Irrationality of the Riemann zeta-function on odd arguments

Question

It is proved that infinitly many of the numbers $\zeta(2n+1)$ are linearly independent over $\mathbb{Q}$. Is there a conjecture such as 'the numbers $\zeta(2n+1)$ are linearly independent over $\mathbb{Q}$'?

Thank you!

Answer 1

If infinitely many $\zeta(2n+1)$ are linearly independent over $\mathbb Q$, you can conclude that infinitely many are irrational.

Because if $x$ and $y$ are linearly independent over $\mathbb Q$, then

$$\forall (r,r')\in\mathbb Q^2\setminus\{(0,0)\}, \quad rx+r'y\ne 0,$$

so $x$ or $y$ is irrational, because if they were both rational, $r=y$ and $r'=-x$ would work.

And yes, there is such a conjecture, widely open so far.

(Because requested in the comments, this paper shows that infinitely many of the $\zeta(2n+1)$ are irrational.)