Question

If $p$ is a prime number, prove that $\sqrt{p}$ is irrational.

Answer 1

Let $\sqrt{p}$ be a rational number and

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

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

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

$\therefore p$ divides $a^2$

But when a prime number divides the product of two numbers, it must divide atleast one of them.

here $a^2=a\times a$

$p$ divides $a$

Let $a=pk$ ......$\left(1\right)$

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

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

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

$\therefore p$ divides $b^2$

But $b^2=b\times b$

$\therefore p$ divides $b$

Thus, $a$ and $b$ have atleast one common multiple $p$

But it arises the contradiction to our assumption that $a$ and $b$ are coprime.

Thus, our assumption is wrong and $\sqrt{p}$ is irrational number.