The result closest to Goldbach conjecture is Chen's theorem [Sci. Sinica 16 157–176], the proposition ``1+2''. It is natural to ask if it is likely that under our arithmetic axioms the Goldbach conjecture is an undecidable proposition.

  • 3
    $\begingroup$ As far as I know, there isn't any such established result. Personally I think it is unlikely that it's undecidable. $\endgroup$ – Balarka Sen Jul 11 '14 at 11:39
  • 3
    $\begingroup$ @downvoters I am not sure why this post is being downvoted. I think it's a perfectly good question and been asked before on mathoverflow. $\endgroup$ – Balarka Sen Jul 11 '14 at 11:40
  • 4
    $\begingroup$ This MO post elaborates on this stuff. It was Knuth who first suspected this possibility. $\endgroup$ – Balarka Sen Jul 11 '14 at 11:41
  • 1
    $\begingroup$ @BalarkaSen I didn't down vote and I don't think what I'm about to say makes it reasonable to down vote the question. But it would have been proved that GC was undecidable, that would be it was solved! $\endgroup$ – Git Gud Jul 11 '14 at 11:51
  • 1
    $\begingroup$ @GitGud I agree with you. I think it'd have been more appropriate to ask for whether it is "likely" instead of a proof. $\endgroup$ – Balarka Sen Jul 11 '14 at 11:55

Suggesting the Goldbach conjecture specifically (or other $\Pi^0_1$ statements like it) is undecidable can make some statements about mathematical Platonism that isn't intrinsically obvious, because undecidability of $\Pi^0_1$ statements like this determine their truth value...

If you can prove the Goldbach conjecture is undecidable in the Peano axioms in some larger axiom system like $\mathsf {ZFC}$, then you've essentially said that you have some model that satifies $\mathsf{PA}$ + (Goldbach) and another model that satisfies $\mathsf{PA}$ + $\lnot$(Goldbach.)

Now what this means is that one of these models is necessarily a non-standard model of arithmetic, because in one of these Goldbach's conjecture will fail... and in another it won't. You've proven not only the undecidability of the Goldbach conjecture with respect to a particular axiom system, but also proven it in a larger one, somewhat analogous to the Goodstein theorem.

But this is just if we somehow manage to stumble on a proof of independence of some statement in a larger axiom system. What we don't know is if any particular unproven $\Pi^0_1$ statement is meaningfully unprovably undecidable, or if it makes sense to assume that "out there" a statement like the Goldbach conjecture is "true" but unprovable in a particular axiom system.

You might be able to prove it's "true" or "independent" (and thus 'true' in the standard model of arithmetic) in some larger axiom system, but at some point you have to start assuming axioms that you have wonder if they are "really" true, like the Continuum Hypothesis, or Large Cardinal Axioms, or just some other unproven random statement... or the Goldbach (or whatever $\Pi^0_1$ statement) itself! Then you need to prove consistency of the system in some yet larger system, which brings us to some problems which should be familiar to anyone who's read about the incompleteness theorems.

As to the likelihood of the undecidability of the Goldbach conjecture under $\mathsf{PA}$ in particular, I find it rather unlikely given the results of theorems of very slightly weaker, similar statements.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.