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Real numbers are a totally ordered set, while complex numbers are not. An intuitive explanation can be given by the fact complex numbers can be represented as points of a plane, while real numbers as point on a line. Pure imaginary numbers can be represented as points of a line, so my question is: can we give some kind of total ordering (I mean defining an arbitrary order relation which satisfies the properties of reflexivity, antisymmetry and transitivity) to pure imaginary numbers? If not, why? The easiest one I can think of is using the ordering of real numbers:

$\quad ai>bi\iff a>b \qquad a, b\in\mathbb{R}$

In real numbers $3>2$ because we defined the order relation in a certain way, but we could also define some order relation such that $2>3$ (after all, > is just a symbol that stands for an order relation). Is that right?

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  • $\begingroup$ Yes, notice that the function $x\mapsto x\cdot i$ is an order morphism and it is a bijection in between $\mathbb{R}$ and $i\cdot \mathbb{R}$ $\endgroup$ – Phicar Jun 11 at 17:54
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    $\begingroup$ Does this help? math.stackexchange.com/questions/487997/… $\endgroup$ – Äres Jun 11 at 17:55
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    $\begingroup$ Part of the special consideration for the complex numbers when it comes to orderings is that if we want it to be an Ordered Field, not only do we need it to have an order and to be a field, but we also require that order satisfy additional requirements such as if $0\prec a\leq b$ then $0\prec a\cdot b$ as well. Your example with the purely imaginary numbers certainly is an order but the purely imaginary numbers with the usual addition/multiplication do not form a field. (You would need to change to a different multiplication) $\endgroup$ – JMoravitz Jun 11 at 18:05
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    $\begingroup$ The punchline is that the complex numbers with the usual addition/multiplication cannot form an ordered field, this despite an order existing and it being a field it will fail these additional requirements. $\endgroup$ – JMoravitz Jun 11 at 18:07
  • $\begingroup$ I understand, so it was related to the operations and not to the ordering itself. What is that symbol between $0$ and $a$ (bear with me, please. I am an undergrad physics student). Also, is the link posted by Wrench in contradiction with lexicographic ordering mentioned by an user below? $\endgroup$ – Feynman_00 Jun 11 at 18:15
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That is correct, yes.

The thing to note is, an order by itself isn't actually useful -- it's only when we can do stuff with the order that it becomes interesting.

A cool thing to note is that if you take Axiom of Choice, any set is well-orderable, and thus totally orderable.

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You can easily have a total order on the complex numbers using the lexicographic ordering. Define $$a+bi \gt c+di \iff \begin {cases} a \gt c\\a=c \ \&\ b \gt d \end {cases}$$ This is a fine total order. The usual order on the reals plays nicely with addition and multiplication on the reals, which this one does not, but that is not part of the definition of a total order.

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  • $\begingroup$ Oh, I read something like this somewhere on a Real Analysis book. I thought there was something wrong about it since that was the only time I've read about ordering complex numbers. I'll do more research then. $\endgroup$ – Feynman_00 Jun 11 at 18:09
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Yes you can totally order imaginary numbers that way. Note though that if you just want an order on $\mathbb{C}$ as a set you can totally order $\mathbb{C}$ with a lexicographic order, for example: $$z<z' \Longleftrightarrow \Re(z)>\Re(z') \lor \big(\Re(z)=\Re(z') \land \Im(z)<\Im(z)' \big)$$

The point with orders on structures is that you usually want them to behave well with other operations.The order defined above for example is such that $a<b \Rightarrow a+c<b+c$ for every $a,b,c$.

The real problem with $\mathbb{C}$ and order is that it cannot be made into an ordered field. That is, you cannot find a total order on $\mathbb{C}$ such that for every $a,b,c$ you have $a<b \Rightarrow a+c<b+c$ and for every $a,b>0$ you have $ab>0$. This is essentially because from the above properties it would follow that $-1<0$ and $a^2\ge 0$ for every $a$.

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