# Complex Numbers cannot be Totally Ordered

## Theorem

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

There exists no total ordering on $\left({\C, +, \cdot}\right)$ which is compatible with the structure of $\struct {\C, +, \cdot}$.

## Proof 1

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:

\(\displaystyle 0\) | \(\prec\) | \(\displaystyle i \cdot i\) | $\quad$ from $(2)$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle -1\) | $\quad$ Definition of Complex Number | $\quad$ |

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

Then:

\(\displaystyle 0\) | \(\prec\) | \(\displaystyle \left({-i}\right) \cdot \left({-i}\right)\) | $\quad$ from $(2)$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle -1\) | $\quad$ Definition of Complex Number | $\quad$ |

Thus by Proof by Cases:

- $0 \prec -1$

Thus it follows that:

\(\displaystyle 0\) | \(\prec\) | \(\displaystyle \left({-1}\right) \cdot \left({-1}\right)\) | $\quad$ from $(2)$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1\) | $\quad$ | $\quad$ |

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.

$\blacksquare$

## Proof 2

Aiming for a contradiction, suppose such a total ordering $\preceq$ exists.

By the definition of a total ordering, $\preceq$ is connected.

That is:

- $0 \preceq i \lor i \preceq 0$

Using Proof by Cases, we will prove that:

- $0 \preceq -1$

- Case 1

Assume that $0 \preceq i$.

By definition of an ordering compatible with the ring structure of $\C$:

- $\forall x, y \in \C: \left({0 \preceq x \land 0 \preceq y}\right) \implies 0 \preceq x \times y$

Substituting $x = i$ and $y = i$ gives:

- $0 \preceq i \times i$

Simplifying:

- $0 \preceq -1$

which is the result required.

- Case 2

Assume that $i \preceq 0$.

By definition of compatibility with addition:

- $\forall x, y, z \in \C: x \preceq y \implies \left({x + z}\right) \preceq \left({y + z}\right)$

Substituting $x = i$, $y = 0$, $z = -i$ gives:

- $i + \left({-i}\right) \preceq 0 + \left({-i}\right)$

Simplifying:

- $0 \preceq -i$

By definition of an ordering compatible with the ring structure of $\C$:

- $\forall x, y \in \C: \left({0 \preceq x \land 0 \preceq y}\right) \implies 0 \preceq x \times y$

Substituting $x = -i$ and $y = -i$ gives:

- $0 \preceq \left({-i}\right) \times \left({-i}\right)$

Simplifying:

- $0 \preceq -1$

This has been demonstrated to follow from both cases, and so by Proof by Cases:

- $0 \preceq -1$

$\Box$

By definition of an ordering compatible with the ring structure of $\C$:

- $\forall x, y \in \C: \left({0 \preceq x \land 0 \preceq y}\right) \implies 0 \preceq x \times y$

Substituting $x = -1$ and $y = -1$:

- $0 \preceq \left({-1}\right) \times \left({-1}\right)$

Simplifying:

- $0 \preceq 1$

By definition of compatibility with addition:

- $\forall x, y, z \in \C: x \preceq y \implies \left({x + z}\right) \preceq \left({y + z}\right)$

Substituting $x = 0$, $y = 1$, $z = -1$ gives:

- $0 + \left({-1}\right) \preceq 1 + \left({-1}\right)$

Simplifying:

- $-1 \preceq 0$

From the definition of ordering:

- $\forall a, b \in \C: \left({a \preceq b \land b \preceq a}\right) \implies a = b$

Substituting $a = -1$ and $b = 0$ gives:

- $-1 = 0$

which is a contradiction.

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

$\blacksquare$

## Proof 3

From Complex Numbers form Integral Domain, $\left({\C, +, \times}\right)$ is an integral domain.

Aiming for a contradiction, suppose that $\left({\C, +, \times}\right)$ can be ordered.

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

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

From Unity of Ordered Integral Domain is Positive:

- $1$ is positive.

Thus by positivity, axiom $3$:

- $-1$ is not positive.

Consider the element $i \in \C$.

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

- $i$ is positive

or:

- $-i$ is positive.

Suppose $i$ is positive.

Then by Square of Element of Ordered Integral Domain is Positive:

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

Similarly, suppose $-i$ is positive.

Then by Square of Element of Ordered Integral Domain is Positive:

- $\left({-i}\right)^2 = -1$ is positive.

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

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

- 1960: Walter Ledermann:
*Complex Numbers*... (previous) ... (next): $\S 1.2$. The Algebraic Theory - 1986: David Wells:
*Curious and Interesting Numbers*... (previous) ... (next): $-1$ and $i$ - 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $-1$ and $i$