$\sqrt[31]{12} +\sqrt[12]{31}$ is irrational

Question

Prove that $\sqrt[31]{12} +\sqrt[12]{31}$ is irrational.

I would assume that $\sqrt[31]{12} +\sqrt[12]{31}$ is rational and try to find a contradiction.

However, I don't know where to start. Can someone give me a tip on how to approach this problem?

Answer 1

Let $\mathbb{Q}(\alpha)$ denote the smallest field containing $\mathbb{Q}$ and $\alpha$.

The theory of field extensions tells us that $\mathbb{Q}(\sqrt[31]{12})$ has degree $31$ over $\mathbb{Q}$, $\mathbb{Q}(\sqrt[12]{31})$ has degree $12$ over $\mathbb{Q}$, and, because $(31,12)=1$, we have $\mathbb{Q}(\sqrt[31]{12})\cap\mathbb{Q}(\sqrt[12]{31}) = \mathbb{Q}$.

If $\sqrt[31]{12} +\sqrt[12]{31}$ were a rational number, we would have $\sqrt[31]{12} \in \mathbb{Q}(\sqrt[31]{12})\cap\mathbb{Q}(\sqrt[12]{31}) = \mathbb{Q}$. But $\sqrt[31]{12}$ is not rational, contradiction.

Answer 2

Here is a simpler variant of Slade's answer:

If $\sqrt[31]{12} +\sqrt[12]{31}$ were a rational number, we would have $\sqrt[31]{12} \in \mathbb{Q}(\sqrt[12]{31})$ and so $\mathbb{Q}(\sqrt[31]{12}) \subseteq \mathbb{Q}(\sqrt[12]{31})$.

But $\mathbb{Q}(\sqrt[31]{12})$ has dimension $31$ over $\mathbb{Q}$ and so cannot be a subspace of $\mathbb{Q}(\sqrt[12]{31})$, which has dimension $12$.

Answer 3

It is known that algebraic integers are closed under addition, subtraction, product and taking roots.

Since $12$ and $31$ are algebraic integers, so does their roots $\sqrt[31]{12}$, $\sqrt[12]{31}$. Being the sum of two such roots, $\sqrt[31]{12} + \sqrt[12]{31}$ is an algebraic integer.

It is also known that if an algebraic integer is a rational number, it will be an ordinary integer. Notice $$2 < \sqrt[31]{12} + \sqrt[12]{31} < \sqrt[31]{2^4} + \sqrt[12]{2^5} = 2^{\frac{4}{31}} + 2^{\frac{5}{12}} < 2\sqrt{2} < 3$$ $\sqrt[31]{12} + \sqrt[12]{31}$ isn't an integer and hence is an irrational number.