This question is motivated by another question posted earlier today.
Let $F$ be a field endowed with an embedding $F\hookrightarrow\Bbb R$. Then the ordering of $\Bbb R$ induces an ordering on $F$.
Is the converse true?
Namely, let $(F,\leq)$ be a characteristic zero ordered field. Is it true that $F$ admits a real embedding such that $\leq$ is induced by the ordering of $\Bbb R$?
I think that if $\Bbb Q$ is dense in $F$ (meaning that for all $x\in F$ and for all $0<q\in\Bbb Q$ there exists $a$, $b\in\Bbb Q$, $a<x<b$ such that $0<b-a<q$)then one can proceed as follows.
Given $x\in F$ let $S_x^+=\{q\in\Bbb Q\mid q>x\}$ and $S_x^-=\{q\in\Bbb Q\mid q<x\}$. Then we can define an embedding $F\hookrightarrow\Bbb R$ sending $\Bbb Q$ to $\Bbb Q$ identically and every $x\in F\setminus\Bbb Q$ to the unique real number $r\in\Bbb R$ bigger than every rational in $S_x^-$ and smaller than any rational in $S_x^+$.
If this is correct, the question is which ordered fields admit $\Bbb Q$ as a dense subset in the above sense. My guess is that all the algebraic extensions of $\Bbb Q$ which are ordered are of this sort.
It is clear that algebraic extension do not exhaust all possibilities. For instance one can order a field of transcendence degree 1 over $\Bbb Q$ in such a way $\Bbb Q$ is dense in it, e.g. $F=\Bbb Q(T)$ can be ordered by fixing an isomorphisms $F\simeq\Bbb Q(\pi)$ and since there are actually uncountably many algebraically independent transcendental numbers there's room for even (way) more complicated examples. But these examples are fields that are already taken, up to isomorphism, inside $\Bbb R$ so the reasoning here is somewhat circular.
I am sure that the answer is actually well-known, but I don't recall having read anything of this sort.