Question

Verify by the method of contradiction that √7  is irrational

Answer 1

Let  p : √7  is irrational 

Let us assume p is not true i.e..,  √7 is rational . 

⇒ √7 = a/b, where a and b are integers having no common factor.

⇒ 7 = a²/b²

⇒ a²  =7b² 

⇒ 7 divides a² 

⇒ 7 divides a 

⇒ a = 7c, for some integer c. 

⇒ a²  = 49c² 

⇒ 7b²  = 49c² 

⇒ b²  = 7c² 

⇒ 7 divides b² 

⇒ 7 divides b 

Thus, 7 is common factor of both a and b. This contradicts that a and b have no common factor. 

So, our assumption is wrong. 

Hence, √7  is irrational is true.