Proof that rational functions are an ordered field, but non-archimedean - Bartle's elements of real analysis I read that the set of rational functions with rational coefficients forms an ordered field, yet it is non-archimedean. I tried googling this, but I don't think I understood the solution.


How does one define an order on rational functions of the form $\mathbb{Q}(x)=p(x)/q(x)$? 
How do you show that $\mathbb{Q}(x)$ is non-archimedean? That there is no natural number $n$, such that $n&gt;\mathbb{Q}(x)$? Do I substitute numerical values of $x$ and show that $\mathbb{Q}(x)$ is unbounded or something?
Consider the following order: let $f(x)=p(x)/q(x)$ be a rational function with

\begin{aligned}p(x)=a\cdot x^n &amp;+ \text{ terms of degree less than $n$}\\
q(x)=b\cdot x^m &amp;+ \text{ terms of degree less than $m$}\end{aligned}

where of course $a,b \in \mathbb{Q}$. Our order says that $f &gt; 0$ if and only if $\frac{a}{b} &gt;0$. Notice this defines the order throughout the field; if one wishes to determine whether $f_1 &gt; f_2$, write the difference $f_1-f_2$ as a single rational function and determine whether it is $&gt;0$, $=0$ or $&lt;0$.

Now, this totally ordered field is not Archimedean. Indeed, consider the rational functions $f(x)=x$ and $g(x) = 1$. No matter how large you choose $n \in \mathbb{N}$, $f(x)&gt;n\cdot g(x)$, because $\big(f-n\cdot g\big)(x)=x-n$ and the leading coefficient of $\big(f-n\cdot g\big)$ is $1$, which is positive.
A comment on rigor.

For polynomials $p,q$ where $q(x)$ is not identically $0,$ the domain of $p/q$ must exclude the finite set of $x$ for which $q(x)=0.$  If these are the only real numbers excluded from dom $(p/q)$ then we must say that $p_1/q_1\ne p_2/q_2$ when $p_1=q_1=id_R$ and $p_2=q_2=1$ because dom$(p_1/q_1)\ne$ dom $(p_2/q_2)$. Which is not what we want to say.

For polynomials $p_1, q_1,p_2,q_2$ with $q_1\ne 0\ne q_2,$ and any finite $S_1,S_2$ such that $q_1^{-1}\{0\}\subset S_1$ and $q_2^{-1}\{0\}\subset S_2,$ let $$(p_1,q_1,S_2)\sim (p_2,q_2,S_2)\iff \forall x\in (R \backslash (S_1\cup S_2)\; (p_1(x)q_2(x)=p_2(x)q_1(x)).$$  Then  for polynomials $p,q$ with $q\ne 0$ we write $p/q$ for the set of all $(p_1,q_1,S_1)$ such that $$(p,q,R \backslash q^{-1}\{0\})\sim (p_1,q_1, S_1).$$