Question

Prove that ^3√2 is an irrational number

Answer 1

let 3√2 be a rational number
3√2 = a/b ( where b not equals to 0 and a and b are co - prime)
√2 = a / 3b

√2 is irrational

our assumption is wrong of taking 3√2 be rational
it is a contradiction therefore 3√2 is irrational
hence, proved

note : In question where you have given root then all the questions will done by the above process ( like - √3, √5 , √7 , etc )

Answer 2

ANSWER :

To Prove : 3√2 is an irrational number.

Let, us assume on the contrary that 3√2 is a rational no. Then , there exist positive integers a and b such that :

3√2 = a/b ( where a and b are co prime therefore their HCF is 1 )  

√2 =  a / 3b

√2 is rational number         [ ∵ 3, a and b are integers ∴ a/3b is a                                                                      rational number. ]

                                             

→ This contradicts the fact that √2 is irrational number.

→ So our supposition is not correct. It's wrong. 3√2 is not a rational number.

→ Hence, 3√2 is an irrational number.

Hence it is Proved...!!!