# Prove that $\alpha:=\sqrt{\pi}+\sqrt 2 \in \Bbb{C}$ is trancendental over $\Bbb{Q}$.

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.

## 3 Answers

If it was algebraic then, since the algebraic numbers form a field and $$-\sqrt2$$ is algebraic,$$\left(\sqrt\pi+\sqrt2\right)-\sqrt2$$would be algebraic too. But this number is $$\sqrt\pi$$, which is transcendental, since $$\pi$$ is transcendental.

• Professor, thank you for your simple and elegant answer. Is there a way to do the proof using the relations above? – Chris Sep 9 '18 at 16:04
• @Chris I don't know how to do that. – José Carlos Santos Sep 9 '18 at 16:06
• Ok. Any other possible ways are welcome at anytime! – Chris Sep 10 '18 at 23:08

You must use the fact that $\pi$ (and so $\sqrt{\pi}$) is transcendental.

Before proving your proposition, I would like to note the fact that if $R$ is a subring of some ring $P$ then the integral elements form a ring (for a proof, go on my post: Proof that the algebraic integers form a subring of $\mathbb{C}$ and replace $\mathbb{Z}$ with $R$ and $\mathbb{C}$ with $P$).

Consider now the evaluation homomorphism $\psi_{\alpha}: \mathbb{Q}[T] \longmapsto \mathbb{Q}(\sqrt{\pi}; \sqrt{2})$, then $$\frac{\mathbb{Q}[T]}{<P_{\alpha}>} \cong \mathbb{Q}(\alpha)$$, where $P_{\alpha}$ is the minimal polynominal of $\alpha$.

From my proof that on the link above you can notice that $$dim_{\mathbb{Q}}(\mathbb{Q}(\alpha)) \leq dim_{\mathbb{Q}}(\mathbb{Q}(\sqrt{\pi};\sqrt{2})) \leq dim_{\mathbb{Q}}(\mathbb{Q}(\sqrt{\pi}) \otimes_{\mathbb{Q}}\mathbb{Q}(\sqrt{2}))$$ but since $$\mathbb{Q}(\sqrt{\pi};\sqrt{2})= \mathbb{Q}(\sqrt{\pi})\mathbb{Q}(\sqrt{2})$$, where by some fields $K$ and $F$, by $KF$ we mean the composite extension also, $$dim_{\mathbb{Q}}(\mathbb{Q}(\sqrt{\pi}) \otimes_{\mathbb{Q}}\mathbb{Q}(\sqrt{2}))= dim_{\mathbb{Q}}(\mathbb{Q}(\sqrt{\pi})\mathbb{Q}(\sqrt{2})) \Rightarrow dim_{\mathbb{Q}}(\mathbb{Q}(\sqrt{\pi}) \otimes_{\mathbb{Q}}\mathbb{Q}(\sqrt{2}))=dim_{\mathbb{Q}}(\mathbb{Q}(\sqrt{\pi};\sqrt{2}))$$.
Since $\sqrt{{\pi}}$ is transcendental, $$dim_{\mathbb{Q}}(\mathbb{Q}(\sqrt{\pi};\sqrt{2}))= \infty$$.
Since $\mathbb{Q}(\sqrt{\pi}) \subseteq \mathbb{Q}(\alpha) \subseteq \mathbb{Q}(\sqrt{\pi};\sqrt{2})$ and $dim_{\mathbb{Q}}(\mathbb{Q}(\sqrt{\pi}))= \infty$ (because $\sqrt{\pi}$ is transcendental), obviously: $$dim_{\mathbb{Q}}(\mathbb{Q}(\alpha))= \infty$$ and hence $P_{\alpha}=0$ and hence $\alpha$ is transcendental. $\,\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \square$

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