TITLE: Proof that rational functions are an ordered field, but non-archimedean - Bartle's elements of real analysis
QUESTION [12 upvotes]: 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 $n>\mathbb{Q}(x)$? Do I substitute numerical values of $x$ and show that $\mathbb{Q}(x)$ is unbounded or something?

REPLY [5 votes]: For polynomials $f(x),g(x)\in\mathbb{Q}[x]$ define $f<g$ if and only if $f\ne g$ and the leading coefficient of $g-f$ is positive.
It's easy to see that this defines a (strict) order relation on $\mathbb{Q}[x]$ compatible with the operations, in the sense that

for every $f\in\mathbb{Q}[x]$ it holds exactly one among $f>0$, $f=0$, or $0>f$;
if $f,g,h\in\mathbb{Q}[x]$ and $f<g$, then $f+h<g+h$
if $f,g,h\in\mathbb{Q}[x]$, $f<g$ and $0<h$, then $fh<gh$.

Now it's easy to see that $0<f$ if and only if $-f<0$, so any element of $\mathbb{Q}(x)$ can be written as a quotient $f(x)/g(x)$ where $0<g$.
Define, for $f_1(x)/g_1(x),f_2(x)/g_2(x)\in\mathbb{Q}(x)$ with $0<g_1$ and $0<g_2$,
$$
\frac{f_1(x)}{g_1(x)}<\frac{f_2(x)}{g_2(x)}
\quad\text{if and only if}\quad
f_1(x)g_2(x)<f_2(x)g_1(x)
$$
and prove that this defines a (strict) order relation with the same properties above, so $\mathbb{Q}(x)$ becomes an ordered field.
Now, of $q\in\mathbb{Q}$, it's obvious that $q<x$ and therefore the order on $\mathbb{Q}(x)$ is not Archimedean. Note also that the order induced on $\mathbb{Q}$ is the usual one.