In $\mathbb{R}(t)$ the rationals are not dense: no rational lies strictly between $0$ and $1/t$

Refuted claim: in every ordered field $F$ the image of $\mathbb{Q}$ is dense, that is, for all $x < y$ in $F$ there is a rational $q$ with $x < q \cdot 1_F < y$.

The witness is $\mathbb{R}(t)$ with the eventual-sign order (Not every ordered field is Archimedean, The rational function field $\mathbb{R}(t)$ ordered by the eventual sign is an ordered field, worked out), and the pair $x = 0$, $y = 1/t$: the interval between them contains no rational at all.

The true statement requires the Archimedean property and is ℚ is dense in every Archimedean ordered field; $\mathbb{R}(t)$ is not Archimedean, and this counterexample is exactly the failure that the Archimedean hypothesis rules out.

• What density really needs. Given $0 < x < y$ in an Archimedean field one finds $n$ with $1/n < y - x$ and then a multiple of $1/n$ in the gap; the Archimedean property is used precisely to make the mesh $1/n$ finer than the gap. In $\mathbb{R}(t)$ the gap $1/t - 0$ is smaller than every $1/n$, so no mesh built from rationals is ever fine enough.

• An element like $1/t$ is called an infinitesimal: positive, and below every positive rational. A non-Archimedean ordered field always has one, since if $x$ exceeds every canonical natural then $1/x$ is below every $1/n$ (Inverses of positives are positive, and reciprocation reverses order). So the failure of density is not special to this field; it happens in every non-Archimedean ordered field, including $\mathbb{R}((t^{-1}))$.

• Density is not the same as completeness. $\mathbb{Q}$ is dense in itself and in $\mathbb{R}$, and $\mathbb{Q}$ is not complete. What this counterexample shows is only that density of $\mathbb{Q}$ needs the Archimedean property, which is also the hypothesis missing from FALSE: the nested interval property alone implies the least-upper-bound property and FALSE: an ordered field in which every Cauchy sequence converges has the least-upper-bound property.