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.

  • 1
    $\begingroup$ "The "inequality" $x>y$ implies that $c_x>c_y$" It doesn't. Consider $$x = \frac{x^2+3}{2x+1};\quad y = \frac{2x-1}{2x+1},$$ then $c_x < c_y$, but $x > y$. $\endgroup$ Sep 5, 2013 at 8:56
  • $\begingroup$ What is the dimension of this field over $\Bbb R$? $\endgroup$ Sep 11, 2016 at 23:36

3 Answers 3


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 $b,d>0$.


You’re making matters more difficult than necessary. Let $P$ be the set of rational functions with positive ratio together with the $0$ function; what you want to show is that $P$ satisfies the definition of a positive cone in the field $F$ of rational functions: it’s closed under addition and multiplication, the square of every $f\in F$ is in $P$, and $-1\notin P$. All of these are very easily verified. Now for rational functions $f$ and $g$ define $f\le_P g$ if and only if $g-f\in P$, and prove that $\langle F,\le_P\rangle$ is an ordered field according to the order-based definition (or just appeal to the easy result that the definition via a linear order is equivalent to the definition via a positive cone).


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.


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