Question

prove that 3√2 is an irrational​

Answer 1

Let us assume, to the contrary, that 3√2 is

3√2 is rational.

Then, there exist co-prime positive integers a and b such that

3√2 = a/b

$\mapsto$√2 = a/3b

$\mapsto$ √2 is rational

. . . . [∵ 3,a and b are integers

∴ a/3b is a rational number]

This contradicts the fact that √2 is irrational.

So, our assumption is not correct.

Hence, 3√2 is an irrational number.