Between any two rationals lies an irrational

Construction. Let $a<b$ be rationals. Set $x=a+\frac{b-a}{\sqrt2}$.

Irrational. $x=a+\frac{(b-a)}{\sqrt2}$. Since $b-a>0$ and $\sqrt2$ is irrational, $\frac{b-a}{\sqrt2}$ is irrational; adding rational $a$ keeps it irrational.

Position. $a<x<b$ because $0<\frac{b-a}{\sqrt2}<b-a$ (since $\sqrt2>1$).

Conclusion. Between any two rationals lies an irrational.