# 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.