Complex Numbers cannot be Totally Ordered/Proof 1

Theorem

Let $\struct {\C, +, \times}$ be the field of complex numbers.

There exists no total ordering on $\struct {\C, +, \times}$ which is compatible with the structure of $\struct {\C, +, \times}$.

Proof

Aiming for a contradiction, suppose there exists a relation $\preceq$ on $\C$ which is ordering compatible with the ring structure of $\C$.

That is:

$(1): \quad z \ne 0 \implies 0 \prec z \lor z \prec 0$, but not both
$(2): \quad 0 \prec z_1, z_2 \implies 0 \prec z_1 z_2 \land 0 \prec z_1 + z_2$

By Totally Ordered Ring Zero Precedes Element or its Inverse, $(1)$ can be replaced with:

$(1'): \quad z \ne 0 \implies 0 \prec z \lor 0 \prec -z$, but not both.

As $i \ne 0$, it follows that:

$0 \prec i$ or $0 \prec -i$

Suppose $0 \prec i$.

Then:

 $\displaystyle 0$ $\prec$ $\displaystyle i \times i$ from $(2)$ $\displaystyle$ $=$ $\displaystyle -1$ Definition 1 of Complex Number

Otherwise, suppose $0 \prec \paren {-i}$.

Then:

 $\displaystyle 0$ $\prec$ $\displaystyle \paren {-i} \times \paren {-i}$ from $(2)$ $\displaystyle$ $=$ $\displaystyle -1$ Definition 1 of Complex Number

Thus by Proof by Cases:

$0 \prec -1$

Thus it follows that:

 $\displaystyle 0$ $\prec$ $\displaystyle \paren {-1} \times \paren {-1}$ from $(2)$ $\displaystyle$ $=$ $\displaystyle 1$

Thus both:

$0 \prec -1$

and:

$0 \prec 1$

This contradicts hypothesis $(1')$:

$(1): \quad z \ne 0 \implies 0 \prec z \lor 0 \prec -z$, but not both

Hence, by Proof by Contradiction, there can be no such ordering.

$\blacksquare$

Historical Note

The fact that Complex Numbers cannot be Totally Ordered was realized by Leonhard Paul Euler.