# Prove that

Question:

Prove that $\sqrt{3}$ is an irrational number.

Solution:

Let $\sqrt{3}$ be rational and its simplest form be $\frac{a}{b}$.

Then, $a, b$ are integers with no common factors other than 1 and $b \neq 0$.

Now $\sqrt{3}=\frac{a}{b} \Rightarrow 3=\frac{a^{2}}{b^{2}}$      [on squaring both sides]

$\Rightarrow 3 b^{2}=a^{2}$      ... (1)

$\Rightarrow 3$ divides $a^{2} \quad$ [since 3 divides $3 b^{2}$ ]

$\Rightarrow 3$ divides $a \quad$ [since 3 is prime, 3 divides $a^{2} \Rightarrow 3$ divides $a$ ]

Let $a=3 c$ for some integer $c$.

Putting $a=3 c$ in equation (1), we get

$3 b^{2}=9 c^{2} \Rightarrow b=3 c^{2}$

$\Rightarrow 3$ divides $b^{2} \quad$ [since 3 divides $3 c^{2}$ ]

$\Rightarrow 3$ divides $b \quad$ [since 3 is prime, 3 divides $b^{2} \Rightarrow 3$ divides $b$ ]

Thus, 3 is a common factor of both $a, b$.

But this contradicts the fact that $a, b$ have no common factor other than 1 .

The contradiction arises by assuming $\sqrt{3}$ is rational.

Hence, $\sqrt{3}$ is rational.