# Is 2i less than 3i? (Part 4: Two other attempted inequalities)

In yesterday’s post, I demonstrated that there is no 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.
• $0 \notin \mathbb{C}^+$

However, it’s instructive (and fun) to try to construct such a set. Yesterday I showed that the following subset satisfies three of the four axioms:

$\mathbb{C}^+ = \{x + iy : x > 0 \qquad \hbox{or} \qquad x = 0, y > 0\}$

Apostol’s calculus suggests two other subsets to try:

$\mathbb{C}^+ = \{x + iy : x^2 + y^2 > 0 \}$

and

$\mathbb{C}^+ = \{x + iy : x > y\}$

Neither of these sets work either, but I won’t spoil the fun for you by giving you the proofs. I leave a thought bubble if you’d like to try to figure out which of the four axioms are satisfied by these two notions of “positive” complex numbers.