# Why does, in theory, you can prove Riemann hypothesis with Godel Incompleteness Theorem, but can't prove the consistency of mathematics?

Some time ago, I watched this fascinating episode of Numberphile about Godel Incompleteness Theorem: https://www.youtube.com/watch?v=O4ndIDcDSGc&t=4s In the video, Professor du Sautoy explains the theorem (with big simplifications) and he also mentions that if we could prove that the Riemann hypothesis is one of those propositions that can't be proved, then it means that the theorem is true, since if it was false we would have been able to find a contradiction. So we would be able to disprove it.

In the first Extra footage video https://www.youtube.com/watch?v=mccoBBf0VDM Professor du Sautoy mention that one of the conclusions from this theorem is that the proposition mathematics \ arithmetic is consistent is one of those propositions that we can't prove. I don't understand why that doesn't also imply that the statement holds, since if it wouldn't have hold we could have proved it by finding a contradiction.

• math.stackexchange.com/q/2305177/622 Jul 4, 2018 at 15:13
• Let me elaborate on Asaf's answer - specifically, the fourth paragraph. Briefly, Godel showed (e.g.) that ZFC proves that PA cannot prove its own consistency. This does imply that ZFC proves that PA is consistent, but that's not surprising. Meanwhile, PA itself does not (we hope!) prove that it can't prove its own consistency; all PA proves is "If PA is consistent, then PA can't prove its own consistency." See also this old answer of mine. Jul 4, 2018 at 16:28

The key point is that "proof" requires you to at least understand what theory you are assuming.

Both the Riemann Hypothesis and "$T$ is consistent" (where $T$ is any recursively enumerable theory) are $\Pi^0_1$ statements. This means that if there is a counterexample, then the statement is disprovable. Or in other words, if the statement is not disprovable, then it is true.

But what does "true" mean? In the context of arithmetic true means "true in the natural numbers". In other contexts, it is usually meaningless, or synonymous for "provable".

Yes. $\sf PA$ is consistent is a $\Pi^0_1$ statement, but we also can prove that it is a true statement, since $\Bbb N$ exists and it is a model of $\sf PA$. Or at least that is the case if your metatheory is sufficiently strong (e.g. set theory à la $\sf ZFC$). The point is that $\sf PA$ itself cannot prove that $\sf PA$ is consistent, and that $\sf ZFC$ cannot prove that $\sf ZFC$ is consistent.

The difference between that and the Riemann Hypothesis is that you don't really talk about consistency of a theory. So there is no natural candidate of a theory which is insufficient for proving the Riemann Hypothesis.

It is possible for a $\Pi^0_1$ statement to turn out equivalent to "$\sf ZFC$ is consistent", in which case it cannot be proved. But it would be a mind shattering surprise to modern mathematics if RH turns out to be such statement. So what we expect is that RH is either outright refutable from a weak theory like $\sf PA$, and if not then it is at least true in $\Bbb N$, which is "the thing we care about" for the Riemann Hypothesis.

• Thank you. To be honest, I didn't fully understand your answer due to some terminology I'm not familiar with, but that's fine. You gave me the feeling for the answer. I'm planning to learn Logic in the future so hopefully, in the future, I'll be able to understand it fully. Jul 5, 2018 at 16:30
• If you want to be more specific, I could try and help clarify or direct you to sources. Jul 5, 2018 at 16:40