The field of rational functions $F_{\mathbb{Ra}}$ in one variable is a classic example of a non-archimidean ordered field.
For an ordered field to be non-archimidean, it must not be dedekind complete (if a field were both ordered and dedekind complete, then it would be $\mathbb{R}$, which is archimidean).
I believe that I understand a counterexample to the claim that rational functions are archimidean:
Consider two members of $F_{\mathbb{Ra}}$, $g(x) = x$ and $f(x) = n$ for $ n \in \mathbb{N}$. When we assess whether $g(x)$ is larger than $f(x)$ -- or any two members of $F_{\mathbb{Ra}}$ -- we assess whether whether for sufficiently large $x$, $g(x)$ is greater than $f(x)$. In this example, it is true that no matter the member of $\mathbb{N}$ I select for $f(x) = n$, sufficiently large $x$ in $g(x) = x$ will always exist that are greater than $n$. Thus, there is no way of ever making $f(x)$ larger than $g(x)$.
This is the "assessment" AFAIK
"We also have trichotomy: if $h \in F_{\mathbb{Ra}}$ and $h(x) \not = 0$, then either $h(x) > 0$ for all sufficiently positive x, or $h(x) < 0$ for all sufficiently positive $x$."
I wanted to learn why rational functions are not dedekind complete.
Dedekind complete: every nonempty subset $S \subset F$ which is bounded above has a least upper bound.
Let $E:=\mathbb{Q}$. Clearly, $E$ is a Riesz space under the natural ordering of $\mathbb{Q}$. Let $A:=\{(1+\frac{1}{n})^n \mid n \in \mathbb{N}\}$, then $\emptyset \not = A \subseteq E$. Furthermore, we know that $\sup A=e$ (we know this result from real analysis; this gives that $A$ is a non-empty subset of $E$ which is bounded above), but $e \not \in E$, so $E$ is not Dedekind complete ($E$ being Dedekind complete requires that every non-empty subset of $E$ which is bounded above has supremum in $E$). This shows that the set of rationals is not a Dedekind complete space. We can use a similar argument, choosing appropriate functions, to show that the field of one-variable rational functions is not Dedekind complete.