Is Riemann Hypothesis provable?

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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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  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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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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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.