algebra precalculus - How to show $sqrt{p}$ is irrational

How to show that $\sqrt{p}$ is an irrational number, given that $p$ is a prime number?

$\sqrt{p}=\frac{a}{b}$

$p=\frac{a^2}{b^2}$

Answer

Suppose $\sqrt p=\frac{a}{b}$ with $(a,b)=1$.

Then

$$\begin{align*} b^2p=a^2&\implies p\mid a^2\\ &\implies p\mid a\\ &\implies \exists k\in\mathbb Z: a=kp\\ &\implies b^2p=k^2p^2\\ &\implies pk^2=b^2\\ &\implies p\mid b^2\\ &\implies p\mid b. \end{align*}$$ Since $(a,b)=1$, we get $p=1$. Contradiction.