5
$\begingroup$

I recently stumbled upon the following article:

This article claims to prove that the values of the Riemann Zeta function at all odd positive integers are irrational. Last time I checked, this problem was categorized as "near to impossible with the current state of number theory". Naturally, I am a bit skeptic on whether this proof is correct. However, I do not have the time (or more precisely the required skill) to verify the proof by myself.

Is the proof in the above paper correct?

$\endgroup$
  • 2
    $\begingroup$ For something so consequential it is a surprisingly short proof, but this would still take at least a full afternoon for a number theorist to check in detail. At least a few will need to do so before we know the proof is sound. Having said that, Apery's proof on the irrationality of $\zeta(3)$ was important enough, so this would be earth-shattering if it turns out to be correct. I myself am tangentially familiar with this area and did my dissertation in analytic number theory, and in a quick scan nothing leaps out as being grossly incorrect. $\endgroup$ – Sputnik May 28 '18 at 13:16
  • 1
    $\begingroup$ Is lemma 3.1 correct? Shouldn't it be an implication only in one direction? $\endgroup$ – Jakobian May 28 '18 at 13:30
  • 4
    $\begingroup$ See arxiv.org/abs/1608.03174v4 . The author comments "An erroneous claim of irrationality of all zeta(2n+1) has been withdrawn, with apologies". $\endgroup$ – Lord Shark the Unknown May 28 '18 at 13:31
  • 1
    $\begingroup$ @Adam I wondered about that and thought I was reading it wrong... You're right, not only does the implication only work forwards, I believe there is a 0 missing, as in $0<|p\alpha - q| < \varepsilon$. Otherwise the statement is false, as it works for rationals too. Re: arXiv, glad that's cleared up. I thought it was strange that it was dated from 2016 and nothing else had been said about it... $\endgroup$ – Sputnik May 28 '18 at 13:43
  • $\begingroup$ @LordSharktheUnknown Thank you for the link, that clears it up. $\endgroup$ – Klangen May 28 '18 at 13:48
1
$\begingroup$

As shown in the comment section, the claim was withdrawn by the author.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.