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A look at the first few zeros $$14.134725,21.022040,25.010858,30.424876,32.935062,37.586178,\dots$$ is in accord with

Numerical evidence suggests that all values of $t$ (the imaginary part of a root of $\zeta$) corresponding to nontrivial zeros are irrational (e.g., Havil 2003, p. 195; Derbyshire 2004, p. 384).

(numbers and quote taken from here). What are the attempts to prove that all values of $t$ are irrational? Would it mean something to the distribution of primes, if one, some or plenty of rational roots $\frac{1}{2}+i\frac{q}{r}$ exist?

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5 Answers 5

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Actually, the rationality or irrationality of the Riemann zeros does have subtle influence on the distribution of primes. In analytic number theory, this sub-subject goes under the name 'Oscillation Theorems'. An example can be found in the (excellent) book "Multiplicative Number Theory" by Montgomery and Vaughan. Corollary 15.7 says that if the ordinates $\gamma>0$ of the Riemann zeros are linearly independent over $\mathbb Q$, then $$ \limsup_{x\to\infty}\frac{M(x)}{x^{1/2}}=+\infty $$ and $$ \liminf_{x\to\infty}\frac{M(x)}{x^{1/2}}=-\infty. $$ Here $M(x)$ is the summatory function of the Möbius $\mu$ function: $$ M(x)=\sum_{n<x}\mu(n). $$ The connection is, of course, the explicit formula.

Edit: As a second example, Rubinstein and Sarnak show (roughly speaking) that under the Riemann Hypotheis and the assumption the zeros are linearly independent over $\mathbb Q$, that $$ \lim_{x\to\infty}\frac{1}{\log(x)}\sum_{\substack{n<x\\\pi(n)\ge \text{Li}(n)}}\frac{1}{n}=0.00000026 $$ as well as other results, in their paper Chebyshev's Bias.

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  • $\begingroup$ +1 Great. Do they say somthing about the converse (the Riemann zeros are not linearly independent over $\Bbb Q$) too? A sketch of the connection would be ever so cool? If you need some german "Umlaute", here you are: Ä ä Ö ö Ü ü ß ;-) $\endgroup$
    – draks ...
    Jan 7, 2013 at 21:06
  • $\begingroup$ As others previously mentioned, the Riemann zeros are conjectured to be linearly independent over $\mathbb Q$. This is sometimes called the 'Grand Simplicity Hypothesis', in analogy with the 'Grand Riemann Hypothesis.' Google this for more applications, in particular a nice paper by Rubinstein and Sarnak on Chebyshev's Bias. $\endgroup$
    – stopple
    Jan 7, 2013 at 21:11
  • $\begingroup$ I will, but if you have anything you can recommend, I would be glad to start with that... $\endgroup$
    – draks ...
    Jan 7, 2013 at 21:14
  • $\begingroup$ ...very interesting. Could you check if the link I added is the one you meant? And further, do Montgomery and Vaughn say something about Numbers $n$ such that Mertens' function is zero in connection with Riemann's roots or Chebyshev Bias, too? $\endgroup$
    – draks ...
    Jan 7, 2013 at 23:40
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It is very easy to make conjectures that some numbers are irrational (as a general principle, it's a good bet that something is irrational unless there is a good reason for it to be rational), but with a few exceptions it's very hard to prove them. AFAIK there is no reason for any of the nontrivial zeros to have rational $t$, but no reasonable hope of proving any of these $t$ to be irrational.

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  • $\begingroup$ So you think, there aren't any and if there were some, this doen't have any impact, no matter how many (rational roots) we find? $\endgroup$
    – draks ...
    Mar 18, 2012 at 19:57
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    $\begingroup$ If it could be shown that there were some rational $t$'s, or even some algebraic ones, I think that would be an amazing result, mainly because it would be so totally unexpected. Unexpected results do occasionally come up in mathematics. But I'm not holding my breath. $\endgroup$ Mar 19, 2012 at 1:29
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    $\begingroup$ For something unexpected (at least for me), have a look at stopple's answer... $\endgroup$
    – draks ...
    Jan 7, 2013 at 21:12
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To the best of my knowledge, rationality of the imaginary part of one/some/all of the non-trivial zeros of zeta would have no consequences for the distribution of primes. On the other hand, irrationality of the real part of even one of these zeros would be another thing entirely!

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    $\begingroup$ See my response below. $\endgroup$
    – stopple
    Jan 7, 2013 at 20:37
  • $\begingroup$ Sorry, Gerry. I hope you don't mind that I accepted stopple's answer above... $\endgroup$
    – draks ...
    Jan 7, 2013 at 23:42
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    $\begingroup$ No worries ... I note for posterity that your "above" is stopple's "below" ... is one of you posting from Australia? $\endgroup$ Jan 9, 2013 at 0:41
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A.M. Odlyzko on the nontrivial zeros of the zeta function:

"We will...write $$\rho_{n}=1/2+i\gamma_{n}$$ ...Nothing is known about the $\gamma_{n}$, but they are thought likely to be transcendental numbers, algebraically independent of any reasonable numbers that have ever been considered."

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  • $\begingroup$ Can you give a more explicit reference, i.e. where did Odlyzko state that? $\endgroup$ Jan 7, 2013 at 21:22
  • $\begingroup$ I found it here, unfortunately the correpsonding link there doesn't work anylonger... $\endgroup$
    – draks ...
    Jan 7, 2013 at 23:04
  • $\begingroup$ @NilsMatthes I put a (currently working) link in the post... $\endgroup$
    – draks ...
    Jan 8, 2013 at 22:54
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There were some recent attempts to see what can be done with the current technology, see

http://arxiv.org/abs/1109.1788

http://arxiv.org/abs/1208.2684

But as you can tell, we are still very far from being able to prove that any zero of $\zeta(s)$ is irrational.

If you had finitely many exceptions to the Riemann Hypothesis and all of them had rational imaginary parts this would have very surprising consequences for the distribution of the primes (from the explicit formula, there would be some very strong periodicity phenomena).

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