The question on Zorich p.63. d) Show that the set $\mathbb{Q}(x)$ of rational fractions $$R_{m,n}=\frac{a_0+a_1x+\cdots+a_mx^m}{b_0+b_1x+\cdots+b_nx^n}$$ with coefficients in $\mathbb{Q}$ or $\mathbb{R}$ becomes an ordered field, but not an Archimedean ordered field, when the order relation $R_{m,n}>0$ is defined to mean $a_mb_n>0$ and the usual arithmetic operations are introduced. This means that the principle of Archimedes cannot be deduced from the other axioms for $\mathbb{R}$ without using the axiom of completeness.
It seems that I have figured out (d): the field is ordered by $\frac{P_m(x)}{Q_k(x)} \leq \frac{P_{m^{\prime}}(x)}{Q_{k^{\prime}}(x)} \Leftrightarrow \frac{P_m(x)}{Q_k(x)} - \frac{P_{m^{\prime}}(x)}{Q_{k^{\prime}}(x)} \leq 0$. And so $\forall n \in \mathbb{N} (n \geq x)$, which means it does not have Archimedean principle.
My question is about the last sentence in (d): Why this example shows that Archimedean principle is dependent on completeness? I think Archimedes principle only relies on the existence of maximal element in any subsets of $\mathbb{Z}$, which can be proved by defining a recursive function to find.
