 \sqrt{12}+\sqrt{31} is irrational? aramutselv 2022-01-06 Answered

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

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I would assume that $$\sqrt{12}+\sqrt{31}$$ 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?

Not exactly what you’re looking for? 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 $$\mathbb Q(\sqrt{12})$$ has degree $$\displaystyle{31}$$ over $$\displaystyle{\mathbb{{Q}}}$$, $$\mathbb Q(\sqrt{31})$$ has degree 12 over $$\displaystyle{\mathbb{{Q}}}$$ , and, because $$\displaystyle{\left({31},{12}\right)}={1}$$, we have
$$\mathbb Q(\sqrt{12})\cap\mathbb Q(\sqrt{31})=\mathbb Q$$
If $$\sqrt{12}+\sqrt{31}$$ were a rational number, we would have
$$\sqrt{12} \in \mathbb Q(\sqrt{12})\cap\mathbb Q(\sqrt{31})=\mathbb Q$$ But $$\sqrt{12}$$ 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 $$\sqrt{12},\sqrt{31}$$. Being the sum of two such roots, $$\sqrt{12}+\sqrt{31}$$ is an algebraic integer.
It is also known that if an algebraic integer is a rational number, it will be an ordinary integer. Notice
$$2<\sqrt{12}+\sqrt{31}<\sqrt{2^4}+\sqrt{2^5}=2^{\frac{4}{31}}+2^\frac{5}{12}<2\sqrt2<3$$
$$\sqrt{12}+\sqrt{31}$$ isn't an integer and hence is an irrational number.