initial subcollections) have. So you admit there is an infinite number of finite prefixes of all digits. But you still think there are *new digit sequences* that it misses. Here is a counterintuitive argument about properties of infinite SETS and their ELEMENTS. In pure functional programming there is no repeated input, there is a single input stream and there is some tricky redirection in order to process that stream interactively. Say you have a hangman program and the input from the keyboard is "abekdsierktzpleo" The first step is to reorganize the input into a stack of increasing inputs. a ab abe abek ... then apply the hangman check function to each element of the stack and produce a list of outputs "_ _ _, _ _ _, _ e _, _ e _,..." Because functional interpreters are LAZY, they can produce the partial output as the partial input is received, and it produces an interactive hangman game, despite the only input being an infinite stream. In this context, 3 31 314 3141 31415 .... and 31415... are 100% EQUIVALENT DATA. So when you consider the whole set, it DOES have a new property. That all digit sequences are computable to infinite length. Herc
From: |-|ercules on 13 Jun 2010 21:59 "Mike Terry" <news.dead.person.stones(a)darjeeling.plus.com> wrote >> >> These are 2 QUANTITATIVE proofs that 1 does not imply 2. >> > > Correct - the digit sequence for Pi is clearly not in either of the lists. But all of pi's digits are! Herc
From: George Greene on 14 Jun 2010 02:04
On Jun 14, 12:43 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: > Do you agree that all the digits of pi are in this set? > > 3 > 31 > 314 Of course we do, but the infinite digit string 111111.... NEVER OCCURS IN THIS SET, DUMBASS. NOR DOES PI ITSELF, DUMBASS, because EVERY string in this set IS FINITE, DUMBASS! THERE ARE NO INFINITE DIGIT STRINGS IN THIS SET, DUMBASS! |