TITLE: Showing that $\sqrt \pi$ is transcendental
QUESTION [19 upvotes]: I'm trying to use the fact that $\pi$ is transcendental to show that $\sqrt \pi$ is also transcendental over $\Bbb{Q}$ . I don't know any theorems about algebraic and non-algebraic numbers so I don't have much in my tool box. Here's what I have done so far :
Let $P(x) \in \Bbb{Q}[x]$ and write $P(x)$ in the following way $$P(x)= \underbrace{ \left ( \sum_{\text{$j$ even}} a_j x^j +a_0 \right ) }_{Q(x)}+\underbrace{\sum_{\text{$i$ odd}} a_i x^i }_{G(x)}$$ Then $P(\sqrt \pi )= Q (\sqrt \pi) + G(\sqrt \pi)$ , $ Q (\sqrt \pi) =\displaystyle \sum_{\text{$j$ even}} a_j  \sqrt\pi ^j+a_0  =\sum_{\text{$j$ even}} a_j \pi^{\frac j 2} +a_0$ hence $Q (\sqrt \pi) \not= 0$ since $\pi$ is transcendental, Similarly I showed $G(\sqrt \pi)$ is also non zero by taking $\sqrt \pi$ as a common factor. I'm now stuck on showing that $G(\sqrt \pi) \not= -Q(\sqrt \pi) $ . If can prove that then $P(\sqrt \pi) \not= 0$.  How can it be done?

REPLY [4 votes]: I found another way to solve it, suppose that $\sqrt \pi$ is a root of $P(x)$ which is defined in the question then we have$$  \left ( \sum_{\text{$j$ even}} a_j \pi^{\frac j 2} +a_0 \right ) +\sqrt \pi \sum_{\text{$i$ odd}} a_i \pi^{i-1} = 0 $$ then $Q(\pi)=-\sqrt \pi G(\pi)$ upon squaring both sides we get $Q^2(\pi)= \pi G^2(\pi)$ then $Q^2(\pi)- \pi G^2(\pi)=0$ is a polynomial in $\pi$ equal to zero which is a contradiction since $\pi$ is transcendental.