Solution: Let $x,y$ be two real numbers such that $x < y$. We need to prove that there exists a rational number $r$ such that $x < r < y$. Since $y - x > 0$, by the Archimedean property, there exists a natural number $n$ such that \[ \frac{1}{n} < y - x \implies ny - nx > 1. \] Since the difference between $nx$ and $ny$ is bigger than $1$, this implies there must be an integer between $nx$ and $ny$, say $m$. Thus we have \[ nx < m < ny \implies x < \frac{m}{n} < y. \] So we proved that there is a rational number $r = \frac{m}{n}$ between $x$ and $y$.

Now to prove that between any two real numbers there exists an irrational number, let $x,y$ be two real numbers with $x < y$. This implies $\sqrt{2} x < \sqrt{2} y$. Thus from the previous part, we can find a rational number $r$ such that \[ \sqrt{2} x < r < \sqrt{2} y \implies x < \frac{r}{\sqrt{2} } < y. \] Since $r$ is a rational number, $\frac{r}{\sqrt{2} }$ must be an irrational number. Thus, we proved that between any two real numbers there exists an irrational number.