Given two real numbers $x,y\in \mathbb{R}$ with $x<y$, then there exists some rational number $r\in \mathbb{Q}$ such that $x<r<y$.

We say the rational numbers are dense in $\mathbb{R}$.

Proof

Without loss of generality, let $0<x<y$.
Then $y-x>0$.
Since The infimum of the set of reciprocals of the natural numbers is zero., there exists some $n\in \mathbb{N}$ such that $y-x>\frac{1}{n}$.
Then $ny-nx>1$, so $ny>1+nx$.
Since Every number is between two natural numbers, there exists some $m\in \mathbb{N}$ such that $m-1\le nx<m$.
Then $m\le nx+1<m+1$,
so $nx<m\le nx+1<ny$
and $x<\frac{m}{n}<y$.
QED.