Ordered Fields and When You Can’t Order Them |

The real numbers $\mathbb{R}$ have an ordering on them–given two numbers $x$ and $y$ , we can tell whether $x = y, x > y,$ or $y<x$ . So as math people, we like to generalize this to other sets–when can we say that a general set is ordered? In this post, we’re going to explain the explicit of idea of what it means for a field to be ordered, and then show that the complex numbers $\mathbb{C}$ can’t be ordered–no matter what ordering you put on it. (If you don’t know what a field is, just think of the real numbers $\mathbb{R}$ or the complex numbers $\mathbb{C}$ .)

In the real numbers, we say that $x > y$ exactly when $x - y > 0$ , that is, $x - y$ is positive. So since we’re trying to generalize the idea of “ordering”, one way is to do that is to figure out how to generalize the idea of “positive” numbers. So (and I’m being a little loose here), let’s say we only have the idea of “adding”, which we’ll denote with +, and multiplication, which we’ll denote by “*”. This is sort of what it means to be a “general field” (You might want to think about which properties of positive numbers can be defined only with +, *, or if this is too vague, just keep reading.)

Here are the two that are most important about the positive numbers–one is that if you add two positive numbers, you get a positive number again. The other is that if you multiply two positive numbers, you get a positive number again. These properties say that whatever set of positive numbers we have, it must be closed under addition and closed under multiplication resepectively.

Oh–and one other point. If we pick a generic number $x$ , we want to say that either $x$ is positive, negative, or zero. Also, has to be exactly one of them (so 10 can’t also be negative, for example). It turns out that’s all we need to make our definition of an ordered field:

Definition: We say that a field $\mathbb{F}$ is an ordered field if it has a set $\mathscr P$ (of “positive numbers”) such that:

( $\mathscr{P}$ is closed under addition) If we have two elements $x \in \mathscr{P}$ and $y \in \mathscr{P}$ , then their sum is also in $\mathscr{P}$ , that is, $x + y \in \mathscr{P}$ .
( $\mathscr{P}$ is closed under multiplication) If we have two elements $x \in \mathscr{P}$ and $y \in \mathscr{P}$ , then their product is also in $\mathscr{P}$ , that is, $x*y \in \mathscr{P}$ .
(All nonzero numbers are positive or negative) For all $x$ in our field, exactly one of the following holds: $x \in \mathscr{P}, x = 0,$ or $-x \in \mathscr{P}$ .

Now we’ll show something pretty cool.

Proposition: The complex numbers $\mathbb{C}$ is not an ordered field.

Proof: To show this, we’re going to use a method called proof by contradiction. We’re essentially going to show that if $\mathbb{C}$ was an ordered field, something bad will happen. So let’s assume $\mathbb{C}$ was an ordered field and see if we can find anything weird happening.

Well one special element in $\mathbb{C}$ that’s not in the real numbers is $i$ , where $i^2 = -1$ . So since $i \neq 0$ , either $i$ or $-i$ is positive, according to (3) above.

If $i$ was positive, then $i*i = -1$ is a positive number, by (2). But again by (2), this says that $1 = (-1)(-1)$ is positive, so $1$ and -1 are both positive. This violates (3).

Okay, so what if $-i$ was positive instead? Well, a pretty similar thing happens, since $(-i)(-i) = -1*(-1)*i*i = -1$ will still be positive, so we’ll get the same contradiction that 1 and -1 are both positive.

So there’s no way to order the complex numbers, at least as a field. Woah! That’s pretty neat. The mathematicians reading this may argue that if you loosen up your definition of just a set ordering, instead of a field ordering, you could put an ordering on $\mathbb{C}$ . But instead of doing that and arguing with your computer screen, you should try to prove to yourself that any finite field can’t be ordered (as a field). It’s more fun that way.

Ordered Fields and When You Can’t Order Them

Categories

Archives

Retired Blogs

Meta