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Someone recently quoted to me this recent article that claims to prove that $\zeta(2n+1) \notin (2\pi )^{2n+1} \mathbb{Q}$.

I always assumed this was well known. More precisely I thought this result followed from the fact that the regulator $$ K_{2n-1}(\mathbb{Z})\otimes \mathbb{Q} = Ext^1_{MT(\mathbb{Z})}(\mathbb{Q}(0),\mathbb{Q}(n)) \longrightarrow Ext^1_{MHS}(\mathbb{Q}(0),\mathbb{Q}(n)) = \mathbb{C}/(2\pi i)^n\mathbb{Q} $$ is injective (this is usually quoted as a consequence of Borel's computations of K-groups "Stable real cohomology of arithmetic groups", "Cohomologie de $SL_n$ et valeurs de fonctions zêta aux points entiers")

Am I mistaken?

PS: corrected a typo thx to Pete L Clark

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Is there a typo in your noninclusion statement in the first line? The zeta value is real, and what's on the right hand side is purely imaginary, so, yes, this is well-known, but probably not what you want to say. –  Pete L. Clark Sep 9 '10 at 16:34
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Leading indicators on the paper: uninformative abstract. "Interesting" search engine results for the author's name (antigravity propulsion, degrees not in math, 60 years old, etc). Breakthrough if correct, but was not publicized. Appears in an obscure journal. –  T.. Sep 9 '10 at 18:35
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@T: It has been discussed on this site before that we are generally not interested in non-expert opinions on whether a paper is correct, and, especially, opinions that are based on anything other than a careful reading of the paper. I am particularly uncomfortable that you cite the author's age as a strike against him -- this seems downright discriminatory. (Note that I cannot get free online access to the paper and therefore am certainly not claiming that it is correct. I edited the question so as to make it sound like less of an endorsement of the paper's correctness.) –  Pete L. Clark Sep 9 '10 at 20:28
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Pete, how do you know T. isn't an expert? Anyway, following the links via Google as T. did led to Journal of Theoretics, a journal whose article titles suggest it is entirely devoted to the work of physics cranks, and in which Musha has multiple articles. The story of Heegner shows that amateurs can make breakthroughs, but just as well-known experts are often given benefit of the doubt when they make announcements without producing details, why shouldn't people who publish crank articles be doubted before seeing the details? –  BCnrd Sep 10 '10 at 2:09
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@B: I don't know who T. is, and that's part of my point: it's hard to place much value in anonymous opinions. But more to the point, even from an expert I don't want to see on a public forum opinions that are not based on the work itself. Anyway, of course you're welcome to doubt the validity of someone's article that you haven't seen, for any reason. [Was there anything in what I wrote that made you think that I have confidence in Musha's result?] It is not always appropriate or wise to publicly express every opinion you have. –  Pete L. Clark Sep 10 '10 at 5:18

1 Answer 1

It is not known (but conjectured) whether the numbers $\zeta(2n+1)/\pi^{2n+1}$ are irrational, $n=1,2,\dots$. It is not even known whether at least one of these numbers is irrational! In fact, the most general (folklore) conjecture states that $\pi$ and all odd zeta values are algebraically independent over $\mathbb Q$. There are natural links between this conjecture and the expected structure of the so-called multiple zeta values; the references I have in mind are papers by A. Goncharov and surveys/talks by M. Waldschmidt.

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