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

aramutselv 2022-01-06 Answered

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

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Elois Puryear
Answered 2022-01-07 Author has 1831 answers

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
Answered 2022-01-08 Author has 1651 answers

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.

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Vasquez
Answered 2022-01-11 Author has 8850 answers

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.

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