Complex Numbers cannot be Totally Ordered/Proof 3

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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