Is less than ?

That’s a very natural question for a student to ask when first learning about complex numbers. The short answer is, “No… there isn’t a way to define inequality for complex numbers that satisfies all the properties of real numbers.”

I demonstrated this in a roundabout way in yesterday’s post. Today, let’s tackle the issue using a proof by contradiction. (I almost said, “Let’s tackle the issue more directly,” but avoided that phrase because a proof by contradiction is sometimes called an indirect proof, which would lead to the awkward sentence “Let’s tackle the issue more directly with an indirect proof.”)

Suppose that there’s a subset of the complex numbers which satisfies the following four axioms:

• If , then
• If , then .
• For every , either or , but not both.
• 

As discussed yesterday, the ordinary properties of inequalities derive from these four order axioms. Assuming these four order axioms are true for the complex numbers, let’s investigate whether is “positive” or not. According to the third axiom, either or , but not both.

Case 1. . Then by the second axiom, , or . Applying the second axiom again, since and , we have , or . But that’s impossible because we assumed that .

Case 2. . Then by the second axiom, , or . Applying the second axiom again, since and , we have , or . But that’s impossible because we assumed that .

Either way, we obtain a contradiction. Therefore, there is no subset of the complex numbers that serves as the “positive” complex numbers, and so there’s no way to define inequalities for complex numbers that satisfies all of the usual properties of inequalities.