1.

Verify by the method of contradiction that “√7 is irrational.” 

Answer»

Prove that √7 is irrational 

Assume √7. is an rational 

 √7 = p/q

Squaring on b.s. 

∴ 7 = p2/q2

∴ 7q2 = p2 i.e., p in a multiple of again put p = 7k (k+1) : 

∴ 7q2 = (7k)2 ⇒ 7q2 = 49k2 ⇒ q2 = 7k2 

⇒ q is also a multiple of q

∴ both “p’ and ‘q’ are the factor of 7 

∴ √7 ≠ p/q (q = 0)

∴ our assumption that √7 is an partial is wrong. 

∴ √7 is an irrational.



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