If I'm not mistaken, one of the things Godel showed with his incompleteness theorems is that things are true in models, not theories. Since ZFC is a theory, not a model, asking whether the Riemann hypothesis is unprovably true or false is meaningless, just like asking "is the continuum hypothesis true or false in ZFC" is meaningless. It's true in some models of ZFC and false in others, which is why it's called "independent" of ZFC.

@Mana not true afaik. There are statements that are true in ZFC but not provable in ZFC. That’s exactly one of Godels results.

As soon as ZFC is shown to be inconsistent, both provably true and provably false would need to be resolved as YES, right?

How could it be unprovably false in ZFC? If it's false, there should be a counter-example.

@OlegEterevsky Same way as how can axiom of choice be false in ZF.

@mathvc I don't think that's the right analogy. Riemann's hypothesis is constructive and can be formulated using only countable sets. For countable sets Axiom of choice can not be false.

@OlegEterevsky RH may be false, it may have a counter-example, but ZFC is not enough to verify that it is indeed a counterexampe. There are plenty of such statements independent of ZFC e.g. Whitehead problem.

@mathvc Since Riemann's zeta-function is analytical (or more precisely meromorphic), we can tell whether a given patch of complex plane contains at least one zero or not by integrating zeta' / zeta over the contour around the patch.

Suppose Riemann's function has a non-trivial zero at the distance eps from the line Re z = 1/2. If we cut the strip 0 < Re z < 1 into small squares by drawing a grid with the step <eps/2, then the zero will lie in one such square that does not intersect with the middle line.

Now follow the following procedure: iterate by natural N. For each N take the part of the strip with -N < Im z < N and deliniate it by the grid with the step 1/N. Then test for each cell whether it contains a zero or not. Thus in a finite number of steps you verify whether there is a zero with abs(Im(z)) < N and at the distance >1/N from the center line.

Returning to the non-trivial root z at the distance eps from the middle line. If we take N that is big enough that N > Im z and 1/N < eps/2, then we are bound to find a small square that contains a zero and doesn't intersect with the middle line, all in the finite number of steps.

Thus, if Riemann hypothesis is wrong, this fact is constructively provable.