Question

show that 3√3 is an irrational no

Answer 1

hence √3 is irrational no.
so 3√3 is also irrational no.

Answer 2

let's assume √3 is rational , therefore √3=a/b where a and b are co primes
3=a^2/b^2
b^2 x3= a^2
a is divisible by 3, so 3 is a factor of a
now,
a=3c
b^2 x 3 = (3c)^2
b^2= 3c^2
b is also divisible by 3
3 is a factor of both a and b but we assumed a and b are co primes, the contradiction arises due to wrong assumption therefore √3 is irrational
similarly let's assume 3√3 is rational
3√3=a/b where a and b are co primes
√3= a/bx3
√3 is irrational but a/bx3 is rational the contradiction arises due to wrong assumption therefore 3√3 is irrational