Ordering of the complex numbers

Question

The complex numbers as a whole cannot be ordered but could you order the complex numbers of the form $a i$ where $a \in \mathbb{R}$?

Answer 1

Yes. The ordering corresponds to that of the reals.

Define an ordering $<_i$ as follows:

$$ai <_i bi \iff a < b$$

where "$<$" is just as in $\mathbb{R}$.

Since $<$ is an ordering of $\mathbb{R}$, $<_i$ is an ordering of the purely imaginary numbers $\{z \in \mathbb{C} \,\mid\, \exists\, a \in \mathbb{R} \,\,\text{ with } \,\,z=ai\}$

EDIT3: Thanks to Asaf, I think I understand now. $\mathbb{C}$ as a set is totally ordered, but if you consider it as a field it is not.

Answer 2

You could define an order, though it may not be useful: $a+bi<c+di$ if $a<c$; and $a+bi<a+ci$ if $b<c$. I think that is called the 'long line' http://en.wikipedia.org/wiki/Long_line_%28topology%29 because you rearrange the complex plane into lines of constant real part. EDIT: see comment, (thanks) it isn't the long line.

Answer 3

The complex numbers can be ordered. It cannot be made into an ordered field with the usual addition and multiplication. And that's different.

We can, for example, define $(a+bi)\preceq(c+di)$ if and only if $a<c$ or $a=c$ and $b\leq d$. This defines a linear ordering of the complex numbers.

Another way would be to note that $\Bbb C$ and $\Bbb R$ have the same cardinality, and therefore there is a bijection $f\colon\Bbb C\to\Bbb R$. Now we can define $z_1\leq_f z_2\iff f(z_1)\leq f(z_2)$. And it is not hard to see that this is a linear order as well.

Now the set $\{ai\mid a\in\Bbb R\}$ is linearly ordered as a subset of a linearly ordered set.

But all these orders are incompatible with the field operations. Namely, $\Bbb C$ cannot be made into an ordered field, where $a<b$ implies $a+c<b+c$ for all $c$. The reason is that in an ordered field, we can prove both these statements:

1. $-1<0$.
2. $x^2\geq 0$ for every $x$.

And in $\Bbb C$, and in fact in any algebraically closed field, $i^2=-1$.