Why can no one in sci.math understand my simple point?
in [Logic]
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From: Peter Webb on 15 Jun 2010 10:07 "Aatu Koskensilta" <aatu.koskensilta(a)uta.fi> wrote in message news:87typ431fz.fsf(a)dialatheia.truth.invalid... > "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? > Well, what I think he is really going on about is whether the diagonal number is computable, hence the reference to a "natural" number n in which the digits vary. Taken at face value, that "there exists a natural number n such that r and r' disagree at the nth decimal place" is simply the statement that r <> r'. So (B) is equivalent to the statement "there exists an uncomputable number". Which is true, of course, although I acknowledge that I cannot provide the decimal expansion of one (by definition).
From: Daryl McCullough on 15 Jun 2010 10:17 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? Another answer to the question is that r is *definable*. Every real number between 0 and 1 can be represented as an infinite series of the form: sum from n=1 to infinity of a_n 2^{-n}. So a real number r can be said to be definable by a formula Phi(x) in the language of arithmetic if r = sum over all n such that Phi(n) of 2^{-n} The formula Phi(x) uniquely determines the value of r. In this sense, the antidiagonal of the list of all computable reals is definable (but not computable). -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 15 Jun 2010 10:18 Peter Webb says... >"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. No, it's computable *relative* to the list of all computable reals. But that list is not computable. -- Daryl McCullough Ithaca, NY
From: Aatu Koskensilta on 15 Jun 2010 10:23 "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> writes: > So (B) is equivalent to the statement "there exists an uncomputable > number". Right. But why then did you say the number was computable? -- 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 10:32
WM says... > >On 15 Jun., 12:39, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > >> That's *all* that matters, for Cantor's theorem. The claim >> is that for every list of reals, there is another real >> that does not appear on the list. > >The claim is only proved for every finite subset of the list. The proof does not make use of any property of infinite lists. The proof establishes: (If r_n is the list of reals, and d is the antidiagonal) forall n, d is not equal to r_n There is no "extrapolation" involved. The way that you prove a fact about all n is this: Prove it about an unspecified n. Use universal generalization. There is no extrapolation involved. -- Daryl McCullough Ithaca, NY |