The square root of two is irrational.

See square root, rational number
Bartle, Sherbert - Introduction to Real Analysis, page 42

If $\mathbb{Q}$ is the set of rational numbers $\mathbb{Q}=\{\frac{p}{q}:p,q\in \mathbb{Z}\mathtt{\text{ AND }}q\ne 0\}$,
there is no $r\in \mathbb{Q}$ such that $r^2=2$.

Proof

Suppose for the sake of contradiction that $p,q\in \mathbb{Z}$ such that ${\left(\frac{p}{q}\right)}^2=2$ and $q\ne 0$.
Assume without loss of generality that $p,q$ are coprime and have the same sign.

Then $\frac{p^2}{q^2}=2$, and $p^2=2q^2$, so $p^2$ is even.
By contrapositive of An odd number squared is odd., the square root of an even number is even.
Then $p$ is even. So $p=2m$ for some integer $m$.

So $p^2=(2m{)}^2=4m^2=2q^2$,
so $q^2=2m^2$, so $q^2$ is even.
Then $q$ is even.

Then $p,q$ both have a factor of 2. However, this contradicts the assumption that they are coprime.

reductio ad absurdum, QED.