Density of the rationals

The proof uses the Archimedean property. Given \(a<b\), choose \(n\in\mathbb N\) with \(1/n<b-a\) and take \(m=\lfloor na\rfloor+1\), the least integer greater than \(na\). Then \(na<m\le na+1<nb\), hence \(a<m/n<b\). With \(a=0.3\), \(b=0.4\) and \(n=11\) we get \(m=4\) and \(q=4/11\).

Density implies there are infinitely many rationals in every interval: apply the result again between \(q\) and \(b\), and so on. For the same reason there is no smallest positive rational. In terms of distance, for every real \(x\) and every \(\varepsilon>0\) there is a rational \(q\) with \(|x-q|<\varepsilon\).

The irrationals are dense too: if \(a<b\), there is a rational \(q\) between \(a/\sqrt2\) and \(b/\sqrt2\), and then \(q\sqrt2\) is irrational (if \(q\ne0\)) and lies between \(a\) and \(b\). Density is not completeness: the set \(\{q\in\mathbb Q: q^2<2\}\) is bounded but has no supremum in \(\mathbb Q\).

A common mistake is to think that, being dense, \(\mathbb Q\) fills the line; yet \(\sqrt2\) is not rational. Density will be used to approximate and to pass from rationals to reals.