# Is Riemann Hypothesis provable?

Loosely speaking, there are three kinds of propositions.

1. Those propositions which are true and can be proved to be true.

2. Those propositions which are false and which can be proved to be false.

3. Those propositions which are true, but it can't be proved that it is true.

Here the word "proof" is used in a strict mathematical sense.

My question simply is that under which category does Riemann Hypothesis belong? Or if a bit specification is more preferred, is RH ZFC-independent?

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if you prove something to be provable, then you proved it. – Ittay Weiss Jun 22 '14 at 7:28
I don't understand how proving that RH is provable is related with the position of the real part of zeros on the critical line. Can you be a bit elaborate? – William Hilbert Jun 22 '14 at 7:51
I am not sure what this question is asking. Do you mean something along the lines: Is it possible that RH is ZFC-independent? – Martin Sleziak Jun 22 '14 at 8:08
@MartinSleziak: My question actually is that: Has RH been proved to be formally decidable proposition? If yes, then I want some reference. – William Hilbert Jun 22 '14 at 12:20
You may find mathoverflow.net/questions/79685/… helpful. – Gerry Myerson Jul 17 '14 at 6:41

## 3 Answers

The question incorporates a point of confusion that is unfortunately common in the popularized literature about these things. There are no propositions of the following form:

1. Those propositions which are true, but it can't be proved that they are true.

The first reason there are no such propositions is that, in order to recognize that a proposition is true, we already need some sort of proof for it. In other words, "proved that it is true" is no different then simply "proved", assuming that we recognize the axioms of the proof as "true". And for mathematicians to widely acknowledge something as true, they need some sort of proof - possibly very informal and intuitive, of course, but some sort of proof nevertheless.

The deeper reason is that "can't be proved" is not a well-defined property of a proposition - it depends on a formal system as well. In other words, as long as we are able to change the meaning of "provable" at any moment, we will never be able to show that something is "unprovable".

As a concrete example, if the Riemann Hypothesis is true, then there is no harm in taking it as an axiom as part of some formal deductive system - and then it would be provable, trivially, in that system. The same holds for any other true proposition; there is always some formal system for which the proposition is provable. We can argue about whether it would make a "good" axiom or not, but that is a different question.

So, in order to talk rigorously about a proposition being unprovable, we need to have a rigorous notion of "provable". The normal natural-language proofs we use are not rigorously specified (even the natural language itself is not). To talk about non-provability in a rigorous way, we need to look at formalized proof systems.

There are many choices for that system Both of the following make sense:

3a. Those propositions which are true, but can't be proved in the formal system of ZFC set theory

3b. Those propositions which are true, but can't be proved in the formal system of Peano Arithmetic

It is perfectly conceivable (although we have no evidence to support it) that the Riemann Hypothesis could be in 3a and/or 3b. However, to recognize that it was in one of these sets, we would need to prove it in some stronger system, in order to know that it is true. So we would know that it is provable, in the natural-language sense, if we knew that it was in 3a or 3b. So no proposition in 3a or 3b is in 3.

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Eppur si muove...... – Piquito May 8 at 14:21
Carl Mummert: I have read your presentation and I want to say you that @CarlMummert (not spaced) is an automatic of StackExchange's website. – Piquito May 8 at 14:32
1. Since it is still a conjecture (a "Millennium Prize Problem") we don't know if it is true or false
2. Therefore we cannot even know whether it is independent from $ZFC$ because it would imply that we know that it is also true (check the replies here).
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This isn't quite accurate - in principle, we could prove that RH was decidable without deciding it (e.g. prove "RH iff R(5, 5) is odd"), although I don't think anyone believes this is a real possibility. You are right, though, that if we proved that RH was independent that would imply that RH is true. Of course, that's all relative to the consistency of the system in which independence is proved - if ZFC proved that PA did not decide RH, all that would mean is that RH is true or ZFC is inconsistent. Of course, most of us strongly doubt that latter possibility, but it's worth mentioning. – Noah Schweber Feb 29 at 15:35
@NoahSchweber What is R? – Dustan Levenstein Apr 6 at 14:36
– Dustan Levenstein Apr 6 at 14:41

Propositions 1, 2 and 3 are very easy to formulate but could be very difficult to explain fully because of the deep gnoseological and epistemological implications that could have.

►Pythagorean theorem belongs to the set involved by 1 but how to describe this set entirely?

►The denial of any unprovable axiom could be say belonging to the set corresponding to proposition 2?

What about of undecidable propositions?

► Some time ago, some people thought Fermat’s last theorem was not provable. How to know a proposition is true without proof? A famous mathematician told me with conviction the Birch and Swinnerton-Dyer conjecture is true; but this almost miraculous conjecture, despite all acreditable belief, will be part of Millennium Prize Problems while not proven.

I think Riemann Hypothesis belongs to category 1 but who cares really someone's opinion here, what really matters is proof.

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At what point did people believe that Fermat's last theorem was unprovable? – Carl Mummert May 8 at 11:59
I have read that in several oportunities. If you don´t believe me I can´t do nothing not having at hand in this moment the references. – Piquito May 8 at 12:42
@CarlMummert: Here one of the several references, this one involving Martin Gardner: "Tales eran las dificultades que se encontraban en su demostración que Martin Gardner (Orden y Sorpresa, Alianza Editorial) se llega a plantear la posibilidad de que “... el teorema fuera cierto pero indemostrable. Pero aún, puede ser imposible demostrar que es indemostrable”. – Piquito May 8 at 12:49

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