Why can no one in sci.math understand my simple point?
in [Logic]
|
From: Aatu Koskensilta on 15 Jun 2010 07:53 WM <mueckenh(a)rz.fh-augsburg.de> writes: > In what form does an uncountable real exist? What is an "uncountable real"? -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Peter Webb on 15 Jun 2010 08:39 "|-|ercules" <radgray123(a)yahoo.com> wrote in message news:87ov0tFu9lU1(a)mid.individual.net... > "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote >> "|-|ercules" <radgray123(a)yahoo.com> wrote in message >> news:87om34FahrU1(a)mid.individual.net... >>> "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote >>>> "|-|ercules" <radgray123(a)yahoo.com> wrote in message >>>> news:87ocucFrn3U1(a)mid.individual.net... >>>>> Consider the list of increasing lengths of finite prefixes of pi >>>>> >>>>> 3 >>>>> 31 >>>>> 314 >>>>> 3141 >>>>> .... >>>>> >>>>> Everyone agrees that: >>>>> this list contains every digit of pi (1) >>>>> >>>> >>>> Sloppy terminology, but I agree with what I think you are trying to >>>> say. >>>> >>>>> as pi is an infinite digit sequence, this means >>>>> >>>>> this list contains every digit of an infinite digit sequence (2) >>>>> >>>> >>>> Again sloppy, but basically true. >>>> >>>>> similarly, as computable digit sequences contain increasing lengths of >>>>> ALL possible finite prefixes >>>>> >>>> >>>> Not "similarly", but if you are claiming that all Reals which have >>>> finite decimal expansions can be listed, this is correct. >>> >>> You didn't follow the similarity. >>> >>> Given the increasing finite prefixes of pi >>> >>> 3 >>> 31 >>> 314 >>> .. >>> >>> This list contains every digit of the infinite expansion of pi. >>> >> >> But pi doesn't appear on the list. >> >> So? > > > that doesn't matter, because that's a convergent sequence. > So what? If this has something to do with Cantor, you don't need to construct pi as some limit of a sequence, just whack pi in as the first in your list of Reals. Yes, Reals form the limit of sequences of rationals, and you can draw up a list of terminating rationals as Cantor's list, and any Real can be derived as the limit of some subsequence of these Rationals, but that still doesn't change the fact that the specific Real does not appear on the list. Cantors proof only applies to a list of Reals, not a sequence of Rational approximations, which are themselves uncountable. > This is what matters. > No, that sequences in list of Reals can define other Reals (as there limit) is interesting, but nothing to do with Cantor's diagonal proof.
From: Peter Webb on 15 Jun 2010 08:45 "WM" <mueckenh(a)rz.fh-augsburg.de> wrote in message news:62ae795b-1d43-4e1f-8633-e5e2475851aa(a)x21g2000yqa.googlegroups.com... > On 15 Jun., 12:26, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > >> (B) There exists a real number r, >> Forall computable reals r', >> there exists a natural number n >> such that r' and r disagree at the nth decimal place. > > > In what form does r exist, unless it is computable too? > Of course its computable. You compute it by changing the nth digit of the nth number on the list to a 7, unless it is already a 7 in which case you make it an 8. Sounds pretty easy to compute, I would have thought. I reckon I could code it in about 2 minutes. > But if r is computable, then this theorem shows that the computable > numbers are uncountable. Contradiction. > Huh? > And if r is not computable, then it is impossible to prove > disagreement with any r'. > Computing the diagonal number is actually very easy. > Regards, WM
From: Aatu Koskensilta on 15 Jun 2010 08:49 "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> writes: > "WM" <mueckenh(a)rz.fh-augsburg.de> wrote in message > news:62ae795b-1d43-4e1f-8633-e5e2475851aa(a)x21g2000yqa.googlegroups.com... >> On 15 Jun., 12:26, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: >> >>> (B) There exists a real number r, >>> Forall computable reals r', >>> there exists a natural number n >>> such that r' and r disagree at the nth decimal place. >> >> In what form does r exist, unless it is computable too? > > Of course its computable. There is a computable real that differs from every computable real? -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Daryl McCullough on 15 Jun 2010 10:06
WM says... > >On 15 Jun., 12:26, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > >> (B) There exists a real number r, >> Forall computable reals r', >> there exists a natural number n >> such that r' and r disagree at the nth decimal place. > > >In what form does r exist, unless it is computable too? r is computable *relative* to the list L of all computable reals. That is, there is an algorithm which, given an enumeration of computable reals, returns a real that is not on that list. In the theory of Turing machines, one can formalize the notion of computability relative to an "oracle", where the oracle is an infinite tape representing a possibly noncomputable function of the naturals. -- Daryl McCullough Ithaca, NY |