Question

$\sqrt{3}$ is proved irrational by _____.

• A. factorisation
• B. rationalisation
• C. contradiction
• D. expansion

Answer 1

The correct option is C contradiction

Let us assume on the contrary that $\sqrt{3}$ is a rational number.
Then, there exist positive integers $a$ and $b$ such that $\sqrt{3}$ = $\frac{a}{b}$ where, $a$ and $b$ are , are co-prime i.e. their HCF is $1$.
Now,
$\sqrt{3}$ = $\frac{a}{b}$
⇒ $3$ = $\frac{a^2}{b^2}$
⇒ $3b^2$ = $a^2$
⇒ $3$ divides $a^2$ [∵ $3$ divides $3b^2$]
⇒ $3$ divides $a$ ...(i)
⇒ $a$ = $3c$ for some integer c
⇒ $a^2$ = $9c^2$
⇒ $3b^2$ = $9c^2$
⇒ $b^2$ = $3c^2$
⇒ $3$ divides $b^2$ [∵ $3$ divides $3c^2$]
⇒ $3$ divides $b$ ...(ii)
From (i) and (ii), we observe that $a$ and $b$ have at least $3$ as a common factor.
But, this contradicts the fact that $a$ and $b$ are co-prime. This means that our assumption is not correct.

Hence, $\sqrt{3}$ is an irrational number.