How to show $\sqrt{p}$ is irrational

Question

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 1

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.

Answer 2

Suppose $a$ and $b$ have no common factors, i.e., the fraction is in its lowest form. $$a^2 = pb^2 \implies p | a^2 \implies p | a \implies a = pa_1 \implies p^2a_1^2 = pb^2 \implies pa_1^2 = b^2 \implies p | b^2 \implies p|b$$

This implies $a$ and $b$ have $p$ as a common factor which contradicts what we assumed without loss of generality.