Complex Numbers cannot be Ordered Compatibly with Ring Structure/Proof 2

Theorem
Let $\left({\C, +, \times}\right)$ be the field of complex numbers.

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

Proof
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:
 * $0 \preceq -1$

Assume that $0 \preceq i$.
 * Case 1

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.

Assume that $i \preceq 0$.
 * Case 2

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:


 * $0 \preceq -1$

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.