Prove that for any prime positive integer p,

Solution:

Let us assume that $\sqrt{p}$ is rational . Then, there exist positive co primes a and $\mathrm{b}$ such that

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

$p=\left(\frac{a}{b}\right)^{2}$

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

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

$\Rightarrow p \mid a^{2}$

$\Rightarrow p \mid a$

$\Rightarrow a=p c$ for some positive integer $c$

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

$\Rightarrow b^{2} p=p^{2} c^{2}(\because a=p c)$

$\Rightarrow p \mid b^{2}\left(\right.$ since $\left.p \mid c^{2} p\right)$

$\Rightarrow p \mid b$

$\Rightarrow p \mid a$ and $p \mid b$

This contradicts the fact that a and $b$ are co primes

Hence $\sqrt{p}$ is irrational