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Conjectures concerning natural numbers which could be settled by a counterexample can, as far as I understand, not be proved to be undecidable without being proved not having a counterexample at the same time. Or?

Lehs
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  • Not sure what answer can be given here except "yes, that's right". – Ben Millwood Dec 07 '14 at 07:29
  • Well @Ben, is it right? There are people thinking that the Goldbach conjecture is unprovable. http://mathoverflow.net/questions/27755/knuths-intuition-that-goldbach-might-be-unprovable – Lehs Dec 07 '14 at 07:31
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    It can be unprovable without it being provable that it's unprovable. – aes Dec 07 '14 at 07:54
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    If Goldbach's conjecture is proved to be undecidable in Peano arithmetic, there are models of arithmetic in which it is true, and models in which it is false. The existence of counterexamples is model-dependent, while provability of Goldbach's conjecture depends on the theory. But finding a counterexample in some model would prove that Goldbach's conjecture is either undecidable or false. – zarathustra Dec 07 '14 at 08:00
  • "The existence of counterexamples is model-dependent..." Not sure I get that, @zara. If it has a counterexample, then it's false, hence, decidable; so if it's undecidable, then it's true. – Gerry Myerson Dec 07 '14 at 08:38
  • In the case it is undecidable, there exists a counterexample in some model, and there is a model in which there is no counterexample, that's what I was saying. Is there something wrong with that? @Gerry – zarathustra Dec 07 '14 at 08:41
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    I just don't see how there could be a counterexample in one model, and not in another. Can there be a number that's prime in one model, but not in another? Can there be two primes that add up to different numbers in two different models? – Gerry Myerson Dec 07 '14 at 08:50
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    Since $\mathbb N$ is a minimal model a counterexample would be a counterexample in every model, but @aes is right of course, that a conjecture could be unprovable unprovable. – Lehs Dec 07 '14 at 10:54
  • @GerryMyerson: Any conditional counterexample would have to be a nonstandard integer. If it were a standard integer it would be a counterexample in all models. – Charles Dec 07 '14 at 16:41
  • Extensely discussed in http://mathoverflow.net/questions/27755/knuths-intuition-that-goldbach-might-be-unprovable and http://math.stackexchange.com/questions/864149/could-it-be-that-goldbach-conjecture-is-undecidable. – Martín-Blas Pérez Pinilla Dec 13 '14 at 14:48
  • Till a perfect 'proving' mechanism is created, there will always be new ways of proving statements to be found, and new ways to use the existing ones. We can not prove it unprovable till such a mechanism is found, and even that might not be implementable in finite time. – ghosts_in_the_code Jan 20 '15 at 17:00

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If the Goldbach conjecture is undecidable (independent) of PA, then it is true for the standard model of arithmetic, i.e. the model $(\mathbb{N}, +, \times, 0, 1, <)$ with their normal interpretations.

Proof: Recall that $\mathbb{N}$ is the prime model of PA, (i.e. it embeds into every model $M$ such that $M\models$ PA). Further more, $\mathbb{N}$ is the initial segment of all models of PA. Assuming that $GC$ is independent of PA, we note that PA + $GC$ and PA + $\neg GC$ are consistent. Then, by Gödel's completeness theorem, there exists models $M_1$ and $M_2$ such that $M_1 \models$ PA + $GC$ and $M_2 \models$ PA + $\neg GC$. Suppose that $\mathbb{N} \models$ PA + $\neg GC$. Then, there exists some element $a \in \mathbb{N}$ such that $a$ cannot be written as the sum of two primes. Since $\mathbb{N}$ is the prime model of PA, we recall the fact that $\mathbb{N}$ embeds into all models of arithemtic. Therefore, if $\varphi(x)\equiv$ "$x$ cannot be written as the sum of two primes", and $\mathbb{N} \models \varphi(a)$.

By above, we showed there exist an $M_1$ such that $M_1 \models $ PA + $GC$. However, since $\mathbb{N}$ embeds into $M_1$, we note that $M_1 \models \varphi(\pi(a))$ (where $\pi$ is the embedding). However this implies that $M_1\models \neg GC$ which is a contradiction.

Thus, if $GC$ is independent of PA, then $\mathbb{N}\models GC$.

Kyle Gannon
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    Are you sure the question isn't asking i Goldbach is independent of ℕ? Almost all references to PA are really to ℕ. – Joshua Dec 27 '14 at 19:59
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    @Joshua: I hope they are referencing PA (or some non-specified theory). When you talk about decidability, it should be in relation to a theory and not a model. – Kyle Gannon Dec 27 '14 at 20:04
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    @Joshua: Someone reference non-standard elements in the comments and so I think they are actually referring to a theory. – Kyle Gannon Dec 27 '14 at 20:05