# Total ordering on complex numbers

Show that there doesn't exist a relation $$\succ$$ between complex numbers such that

(i) For any two complex numbers $$z,w$$, one and only one of the following is true: $$z\succ w,w\succ z,$$ or $$z=w$$

(ii) For all $$z_1,z_2,z_3\in\mathbb{C}$$ the relation $$z_1\succ z_2$$ implies $$z_1+z_3\succ z_2+z_3$$.

(iii) For all $$z_1,z_2,z_3\in\mathbb{C}$$ with $$z_3\succ 0$$, then $$z_1\succ z_2$$ implies $$z_1z_3\succ z_2z_3$$.

Suppose $$i\succ 0$$. From (iii) we have $$i^2\succ 0$$, so $$-1\succ 0$$, so applying (ii) we get $$0\succ 1$$. But repeating (iii) on $$-1\succ 0$$ we get $$1\succ 0$$, a contradiction. So either $$i=0$$ or $$0\succ i$$.

How can I proceed from here?

• If $i\prec0$ then $0\prec-i$. Commented Sep 9, 2013 at 3:01
• Note that a relation $\prec$ which satisfies (i), (ii) and (iii) is stronger than a total order the way it is typically defined. Every nonempty set admits a total ordering by Zermello's theorem. Commented Apr 7, 2017 at 20:25

If we had an order on the complex numbers, then either $i \prec 0$ or $0 \prec i$.

If $0 \prec i$, then $$0i \prec ii \implies 0 \prec -1$$

Then since $0 \prec -1$, we see that $0 \prec (-1)^2 = 1$. Using (iii) we get

$$0 \prec -1 \implies 1 = 0 + 1 \prec -1 + 1 = 0 \implies 1 \prec 0 \prec 1$$

contradicts (i). The case that $i \prec 0$ is similar. Just use (ii) and add $(-i)$ both sides.

There is a well-known geometric interpretation of complex multiplication as a scaling followed by a rotation on a vector in the plane.

In brief, if $$z_1$$ is a complex number represented in the polar form $$|z_1|e^{i \theta}$$, then multiplying $$z_2$$ by $$z_1$$ is equivalent to scaling $$z_2$$ by $$|z_1|$$ and rotating it counterclockwise by $$\theta$$.

Here is a proof using this idea. Suppose $$0 \prec 1$$.

In the picture links 'red' means '$$\succ$$' and 'blue' means '$$\prec$$'.

Multiplying the values on sides by the complex unit vectors $$e^{i \theta}$$ is equivalent to rotating them by $$\theta$$. When we do so for $$\theta \in [0, 2 \pi]$$ 1 becomes the unit circle, while 0 remains the same. Because our ordering is preserved by complex multiplication, everything on the unit circle $$\succ$$ 0.

On the other hand, adding a real number $$x$$ to both sides is equivalent to translating the values along the x-axis. When we translate 0 and 1 by all of the real numbers $$x$$, we find that everything to the 'right' of zero $$\succ$$ 0, and everything to the 'left' of zero $$\prec$$ 0.

By inspecting the pictures, we see that complex multiplication forces $$-1 \succ 0$$, while complex addition forces $$-1 \prec 0$$. A contradiction!

With a little work this can be made rigorous. The idea also informs a variety of alternative proofs for this theorem. For example, the accepted answer uses rotation by unit vectors and translation along the imaginary axis to provide the contradiction.

• During the rotation when you multiply $1$ by $\exp(i\theta)$, do you explicitly need to assume $\exp(i\theta)$ $\succ$ $0$?
– Aman
Commented Apr 30, 2023 at 2:13

$i\ne 0$ since $i$ has an inverse but $0$ does not. Now just note that all squares are positive, and thus that the argument you gave stays true word for word if you start by assuming that $i\prec 0$. The point is really, that in $\mathcal C$ the number $-1$ is a square. In any ordered field, $1\ge 0$ and all squares are positive. It does not matter which of the two square roots of $-1$ you use, you'll get the same contradiction.

• If $i= 0$, how is it a contradiction with one of the given three properties? Commented Sep 9, 2013 at 3:07
• i has an inverse, but 0 does not. So i=1 is impossible. Commented Sep 9, 2013 at 3:07
• Yes, I realize that $i$ has an inverse but $0$ does not. I'm asking how it contradicts one of (i), (ii), (iii). Commented Sep 9, 2013 at 3:11
• It doesn't. It's already a contradiction on its own. Commented Sep 9, 2013 at 3:24