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.
Best Answer
It's easier to order first $\mathbb{R}[X]$ (the ring of polynomials) and then use the fact that an ordering on a domain extends uniquely to its field of fractions.
The ordering on $\mathbb{R}[X]$ can be defined by
$$ f < g \quad\text{if and only if}\quad \lim_{x\to\infty}(g(x)-f(x))>0 $$ Where the limit can be $\infty$. This is is the same as saying that the leading coefficient of $g-f$ is positive. Checking the order properties is easy.
When $D,\le$ is an ordered domain and $F$ is its field of fractions, then there's a unique extension of $\le$ to $F$ making $F,\le$ an ordered field, by defining $$ \frac{a}{b}\le\frac{c}{d} \quad\text{if and only if}\quad ad\le bc $$ where $b,d>0$.