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Suppose someone proved that the Goldbach conjecture was undecidable in an axiomatic system that is consistent as far as we know. Then in some sense we know that Goldbach conjecture must be "true", because if we could find a counter-example, then we could prove it false, and it wouldn't be undecidable.

How widely accepted is this kind of "proof"? I don't have very much experience, but outside of set theory and logic, I have rarely seen anybody specify in what axiomatic system they are working in.

For instance, would such a proof be accepted in

  • The Putnam exam?
  • Science?
  • Engineering?
  • General Mathematics (e.g. the Monthly)?
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    Outside of set theory and logic, unless otherwise stated, it is usually understood that people work with ZF, possibly with choice. So nobody bothers to specify this explicitly.2012-01-31
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    "Engineering"? I am not aware of engineers being concerned about proofs of formal mathematical statements at all, let alone one whether a given argument is an acceptable proof or not...2012-01-31
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    @ArturoMagidin I would still imagine engineers have some method of justification. I was mostly curious as to whether such proofs/approach to problems are uncommon because they end up being contrived and really lack any benefit, or because they are simply neglected.2012-01-31
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    @math4tots: If you could prove that Goldbach's conjecture is independent of PA in ZFC, then that would in fact be a proof that Goldbach's conjecture is _true in ZFC_. Since most mathematicians don't work in pure PA (where would the "true" real numbers be, for a start?), this would certainly be considered a valid proof. For more discussion on the specific case of Goldbach, see [this MO question](http://mathoverflow.net/questions/27755/knuths-intuition-that-goldbach-might-be-unprovable).2012-01-31
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    Consistent "as far as we know"? If you could prove that Goldbach (or any other statement that can be expressed in the system) is unprovable in a given system, that would prove that the system is consistent, because from a contradiction you can prove anything.2012-01-31
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    I have heard of an engineering proof "The graph plotted in Mathematica crosses the $x$-axis therefore the function has a root", and no - there was no proof of the continuity of the function before this argument.2012-01-31
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    The proof would be valid, but this scenario is very unlikely to occur in practice: typically, proving unprovability is very difficult. If there was a proof that Goldbach's conjecture is undecidable in, let's say, PA, it would almost certainly be easier to prove Goldbach's conjecture directly.2012-01-31
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    Note that the hypothesized proof would be essentially isomorphic to the usual argument that Gödel's undecidable sentence is in fact true for the standard integers -- and nobody seem to take issue with _that_ conclusion in particular.2012-02-01
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    One should also add the caveat that they are defining truth in the intuitionistic, as opposed to constructivist, sense. You know, just in case there are philosophers around...2012-02-01
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    Since you're asking about the "acceptance" of such a "proof" in several areas, I'm not so sure that you've asked any sort of mathematical question here. This seems more a question about the standards of Putnam Exam, Science, Engineering, and General Mathematics, which I don't think a question about mathematics, but rather a question about how people approach those fields. It might fit better on the meta-math stack exchange.2012-02-02
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    @Doug: [As the FAQ indicates](http://meta.math.stackexchange.com/faq), meta.math.SE is for discussion about math.SE.2012-02-02
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    The Goldbach conjecture is undecidable in the system with *no* axioms, which is certainly consistent. You need more than just consistency to conclude that the conjecture must be true.2012-10-06

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We do not know it in some sense, but taking Platonic stance we actually know it for sure. Assume that there exists a totality of standard natural numbers, say $\mathfrak{N}$. Suppose $\varphi$ is the Goldbach conjecture, which is $\Pi_1$ sentence (begins with a universal unbounded quantifier followed be a formula expressing recursive property or realtion). Finally suppose that you managed to show that neither $PA\vdash\varphi$ nor $PA\vdash\neg\varphi$ (where $PA$ is first-order Peano arithmetic).

However, as a Platonist I believe that either $\varphi$ is true about $\mathfrak{N}$ or it is false about it. Assume it is false, which entails that its negation must be true. But $\neg\varphi$ is equivalent to a $\Sigma_1$ formula (truly) asserting existence of a natural number which is not a sum of two primes: $\exists_x\psi(x)$. Take this (standard) number, say $n$, and substitute $\overline{n}$ for $x$ to obtain $\psi(\overline{n})$ which asserts a recursive property of $n$. From representability of recursive properties you obtain that $PA\vdash\psi(\overline{n})$, so $PA\vdash\exists_x\psi(x)$, so $PA\vdash\neg\varphi$, contrary to the fact that $\varphi$ is undecidable. (The mehtod described applies to all $\Pi_1$ sentences and uses so called $\Sigma_1$ completeness of $PA$)

Of course the hard part is proving undecidability of $\varphi$, and as it was said by Robert Israel in the comments above, this would probably be a longer way than challenging Goldbach conjecture directly. Moreover being Platonist is not enough as well, since you must engage some (stronger than $PA$) theory in which you can carry a proof of independence of $\varphi$.

Thus reception of a proof of this kind would probably rely on nature of methods applied in showing undecidability of the conjecture. Last but not least, constructivists could attack the assumption about definite truth value of the conjecture or (maybe even more likely) the part in which one infers existence of $n$ from a premiss saying: not every natural number... .