# The square root of 2 is not rational

## Proof by contradiction

Let's assume that there exists a rational number $q$ equal to $\sqrt{2}$. Therefore, by definition,

\begin{equation} \exists a \in \mathbb{N}, b \in \mathbb{N} \setminus \{ 0 \}, \sqrt{2} = q = \frac{a}{b} \end{equation}We can also write it as an irreducible fraction, so that $q = c/d$, such that $d$ does not divide $c$. From the properties of the square root, we have

\begin{equation} q^2 = \frac{c^2}{d^2} = 2 \end{equation}We can multiply both sides by $b^2$ to obtain

\begin{equation} c^2 = 2 d^2 \end{equation}