Complex Numbers Cannot be Totally Ordered

Theorem

There exists no total ordering on the field of complex numbers $\left({\C, +, \times}\right)$ compatible with the structure of $\left({\C, +, \times}\right)$.

Proof

Suppose to the contrary that the relation $\preceq$ is such that:

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

$(2): \quad 0 \prec z_1, z_2 \implies 0 \prec z_1 z_2, 0 \prec z_1 + z_2$

Then since $i \ne 0$, it follows that $0 \prec i$ or $0 \prec -i$.

Suppose $0 \prec i$.

Then $0 \prec i \cdot i = -1$.

Otherwise, suppose $0 \prec \left({-i}\right)$.

Then $0 \prec \left({-i}\right) \cdot \left({-i}\right) = -1$.

So whichever, we have $0 \prec -1$.

But at the same time, $0 \prec \left({-1}\right)^2 = 1$.

So we have both $0 \prec -1$ and $0 \prec 1$ which contradicts the hypothesis that $\prec$ is an compatible with the structure of $\left({\C, +, \times}\right)$.

Hence there can be no such ordering.

$\blacksquare$

Sources

• C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 2.7$: Example $12$