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# 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,

$$\exists a \in \mathbb{N}, b \in \mathbb{N} \setminus \{ 0 \}, \sqrt{2} = q = \frac{a}{b}$$

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

$$q^2 = \frac{c^2}{d^2} = 2$$

We can multiply both sides by $b^2$ to obtain

$$c^2 = 2 d^2$$