Wikipedia:Reference desk/Archives/Mathematics/2012 July 29

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July 29

under a coordinate transformation

$a(x,y){\frac {\partial ^{2}u}{\partial x^{2}}}+2b(x,y){\frac {\partial ^{2}u}{\partial x\partial y}}+c(x,y){\frac {\partial ^{2}u}{\partial y^{2}}}+d(x,y,u,{\frac {\partial u}{\partial x}},{\frac {\partial u}{\partial y}})=0$.

Under a coordinate transform $\xi =\xi (x,y),\eta =\eta (x,y)$ we magically get the following:

$\alpha (\xi ,\eta ){\frac {\partial ^{2}u}{\partial \xi ^{2}}}+2\beta (\xi ,\eta ){\frac {\partial ^{2}u}{\partial \xi \partial \eta }}+\gamma (\xi ,\eta ){\frac {\partial ^{2}u}{\partial \eta ^{2}}}+\delta (\xi ,\eta ,u,{\frac {\partial u}{\partial \eta }},{\frac {\partial u}{\partial \eta }})=0$ where $\gamma (\xi ,\eta )=a\eta _{x}^{2}+2b\eta _{x}\eta _{y}+c\eta _{y}^{2}$.

Could somebody please explain. --150.203.114.14 (talk) 07:20, 29 July 2012 (UTC)

I don't think you should be able to prove it.

So, someone here refused to suppose that a prime, j, was the sum of a certain number of perfect numbers. This makes sense if the perfect numbers are all even, as, with the first one being 6, you could never get to an odd prime by adding even numbers. (I say "odd prime" because above the first perfect number they're all odd.)

So, I wonder if he tacitly was refusing to suppose that there was an odd perfect number. I looked at the article and there is no proof - only strong heuristics that there wouldn't be.

I personally think that it can't ever be proven. 84.3.160.86 (talk) 08:38, 29 July 2012 (UTC)

What is your question? --Tango (talk) 14:55, 29 July 2012 (UTC)

as stated. I've given you my reasoning. 84.3.160.86 (talk) 23:51, 29 July 2012 (UTC)

well, I guess I haven't given my reasoning. The question is I personally think that it can't ever be proven? 84.3.160.86 (talk) 23:54, 29 July 2012 (UTC)

That isn't a question. Can you rephrase? Are you asking if we think it can be proven? I believe consensus is that it will someday be proven that there is no odd perfect number, but this is just a (educated) guess. Staecker (talk) 00:47, 30 July 2012 (UTC)

Why? Don't mathematicians expect loads of statements to turn out to be true-but-only-statistically-provable, being able to produce a proof that it's not actually provable one way or another, while having an actual reality and not being independent? 84.3.160.86 (talk) 01:35, 30 July 2012 (UTC)

Yes, but perfect numbers aren't the kind of thing we would expect to run into logical difficulties with. They aren't weird enough for that! --Tango (talk) 20:04, 31 July 2012 (UTC)

Now that I've gotten the answers below, isn't "while having an actual reality and not being independent?" wrong - shouldn't it read "while having an actual reality despite being independent?" 84.3.160.86 (talk) 13:05, 1 August 2012 (UTC)

I've also asked another question below which you may be interested in. 84.3.160.86 (talk) 00:01, 30 July 2012 (UTC)

is "can't be proof" synonymous with independent?

Suppose I produce this fine glowing proof on one sheet of crisp high-quality slightly rosy-colored paper that it is impossible to prove that there are no odd perfect numbers. Mathematicians look over the proof but find no flaw and accept it. (And not because you can't argue with high-quality paper.)

So, we now have an accepted proof that we can't prove that there are no odd perfect numbers. Does the existence of this correct proof ==> that suddenly this question has become independent and ==> now I can make a mathematics that is completely the same as the standard mathematics accept that it assumes there is an odd perfect number?

Or, even though there is a proof that we can't prove that there are no odd perfect numbers, is it still possible for it to be impossible for it to be independent? (for example due to the incredibly strong heuristics.)

I guess what I am saying is, is a proof that something can't be proved at once a proof that under the same system it cannot be wrong (i.e. must be independent) to assume that it is true? 84.3.160.86 (talk) 23:59, 29 July 2012 (UTC)

Well, you need to specify what axioms are incapable of proving it. I can construct a very short proof that there are no odd perfect numbers, by taking it as an axiom. Once you've specified your axiom system, then yes, if you can prove a statement unprovable, you can add its negation to your axiom system and have a fine new system. This is either a consequence of Godel's completeness theorem, or a consequence of the fact that classical logic supports proof by contradiction, depending on what you require of an axiom system.--119.225.45.242 (talk) 00:09, 30 July 2012 (UTC)

Can you explain more about the last part? Let's take our normal axioms, nothing extra. Now we examine the "no odd perfect numbers" (nopn) statement, which there is incredibly compelling heuristics for believing. It is vanishingly unlikely to be false. Suppose that we have a proof, P, that the nopn is unprovable. Now might it not still be wrong to assume either nopn or to assume aopn, since, in fact, there could be a reality? (the reality for which we have incredibly statistical certainty or its opposite reality).

(I guess what I'm asking is that if you prove that nopn is unprovable, but the reality of the system - if you could do an infinite check real quick - is that the numbers that match opn are {} - then you, thinking nopn is unprovable, might feel free to add an axiom that is it is false -- as you've just said is allowed -- but in the physical sense under that axiom you've just taken a false statement as an axiom!!! which is obviously horrifically bad.

(it's like this: if under some weak system it is impossible to prove something trivial - but you can prove this non-provability - that doesn't make it independent - it could still have a truth-value, couldn't it? Therefore, if it has a truth value, then if you pick the wrong side you've just made your system include an axiom that is false under the other axioms. So it's not really independent at all, even though you can't prove it. Only by not including it do you have a consistent system.

(So, assuming our current system is consistent, couldn't it be made inconsistent by trying to add the axiom as you suggest, after proving it's unprovable?) 84.3.160.86 (talk) 01:32, 30 July 2012 (UTC)

You're confusing several different notions here. When you talk about "a reality", or about statements "having a truth-value", you're talking about whether or not statements are true in the standard model of arithmetic. This is the setting we all know and love--the natural numbers with the usual arithmetic operations on them.

Suppose you've proven that nopn can neither be proven nor disproven in PA. Then you've shown that nopn is independent of PA, and so PA + aopn is consistent. This new system doesn't describe the standard model of arithmetic, but that doesn't make horrifically bad or not really independent. The axioms of group theory don't describe the standard model of arithmetic, but that's no reason to condemn them.--119.225.45.242 (talk) 02:09, 30 July 2012 (UTC)

How do you go from "Suppose you've proven that nopn can neither be proven nor disproven in PA. Then you've shown that nopn is independent of PA" is what I want to know.

Definition. Independence (mathematical logic).--119.225.45.242 (talk) 02:35, 30 July 2012 (UTC)

This is interesting. So, as long as you can't prove a false statement or it's opposite, you can assume the false statement as an axiom and nothing bad will happen? 84.3.160.86 (talk) 08:58, 30 July 2012 (UTC)

No, that's not correct. You won't be able to prove a contradiction. But that doesn't mean "nothing bad will happen". You will be able to prove false things (the false axiom, as one example, will have a trivial proof), and it is bad to be able to prove false things, because that means your proof system is not reliable. --Trovatore (talk) 10:26, 30 July 2012 (UTC)

So you're saying independence is independent of truth, so to speak. So, is the axiom of choice actually probably true under the rest of zfc or actually probably false under the rest of zfc. (not that I'm not asking about our world,but the world defined only by zfc and nother else). 84.3.160.86 (talk) 14:04, 30 July 2012 (UTC)

It might very much be dependent on PA - after all, PA is what produces all the statistical evidence we have for it!!!! - and have one monstruously definite truth value. However PA might be just weak enough that that truth value cannot be proven. How do you then go and say it could be the opposite of what it would be if PA were slightly stronger and could prove it?

Basically, you are saying that the moment I make a system just dumb enough to not be able to prove that Pi is irrational - but we do have pi in that system, which has the same value there as anywhere else - then you might as well assume the axioms that lead to the normal value of pi, and also that it terminates. But this is now inconsistent.

What I'm saying is just because you're a shade too weak to prove something doesn't mean it can't still be true. So there could be a special case of you being just strong enough to prove that you're just weak enough not to be able to prove something, but also that you're strong enough for these axioms to definitely imply a truth-value under the same axioms.

It's like saying that if I prove to myself that I'm not smart enough to prove whether there is an integer that is both divisible by 2 and divisible by 3, but not divisible by 6, then suddenly this is independent. Well, no. There is no such number, regardless of whether I've just proven to myself that I'm not smart enough to prove it. 84.3.160.86 (talk) 02:22, 30 July 2012 (UTC)

This looks suspiciously like it's veering off into a discussion of the nature of truth, but I'll try to explain the point in the third paragraph. The point is you have "you're strong enough for these axioms to definitely imply a truth-value under the same axioms". The question is what exactly is meant by "imply". Assuming we're (sufficiently strong - note that strength here has got nothing to do with how much you know about arithmetic, and is dealing with things like A->B and B->C implying A->C) in first-order logic of some kind (such as weak PA), Godel's completeness theorem says that imply in the sense of "can be proved from" (syntactic) and in the sense of "must be true if _ are" (semantic) are exactly the same. Now I've written that, I've realised you might not be in weak PA but in PA with a full induction axiom (all subsets containing 0 and successors of everything they contain are the whole model): our article on the Peano axioms#Models says that this any model of this is ordinary arithmetic. For the record I also ought to point out that strong heuristic evidence is not proof, especially in number theory. Does that help at all? Straightontillmorning (talk) 18:35, 30 July 2012 (UTC)

I'm having trouble understanding what you're saying. Let me tell you more about my thinking. Suppose I the following axiom: a sibling is a boy or a girl but not both. Now suppose I take as axiomatic that "I have two siblings". Can you prove (from this and the other axiom, plus the rest of zfc+) they are both girls? No. Can you prove they are both boys? No. Can you prove that one is a girl, one is a boy? No. But can you prove that at least one is a boy or a girl? Yes. So returning to the first two statements "Both are girls" is neither false nor true under the axiomatic system - it really is totally independent and has NO truth value under just those axioms.

So this is one kind of "independence" - lack of enough "implicative" power in the axioms to pigeonhole the thing to a truth value. But there is the other kind -> where the axioms DO pigeonhole the thing into having an actual truth-value. Thus if some axiomatic system comes up with Pi but can't determine whether it's rational or irrational - well, despite the fact that they can't determine it, if it's the same pi we are talking about then the answer is "it's irrational". So there is a truth-value there, JUST from the rest of the axioms. It's impossible to come up with a true system (i.e. one where you can't prove statements that are actually false in that system) where Pi has the same value it has in our system but is proved to be rational. Sure, with the right weak system of axioms maybe they can come up with a non-constructive proof of hte false statement that there is an exact numerator and denominator that is the reduced value of Pi - but it will be false despite being consistent there. So whereas I can have two sisters or I can have two brothers, and you can't prove it from what I've told you and EITHER could be correct, if pi could be rational and pi could be irrational and "EITHER" could be correct, in fact one "can't". One is really false. Do you see my thinking? That there are two kinds of independent axioms: ones that actually have a pigeon-holed value (such as the truth that pi is irrational in any system that produces its true - our - value) and others that actually are free to be one way or another (such as having two brothers or two sisters in the axiomatic system that says siblings are just male or female. likewise in that system it is simply "false" to say someone could have 1 sibling, and at least one sister and one brother - even if they can't prove it because their axiomatic system is too weak for the pigeonhole principle.) DO you get where I'm coming from? 84.3.160.86 (talk) 20:56, 30 July 2012 (UTC)

I understand what you're saying - you're talking about the difference between proving and entailment (our article on this doesn't make a lot of sense to me but that's the right concept) - but Godel's completeness theorem says that in the cases we normally care about these are the same - if you know what pi is, you know it is irrational. If you can't establish it's irrational, you certainly can't establish that it's the same pi. Straightontillmorning (talk) 12:37, 31 July 2012 (UTC)

A first order theory is consistent iff it has a model, a statement is independent if it and its negation is consistent with the theory; so if p is independent of T, then there should be a model for T+p and T+~p. This doesn't seem to jive with the above at all; then again, I'm at work and having a dreadfully unpleasant day, maybe I missed something or am having a moment:-)Phoenixia1177 (talk) 11:38, 1 August 2012 (UTC)