8
$\begingroup$

I know this may very well be a silly question. I always hear that Complex numbers cannot be ordered. But there's something I'm missing... Why can't we just compare two complex numbers $z_1,z_2$ as follows:

A number is less then the other if the modulo is less or equal modulo but angle less from the x axis, in other words $z_1$<$z_2$ if $\rho_1<\rho_2$; or $\rho_1=\rho_2$ and $\theta_1<\theta_2$ where obvously $$z_1=\rho_1 e^{i\theta_1}, z_2=\rho_2 e^{i\theta_2}, \mathrm{with\ }\theta_1,\theta_2 \in [0,2\pi)$$

Why don't we do it? And if we do, why everyone always says that the Complex numbers cannot be ordered?

| cite | improve this question | |
$\endgroup$
  • 2
    $\begingroup$ In this way, the number $(1,0)$ is less than $(-1,0)$... $\endgroup$ – Mauro ALLEGRANZA May 15 '16 at 19:50
  • $\begingroup$ So there are no "negative" numbers? And the sum of two "positive" numbers, $z+(-z),$ can be zero? $\endgroup$ – bof May 15 '16 at 23:15
  • $\begingroup$ In this case, all the positive Real numbers are equivalent which may not go over well. $\endgroup$ – JB King May 15 '16 at 23:38
  • $\begingroup$ @JBKing What do you mean with "equivalent"? I mean if $z_1=1\ \mbox{and} \ z_2=2$ then $z_1<z_2$. $\endgroup$ – Ixion Sep 4 '18 at 17:08
17
$\begingroup$

You are right. Your order will work perfectly well, along with infinitely many others. The problem is that this order, and any others that might be defined, is not mathematically interesting; that is, such orders do not interact with the algebraic, geometric, or analytic structures of the complex numbers in any way that has a reasonably simple or satisfying description. They are all just arbitrary. It might be more precise to say that there is no canonical ordering of $\Bbb C$.

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.