Complex Numbers cannot be Totally Ordered

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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.


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