Proof of irrationality of square root of 5

Assume to the contrary that $\sqrt{5}$ is rational. Let $\sqrt{5} = \frac{a}{b}$, where $a, b$ are coprime integers and $b \neq 0$. $5 = \frac{a^2}{b^2} \implies a^2 = 5b^2$ Thus, 5 divides $a^2$, which implies 5 divides $a$. Let $a = 5c$ for some integer $c$. $(5c)^2 = 5b^2 \implies 25c^2 = 5b^2 \implies b^2 = 5c^2$ Thus, 5 divides $b^2$, which implies 5 divides $b$. Therefore, 5 is a common factor of both $a$ and $b$, contradicting the assumption that $a$ and $b$ are coprime. Hence, $\sqrt{5}$ is irrational.