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Certain statements are known to be unprovable within a given axiomatic system; the continuum hypothesis within ZFC is an example. We can either add the continuum hypothesis, or its negation, to ZFC, and proceed with the new consistent set of axioms.

My question is: how do we know whether a given statement (say, within ZFC) is provable or unprovable? I'm intrigued because I read that for many years people sought a proof of the continuum hypothesis within ZFC (e.g. it was no. 1 on Hilbert's list of pressing problems), which turns out to be impossible. Could it be that there are "unsolved" problems out there for which people are searching solutions, while in fact those statements are unprovable? Must a mathematician always live with the fear of doing Sisyphus' labour? As specific examples, could the Riemann hypothesis or the Goldbach conjecture actually be unprovable within ZFC?

Latrace
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    We , in general , cannot know it. – Peter Feb 02 '23 at 13:52
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    Your last question was already discussed in great detail, see this post. Here is the spoiler from Asaf Karagila's great answer:"This means that if you cannot disprove the Riemann hypothesis, it has to be true. The same can also be said on the Goldbach conjecture." – Dietrich Burde Feb 02 '23 at 13:55
  • These questions and the links within seem relevant: https://math.stackexchange.com/questions/2305177/decidability-of-the-riemann-hypothesis-vs-the-goldbach-conjecture, https://math.stackexchange.com/questions/4197644/how-could-the-riemann-hypothesis-be-independent-of-zfc . – Aphelli Feb 02 '23 at 13:58
  • "Must a mathematician always live with the fear of doing Sisyphus' labour?" Basically , everything we try could turn out to be a useless effort , unless we KNOW it is feasible. No reason to not investigate such problems. But I advice everyone not to try to prove such super-difficult theorems. Even, if there is such a proof , the odds for a success are extremely against everyone trying it. – Peter Feb 02 '23 at 14:18
  • Both the Riemann hypothesis and the Goldbach conjecture (assuming they are true) can be unprovable in ZFC. We ASSUME that ZFC is powerful enough to do this, but there is no guarantee. If we can prove them to be independent of ZFC within a stronger theory (in other words undecidable in ZFC) , then we have proven them because in the case they are FALSE ZFC can prove this ! – Peter Feb 02 '23 at 14:35
  • A false statement cannot be proven in ZFC anyway (if ZFC is consistent) , so the unprovability does NOT prove the claim. – Peter Feb 02 '23 at 14:37
  • Short answer: a statement is undecidable iff its truth value is model-dependent, so you just need one model for each. Long answer: we obtain these models with techniques such as this. – J.G. Feb 02 '23 at 15:21

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What you are interested in here is the study of Consistency Results.

Definition:

A statement $\phi$ is independent of ZFC if neither $\phi$ nor $\neg\phi$ are provable from ZFC.

Kunen's "Set Theory An Introduction to Independence Proofs" goes into great detail on how to show consistency results.

If those methods ( or any valid method) can be applied to $ \phi$ and $\neg \phi$ then $ \phi $ is independent of ZFC i.e. not provable from ZFC, and not disprovable.

A classical example:

One method is the careful construction of models. It is possible to construct a model of ZFC where the Continuum Hypothesis is true; and it is possible to construct a model where the Continuum Hypothesis is false.

You can conclude, the Continuum Hypothesis is indepedent ( not provable from) ZFC.

The method above uses Forcing, which is the basis for many consistency results. However, it is quite technical and so I reccomend you refer to Kunen for details.

Note: Of course, this is all under the assumption that ZFC is consistent. That is, for all $\phi $, ZFC cannot prove both $\phi$ and $ \neg \phi $

Michael Carey
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