Is it known if $\frac{\zeta(3)}{\pi^3}\in\mathbb{Q}$?
It is obvious that $\frac{\zeta(2n)}{\pi^{2n}}\in\mathbb{Q}$, but since there is no closed form for the odd values, are we left to be unable to determine if the solution could have such a form?
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Simply Beautiful Art
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4http://mathoverflow.net/questions/60595/is-zeta3-pi3-rational – Wojowu May 05 '16 at 11:46
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@Wojowu Thanks. – Simply Beautiful Art May 05 '16 at 11:47
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1I'm not sure it is that "obvious" that $\frac{\zeta(2n)}{\pi^{2n}}\in\mathbb{Q}$... – fretty May 05 '16 at 12:14
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@fretty Did you look at the general solution to $\zeta(2n)$? Every component with the exception of the $\pi^{2n}$ is rational, and multiplied and divided together, ${\mathbb{Q}\over\mathbb{Q}}\in\mathbb{Q}$ – Simply Beautiful Art May 05 '16 at 20:54
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But the formula itself is hardly obvious... – fretty May 05 '16 at 22:29
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@fretty Well, it was rather obvious when I looked at it. What gave you doubts? – Simply Beautiful Art May 05 '16 at 22:32
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The formula is non-trivial...sure once you KNOW it then it is obvious that $\frac{\zeta(2k)}{\pi^{2k}}\in\mathbb{Q}$ but the formula itself isn't obvious, it requires some non-trivial maths to find/prove it! – fretty May 05 '16 at 22:37
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It's only obvious ONCE you know the formula. It isn't obvious just from the definition of the zeta function. – fretty May 05 '16 at 22:55
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@fretty Oh, of course not. When you said formula, I didn't think you meant the definition. Sorry. – Simply Beautiful Art May 05 '16 at 23:33
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Well $\zeta(3)$ is what known as Apery's constant, and it is proved to be irrational. But whether $\frac{\zeta(3)}{\pi^3}$ is rational or not is not known. For little more info you can check this link
Kushal Bhuyan
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