Is $-\frac{\pi^5}{2\sqrt2}\times\cot(\frac{\pi}{\sqrt2})$ irrational? I have an idea, and this is the last question for this idea. Pi, and square root of 2 is irrational numbers, but these expression, i don't know. Please, help! Thanks for the answers!
2026-03-29 13:27:40.1774790860
$-(\pi^5)/(2\sqrt{2})\cdot(\cot(\pi/\sqrt{2}))$ is irrational? How to prove it?
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COMMENT.- It is known from Lindemann's paper in which he proved the transcendance of $\pi$ that, among other results, the value in $x$ of the trigonometric function $\cot(x)$ is trascendental if $x$ is non-zero algebraic. But if $x$ is trascendental then $\cot (x)$ can be algebraic (even rational), or transcendental.
On the other hand, the product of two transcendental numbers is not necessarily transcendental (for example until today it is ignored if $e\times \pi$ is algebraic or transcendental).
If $\cot\left(\dfrac{\pi}{\sqrt2}\right)\approx-0.7612229$ is algebraic then we can say that the proposed number is transcendental but if $\cot\left(\dfrac{\pi}{\sqrt2}\right)$ is transcendental then we cannot answer.
The true problem is to know the nature of $\cot\left(\dfrac{\pi}{\sqrt2}\right)$ and this is not known as far as I know.