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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?

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What you describe isn't a total order (in fact, abuse of notation aside, it's a preorder).

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  • $\begingroup$ I see, but actually I notice that Rudin simply says 'an order' and not 'total order' and still later asserts the complex field cannot be ordered. $\endgroup$ – user137794 Nov 14 '15 at 2:35
  • $\begingroup$ @user137794 "Order" means "total order," in this context - look for where he defines "order." $\endgroup$ – Noah Schweber Nov 14 '15 at 3:13
  • $\begingroup$ (Actually, I don't have Rudin on hand, so maybe he doesn't define "order" - but if he doesn't, he should, and what he means is "total order.") $\endgroup$ – Noah Schweber Nov 14 '15 at 3:14
  • $\begingroup$ He does define an order and simply states rules 1 and 2. $\endgroup$ – user137794 Nov 14 '15 at 3:15
  • $\begingroup$ @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 $y\le x$." $\endgroup$ – Noah Schweber Nov 14 '15 at 3:19

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