Complex Numbers cannot be Ordered Compatibly with Ring Structure/Proof 1
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
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:
| \(\ds 0\) | \(\prec\) | \(\ds i \times i\) | from $(2)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds -1\) | Definition 1 of Complex Number |
Otherwise, suppose $0 \prec \paren {-i}$.
Then:
| \(\ds 0\) | \(\prec\) | \(\ds \paren {-i} \times \paren {-i}\) | from $(2)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds -1\) | Definition 1 of Complex Number |
Thus by Proof by Cases:
- $0 \prec -1$
Thus it follows that:
| \(\ds 0\) | \(\prec\) | \(\ds \paren {-1} \times \paren {-1}\) | from $(2)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds 1\) |
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.
$\blacksquare$
Historical Note
The fact that Complex Numbers cannot be Ordered Compatibly with Ring Structure was realized by Leonhard Paul Euler.