Prove that the square root of any irrational number is irrational.
Said shortly, $$\left(\frac pq\right)^2=\frac{p^2}{q^2}$$ is rational.
The square of any rational is rational, hence no rational is the square root of an irrational.
Unnecesarily sophisticated proof: $$ m^2\not\in{\Bbb Q}\implies[{\Bbb Q}(m^2):{\Bbb Q}]>1\implies [{\Bbb Q}(m):{\Bbb Q}] = [{\Bbb Q}(m):{\Bbb Q(m^2)}]\,[{\Bbb Q}(m^2):{\Bbb Q}] > 1. $$