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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?

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    $\begingroup$ If $i\prec0$ then $0\prec-i$. $\endgroup$ Commented Sep 9, 2013 at 3:01
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    $\begingroup$ 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. $\endgroup$ Commented Apr 7, 2017 at 20:25

3 Answers 3

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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.

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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.

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

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

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