\[\textbf{Prove square root of 2 is irrational}\]

$\text{Assume } \sqrt{2} \in \mathbb{Q}$

$\text{let }n = min\{ n \in \mathbb{N} \mid n\sqrt{2} \in \mathbb{N}\}$

$\Rightarrow n(\sqrt{2} - 1)\sqrt{2} \in \mathbb{Q}$

$\because \sqrt{2} - 1 \le 1$

$\Rightarrow n(\sqrt{2} - 1)\sqrt{2} \le n\sqrt{2}$

$\Rightarrow n(\sqrt{2} - 1) \le n \text{ such as } n(\sqrt{2} - 1)\sqrt{2} \in \mathbb{N}$

$\Rightarrow \text{This contracts our assumption } \quad \square$

\[ \text{Prove Square root of 2 is irrational} \text{ [ Geometric proof ] }\] $\text{Assume } \sqrt{2} \in \mathbb{Q}$
$\Rightarrow \frac{a}{b} = \sqrt{2} \quad a, b \in \mathbb{N} \text{ and } \gcd(a, b) = 1, a > b$

Given a right issoseles triangle [above figure]
$AB = AC, \quad FE \text{ tangles to arc at point } E $
$\Rightarrow AF = EF$
$\text{Let } AB = AC = 1$
$\Rightarrow BC = \sqrt{2}$
$\Rightarrow FB = 1-EB = 1-(\sqrt{2} - 1) = 2-\sqrt{2}$
$\because \frac{AC}{CB} = \frac{EB}{FB} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}-1}{2-\sqrt{2}}$
$\therefore \frac{1}{\sqrt{2}} =\frac{1}{\frac{a}{b}} = \frac{\sqrt{2}-1}{2-\sqrt{2}} = \frac{\frac{a}{b} - 1}{2-\frac{a}{b}}$
$\therefore \frac{b}{a} = \frac{ \frac{a-b}{b} } {\frac{2b - a}{b}} = \frac{a-b}{2b-a}$
$\because \sqrt{2} \lt \sqrt{4} = 2 \therefore a \lt 2b $
$\Rightarrow \color{red}{a-b \lt b} $
$\because 2a \gt 2b$
$\Rightarrow \color{red}{2b - a \lt a}$
$\frac{b}{a} = \frac{\color{red}{a-b}}{\color{red}{2b-a}}$
$\text{That contracts our assumption } \gcd(a, b) = 1$
$\Rightarrow \frac{b}{a} \notin \mathbb{Q}$

The geometric proof is similar to the first proof, in fact both proofs use the idea which rational number can be written in $\frac{a}{b} \text{ and }\gcd(a,b)=1$ and contraction will be merged once the donominator/numerator are reducted