# Is 2i less than 3i? (Part 1: Order Axioms)

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.”

However, for this answer to make sense, we need to talk about how inequalities are defined in the first place.

Following Apostol’s calculus, there are four axioms from which the ordinary notions of inequality follow. We shall assume that there exists a certain subset $\mathbb{R}^+ \subset \mathbb{R}$, called the set of positive numbers, which satisfies the following four order axioms:

• If $x, y \in \mathbb{R}^+$, then $x+y \in \mathbb{R}^+$
• If $x, y \in \mathbb{R}^+$, then $xy \in \mathbb{R}^+$.
• For every real $x \ne 0$, either $x \in \mathbb{R}^+$ or $-x \in \mathbb{R}^+$, but not both.
• $0 \notin \mathbb{R}^+$

We then define the symbols $<$, $\le$, $>$, and $\ge$ in the obvious way:

• $x < y$ means that $y-x$ is positive.
• $x > y$ means that $y < x$.
• $x \le y$ means that either $x < y$ or $x=y$.
• $x \ge y$ means that $y \le x$

From only these four axioms, many familiar theorems about inequalities can be proven. For what it’s worth, when I was a student in Algebra I, I had to prove nearly all of these theorems.

• If $a, b \in \mathbb{R}$, then exactly one of the following three relations is true: $a < b$, $a > b$, $a = b$.
• If $a < b$ and $b < c$, then $a < c$.
• If $a < b$, then $a + c < b + c$.
• If $a < b$ and $c > 0$, then $ac < bc$.
• If $a \ne 0$, then $a^2 > 0$.
• $1 > 0$.

We interrupt this list with a public-service announcement: Yes, there’s a proof that $1$ is greater than $0$. When I tell this to students, I can usually see their heads start to spin, as they think, “Of course we know that!” Then I ask them what the definitions of $1$ and $0$ are. Usually, they have no idea. Then I’ll remind them that $0$ is defined to be the additive identity (so that $x + 0 = x$ and $0 + x = x$ for all real numbers $x$), while $1$ is defined to be the multiplicative identity (so that $x \cdot 1 = x$ and $1 \cdot x = x$ for all real numbers $x$). Based on those definitions alone, I then ask my students, is it obvious that the multiplicative identity has to be larger than the additive identity? The answer is no, which is why the above order axioms are needed.

Here are some more familiar theorems about inequalities that derive from the four order axioms.

• If $a < b$ and $c < 0$, then $ac > bc$.
• If $a < b$, then $-a > -b$.
• If $a < 0$, then $-a > 0$.
• If $ab > 0$, then $a$ and $b$ are either both positive or both negative.
• If $a < c$ and $b < d$, then $a + b < c + d$.
• If $a < 0$ and $b < 0$, then $a + b < 0$.
• If $a > 0$, then $1/a > 0$.
• If $a < 0$, then $1/a < 0$.
• If $0 < a < b$, then $0 < 1/b < 1/a$.
• If $a \le b$ and $b \le c$, then $a \le c$.
• If $a \le b \le c$ and $a = c$, then $b = c$.

These last two theorems are less familiar. They basically state that (1) there is no “biggest” real number and (2) there is no positive number that’s immediately to the right of 0.

• There is no real number $a$ so that $x \le a$ for all real numbers $x$.
• If $0 \le x < h$ for every positive real number $h$, then $x =0$.

Here’s another important theorem that ultimately derives from the four order axioms, proving that a number system including $\sqrt{-1}$ is incompatible with the four order axioms.

• There is no real number $x$ so that $x^2 + 1 = 0$.

The proof of this theorem is simple, given the theorems above. If $x = 0$, then $x^2 + 1 = 1$, which is greater than 0. If $x \ne 0$, then $x^2 > 0$, and so $x^2 + 1 > 0 + 1 > 1$. Since $1 > 0$, it follows by transitivity that $x^2 > 0$. Either way, $x^2 + 1 > 0$, and so $x^2 + 1 \ne 0$.