The correct option is C contradiction
Let us assume on the contrary that √3 is a rational number.
Then, there exist positive integers a and b such that √3 = ab where, a and b are , are co-prime i.e. their HCF is 1.
Now,
√3 = ab
⇒ 3 = a2b2
⇒ 3b2 = a2
⇒ 3 divides a2 [∵ 3 divides 3b2]
⇒ 3 divides a ...(i)
⇒ a = 3c for some integer c
⇒ a2 = 9c2
⇒ 3b2 = 9c2
⇒ b2 = 3c2
⇒ 3 divides b2 [∵ 3 divides 3c2]
⇒ 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, √3 is an irrational number.