Question

Prove that $\sqrt{n}$ is not a rational number, if $n$ is not perfect square.

Answer 1

Let $\sqrt{n}$ be a rational number.

$\therefore \quad \sqrt{n}=\frac{p}{q}$, where $p$ and $q$ are co-prime

Integers, $q \neq 0$

On squaring both sides, we get

$\Rightarrow \quad n=\frac{p^2}{q^2}$

$p^2=n q^2$ ........(i)

$\Rightarrow n$ divides $p^2$ 

[Let $p$ be a prime number. If $p$ divided $a^2$ then $p$ divides $a$, where $a$ is a positive integer]

$\Rightarrow n$ divides $p$ ........(ii)

Let $p=n m$, where $m$ is any integer.

$\Rightarrow \quad p^2=n^2 m^2$

$\Rightarrow \quad n^2 m^2=n q^2$

$\Rightarrow \quad q^2=n m^2$

$\Rightarrow n$ divides $q^2$

$\Rightarrow \quad n$ divided $q$ ........(iii)

[Let $p$ be a prime number. If $p$ divided $a^2$ then $p$ divides a where $a$ is a positive integer]

From (ii) and (iii), $n$ is a common factor of both $p$ and $q$ which contradicts the assumption that $p$ and $q$ are co-prime integer.

So, our supposition is wrong, $\sqrt{n}$ is an irrational number.