Prove that root n is not a rational number, if n is not perfect square

Solution:

Let √n be a rational number.

∴ $\sqrt{n} = \frac{p}{q}$ , where p and q are co-prime integers q≠ 0.

On squaring both side we get

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

$\Rightarrow p^2 = nq^2$ ….(i)

⇒ n divides p²

⇒ n divides p … (ii)

[If n divides p², then n divides p]

Let p = nm, where m is any integer.

⇒ $p^2 = n^2m^2$

From eq (i)

⇒ $n^2m^2 = nq^2$

$\Rightarrow q^2=nm^2$

⇒ n divides q²

⇒ n divides q …..(iii)

[If n divides q², then n divides q]

From (ii) and (iii), n is common factor of both p and q which contradicts the assumption that

So, our assumption is wrong.

Hence, √n is an irrational number.