# Proof by cases that there is an irrational number $a$ such that $a^{\sqrt{3}}$ is rational?

My textbook has established the classic proof by cases that there exist irrational $a$ and $b$ such that $a^b$ is rational (namely, assuming $\sqrt{2}^\sqrt{2}$ is irrational, let $a=\sqrt{2}^\sqrt{2}$ and $b=\sqrt{2}$, then $a^b=2$; then, assuming $\sqrt{2}^\sqrt{2}$ is rational, $a=b=\sqrt{2}$ satisfies the equation), but then presents the following twist on the problem that I am having difficulty with.

Prove that there is an irrational number $a$ such that $a^\sqrt{3}$ is rational. Hint: Consider $\sqrt[3]{2}^\sqrt{3}$ and argue by cases.

So, assuming I am being prompted to structure the proof by cases like the classic example, I am not supposed to have any information on whether $\sqrt[3]{2}^\sqrt{3}$ is rational or not. For the case where $\sqrt[3]{2}^\sqrt{3}$ is irrational, it's trivial:

Assume $\sqrt[3]{2}^\sqrt{3}$ is irrational. Then, $a = \sqrt[3]{2}^\sqrt{3}$, and $a^\sqrt{3} = (\sqrt[3]{2}^\sqrt{3})^\sqrt{3} = 2$.

Moving on to the rational case, I played around with multiplication with a known irrational number $x$, so $a = x\cdot\sqrt[3]{2}^\sqrt{3}$, so $a^\sqrt{3} = (x\cdot\sqrt[3]{2}^\sqrt{3})^\sqrt{3}$. But then found I would have to demonstrate the rationality of $x^\sqrt{3}$, which seemed like it was getting away from the point of the problem, in particular if I am not supposed to know anything about the rationality of $\sqrt[3]{2}^\sqrt{3}$.

How do I use $\sqrt[3]{2}^\sqrt{3}$ to prove the rational case?

• Whoops, that was a careless transcription error. Thanks! – Kanpachi May 27 '18 at 8:16

## 1 Answer

If $\sqrt[3]2^{\sqrt3}\in\mathbb Q$, then take $a=\sqrt[3]2$. Then $a$ is an irrational number such that $a^{\sqrt3}$ is rational.

• Well, don't I feel silly. Thank you! – Kanpachi May 27 '18 at 8:02