(related to Proposition: Complex Numbers Cannot Be Ordered)

We provide a proof by contradiction. * Assume, the field of complex numbers $(\mathbb C,+,\cdot)$ is an ordered field, i.e. there is a linear order „\(\geq \)“ on \(\mathbb C\), which fulfills the two properties: * from \(z_1\geq z_1\) it follows \(z_1+z_3\geq z_1+z_1\) (for arbitrary \(z_1,z_2,z_3\in \mathbb C\)), and * from \(z_1\geq 0\) and \(z_2\geq 0\) it follows \(z_1\cdot z_2\geq 0\) (for \(z_1,z_2\in F\)). * It suffices to show that the two complex numbers, the imaginary unit $i\in\mathbb C$ and the complex zero $0\in\mathbb C,$ cannot be ordered to fulfill the second property. * Case $i \ge 0.$ * Then $i\cdot i\ge 0\cdot i=0,$ or $-1 \ge 0$ which is a contradiction in the embedded real numbers as a special case of complex numbers. * Case $i < 0.$ * Then $(-i)\cdot (-i) > 0,$ or $-1 > 0$ which is again a contradiction.

∎