Why can no one in sci.math understand my simple point?
in [Math]
|
Prev: power of a matrix limit A^n
Next: best way of testing Dirac's new radioactivities additive creation Chapt 14 #163; ATOM TOTALITY
From: WM on 15 Jun 2010 11:44 On 15 Jun., 16:23, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > "Peter Webb" <webbfam...(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? And in what form does it exist? Regards, WM
From: WM on 15 Jun 2010 11:50 On 15 Jun., 16:32, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > 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 As every n is finite, it belongs to a finite initial segment of the infinite list. > > There is no "extrapolation" involved. The way that you prove > a fact about all n is this: > > Prove it about an unspecified n. Specified or not. n is finite anyhow and belongs to a finite initial segment of the list. Only for that always finite segment the proof is correct. > Use universal generalization. > > There is no extrapolation involved. Similarly we see that Hercules' list contains no line that is the last one. Hence there is no digit of the decimal expansion of pi that is not contained together with all its predecessors in one single line. If none is not contained, then all are contained, arn't they? Regards, WM
From: Aatu Koskensilta on 15 Jun 2010 11:59 WM <mueckenh(a)rz.fh-augsburg.de> writes: > We should not use oracles in mathematics. Nonsense. Using orcales we can show for example that the P = NP problem can't be solved using any technique that relativizes. This is a useful result. > A real is computable or not. My list contains all computable numbers: > > 0 > 1 > 00 > ... Your list doesn't appear to contain any real at all, just finite binary sequences. -- 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 12:46 WM says... > >On 15 Jun., 16:32, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: >> 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 > >As every n is finite, it belongs to a finite initial segment of the >infinite list. I'm not sure what you are saying. The fact is, we can prove that for every real r_n on the list, d is not equal to r_n. That means that d is not on the list. There is no extrapolation involved. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 15 Jun 2010 12:49
WM says... > >On 15 Jun., 16:18, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: >> Peter Webb says... >> >> >"WM" <mueck...(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. > >That is nonsense! > >The list of all definitions is possible and obviously contains all >definitions of real numbers. I was talking about the list of all *computable* reals. There are definable reals that are not computable, and Cantor's proof shows how to define one. You can similarly get a list of all definable reals for a specific language L. Then Cantor's proof allows us to come up with a new real that is not definable in language L. (It is definable in a new language that extends L). -- Daryl McCullough Ithaca, NY |