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 . It’s straightforward to show that . Let and , where . Then , and so .
- 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 , where . Then or . We now show that, no matter what, or , 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 , then since . Also, if , then , so that and .
  - 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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Is 2i less than 3i? (Part 3: An inequality that almost works)

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Published by John Quintanilla

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