# Why is there no 'trivial' ordering on the complex numbers

The ordering I'm trying to consider is simply for $x,y \in \mathbb{C}$ then $x=y$.

I'm going through Rudin's Principles of Mathematical Analysis and the only restrictions he gives for an ordered field are that it has some total order and follows a few more rules.

1. For $x,y$ exactly one of the following holds: $x<y$, $x=y$, $x>y$.
2. If $x<y$ and $y<z$ then $x<z$.
3. If $y<z$ then $x+y < x+z$.
4. If $x,y > 0$ then $xy > 0$.

As I see it, this trivial ordering satisfies 1 and vacuously satisfies 2-4. Thus $\mathbb{C}$ is ordered and ever field can be ordered likewise.

When I looked on Wikipedia, they note that since $-1$ is a square, it must be positive thus no ordering is possible. But I believe that should say $-1$ is nonnegative, and that all numbers or nonnegative. With all numbers equal to each other and $0$, there is no issue.

Am I missing something?

• @user137794 Ah, I see the issue: "=" isn't an arbitrary symbol, it really is equality. So you can't just say "$x=y$ for all $x,y\in\mathbb{C}$" - for instance, $1\not=7$. In particular, "$x=y$" is not a priori the same as "$x\le y$ and $y\le x$." – Noah Schweber Nov 14 '15 at 3:19