Prove that the complex numbers cannot be ordered

The three axioms for an order relation are (see page 20 of Apostol):

If $x$ and $y$ are in $\mathbb{R}^+$ , so are $x+y$ and $xy$ .
For every real $x \neq 0$ , either $x \in \mathbb{R}^+$ or $-x \in \mathbb{R}^+$ , but not both.
$0 \notin \mathbb{R}^+$ .

Prove that the complex number system cannot be equipped with an ordering relation satisfying all of these axioms.

Proof. Suppose otherwise, that such an ordering exists. Denote the positive complex numbers by $\mathbb{C}^+$ . Then, since $i \neq 0$ , we must have either $i \in \mathbb{C}^+$ or $i \in \mathbb{C}^-$ , but not both by the second axiom.

But, if $i \in \mathbb{C}^+$ then $i^3 \in \mathbb{C}^+$ (since the product of positive elements must also be positive by applying the first axiom twice). Since $i^3 = -i$ this is a contradiction; therefore, $i \notin \mathbb{C}^+$ .

Similarly, if $-i \in \mathbb{C}^+$ then $(-i)^3 \in \mathbb{C}^+$ which is also a contradiction since $(-i)^3 = i$ . Hence, there can be no such ordering $. \qquad \blacksquare$

Stumbling Robot

Prove that the complex numbers cannot be ordered

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