How many orders on the rational numbers are compatible with the field operations?

Question

I know that the real numbers have only one linear order which is compatible with the standard field operations. What about the rational numbers? How many orders on $\mathbb{Q}$ are compatible with the standard field operations? Is it a finite number? Or is it infinite, even uncountably infinite?

Answer 1

An order compatible with the field structure can be defined in terms of the set $P$ of positive numbers, characterized by the following properties:

• $\Bbb Q$ is the disjoint union of $P$, $\{0\}$, and $-P$
• $P+P\subseteq P$
• $P\cdot P\subseteq P$

(Indeed, give $\prec$, we can define $P=\{\,x\in\Bbb Q\mid 0\prec x\,\}$, and vice versa given $P$, we can defined $x\prec y\iff y-x\in P$)

In particular, squares of non-zero numbers are positive (and in case of $\Bbb R$ we could stop here). In particular, $1\in P$ and $-1\notin P$. Assume $-\frac1n\in P$ for some $n\in\Bbb N$. Then by repeated adding, $-1=(-\frac1n)+\cdots+(-\frac1n)\in P$, contradiction. We conclude that $\frac1n\in P$. By repeated adding, $\frac mn\in P$ for $m,n\in\Bbb N$. Ultimately, the set $P$ must consist precisely of the "traditionally" positive numbers and hence there is only one order on $\Bbb Q$ that is compatible with the field structure.