# Complex Numbers cannot be Totally Ordered/Proof 3

## Contents

## 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

From Complex Numbers form Integral Domain, $\struct {\C, +, \times}$ is an integral domain.

Aiming for a contradiction, suppose that $\struct {\C, +, \times}$ can be ordered.

Thus, by definition, it possesses a (strict) positivity property $P$.

Then from Strict Positivity Property induces Total Ordering, let $\le$ be the total ordering induced by $P$.

From Unity of Ordered Integral Domain is Strictly Positive:

- $1$ is strictly positive.

Thus by strict positivity, axiom $3$:

- $-1$ is not strictly positive.

Consider the element $i \in \C$.

By definition of strict positivity, axiom $3$, either:

- $i$ is strictly positive

or:

- $-i$ is strictly positive.

Suppose $i$ is strictly positive.

Then by Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive:

- $i^2 = -1$ is strictly positive.

Similarly, suppose $-i$ is strictly positive.

Then by Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive:

- $\paren {-i}^2 = -1$ is strictly positive.

In both cases we have that $-1$ is strictly positive.

But it has already been established that $-1$ is not strictly positive.

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.

## Sources

- 1969: C.R.J. Clapham:
*Introduction to Abstract Algebra*... (previous) ... (next): Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Example $12$