Irska

Let $A={\tan }^{-1}(\sqrt{2})$ and assume $\frac{A}{\pi }=\frac{m}{n}$ with integers $m,n$. This gives $nA=m\pi$ and $\tan (A)=\sqrt{2}$. The strategy is to study the sequence $\tan (kA)$ for $k=1,2,3,\dots$. We first observe that $nA=m\pi$ implies that this sequence contains only finitely many different values, namely $\tan (A),\tan (2A),\dots ,\tan (nA)$. The reason is the periodicity $\tan (x+\pi )=\tan (x)$. The required contradiction is now achieved by using $\tan (A)=\sqrt{2}$ and the double angle formula $$\tan (2B)=\frac{2\tan (B)}{1-{\tan }^2(B)}$$ If $\tan (B)=\frac{a}{b}\sqrt{2}$, where $a,b$ are co-prime integers and $a$ is even, then $$\tan (2B)=\frac{2ab}{a^2-2b^2}\sqrt{2}$$ and $2ab$ is even, $b^2-2a^2$ is odd. Moreover, $\gcd (ab,b^2-2a^2)=1$, because a prime $p$ which divides $ab$ and $b^2-2a^2$ has either to divide $a$ and so also $b$ or has to divide $b$ (which is odd) and $2a^2$, so it divides $a$. But $\gcd (a,b)=1$ implies that no such prime $p$ can exist. Therefore, we again obtained a fraction in lowest terms and with even numerator. As we have $|2ab|>|a|$ the numerator of $\frac{1}{\sqrt{2}}\tan (2B)$ is bigger than the numerator of $\frac{1}{\sqrt{2}}\tan (B)$. Because $\frac{1}{\sqrt{2}}\tan (2A)=-2$ has even numerator, we can conclude that in the sequence $\tan (2A),\tan (4A),\tan (8A),\dots ,\tan (2^{k}A),\dots$ no two numbers are equal in contradiction to the finiteness shown above. This contradiction proves that $\frac{A}{\pi }$ cannot be rational.