This question was sparked by this Numberphile video: https://www.youtube.com/watch?v=O4ndIDcDSGc. Near the end, (12:05), he speaks about the Riemann Hypothesis. He describes that if Riemann is shown to be an unprovable statement, then this proves that Riemann is true. For if it were false then it would be provably false (demonstrable by the existence of a non-trivial zero not at $\frac{1}{2}$) and thus decidable, contradiction.

My question is about the model theoretic interpretation of this. From my understanding, undecidable sentences are not valid and their negations are not valid. This is from the completeness of the logic i.e. $\{\vdash{A}\}$ if and only if $\{\vDash{A}\}$. Thus, if $A$ is undeceivable then for some models of the theory in predicate calculus, $A$ is $\mathscr{T}$ and other models where $A$ is $\mathscr{F}$.

If we show Riemann to be undecidable, how can we really say it is true if it is only sometimes true as described above?


2 Answers 2


Edit- Added more thorough explanation

When they say the Riemann Hypothesis is true (or anyone says any mathematical assertion is true), they don't mean it is 'logically valid', they mean that it is true, i.e. that there are no nontrivial zeros of the zeta function.

It's not even clear without further elaboration what precisely it would mean for the Riemann Hypothesis to be logically valid. First we must cast the Riemann hypothesis (or some equivalent statement) as a sentence in some formal language, and only then we can start talking about interpretations and validity. Once we've done this, the idea of the sentence being "true" becomes a statement about one interpretation, usually called the 'standard' or 'intended' interpretation / model.

It is certainly not saying that it is valid in predicate logic, i.e. true in every interpretation, which is a very strong statement. More reasonable would be that it is a logical consequence of some effectively describable collection of axioms (perhaps PA if the language is arithmetic... more on that below), i.e. it is true in any model of those axioms. But even in this case there are generally many unintended models that don't have much bearing on ordinary mathematics, so they don't mean that either.

So the way to think about it is just like they say: if RH is false (in the standard interpretation), then its negation is provable, and hence it is not undecidable. So undecidable implies true (in the standard interpretation). If we want to be more model theoretic, we could say that if RH is false in the standard model, then it is false in all models, but really it's easiest to see that it is refutable directly, which implies by soundness that it is false in all models.

(And let me emphasize again, we've been pretty noncommittal about questions like 'what version of RH?', 'expressed in what language?', and 'model of what axioms?'... all of which needs to be fleshed out).

So the assertion in the video makes sense, and as they say, the reasoning is that if RH were false, it would be refutable. However, they gloss over the explanation, perhaps leaving you with the incorrect impression that it is trivial. Even the basic reasoning that if there's a counterexample (i.e. a nontrivial zero), we can check it, is annoyingly very close to being correct while still being very misleading.

To understand why this is an oversimplification, think about what would naively go into 'checking a zero'. The value of the zero itself may in principle contain an infinite amount of information (it's a complex number after all). There's no obvious reason why checking that it gives zero when plugged into the zeta function would be an effective procedure that translates to an effective proof. So its falsity is not "demonstrable by the existence of a non-trivial zero".

However, it has been known for some time that RH is equivalent to a $\Pi_1$ sentence of arithmetic, which is a sentence of the form "for all natural numbers $n$ the effectively computable property $P(n)$ holds" (and this fact isn't obvious unless you are a analytic number theorist well-versed in formalizing things in arithmetic). In this form it is obvious that it is refutable if false, since it is now clear that the procedure for verifying a given counterexample to $P(n)$ is effective, so any reasonable system will be able to turn this into a proof of $\exists n \lnot P(n)$

Original answer:

The truth they refer to is truth in the standard model, not validity. So the way to look at it is that if the Riemann Hypothesis is false in the standard model, then it is false in all models and hence not undecidable. The reason, as it says in the numberphile video, is that if it is false, then its negation is provable, and thus it is false in all models. This is due to the nontrivial theorem that the RH is equivalent to a $\Pi_1$ arithmetical sentence, so a (standard) counterexample to the Riemann Hypothesis will be witnessed by a finite arithmetical computation. (I haven’t watched the numberphile video (I’m traveling) but I’m guessing they oversimplified this... it would not be enough for a counterexample to exist... it needs to be provable.)

  • 1
    $\begingroup$ I am glad you expanded the answer. I wrote a comment on the nature of the counterexample, overlapping with your edit. $\endgroup$ Mar 22, 2019 at 12:35

This is too long for a comment, so I have made it a community wiki post. It is correct that, if the Riemann Hypothesis is false, there is a concrete example. But that example will almost certainly not be "demonstrable by the existence of a non-trivial zero not at $1/2$".

Assume there is a non-trivial zero not at $1/2$. There may be no nice analytic expression for it. We could begin to approximate it, but there is no reason to think that we would actually see "zero" from that approximation. We might see that the results are getting closer and closer to zero, but there is no reason to be certain that we could prove that some particular complex number is a counterexample. There is no reason to think that a root-finding process would produce a counterexample in a finite amount of time. This is closely tied to the issue that the usual statement of the Riemann Hypothesis involves complex numbers, rather than natural numbers.

However, Lagarias proved in 2002 [1] that there is a statement equivalent to the Riemann Hypothesis that, when stated carefully, can be stated as a formula that quantifies only over natural numbers and only with leading universal quantifiers. For this particular kind of statement, if it is false then there is a concrete counterexample consisting of natural numbers only. This kind of statement is called $\Pi^0_1$ in logic. So the claim being made on the video is correct, but the reasoning that was provided in the video is more questionable.

1: An Elementary Problem Equivalent to the Riemann Hypothesis, Jeffrey C. Lagarias, The American Mathematical Monthly Vol. 109, No. 6 (Jun. - Jul., 2002), pp. 534-543. ArXiV preprint.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.