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The question is contained in the title; I mean the standard axioms ZFC. The wiki link: Riemann hypothesis. There are finite algorithms allowing one to decide if there are non-trivial zeroes of the $\zeta$-function in the domains whose union exhausts the whole strip $0<\Re z<1$, but this does not seem to be the obstacle for undecidability. Are there other arguments?

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@Shaqq: undecidable means “neither provable nor refutable”. In particular, an undecidable statement may also be false (but not refutable). OTOH, in principle, a true but unprovable statement may be refutable and therefore decidable (if ZFC is not a sound theory). Finally, it is much better to say “independent” instead of “undecidable”, since the latter word is also used to mean “not computable by an algorithm” in recursion theory, and an awful lot of people get thoroughly confused by mixing these two unrelated terms. –  Emil Jeřábek Nov 1 '11 at 11:58
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related: Is the RH equivalent to a $\Pi_1$ sentence? –  Kaveh Nov 1 '11 at 16:50
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A working link: mathoverflow.net/questions/31846 . –  Emil Jeřábek Nov 1 '11 at 17:41
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Whether or not it is logically possible that the Riemann hypothesis is undecidable, I submit that there is no good reason to believe it is undecidable. "Gosh, people have been trying to prove it for a while without success" is not a good reason - that philosophy has been incorrect on a bazillion theorems so far. –  Greg Martin Nov 1 '11 at 20:21
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@Kofi: I suppose you meant to say that the negation of the Riemann hypothesis was unprovable. Yes, this would imply that RH is true, but it is quite a mess to verify this directly using definition: you have to prove first that “find a [nontrivial] zero on the critical strip” is something you can certify by a finite object. This can be done for example by fixing a rectangle $c$ with Gaussian rational endpoints, disjoint from the critical line, such that the integral of $\zeta'/\zeta$ over $c$ is nonzero, provided you show that you can approximate the integral with sufficient accuracy, ... –  Emil Jeřábek Nov 2 '11 at 11:15

4 Answers 4

I do not know anything about zero-finding algorithms for $\zeta$, so I will make only one small remark which doesn't require such knowledge: If the Riemann Hypothesis is false, then it is provably false (in ZFC, or any similar system).

This is because Robin's theorem tells us that the Riemann hypothesis is equivalent to the assertion that, for every natural $n \geq 5041$, the sum of the divisors of $n$ is less than $e^{\gamma} n \log{\log{n}}$; since there are programs which calculate this latter quantity to arbitrary precision, and thus can verify whether this inequality holds for any given $n$, we find that the Riemann hypothesis is a $\Pi_1$ statement: it is equivalent to the assertion that some computer program never outputs "NO" on any input. (Although not familiar with the proofs of Robin's theorem, etc., I assume they can be carried out in ZFC, and thus establish the relevant equivalence within ZFC.). There may be more direct ways to establish that the Riemann hypothesis is a $\Pi_1$ statement, such as by knowledge of algorithms which enumerate to arbitrary precision the zeros of $\zeta$, but at any rate, there is this one.

Accordingly, if the Riemann hypothesis is false, then the relevant computer program does output "NO" on some input, from which it would follow that ZFC proves that that computer program outputs "NO" on that input, and thus ZFC would prove the Riemann hypothesis to be false.

The possibility still remains, however, as far as I know, that the Riemann hypothesis may be true but unprovable in ZFC.

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I have heard that the Riemann hypothesis cannot be true but unprovable in ZFC. But I have no access to the source anymore, that is why I ask my question here. –  Shaqq Nov 1 '11 at 8:31
    
Lightface $\Pi_1$? –  Michael Blackmon Nov 1 '11 at 8:56
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@Michael: it is $\Pi^0_1$, period. This is a classification of sentences up to provable equivalence, not a classification of sets of reals, therefore the concept of boldface does not make sense here. As a set of reals, any statement has trivial complexity (since it is either the empty set or the full set, depending on whether the statement is true). –  Emil Jeřábek Nov 1 '11 at 11:51
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@Shaqq: It seems likely that you’ve heard (but misremembered) that the negation of the Riemann hypothesis cannot be true but unprovable (which is what Sridhar’s answer is about). –  Emil Jeřábek Nov 1 '11 at 12:00
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@Emil: probably Shaqq heard it in this form: if the Riemann hypothesis is undecidable in ZFC then it is true. That's the most common way I've heard it bandied about. –  Timothy Chow Nov 1 '11 at 14:13

In an interview with Martin Davis that appeared in the notices of the AMS, he speculates by the end (pp. 570) with the possibility of RH being undecidable. He was explaining that every $\Pi_1^0$ is equivalent to a statement asserting about a particular polynomial equation with integer coefficients that that equation has no natural number solutions, and that RH was a statement of that kind, as worked out in "Hilbert’s tenth problem: Diophantine equations: positive aspects of a negative solution", by Martin Davis, Yuri Matijasevic, and Julia Robinson, in Mathematical developments arising from Hilbert problems, Proc. Sympos. Pure Math., AMS, 1976. He then goes on:

"I am certainly no analyst, but the reason I think the Riemann Hypothesis is a good candidate for undecidability by elementary methods is that it is sitting right in the middle of classical analysis, and it has been attacked by brilliant mathematicians—Paul Cohen spent a lot of time on it—and the existing methods just don’t seem to resolve it. It’s hard to believe it isn’t true. And why shouldn’t it be one of those propositions that require set theoretic methods? That would be great!"

As he previously explained, there are some mathematical propositions that "have a very simple form involving solvability of specific Diophantine equations and that require set-theoretic methods for their resolution". The fact that RH might be of that character, as he remarks, had already been conjectured by Gödel at the Gibbs lecture. "And that wouldn’t surprise me in the least", he claims.

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When Davis mentions set-theoretic methods, this suggests that he means undecidability in Peano arithmetic rather than undecidability in ZFC. (The text is unclear, since first he discusses algorithmic decidability but then he mentions measurable cardinals.) RH may well be unprovable in PA and yet provable in ZFC. However, the same basic ideas apply. –  Toby Bartels Nov 1 '11 at 16:58
    
I believe Davis talks about both aspects because he's thinking about what Gödel said at the lecture. There, Gödel was talking about different levels of set theoretic constructions (sets of integers, then sets of sets of integers and so on, eventually iterating this through transfinite ordinals and then using this operation to define further sets...). Gödel emphasized that there might be some set theoretic axioms describing the behaviour of sets of a certain level in the hierarchy which actually have consequences in the 0 level. but he remarks that there is no end in this hierarchy of axioms... –  godelian Nov 1 '11 at 21:05

Yes.

That being said, it's pure speculation. We can as well talk about the decidability of any other famous open problem, but in my experience that usually doesn't lead to anything new. To me, it's highly unlikely that is undecidable, but of course, we can't exclude it. Compare it for example to the rationality of $2*\pi^{4/3} - e^{3/2}$.

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Is the question of the rationality of $2 \pi^{4/3}- e^{3/2}$ undecidable? Or is this just an example of a difficult question? –  Bart Snapp Dec 18 '11 at 17:19
    
It just seems to be extremely hard :) –  Woett Dec 19 '11 at 14:30
    
I think Woett's answer wants to point out the following analogy: it is relatively easy to prove that there are many transcendental numbers (Cantor) or many undecidable sentence (Godel) but to prove that a given, mathematically interesting number/sentence is transcendental/undecidable is much much harder. –  Joël Mar 21 '12 at 16:02
    
Interesting! But what I was going for was the following analogy: in both cases something is extremely hard (proving RH, or proving that some number is irrational), which may lead to some speculation that it just can't be done (that either RH is undecidable or the number is actually rational). While it is OK, I guess, to keep that possibility in the back of your mind, the chances of that happening are, in my belief system, very very slim. –  Woett Mar 22 '12 at 13:38

If negation of the Riemann hypothesis is not provable in Peano arithmetic, then the Riemann hypothesis is true.

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