The complex numbers as a whole cannot be ordered but could you order the complex numbers of the form ai where a is a real number?
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$\begingroup$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:
- $-1<0$.
- $x^2\geq 0$ for every $x$.
And in $\Bbb C$, and in fact in any algebraically closed field, $i^2=-1$.
$\endgroup$ 2 $\begingroup$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' because you rearrange the complex plane into lines of constant real part. EDIT: see comment, (thanks) it isn't the long line.
$\endgroup$ 1 $\begingroup$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.
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