Proving $\sqrt{p}$ is irrational number

Exercise:

Show that , $\sqrt{p}$ is irrational number whenever $p$ is prime number

Solution: suppose that $\sqrt{p}$ is rational then by the definition of rational

We can suppose $\sqrt{p}=\frac{a}{b}$ where $\frac{a}{b}$ is irreducible fraction hence $\gcd \left( a,b \right)=1$

We have $\sqrt{p}=\frac{a}{b}\Leftrightarrow p=\frac{{{a}^{2}}}{{{b}^{2}}}\Leftrightarrow {{b}^{2}}p={{a}^{2}}$ so $p|{{a}^{2}}\Leftrightarrow p|a$

But ${{a}^{2}}$ is multiple of $p$ hence $a$ is multiple of $p$ i.e $a=pk$ where $k\in \mathbb{Z}$

So ${{b}^{2}}=\frac{{{a}^{2}}}{p}=\frac{{{p}^{2}}{{k}^{2}}}{p}=p{{k}^{2}}=pk\Leftrightarrow p|{{b}^{2}}\Leftrightarrow p|b$

So $p|\gcd \left( a,b \right)$ hence $\gcd \left( a,b \right)\ne 1$ thus $\sqrt{p}$ is irrational number