We discussed fields here. Today we’re going to continue by talking about a more specific kind of field.

To recap, here are the basic field axioms:

A1: Addition is commutative: for all x and y, x + y = y + x
A2: Addition is associative: for all x and y, (x + y) + z = x + (y + z)
A3: The field contains an additive identity: there exists a member 0 such that for all x, x + 0 = x
A4: Each member has an additive inverse: for each x there is a -x such that x + -x = 0

M1: Multiplication is commutative: for all x and y, x * y = y * x
M2: Multiplication is associative: for all x and y, (x * y) * z = x * (y * z)
M3: The field contains an multiplicative identity not equal to the additive identity: there exists a member 1 ≠ 0 such that for all x, x * 1 = x
M4: Each member other than the additive inverse has an multiplicative inverse: for each x ≠ 0 there is an x-1 such that x * x-1 = 1

D1: Multiplication distributes over addition: for all x, y, and z, x * (y + z) = (x * y) + (x * z)

Today, we’ll talk about an optional property of fields. A field is called ordered if it has a “greater than” operator that works as we expect. As always, we’ll make this precise. The ordering axioms describe a subset of the field named P, which we usually call the positive numbers, and go on to say:

O1: If p and q are both members of P, then so is p + q
O2: If p and q are both members of P, then so is p * q
O3: For every x, exactly one of the following is true:

  • x is a member of P
  • -x is a member of P
  • x is 0

We then define greater than (>) as

x > y if and only if there is a p in P such that x = y + p

As a simple consequence, x > 0 if and only if x is a member of P, which is what we wanted. Also, clearly 1 > 0, since:

  • We know 1 ≠ 0
  • If -1 were greater than 0, then -1 * -1 = 1 would be also be greater than 0. which is a contradiction.
  • All that remains is 1 > 0

We can check that > acts the way we expect:

  • if x > y then x + z > y + z

x = y + p, so x + z = y + p + z, thus x + z > y + z

  • if x > y and z > 0 then x * z > y * z

x = y + p, so x * z = (y + p) * z = y * z + p * z
p * z > 0, so x * z > y * z

  • if x > y and 0 > z then y * z > x * z

x = y + p, so x * z = (y + p) * z = y * z + p * z, or
x * z + (p * -z) = y * z
0 > p * -z, so x * z < y * z

  • if x > y and y > z then x > z

x = y + p and y = z + q, so x = z + p + q.
Since p + q > 0, x > z

Looks good.

The usual ordering on the rationals and reals satisfies O1-O3. It’s easy to see that there is no such ordering for the complex numbers. By O3, either i or -i would have to be > 0, but i * i = -i * -i = -1, which contradicts O2. So, the complex numbers are not an ordered field. Nor are the finite fields we looked at last time. (Consider the field with members 0, 1, 2, 3, and 4. 1 = -4, but since 1 is positive, 1 + 1 + 1 + 1 = 4 would also need to be positive. This can be generalized to all finite fields easily enough.)

So, among our favorite examples of fields, only the rational and reals are ordered.

Next time, we’ll discuss a more complex property of some fields, completeness.

UPDATE: A challenge, for anyone who cares to take it.

Prove: if a > b and 0 > c, then b * c > a * c

Mike Schilling

Mike has been a software engineer far longer than he would like to admit. He has strong opinions on baseball, software, science fiction, comedy, contract bridge, and European history, any of which he's willing to share with almost no prompting whatsoever.


  1. I’m unclear on what makes the complex numbers un-orderable.

    After all, couldn’t we just adopt a total order relation that can handle complex numbers, such as

    a+bi > c+di iff a > b or (a=b and c>d)

    • whoops. I just realized that under that ordering, i is positive, but i^2 = -1 is negative. I assume any other ordering would similarly result in a situation that violates the properties of positive numbers you’ve given above.

      • Complex numbers don’t fit onto a number line. There are ways of considering sets of complex numbers relative to each other — but they won’t be order-able, not until they fit on a number line.

        • I’m not sure that makes a difference, Blaise. After all, the finite fields definitely fit on a number line, and they’re not an ordered field.

      • Under any ordering that obeys all the axioms, 1 is positive and -1 negative. (I gave the proof in the main post.) So neither i nor -i can be positive, since either one times itself is -1. Thus, no ordering can work.

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