February 11th, 2020

$\mathbf{Theorem}$

$\sqrt{2}$ is irrational.

$\mathbf{Proof}$

Let $a,b \in \mathbb{Z} \quad, \frac{a}{b}=\sqrt{2},\quad b \neq 0, \quad (a,b)=1$

Then $a=\sqrt{2}b$

\begin{align*} \implies & a^2=2 \times b^2 \quad (\Xi) \\ \implies & 2|a^2 \\ \implies & 2|a \\ \end{align*}

Then if $a$ is even, $\exists k \in \mathbb{Z} \ni a = 2k$

Then substitute into $(\Xi)$, we get $(2 \times k)^2=2\times b^2$

\begin{align*} \implies & 4 \times k^2 = 2 \times b^2 \\ \implies & 2 \times k^2 = b^2 \\ \implies & 2|b^2 \\ \implies & 2|b \end{align*}

If $a$ and $b$ are both even, it contradicts the initial condition $(a,b)=1$.

$\therefore$ By Law of Contradiction, $\sqrt{2}$ is irrational ◼︎