# \sqrt[31]{12}+\sqrt[12]{31} is irrational?

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

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Elois Puryear

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?

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Jillian Edgerton

Let $$\displaystyle{\mathbb{{Q}}}{\left(\alpha\right)}$$ denote the smallest field containing $$\displaystyle{\mathbb{{Q}}}$$ and $$\displaystyle\alpha$$
The theory of field extensions tells us that $$\mathbb Q(\sqrt[31]{12})$$ has degree $$\displaystyle{31}$$ over $$\displaystyle{\mathbb{{Q}}}$$, $$\mathbb Q(\sqrt[12]{31})$$ has degree 12 over $$\displaystyle{\mathbb{{Q}}}$$ , and, because $$\displaystyle{\left({31},{12}\right)}={1}$$, 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.

Vasquez

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
$$2<\sqrt[31]{12}+\sqrt[12]{31}<\sqrt[31]{2^4}+\sqrt[12]{2^5}=2^{\frac{4}{31}}+2^\frac{5}{12}<2\sqrt2<3$$
$$\sqrt[31]{12}+\sqrt[12]{31}$$ isn't an integer and hence is an irrational number.