9

I don't know much about foundations and logic, so I ask forgiveness if my question is just plain stupid.

Assume we have a statement of the form:

There exist no $x\in X$ such that $P(x)$.

where $X$ is some kind of set (or class, or similar stuff) and $P$ is a set of properties. An example of such a statement could be the Riemann hypothesis:

There exist no $x\in\mathbb{C}$ such that $\Re(x)\neq\frac{1}{2}$ such that $x$ is not a negative even integer and $\zeta(x)=0$.

Can such a statement be provably unprovable?

Intuitively, I would say no, because to show that it is unprovable we should show that we cannot find $x\in X$ such that $P(x)$ (else finding such an $x$ would be a proof that the statement is false), but doing so would prove the statement to be true.

Is this correct, or am I missing something?

Edit: Please read the question correctly: it is not properly a question on the RH, but more a question on logic!

Daniel Robert-Nicoud
  • 29,795
  • If the Riemann hypothesis is wrong, then it is provable. Just find a contradicting x. But there could be a proof that shows under the condition that the hypothesis is true, there can not exist a derivation of a proof from the axioms of set theory, or similar... – Loreno Heer Nov 06 '14 at 23:36
  • @user2345215 I meant an $x \in \mathbb{C}$ – Loreno Heer Nov 06 '14 at 23:38
  • @user2345215 Finding it is a computational aspect, it is not relevant. If the hypothesis is wrong, such an $x$ exists, regardless if it is found or not. And that $x$ proves the statement is wrong. – Loreno Heer Nov 06 '14 at 23:43
  • @sanjab I don't believe your explanation, but I just realized there are algorithms which can verify there are no sufficiently small zeroes. That proves it, not your "argument", but I'll delete my previous comments as this is getting lengthy. – user2345215 Nov 06 '14 at 23:47
  • no need to delete the comments. try to prove me wrong. – Loreno Heer Nov 06 '14 at 23:49
  • 3
    @sanjab There are real numbers which you can't even define (uncountably many of them, but only countably many formulas). So how can you computationally verify there are not solutions? – user2345215 Nov 06 '14 at 23:50
  • 2
    @user2345215 Jose Arnaldo Dris's Math Overflow link demonstrates that the falsehood of RH implies that it is provably false. – Dustan Levenstein Nov 06 '14 at 23:55
  • @user2345215 yes you are correct about this. I can't make my argument like this. – Loreno Heer Nov 07 '14 at 00:00
  • Your terminology is a bit misleading. Every false statement is unprovable (assuming the consistency of the axioms that you are using). So if the Riemann Hypothesis is false, it is also unprovable. What you should be asking is, could the Riemann Hypothesis be provably undecidable? – TonyK Nov 27 '14 at 17:08
  • @TonyK By unprovable I mean that it cannot be proven to be either true or false (it should be the same as undecidable, right?) – Daniel Robert-Nicoud Nov 27 '14 at 18:36
  • Yes, I know that's what you meant. But that's not what 'unprovable' means. – TonyK Nov 27 '14 at 18:53

5 Answers5

6

If $p$ is a zero off the critical line, there are rational numbers $A<B, C<D$ such that the square with corners $A+iC, B+iC,B+iD, A+iD$ does not intersect the critical line and has a zero (namely $p$) inside it. Thanks to the Argument Principle, this fact can be proven by computing a contour integral numerically with sufficient precision. So if RH is false, it must be provably false. And so a proof of undecidability would be a proof that it is true.

Robert Israel
  • 448,999
  • In ZFC, couldn't the proof of the negation of the Riemann hypothesis be a theorem along the lines of "There is a non-trivial zero not on the critical line, but it's not possible to construct (find) it."? – Nikolaj-K Nov 07 '14 at 11:58
  • 2
    If it is impossible to find it, then it doesn't exist. Now it could be that there is a proof in ZFC of the negation of RH, even though RH is actually true. That would mean ZFC is not $\omega$-consistent. But you couldn't have the theorem you suggested unless ZFC is inconsistent. – Robert Israel Nov 07 '14 at 15:45
5

You can check out the answers to this related MO question:

"Can the Riemann hypothesis be undecidable?"

Community
  • 1
Jose Arnaldo Bebita Dris
  • 1
0

Let us say that there exists some $x \in \Bbb{C}$ such that $x$ is neither a negative integer nor on the line of $\Re(x) = \frac{1}{2}$. Let us further suppose that some inspired mathematician conjectured that $x$ could be expressed as the sum of some convergent but some god-awful complicated series.

Now "experiemental mathematicians" might well look at this $x$ and determine that $\zeta(x)$ is zero to a thousand decimal places; but of course that does not constitute a proof that $\zeta(x)=0$. And it is certainly logically plausible that the theorem "$\zeta(x)$, with $x$ defined as $\sum_\infty (\mbox{this horribly complicated mess})$, is zero" could be true but unprovable.

So it is plausible that the Riemann hypothesis could be undecidable within the usual mathematical axiomatic framework. Contrast this with Fermat's last theorem (ignoring Wile's proof): there, any given purported counterexample can be tested algorithmically in a finite number of steps, so if you could prove it was undecidable, you would be proving it false. (This argument would not say it can't be undecidable, just that it cannot be proven to be undecidable.)

Mark Fischler
  • 41,743
-1

No. The problem is in the definition of zeros. If we would not have 1,2,3,4,5,6... on the left side of the equation, which is the sum

Left side

we could argue that we do not have enough information about zeros, hence some of them are not defined in some particular sense, hence Riemann cannot be decided. However, although we cannot calculate all of the zeros, there is nothing ambiguous about them. They are all perfectly defined objects on their own accord, so if the Riemann hypothesis is false there is the first zero off 1/2 mark.

Although I said this, hypothetically speaking it could be true that the set of zeros is not defined completely, by simply stating that they are zeros of the extended sum as above. This might mean that we would need to add something to the definition of natural numbers, some minuscule and yet not discovered condition that would change their nature in such a way that with one condition Riemann is true, yet with another false.

This last is not likely because we can already define the way of calculating any zero anywhere, although we do not have time and space to collect all of them. So it is definitely true or false with Riemann.

Community
  • 1
  • Why isn't $S= {s \in \mathbb{C} | \sum\limits_{1}^{n}\frac{1}{n^s}=0}$ a complete definition of the set of zeros? – h34 Jul 15 '15 at 11:09
  • The sum is not convergent for s having a real part 1/2. That is why it has to be extended first. –  Jul 16 '15 at 11:36
-1

The answer to the question is no: the Riemann hypothesis couldn't be provably unprovable. A quick way to show this is to start by observing that it cannot be false and provably unprovable, because if it is false there must be a counterexample that can be shown to be a counterexample, by which we could prove it false. Contradiction. So if we assume it's provably unprovable then it must be true, which is also a contradiction.

But it might be unprovably unprovable.

Note that it must be either true or false: either there is a counterexample or there isn't.

h34
  • 119
  • What if it's false but the additional zeros are of the "indescribable" kind, i.e. their real and imaginary components are neither whole, nor rational, nor algebraic, nor lending themselves to any other kind of finite symbolic description? – Ron Inbar Apr 02 '21 at 12:06