"The first incompleteness theorem states that no consistent system of axioms whose theorems can be listed by an effective procedure (i.e., an algorithm) is capable of proving all truths about the arithmetic of the natural numbers. For any such formal system, there will always be statements about the natural numbers that are true, but that are unprovable within the system. The second incompleteness theorem, an extension of the first, shows that the system cannot demonstrate its own consistency." Could Goldbach's conjecture be seen as a statement that is true but not be provable within that consistent system? Every even number can be written as a sum of two primes. This seems pretty obvious if we just think of it as another axiom. And until we can find an even number that can't be written as a sum of two primes Goldbach's conjecture is as true as 2+2=4.

  • $\begingroup$ Can you show that adding Goldbach's conjecture as an axiom does not create any contradictions? $\endgroup$ – mathreadler Sep 5 '17 at 9:09
  • $\begingroup$ I always thought the phrase "true but unprovable" strange. If a thing is unprovable, then we introduce no contradictions by assuming it is false as an axiom (since proof by contradiction is a valid proof in regular logic), so how does it make sense to say that it is true? Of course, in some cases, looking in from the outside, it ought to be true, but that's a different matter, $\endgroup$ – Arthur Sep 5 '17 at 9:12
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    $\begingroup$ As an aside, "truths about the arithmetic of natural numbers" is not an absolute notion -- truth is relative to a choice of model. Often people have in mind, whether implicitly or explicitly, some specific model of natural numbers they mean all such statements to be taken relative to, which is why they make such statements without qualification. $\endgroup$ – Hurkyl Sep 5 '17 at 9:16
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    $\begingroup$ @Arthur Think "true, but unprovable in this given system". Mathematicians are not bounded to a single system, they are able to invent new ones where the prove might (or might not) be easy. Of course they then also need to show that said system can be applied in this case, but that's another topic... $\endgroup$ – Dirk Sep 5 '17 at 9:20
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    $\begingroup$ @Arthur No; if we could prove that, then in Peano arithmetic GC is provable. My point's that, if GC is false, then there's a counterexample $N$, so refuting GC is a finite task: show $N$, & prove that for any $N=a+b$ ($a, b>1$), $a, b$ are not both prime. Even the task of seeking (not just showing) a proof is finite: you need only test all even integers in turn. All this is known even though nobody knows of such an $N$. Is it known that, if there were a cardinality strictly between $\aleph_0$ & $\aleph_1$, then to show one & prove that it is so would be a finite task? $\endgroup$ – Rosie F Sep 6 '17 at 10:38

Almost certainly no. The only way Gödel's incompleteness theorems could possibly be used in such a proof is if you managed to first prove a lemma such as

  • if there is an undecideable statement, then Goldbach's conjecture is undecideable
  • any proof or disproof of Goldbach's conjecture can be modified to prove the consistency of Peano arithmetic

or other similar thing. But it strains credulity to think that one could prove such a lemma without first having a proof that Goldbach's conjecture is undecideable.

If all you really mean to ask is whether it's possible that Goldbach's conjecture is undecideable, then yes — (as far as I know) there does not exist any proof that Goldbach's conjecture is decideable, so the current state of mathematical knowledge leaves all three of the following options open:

  • that Goldbach's conjecture can be proven from Peano arithmetic
  • that Goldbach's conjecture can be disproven from Peano arithmetic
  • that Goldbach's conjecture is undecideable in Peano arithmetic
  • $\begingroup$ Thank you for your answer. I'm a beginner, I mean they've only started trigonometry in my school. But I like calculus, number theory, quantum physics etc. So excuse my incompetence. $\endgroup$ – user167920 Sep 5 '17 at 10:06

As many other conjectures dealing with prime numbers,we can't say yet because we don't have enough understanding of prime numbers yet. There might be an easy and obvious proof once we have better understanding of prime numbers, or there might not be.

However, you can't just assume the conjecture to be true until proven false. If you do, other results will be build on it and you might end up loosing the work of many decades should the conjecture ever turn out to be false. If you are in a more practical field, you might say "it is true for all numbers I will ever encounter during my work", but this line of thought is mostly reserved for engineers, mathematicians tend not to think that way.

  • $\begingroup$ Thanks for your answer. I'm currently 15 and want to become a mathematician. So please excuse for my stupid questions about number theory I'm a beginner really. $\endgroup$ – user167920 Sep 5 '17 at 10:04
  • $\begingroup$ There are no stupid questions, only stupid answers. :) (Stupid question: Who said that? I forgot...^^) $\endgroup$ – Dirk Sep 5 '17 at 10:57

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