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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.

Joe
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  • Can't you just prove that ${y\in\mathbb{R} : y = x^2}$ is the set of non-negative elements of both fields, so they are equal? – Jakobian Jul 04 '23 at 17:34
  • See also https://math.stackexchange.com/questions/449404/is-an-automorphism-of-the-field-of-real-numbers-the-identity-map – lhf Jul 05 '23 at 10:54
  • @Ihf: Can the fact that $(\mathbb R,+,\cdot)$ only admits one order be derived from the fact that only has one automorphism? – Joe Jul 05 '23 at 13:16

2 Answers2

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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.

Noah Schweber
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  • @JyrkiLahtonen - just to clarify, you're saying "not only is there more than one, there's at least one for every transcendental real", yes? (At first read I thought you were disagreeing with there being a second ordering.) – JonathanZ Jul 04 '23 at 18:10
  • @JyrkiLahtonen Derp, not my finest moment - fixed! – Noah Schweber Jul 04 '23 at 18:59
  • @JonathanZonstrike FYI Jyrki was absolutely right, I made a very silly mistake (now corrected). – Noah Schweber Jul 04 '23 at 19:06
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    Happens :-) Adding my upvote. – Jyrki Lahtonen Jul 04 '23 at 19:26
  • @NoahSchweber - Hmm, I must be missing something. I've imagined that all these orders arise from treating $\mathbb Q(\pi)$ like $\mathbb Q(x)$, and then inducing an order by "evaluating" at some transcendental. In which case I don't see a huge difference between letting $x$ be $e/1000000$ and letting $x$ be $-e/1000000$. Is there some algebraic distinction that would put them in separate "categories"? I don't immediately see one. – JonathanZ Jul 04 '23 at 21:34
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    @JonathanZonstrike Nope, as ordered fields they're isomorphic. But I was thinking of them as orderered fields with the point $\pi$ distinguished, in which case we can tell if it is positive or negative. But it was just meant to be informal. – Noah Schweber Jul 04 '23 at 21:35
  • @NoahSchweber: As far as I understand, $\mathbb Q(x)$ is the set of rational functions with coefficients in $\mathbb Q$, and $\mathbb Q(\pi)$ is the set obtained by evaluating all of these rational functions at $\pi$. Is that correct? – Joe Jul 05 '23 at 13:12
  • @Joe When we have fields $k\subset K$ and an element $\alpha\in K\setminus k$, we write $k(\alpha)$ for the smallest subfield of $K$ containing $k$ (as a subset) and $\alpha$ (as an element). So there's some conflation of notation here. Now it's benign - what you've written will coincide with this - but I think the phrasing here is more intuitive. – Noah Schweber Jul 05 '23 at 13:17
  • @NoahSchweber: Right, the definition you have written does seem more intuitive. Is my definition still equivalent to yours if we replaced $\pi$ with $\sqrt 2$, say? Then, "evaluating" some of those rational functions might lead to division by zero, and so I'm not sure what the correct version of my definition should be. Perhaps $\mathbb Q(\sqrt 2)$ is the set obtained by evaluating the rational functions in $\mathbb Q(x)$ whose evaluation maps are defined at $\sqrt 2$? Is that equivalent to the usual definition? – Joe Jul 05 '23 at 13:25
  • @Joe Yes, that's the remaining subtlety and the correct fix. – Noah Schweber Jul 05 '23 at 13:38
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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).

Joe
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