'When Are Relations Neither True Nor False?
in [Math]
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From: Jesse F. Hughes on 23 Mar 2010 07:55 Newberry <newberryxy(a)gmail.com> writes: > It is meaningless only if ~(Ex)Px is necessarly true. If ~(Ex)Px is > contingent then > > (Ax)(Px -> Qx) > > is merely neither true nor false. In arithmetic it would indeed be > meaningless. Okay. I don't recall you mentioning this distinction previously (and I may well forget it hence). Thus. the sentence ~(E a,b)( a^2/b^2 = 2 and gcd(a,b) = 1 ) is meaningless, right? You don't have any idea what it means? Despite the fact that the classical proof that sqrt(2) is irrational proceeds thus: |- (E a,b)( a^2/b^2 = 2 ) -> (E a,b)( a^2/b^2 = 2 and gcd(a,b) = 1 ) |- ~(E a,b)( a^2/b^2 = 2 and gcd(a,b) = 1 ) -------------------------------------------------------------------- So, ~(E a,b)( a^2/b^2 = 2 ) -- "I AM serious about this being a short route to a Ph.d for some of you, but just remember, I'm the guy who proved Fermat's Last Theorem in just a bit over 6 years [...] My standards are kind of high." --James Harris, founding a new mathematical school
From: Aatu Koskensilta on 23 Mar 2010 08:36 "Jesse F. Hughes" <jesse(a)phiwumbda.org> writes: > Thus. the sentence > > ~(E a,b)( a^2/b^2 = 2 and gcd(a,b) = 1 ) > > is meaningless, right? Newberry said that (x)(Px --> Qx) is meaningless if ~(Ex)Px is necessarily true. How do you get from this the meaninglessness of ~(E a,b)( a^2/b^2 = 2 and gcd(a,b) = 1 ) which is not of the form (x)(Px --> Qx)? -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Michael Stemper on 23 Mar 2010 08:37 In article <87iq8op9if.fsf(a)dialatheia.truth.invalid>, Aatu Koskensilta <aatu.koskensilta(a)uta.fi> writes: >mstemper(a)walkabout.empros.com (Michael Stemper) writes: >> In article <ho40c402sgc(a)news5.newsguy.com>, "J. Clarke" <jclarke.usenet(a)cox.net> writes: >>> it establishes some rule by which elements of one set may be >>> associated with elements of another set. >> >> Relations *do* that, yes. > >On the set theoretic conception, most relations don't establish any rule >in the ordinary sense of the word. When you say this, do you mean "a rule that can be expressed more simply than by just listing all of the members of the set"? Would { (0,c), (0,e), (3,a), (2,b), (2,d), (1,c) } not establish a rule by which the elements of {0, 1, 2, 3} can be related with the elements of {a, b, c, d, e}? -- Michael F. Stemper #include <Standard_Disclaimer> This message contains at least 95% recycled bytes.
From: Aatu Koskensilta on 23 Mar 2010 08:40 mstemper(a)walkabout.empros.com (Michael Stemper) writes: > In article <87iq8op9if.fsf(a)dialatheia.truth.invalid>, Aatu Koskensilta > <aatu.koskensilta(a)uta.fi> writes: > >> On the set theoretic conception, most relations don't establish any >> rule in the ordinary sense of the word. > > When you say this, do you mean "a rule that can be expressed more > simply than by just listing all of the members of the set"? In general we can't list the elements of a set at all. > Would { (0,c), (0,e), (3,a), (2,b), (2,d), (1,c) } not establish a > rule by which the elements of {0, 1, 2, 3} can be related with the > elements of {a, b, c, d, e}? Sure. -- 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 23 Mar 2010 08:42
Newberry says... >It is meaningless only if ~(Ex)Px is necessarly true. If ~(Ex)Px is >contingent then > > (Ax)(Px -> Qx) > >is merely neither true nor false. In arithmetic it would indeed be >meaningless. The problem with this talk about statements being "meaningless" is that often the only way that you *know* that ~(Ex)Px is necessarly true is by proving a pair of statements: 1. (Ax) Px -> Qx 2. (Ax) Px -> ~Qx together, these imply 3. ~(Ex) Px which you claim implies that 1&2 are meaningless. Before you start declaring that certain statements are meaningless, you need a proof-theoretic result: that every meaningful true statement can be proved without using meaningless statements in an intermediate step. Another criterion that seems like it would be important is that you should either be able to recognize meaningless statements, or at the least, you should be able to prove that statements are meaningless without using meaningless statements in the proof. I think it makes more sense to say that every statement of arithmetic is meaningful. -- Daryl McCullough Ithaca, NY |