If only the first compatibility condition of an ordered field is satisfied, then can R have other orders?

Question

The two compatibility conditions for an ordered field F are that:

$ \begin{align} x, y, z\in F, & y < z \implies x + y < x + z\\ x, y\in F, & 0 < x, y \implies 0 < xy \end{align} $

Suppose we create an order on the field of real numbers that is somewhat compatible, in the sense that the first condition is satisfied.

Then must this order be our usual order (or the opposite)? Or could some really bizarre order work?

Answer 1

It is easy to show that the first condition implies that $<$ on $\Bbb Q$ is either the usual order or the reverse of the usual order.

Consider $\Bbb R$ to be a vector space over the ordered field $\Bbb Q.$ Let $B$ be a vector-space (Hamel) basis for $\Bbb R$ over $\Bbb Q$. For each $x\in \Bbb R$ there exists a unique $f_x:B\to \Bbb Q $ such that (i) $\{b\in B: f_x(b)\ne 0\}$ is finite, and (ii) $x=\sum_{b\in B}bf_x(b).$

Let $<$ be an order on $\Bbb Q$ that obeys the first condition. Let $<_W$ be any well-ordering of $B$. Now for $x,y\in \Bbb R$ with $x\ne y$ let $b_0$ be the $<_W$-least $b\in B$ such that $f_x(b)\ne f_y(b).$ And define $x<'y \iff f_x(b_0)<f_y(b_0).$ Then $<'$ obeys the first condition. Since there are many different well-orders on the set $B$ there are many different linear orders on $\Bbb R$ that obey the first condition.

BTW. Some proper sub-fields of $\Bbb R$ can be ordered fields with orders different from the usual order, but obeying the first and second conditions. E.g. $\{a+b\sqrt 2\;: a,b\in \Bbb Q\}$ with $1>0>\sqrt 2\;.$

Answer 2

No, there are lots of other orders if we drop the second requirement.

When we forget the second requirement, what we're really asking about is the structure $(\mathbb{R}; +)$ as an ordered group (since now the multiplicative structure is irrelevant). One obvious difference is that now the reverse order is acceptable, but going beyond this we can immediately see that there are lots of orderings since $(\mathbb{R}; +)$ has lots of automorphisms: if $a, b$ are irrational numbers then there is an automorphism of $(\mathbb{R};+)$ sending $a$ to $b$. So there are plenty of "strange" orderings of $(\mathbb{R}; +)$, gotten by applying group automorphisms to the usual one.