$\sqrt{5} \in \mathbb{R}$ is algebraic over $\mathbb{Q}$

Question

I have to show that $\sqrt{5} \in \mathbb{R}$ is algebraic over $\mathbb{Q}$ and then I have to find $Irr(\sqrt{5}, \mathbb{Q})$.

How can I show that $\sqrt{5} \in \mathbb{R}$ is algebraic over $\mathbb{Q}$ ??

Answer 1

Hint: Try find a nontrivial polynomial $f(x)$ over $\mathbb{Q}$ such that $f(\sqrt{5})=0.$

Answer 2

$\sqrt 5$ is a solution of $X^2-5\in\mathbb Q[X]$, moreover, by Eisenstein criterion, this polynomial is irreducible !