Consider that $3\sqrt{2}$ is a rational number. Then,

$3\sqrt{2},=\frac{\mathrm{p}}{\mathrm{q}}$

Where p and q are co-prime numbers and q ≠ 0.

Squaring both sides,

$18=\frac{{\mathrm{p}}^2}{{\mathrm{q}}^2}$

18q² = p²            (1)

This shows that p² is divisible by 18 and thus p is divisible by 18. Therefore,

p = 18k

p² = 324k²                  (2)

From equation (1) and (2),

18q² = 324k²

q2 = 18k²

It can be observed that p and q both are divisible by 18 which is a contradiction to the fact that p and q are co-primes.

Therefore, the assumption is wrong.

Hence, $3\sqrt{2}$ is an irrational number.