Is 2i less than 3i? (Part 2: Proof by contradiction)

Is 2i less than 3i?

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 \mathbb{C}^+ \subset \mathbb{C} of the complex numbers which satisfies the following four axioms:

  • If z_1, z_2 \in \mathbb{C}^+, then z_1+z_2 \in \mathbb{C}^+
  • If z_1, z_2 \in \mathbb{C}^+, then z_1 z_2 \in \mathbb{C}^+.
  • For every z \ne 0, either z \in \mathbb{C}^+ or -z \in \mathbb{C}^+, but not both.
  • 0 \notin \mathbb{C}^+

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 i is “positive” or not. According to the third axiom, either i \in \mathbb{C}^+ or -i \in \mathbb{C}^+, but not both.

Case 1. i \in \mathbb{C}^+. Then by the second axiom, i \cdot i \in \mathbb{C}^+, or -1 \in \mathbb{C}^+. Applying the second axiom again, since i \in \mathbb{C}^+ and -1 \in \mathbb{C}^+, we have i \cdot (-1) \in \mathbb{C}^+, or -i \in \mathbb{C}^+. But that’s impossible because we assumed that i \in \mathbb{C}^+.

Case 2. -i \in \mathbb{C}^+. Then by the second axiom, (-i) \cdot (-i) \in \mathbb{C}^+, or -1 \in \mathbb{C}^+. Applying the second axiom again, since -i \in \mathbb{C}^+ and -1 \in \mathbb{C}^+, we have (-i) \cdot (-1) \in \mathbb{C}^+, or i \in \mathbb{C}^+. But that’s impossible because we assumed that -i \in \mathbb{C}^+.

Either way, we obtain a contradiction. Therefore, there is no subset \mathbb{C}^+ 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.

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  1. Is 2i less than 3i? (Index) | Mean Green Math

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