Question

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

Answer 1

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 $b$ ≠ 0.
Now $\sqrt{3}=\frac{a}{b}$ ⇒ $3=\frac{a^2}{b^2}$ [on squaring both sides]
⇒ $3b^2=a^2$ ... (1)

⇒ $3$ divides $a^2$ [since 3 divides 3$b^2$]
⇒ $3$ divides $a$ [since 3 is prime, 3 divides $a^2$ ⇒ 3 divides $a$]
Let $a=3c$ for some integer $c$.
Putting $a=3c$ in equation (1), we get
$3b^2=9c^2$ ⇒ $b=3c^2$
⇒ 3 divides $b^2$ [since 3 divides 3$c^2$]
⇒ 3 divides $b$ [since 3 is prime, 3 divides $b^2$ ⇒ 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.