Let us assume on the contrary that √2 is a rational number. Then, there exist positive integers a and b such that
√2=ab where, a and b, are co-prime i.e. their HCF is 1
⇒(√2)2=(ab)2
⇒2=a2b2
⇒2b2=a2
⇒2|a2[∵2|2b2 and 2b2=a2]
⇒2|a...(i)
⇒a=2c for some integer c
⇒a2=4c2
⇒2b2=4c2[∵2b2=a2]
⇒b2=2c2
⇒2|b2[∵2|2c2]
⇒2|b...(ii)
From (i) and (ii), we obtain that 2 is a common factor of a and b. But, this contradicts the fact that a and b have no common factor other than 1. This means that our supposition is wrong.
Hence, √2 is an irrational number.