If $\alpha\in(0,\pi/4)$ and $\tan(\alpha)$ is rational, then $\sqrt{2\tan(2\alpha)}$ is irrational

Question

Let $ \alpha\in(0,\frac{\pi}{4})$.

prove that

$$\tan(\alpha) \in \Bbb Q \;\implies \;\sqrt{2\tan(2\alpha)}\notin \Bbb Q$$

I tried to prove the contrapositive, but it is still not obvious. Any help will be appreciated.

Answer 1

Denote $a = \text{tan}(\alpha)$. Since $\alpha \in (0, \frac{\pi}{4})$, it follows that $a \in (0,1)$.

Since $a \in \mathbb{Q} \cap (0,1)$, there exist some coprime numbers $p<q \in \mathbb{N}$ such that $a = \frac{p}{q}$.

Now, the idea is to use the identity: $$\text{tan}(2\alpha) = \frac{2\cdot\text{tan}(\alpha)}{1-\text{tan}^2(\alpha)}$$

Suppose by way of contradiction that $\sqrt{2\cdot \text{tan}(2\alpha)} \in \mathbb{Q}$. Then, there are some $a,b \in \mathbb{N}$, $(a,b)=1$, such that: $$\frac{\frac{p}{q}}{1-\frac{p^2}{q^2}}=\frac{a^2}{b^2} \iff \frac{pq}{q^2-p^2}=\frac{a^2}{b^2}$$ Note that $(pq,q^2-p^2)=1$ (becacuse $p$ and $q$ are coprime), hence $pq=a^2$ and $q^2-p^2 = b^2$. Also, using again the fact that $(p,q)=1$ we get that $p = c^2$ and $q = d^2$ for some $c,d \in \mathbb{N}$ with $(c,d)=1$. Thus, we obtain: $$d^4 - c^4 = b^2$$

The last equality is impossible. See here why.