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 nontrivial 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?

5$\begingroup$ @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. $\endgroup$– Emil JeřábekNov 1, 2011 at 11:58

1$\begingroup$ related: Is the RH equivalent to a $\Pi_1$ sentence? $\endgroup$– KavehNov 1, 2011 at 16:50

1$\begingroup$ A working link: mathoverflow.net/questions/31846 . $\endgroup$– Emil JeřábekNov 1, 2011 at 17:41

21$\begingroup$ 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. $\endgroup$– Greg MartinNov 1, 2011 at 20:21

2$\begingroup$ @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, ... $\endgroup$– Emil JeřábekNov 2, 2011 at 11:15
5 Answers
I do not know anything about zerofinding 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.

5$\begingroup$ @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). $\endgroup$ Nov 1, 2011 at 11:51

5$\begingroup$ @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). $\endgroup$ Nov 1, 2011 at 12:00

4$\begingroup$ @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. $\endgroup$ Nov 1, 2011 at 14:13

2$\begingroup$ @Ed: Not really, since Robinson arithmetic most likely does not prove the equivalence of RH to the relevant $\Sigma^0_1$sentence. In fact, the RH as such is not even an arithmetical statement (it speaks about complex zeroes). I do not know whether Robin’s theorem is provable in ZF, but I note that a good deal of elementary analysis relies on DC (or at least, countable AC). $\endgroup$ Nov 1, 2011 at 16:12

3$\begingroup$ Is it possible, for a given $n$ and $K=\sigma_1(n),$ for the statement "$e^\gamma n\log\log n=K$" to be undecidable? Our arbitraryprecision approximations just showing values closer and closer to $K$ and that's all we know? $\endgroup$ Dec 18, 2011 at 17:33
Without worrying about reductions, if RH is false, it is provably false: suppose $\rho\in\mathbb{C}$ is a zero in the critical strip but off the critical line (say, $\Re(\rho)>\frac12$). Then for a little rectangle $R$ with (say) rational corners, containing $\rho$ in its interior but not containing the pole at $s=1$ or intersecting the critical line we'd have $$\frac{1}{2\pi i}\oint_{\partial R} \frac{\zeta'}{\zeta}(s)\mathrm{d}s \geq 1$$ by the argument principle.
But we can approximate $\zeta(s),\zeta'(s)$ to arbitrary precisition by a finite computation, and similarly we can approximate the integral to arbitrary precision by a numerical computation. In other words, there is a finite computation which provably approximates the integral above to within $\frac{1}{2}$. Then the nonvanishing of the approximation proves the falsity of RH.
Similarly, RH is equivalent to estimates on the primecounting function, for example to $$\psi(x)(x)<\sqrt{x}\log^2 x$$ where $\psi(x) = \sum_{p^r<x}\log p$.
While the models can disagree about some aspects of the real numbers, I don't think they can disagree about the logarithm of an integer (in the sense that $\log(n)=\log(1/n) = \sum_{k=1}^\infty \frac{(n1)^k}{kn^k}$), and it's enough to consider integer $x$ in the inequality. This means that if RH is false in a model, RH will be false in any model whose integers contains those of the original model.
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Edited to clarify the last comment: in the primecounting inequality, replace the term $\log p$ associated to the prime power $p^r$ with $\sum_{k=1}^{100p^{2r}} \frac{(p1)^k}{kp^k}$. Similarly, replace the $\log x$ with a truncation of the logarithm series. RH is still equivalent to the inequality using the truncated logarithm, and in the new formulation both sides of the inequality are explicit rational numbers! In other words, RH is equivalent to a countable sequence of inequalities about explicit rational numbers.
Second edit: I realize that the direct formulation can also be converted to a statement about rationals. Let $R_n$ be the axisparallel rectangle with corners at $\frac{1}{2}+\frac{1}{n}+i$ and $1+ni$. Let $I_n$ be a rational approximation to the zerocounting integral above around $R_n$, accurate to better than $\frac{1}{2}$ (obtained by dividing the boundary of $R$ into many points, at each point approximating $\frac{\zeta'}{\zeta}$ by a rational number, and then integrating numerically). This is an explicit finite computation, and RH is equivalent to the infinite system of inequalities $I_n<0.5$.
Third edit: removed false statement about submodels.

$\begingroup$ “$\cal U$ contains an isomorphism between the natural numbers of $\cal U$ and the natural numbers in a model of ZFC inside $\cal U$” — this is not necessarily the case; even if $\cal U$ is a transitive model of ZFC, the submodel inside $\cal U$ might have nonstandard natural numbers. $\endgroup$– YuryDec 23, 2015 at 22:46

$\begingroup$ Accordingly, there might be a “counterexample” $x$ to RH in the model in $\cal U$, which is a nonstandard natural number. $\endgroup$– YuryDec 23, 2015 at 22:50

1$\begingroup$ Ah I see my error: the "set of standard natural numbers" in the submodel will satisfy the Peano axioms, but need not be a set in the submodel. $\endgroup$ Dec 23, 2015 at 23:00

$\begingroup$ But still the embedding of the natural numbers of $\cal{U}$ in the natural numbers of the submodel will show that if RH is false in $\cal{U}$ it's false in the submodel. $\endgroup$ Dec 23, 2015 at 23:03

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


$\begingroup$ Consider the contrapositive: if the RH is false, then ¬RH is provable in PA. It just means that if there is a counterexample to the RH (a nontrivial root of the Riemann ζfunction off the critical line), it can be demonstrated by a straightforward calculation. It can be done even in a weaker theory than PA. $\endgroup$ Aug 23, 2022 at 23:21
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 settheoretic 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.

5$\begingroup$ When Davis mentions settheoretic 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. $\endgroup$ Nov 1, 2011 at 16:58

1$\begingroup$ 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... $\endgroup$– godelianNov 1, 2011 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 $2*\pi^{4/3}  e^{3/2}$.

$\begingroup$ 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? $\endgroup$ Dec 18, 2011 at 17:19

3

1$\begingroup$ 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. $\endgroup$– JoëlMar 21, 2012 at 16:02

2$\begingroup$ 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. $\endgroup$– WoettMar 22, 2012 at 13:38

2$\begingroup$ I believe even the rationality of $\pi\cdot e$ is open. $\endgroup$ Jun 2, 2017 at 0:58