TITLE: Prove that $\alpha:=\sqrt{\pi}+\sqrt 2 \in \Bbb{C} $ is trancendental over $\Bbb{Q}$.
QUESTION [5 upvotes]: We want to prove that $\alpha:=\sqrt{\pi}+\sqrt 2 \in \Bbb{C} $ is
  trancendental over $\Bbb{Q}$.

Attempt. We use proof by contradiction and so assume that $\alpha \in \Bbb{C}$ is algebraic over $\Bbb{Q}$. Then we have that $$\left[\Bbb{Q}\left(\sqrt{\pi}+\sqrt 2\right):\Bbb{Q}\right]<\infty.$$
It is true that $[K(\sqrt \pi ):K]=\infty$, where $K$ is a field and the proof is the same with this post. So, one can think to use the following relations:

$\sqrt \pi,\sqrt2 \in \Bbb{Q}(\sqrt \pi, \sqrt 2) \implies \sqrt \pi+\sqrt2 \ \in \Bbb{Q}(\sqrt \pi, \sqrt 2) \implies \Bbb{Q}(\sqrt \pi+ \sqrt 2)\subseteq \Bbb{Q}(\sqrt \pi, \sqrt 2) $
$\Bbb{Q} \leq \Bbb{Q}(\sqrt2)\leq  \Bbb{Q}(\sqrt \pi+ \sqrt 2)\leq \Bbb{Q}(\sqrt \pi, \sqrt 2)$ and one logical thought is to apply the Tower Law, but I m not sure if this helps. 

Any ideas please?
Thank you.

REPLY [3 votes]: Continuing your argument:
$$
\pi \in \Bbb{Q}(\sqrt \pi) \subseteq \Bbb{Q}(\sqrt \pi, \sqrt 2) = \Bbb{Q}(\alpha)
\implies 
\Bbb{Q}(\pi) \subseteq \Bbb{Q}(\alpha)
\implies 
[\Bbb{Q}(\alpha ):\Bbb{Q}] \ge [\Bbb{Q}(\pi):\Bbb{Q}]=\infty
$$