Can the field of real numbers be ordered in more than one way?

Question

Consider the field $\mathbb R$ of real numbers under the usual addition and multiplication operations. Can this field be equipped with a total order $\prec$ which is not equal to the usual one $<$, and which gives $\mathbb R$ the structure of an ordered field? If so, is it possible that $(\mathbb R,+,\cdot,\prec)$ is not even isomorphic to $(\mathbb R,+,\cdot,<)$? I believe the second question is equivalent to asking whether there is an ordered field with the same set and binary operations as the real numbers, but which is not Dedekind-complete.

Some progress

Any total order $\prec$ must satisfy $0\prec n$ for every $n\in\mathbb Z^+$. From this, it follows that any total order must satisfy $0\prec r$ for every positive rational (here, I mean "positive" in the sense of the usual ordering). Moreover, if $p,q\in\mathbb Q$, then $p\prec q$ is equivalent to $0\prec q-p$, which in turn is equivalent to $0<q-p$ and thus $p<q$. This means that any total order on $\mathbb R$ agrees with the usual one, as far as rational numbers are concerned.

Answer 1

Perhaps surprisingly, the answer is no - the ordering on $\mathbb{R}$ is unique! Or, more precisely, it's the only ordering which "plays well" with $+$ and $\times$.

This is because in any ordered field, every square is positive. Now just use the fact that for each (nonzero) real $r$ exactly one of $r$ and $-r$ is a square. Note that the ordering is determined entirely by the "positive part," since $x\prec y$ iff $0\prec y-x$.

Note that this relies on $\mathbb{R}$ having "enough" square roots. If we looked instead at something like $\mathbb{Q}(\pi)$, the argument above would break down and we might indeed have multiple compatible orderings. In particular, for $\mathbb{Q}(\pi)$ there are lots of compatible orderings (I made a silly mistake here, now fixed)! They fall into two broad "categories" depending on whether $\pi$ is considered positive or negative, but even within each category there's still lots of freedom.

Answer 2

Minutes after I posted the question, I came up with the following answer: There is only one way to order the field of real numbers if we want to give it the structure of an ordered field.

Suppose $x,y\in\mathbb R$. If $x<y$, then $y-x>0$, so there is a $z\neq0$ such that $z^2=y-x$. Since $z\neq0$, it follows that $z^2\succ 0$, and so $x\prec y$. The converse holds because if $x\ge y$, then $x\succeq y$ (consider the cases $x=y$ and $y<x$ separately).