Question

27. Prove that √2 is irrational.​

Answer 1

Let us assume on the contrary that 2 is a rational number. Then, there exist positive integers a and b such that

2=ba where, a and b, are co-prime i.e. their HCF is 1

⇒(2)2=(ba)2 

⇒2=b2a2 

⇒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∣b

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.