vega12 wrote:skeptical scientist wrote:Perhaps, but we'd know it's true the same way we know Godel's sentence G is true, whether or not you accept choice as valid. We know it's not refutable, and if false it would be refutable, so it's true.

True doesn't mean theorem (provable), and undecidable doesn't mean true (whatever that means

).

I am well aware of the difference between "true", "provable", and "undecidable". "Truth" is determined by your model. For example, the statement "1+1=0" is true in some models of the theory of fields (fields of characteristic 2) and not in others. With PA, we have a standard model in mind, (

N,0,

S,+,*,<). So when I say "true", I mean true in that model. We know that Godel's sentence G is true in the standard model, because in the standard model it can be interpreted as saying "there is no formal derivation of G from the axioms of PA", and if it were false, there would be a formal derivation of G, contradicting the fact that the axioms of PA are true, and can therefore only be used to derive true statements. We also know that there is no formal derivation of G, and thus (by the

completeness theorem), G must be false in some models of the axioms of PA. One consequence of this is that there is no finite (or even computably enumerable) collection of axioms which characterize the theory of natural numbers, and there will always be nonstandard models of such a collection of axioms. (Actually if this was all we cared about there would be a much easier proof using compactness; Godel's theorem tells us a lot more.)

Godel's sentence wasn't proven true because you can accept it or deny it, but rather a much more clever self-referential argument (which actually showed why it can't be true).

I think you mean "provable" here, rather than true. The self-referential argument showed that G could not be formally derived from the axioms of number theory. It also, as I explained above, showed that it must be true (since what it says is exactly that it can't be formally derived from the axioms of number theory; or rather, that's what it says if your background model is actually the natural numbers.)

And how did we know that Fermat's last theorem was not refutable before it was proved true? I thought that we were unsure whether it was undecidable or provable.

We didn't. My point is that by proving FLT by assuming the axiom of choice, we prove that FLT is consistent (in ZF, i.e. set theory without choice), since choice is known to be consistent with ZF. Since if false, FLT would be refutable (by providing a counterexample a

^{n}+b

^{n}=c

^{n}), this implies that it must be true, whether or not you accept choice.

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dadaMathematician wrote:I thought that in GEB he constructed a Godel sentence in peano arithmetic and then demonstrated that you can either assume that the statement is formally true or formally false. When you assume that it is formally false (i.e. there exists a truth) you can construct infinite numbers that represent proofs of G.

Or am I remembering that wrong?

No you are correct. In the latter case, you are no longer working in the natural numbers, but in some nonstandard model of number theory, where you can construct "infinite" natural numbers. (I.e. there is a "natural number" x with x>1, x>2, x>3, x>4, etc. In other words, your model is "

non-Archimedean".)

I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson