Question

prove that √2 is an irrational number​

Answer 1

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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 2b^2=a^2] 

⇒2∣a...(i) 

⇒a=2c for some integer c

⇒a^2=4c^2 

⇒2b^2=4c2[∵2b^2=a^2] 

⇒b^2=2c^2 

⇒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.