Irrationality of the Squareroot of n

Theorem 0.3 Irrationality of $\sqrt{n}$

Let $n \in \mathbb{N}$ and $n$ is not a perfect square (i.e. $n \neq m^2$. Then $\sqrt{n}$ is irrational.

Suppose that $\sqrt{n}$ is rational. $\sqrt{n} = \frac{p}{q}.$ Then let

$$ E = \{k \in \mathbb{N}: k\sqrt{n} \in \mathbb{N}\}. $$

Then $E$ is non-empty since $q\sqrt{n}=p$. This means that $E$ has a least element by the well ordering principle. Let this element be $n_0$.

$\ \blacksquare$

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