From: Jesse F. Hughes on 23 Mar 2010 09:17 "Jesse F. Hughes" <jesse(a)phiwumbda.org> writes: > Aatu Koskensilta <aatu.koskensilta(a)uta.fi> writes: > >> "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)? > > He also says that > > ~(Ex)(Px & Qx) > > is meaningless in exactly the same situations that > > (Ax)(Px -> Qx) > > is meaningless. He wants the two formulas to remain equivalent. This is mistaken, of course. ~(Ex)(Px & Qx) is equivalent to (Ax)(Px -> ~Qx), but aside from that silly mistake, the point is hopefully clear. According to Newberry, ~(Ex)(Px & Qx) is meaningless if ~(Ex)Px is necessarily true. > (Surely, you're not taking issue with the fact that I've used two > existential statements rather than one?) > > As usual, Newberry can correct me if I'm mistaken on his claims. -- "I'm the theory guy. Other people are the experimental people. If you push me on details I get annoyed, as I'm the theory guy. I'm the theoretical amateur mathematician." --James S. Harris, poet
From: Jesse F. Hughes on 23 Mar 2010 09:39 Aatu Koskensilta <aatu.koskensilta(a)uta.fi> writes: > "Jesse F. Hughes" <jesse(a)phiwumbda.org> writes: > >> He also says that >> >> ~(Ex)(Px & Qx) >> >> is meaningless in exactly the same situations that >> >> (Ax)(Px -> Qx) >> >> is meaningless. > > I see your knowledge in this field far surpasses mine! You have to keep up, Aatu. Otherwise, progress will pass you by, old man. -- Scissors and string, scissors and string, When a man's single, he lives like a king. Needles and pins, needles and pins, When a man marries, his trouble begins. --- Mother Goose
From: Jesse F. Hughes on 23 Mar 2010 09:44 Aatu Koskensilta <aatu.koskensilta(a)uta.fi> writes: > stevendaryl3016(a)yahoo.com (Daryl McCullough) writes: > >> Well, Jesse's statement can be rewritten in the form >> >> (forall a,b)( a^2/b^2 = 2 --> gcd(a,b) > 1 ) > > But does classical equivalence preserve meaninglessness? Not all classical equivalences. I think that Newberry has said that (Ax)~P may be meaningful when (Ax)(P -> 0=1) is meaningless He better say that, since otherwise (Ax)(~Px) would be meaningless unless (Ex)Px. Also, (Ax)P is meaningful iff ~(Ex)~P is meaningful. In a related matter, he's said that P is meaningless iff ~P is meaningless. But aside from these particular claims, I don't know much about his desiderata. -- Jesse F. Hughes "If the world weren't rather strange, by now I should at least be with some research group talking about my number theory research." -- James S. Harris learns the world is a funny place
From: Nam Nguyen on 23 Mar 2010 10:18 Aatu Koskensilta wrote: > Nam Nguyen <namducnguyen(a)shaw.ca> writes: > >> Is G(T), to you, a formula written in L(T) or in the language of >> arithmetic [i.e. L(PA)]? > > To me? Surely it's up to you to explain your notation. > "encoded(G(T))" is my notation, "G(T)" is actually not mine. Let me rephrase the question: >> Outside Nam's paragraph, is G(T), to you, a formula written in >> L(T) or in the language of arithmetic [i.e. L(PA)]? How I explain "encoded(G(T))" to you would depend on your answering this question.
From: James Burns on 23 Mar 2010 11:06
MoeBlee wrote: > On Mar 22, 6:30 pm, Transfer Principle > <lwal...(a)lausd.net> wrote: >>I believe so -- even if 0.999...=1 isn't such a >>statement. To me, it seems hard to believe that >>a supermajority of physicists -- who have never >>even heard of ZFC -- would be in complete >>agreement with _all_ of the theorems of ZFC that >>are related to physics (including real numbers). > > A more interesting question would be to look for > a theorem concerning only (in some sense of > "concerning") real numbers that physicists apply > but that is not a theorem of Z-"regularity". I believe that physics folklore claims that the Dirac delta function, representing a perfectly sharp peaked function, was used (by physicists) prior to the development of a justification for using it, generalized functions. Would the Dirac delta function -- prior to the development of generalized functions -- be an example of what you are asking for? The naive (physicists'?) description of the Dirac delta function is that it is zero everywhere except at zero, but when integrated yields one. This is provably impossible, unless one understands the less-than-obvious meaning of "integrate the Dirac delta function", just as one needs to know what the "..." in "0.999..." means. I believe the justification (by physicists) for using the Dirac Delta function was "It works. Use it." This suggests to me that a near-universal fraction of physicists confronted by the statement "0.999... < 1" would reply "It doesn't work. Don't use it." If 0.999... and 1 are supposed to be the result of some measurement (not too unreasonable an assumption, I hope, given that we are talking about physicists), then the physicists' view would be that they must have error bars larger than zero. Clearly, 0.999... and 1 do not differ significantly, no matter what non-zero error bounds one names. However, differing /significantly/ is part of what is implied by "0.999... < 1", in the physicist's view. Or, at least, this is how it seems to me. Jim Burns |