# Is 2i less than 3i? (Part 3: An inequality that almost works)

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. One way of attempting this is defining

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

This set $\mathbb{C}^+$ leads to the lexicographic ordering of the complex numbers: if $z_1 = a_1 + i b_1$ and $z_2 = a_2 + i b_2$, where $a_1, a_2, b_1, b_2 \in \mathbb{R}$, we say that $z_1 \prec z_2$ if

$a_1 < a_2 \qquad \hbox{or} \qquad a_1 = a_2, b_1 < b_2$

I used the symbol $z_1 \prec z_2$ because, as we’ll see, $\prec$ satisfies some but not all of the usual properties of an inequality. This ordering is sometimes called the “dictionary” order because the numbers are ordered like the words in a dictionary… the real parts are compared first, and then (if that’s a tie) the imaginary parts are compared. See Wikipedia and Mathworld for more information.

In any case, defining $\mathbb{C}^+$ in this way satisfies three of the four order axioms.

• Suppose $z_1, z_2 \in \mathbb{C}^+$. It’s straightforward to show that $z_1 + z_2 \in \mathbb{C}^+$. Let $z_1 = a_1 + i b_1$ and $z_2 = a_2 + i b_2$, where $a_1, a_2, b_1, b_2 \in \mathbb{R}$. Then $a_1, a_2 \ge 0$, and so $a_1 + a_2 \ge 0$.
• Case 1: If $a_1 + a_2 > 0$, then clearly $z_1 + z_2 \in \mathbb{C}^+$.
• Case 2: If $a_1 + a_2 = 0$, that’s only possible if $a_1 = 0$ and $a_2 = 0$. But since $z_1, z_2 \in \mathbb{C}^+$, that means that $b_1 > 0$ and $b_2 > 0$. Therefore, $b_1 + b_2 > 0$. Since $a_1 + a_2 = 0$, we again conclude that $z_1 + z_2 \in \mathbb{C}^+$.
• Suppose $z = a + bi \ne 0$, where $a, b \in \mathbb{R}$. Then $a \ne 0$ or $b \ne 0$. We now show that, no matter what, $z \in \mathbb{C}^+$ or $-z \in \mathbb{C}^+$, but not both.
• Case 1: If $a > 0$, then $-a < 0$, and so $z \in \mathbb{C}^+$ but $-z \notin \mathbb{C}^+$.
• Case 2: If $a < 0$, then $-a > 0$, and so $-z \in \mathbb{C}^+$ but $z \notin \mathbb{C}^+$.
• Case 3: If $a = 0$, then $b \ne 0$ since $z \ne 0$. Also, if $a = 0$, then $-a = 0$, so that $z = bi$ and $-z = -bi$.
• Subcase 3A: If $b > 0$, then $-b < 0$, and so $z \in \mathbb{C}^+$ but $-z \notin \mathbb{C}^+$.
• Subcase 3B: If $b < 0$, then $-b > 0$, and so $-z \in \mathbb{C}^+$ but $z \notin \mathbb{C}^+$.
• By definition, $0 = 0 + 0i \notin \mathbb{C}^+$.

However, the fourth property fails. By definition, $i = 0 + 1i \in \mathbb{C}^+$. However, $i \cdot i = -1 + 0i \notin \mathbb{C}^+$.

Because this definition of $\mathbb{C}^+$ satisfies three of the four order axioms, the relation $\prec$ satisfies some but not all of the theorems stated in the first post of this series. For example, if $z_1 \prec z_2$ and $z_2 \prec z_3$, then $z_1 \prec z_3$. Also, if $z_1 \prec z_2$ and $w_1 \prec w_2$, then $z_1 + w_1 \prec z_2 + w_2$.

I’ll leave it to the interested reader to determine which of the theorems are true, and which are false (and have counterexamples).

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