TITLE: Is $\pi/\sqrt{2}$ transcendental?
QUESTION [8 upvotes]: I believe that $\frac{\pi}{\sqrt{2}}$ is transcendental but I'm not sure about how to prove it. If $\frac{\pi}{\sqrt{2}}$ was algebraic, there would exist a polynomial $P \in \mathbb{Q}[X]$ such that $P\big( \frac{\pi}{\sqrt{2}} \big) = 0$. By writing $P = \displaystyle \sum_{k=0}^{N} a_{k}X^{k}$, we have :
$$ \sum_{\substack{\text{k even} \\[1mm] k = 2p}} a_{2p} \frac{\pi^{2p}}{2^{p}} +  \frac{\pi}{\sqrt{2}} \sum_{\substack{\text{k odd} \\[1mm] k = 2p+1}} a_{2p+1} \frac{\pi^{2p}}{2^p} = 0. $$
But I'm not sure that helps..

REPLY [5 votes]: An elementary argument that $\pi$ transcendental implies $\frac{\pi}{\sqrt{2}}$ transcendental goes like this:
If $\;\frac{\pi}{\sqrt{2}}\;$ is algebraic, then it is a root for some polynomial $P(x)$ over $\mathbb{Q}$.  
This in turn implies $\pi$ is a root of polynomial $\displaystyle\;Q(x) = P\left(\frac{x}{\sqrt{2}}\right)P\left(-\frac{x}{\sqrt{2}}\right)$
Splitting $P(x)$ into its even and odd parts, there exists $E(x), O(x) \in \mathbb{Q}[x]$ such that
$$P(x) = E(x^2) + x O(x^2)
\quad\implies\quad Q(x) = E\left(\frac{x^2}{2}\right)^2 - \frac{x^2}{2}O\left(\frac{x^2}{2}\right)^2$$
This means $Q(x)$ also belongs to $\mathbb{Q}[x]$ and this contradicts with the assumption that $\pi$ is transcendental!