Why can no one in sci.math understand my simple point?
in [Logic]
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From: Virgil on 16 Jun 2010 01:34 In article <874oh3ir6m.fsf(a)dialatheia.truth.invalid>, Aatu Koskensilta <aatu.koskensilta(a)uta.fi> wrote: > Virgil <Virgil(a)home.esc> writes: > > > But until you can determine which of those 10 cases, how can you > > compute the number? > > You can't. The number is computable nonetheless, in the sense that there > exists an effective procedure for churning out its decimal expansion. Except for that one digit. > > As noted, computability is a purely extensional notion. Recall the > classical recursion theory exercise, which we find, in some form or > other, in pretty much any text on the subject: > > Let f : N --> N be a function such that > > f(x) = 0 if Goldbach's conjecture is true, and 1 otherwise. > > Is f computable? Not yet!
From: WM on 16 Jun 2010 01:56 On 15 Jun., 21:03, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > WM says... > > >On 15 Jun., 18:46, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > >> 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. > > >Of course. Every real r_n belongs to a finite initial segment of the > >list. > >That does not yield any result about the whole list > > On the contrary, the definition of "d is on the list" > is that "there exists a natural number n such that r_n = d". You make it too easy. r_n = d would mean that there is a finite list containing d. > We proved "forall n, r_n is not equal to d". So that > means "there does not exist a natural number n such that r_n = d", > so that means "d is not on the list". > > We have thus proved something about the whole list. Wrong. You proved something for every finite list. Like for every finite set: |{2, 4, 6,..., 2n}| is not larger than each of its elements. In the same way you may prove that Hercules' list does not contain pi and (!!!) that every single line that contains all digits of Hercules' list 3.1415 and so on in infinity does not contain pi. You are very inconsequent, always changing the meaning of for all. > You have to actually learn logic to be able to tell the difference. No, one has only to become a set theorist and parrot every nonsense of that sad "theory". Regards, WM
From: Peter Webb on 16 Jun 2010 02:22 "Virgil" <Virgil(a)home.esc> wrote in message news:Virgil-66F8AC.23332415062010(a)bignews.usenetmonster.com... > In article <4c1841c6$0$17174$afc38c87(a)news.optusnet.com.au>, > "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: > >> "Virgil" <Virgil(a)home.esc> wrote in message >> news:Virgil-966F99.20493515062010(a)bignews.usenetmonster.com... >> > In article <4c181d87$0$17178$afc38c87(a)news.optusnet.com.au>, >> > "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote: >> > >> >> > Nevertheless your "definition" belongs to a countable set, hence it >> >> > is >> >> > no example to save Cantors "proof". >> >> > >> >> > Either all entries of the lines of the list are defined and the >> >> > diagonal is defined (in the same language) too. >> >> >> >> Yes. If you provide a list of Reals, then the diagonal is computable >> >> and >> >> does not appear on the list. >> >> >> >> > Then the proof shows >> >> > that the countable set of defined reals is uncountable. >> >> >> >> No, it shows that all "definable" (computable) Reals cannot be >> >> explicitly >> >> listed. This is *not* the same as being uncountable. >> >> >> >> >> >> > Or it does not >> >> > show anything at all. >> >> > >> >> >> >> It shows that all "definable" (computable) Reals cannot be explicitly >> >> listed. This is a well known proof in set theory. This is *not* the >> >> same >> >> as >> >> being uncountable. >> >> >> >> >> >> > To switch "languages" is the most lame argument one could think of. >> >> > The diagonal argument does not switch languages. And it cannot be >> >> > applied at all because the list of all finite defiitions does not >> >> > contain infinite entries. Those however are required for the >> >> > diagonal >> >> > argument. >> >> > >> >> >> >> No, that paragraph above is close to gibberish. Cantor said and proved >> >> that >> >> any purported list of all Reals cannot contain all Reals. His proof is >> >> simple and clear, provides an explicit construction of at least one >> >> missing >> >> Real, and does contain or require any concepts of uncomputable >> >> numbers, >> >> or >> >> use of the Axiom of Choice. >> > >> > Isn't there a missing "not" in that last sentence between "does" and >> > "contain"? >> >> Oops. > > When I was younger, one excused such thing by saying "Even Homer nods", > but the young today don't seem to know about the Iliad and Odyssey. Well, I am old, and whilst I am familiar with Homer Simpson, I don't know these other two characters. Perhaps I should have said "Do'h".
From: Herman Jurjus on 16 Jun 2010 04:10 On 6/16/2010 5:37 AM, Aatu Koskensilta wrote: > Virgil<Virgil(a)home.esc> writes: > >> But until you can determine which of those 10 cases, how can you >> compute the number? > > You can't. The number is computable nonetheless, in the sense that there > exists an effective procedure for churning out its decimal expansion. > > As noted, computability is a purely extensional notion. Recall the > classical recursion theory exercise, which we find, in some form or > other, in pretty much any text on the subject: > > Let f : N --> N be a function such that > > f(x) = 0 if Goldbach's conjecture is true, and 1 otherwise. > > Is f computable? To Virgil: Also many classical mathematicians appreciate this as an example showing that the extensional notion 'Turing computable' is a slight distortion of the intuitive notion 'computable'. -- Cheers, Herman Jurjus
From: WM on 16 Jun 2010 04:37
On 16 Jun., 05:37, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > Virgil <Vir...(a)home.esc> writes: > > But until you can determine which of those 10 cases, how can you > > compute the number? > > You can't. The number is computable nonetheless, in the sense that there > exists an effective procedure for churning out its decimal expansion. > > As noted, computability is a purely extensional notion. Recall the > classical recursion theory exercise, which we find, in some form or > other, in pretty much any text on the subject: > > Let f : N --> N be a function such that > > f(x) = 0 if Goldbach's conjecture is true, and 1 otherwise. > > Is f computable? All computable, Turing-computable, nonetheless computable, definable (in any useful, i.e., finite language), and by other means determinable numbers form a countable set. If Cantor's diagonal proof results in any such a number, then it proves in effect the uncountability of a countable set. Otherwise it proves nothing, because an infinite sequence of digits without other information does not determine anything. Regards, WM |