Use contradiction method to prove that

Solution:

Let us assume that $\sqrt{3}$ is a rational number.

For a number to be rational, it must be able to express it in the form $p / q$ where $p$ and $q$ do not have any common factor, i.e. they are co-prime in nature.

Since $\sqrt{3}$  is rational, we can write it as

$\sqrt{3}=\frac{p}{q}$

$\rightarrow \frac{p}{\sqrt{3}}=q$

[ squaring both sides ]

$\rightarrow \frac{p^{2}}{3}=q^{2}$

Thus, $\mathrm{p}^{2}$ must be divisible by 3 . Hence $\mathrm{p}$ will also be divisible by 3 .

We can write $p=3 c$ ( $c$ is a constant ), $p^{2}=9 c^{2}$

Putting this back in the equation,

$\frac{9 c^{2}}{3}=q^{2}$

$\rightarrow 3 c^{2}=q^{2}$

$\rightarrow$$\mathrm{C}^{2}=\frac{\mathrm{q}^{2}}{3}$ 

Thus, $q^{2}$ must also be divisible by 3, which implies that q will also be divisible by $3 .$

This means that both $\mathrm{p}$ and $\mathrm{q}$ are divisible by 3 which proves that they are not co-prime d hence the condition for rationality has not been met. Thus, $\sqrt{3}$ is not rational.

$\therefore \sqrt{3}$is irrational.

Hence, the statement p: $\sqrt{3}$ is irrational, is true.