# 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.