TITLE: Ordering the field of real rational functions
QUESTION [11 upvotes]: Perusing the Wikipedia article on ordered fields, I encountered an interesting statement, a variation of which I am now trying to prove.  Basically, I am trying to show that the set of real rational functions can be made into an ordered field by defining $\frac{p(x)}{q(x)} > 0$ whenever $\frac{a_0}{b_0} > 0$, where $a_0$ is the leading coefficient of $p(x)$ and $b_0$ is the leading-coefficient of $q(x)$.  (Now, we assume that > behaves in the usual way on $\mathbb{R}$.)  I was wondering if I am on any sort of "right track" with this thinking:

Since the relation > on any "pair" of real rational functions $r, r'$  essentially depends on the comparability of the leading-coefficient ratios of $r$ and $r'$ - which we'll call $c$ and $c'$ respectively for ease of reference - look at $c$ and $c'$.
Since $c, c' \in \mathbb{R}$, and $\mathbb{R}$ is a totally ordered field in its own right, $c$ and $c'$ are always comparable and we can either have $c > c'$ or $c \ngtr c'$, which is equivalent to $c \leq c'$ by virtue of total order ($\leq$ being the negation of >).
Since $c$ and $c'$ are always comparable, then always either $r > r'$ or $r \ngtr r'$.
This is just sort of the general "logic" behind the following argumentation.

I'd do something like (for the property "if $a \leq b$ then $a + c \leq b + c$"):
Let $x, y, z$ be real rational functions, and let $c_x, c_y, c_z$ denote their leading-coefficient ratios respectively.  Assume $x > y$.  The "inequality" $x > y$ implies that $c_x > c_y$, which, since $c_x$ and $c_y$ are themselves in the field $\mathbb{R}$ totally ordered by >, $c_x + c_z > c_y + c_z$ holds and implies that $x + z > y + z$.
I'm not going to do anything for the second property - this was just a demonstration, and I'm wondering if this strategy/reasoning is valid.  Thanks in advance and sorry in advance for the long, long post.

REPLY [2 votes]: You write "Since the relation $>$ on any "pair" of real rational functions $r$, $r′$ essentially depends on the comparability of the leading-coefficient ratios...", but this is not really true. It's your choice to desire that method of ordering real rational functions, but it's not mandatory. While $\mathbf R$ has only one possible ordering, there are many orderings on $\mathbf R(x)$ besides the one you're asking about, which could be called ordering by growth at $\infty$. There is also ordering of  rational functions by growth at $-\infty$ or by growth just to the right of a real number or just to the left of a real number (e.g., ordering by growth just to the right of $1$ is the rule $f > g$ when $f(1+\varepsilon) > g(1+\varepsilon)$ in $\mathbf R$ for all small positive $\varepsilon$). Since any rational function has only a finite number of poles (numbers where the denominator blows up), these orderings all make sense: for each $a \in \mathbf R$, the domain of each rational function includes $(a,a+\varepsilon)$ and $(a-\varepsilon,a)$ for small enough $\varepsilon > 0$.  
The orderings I described on $\mathbf R(x)$ exhaust all possibilities, but I don't have the time to write out an explanation of that here.