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}^+.

green line

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).

Leave a comment

1 Comment

  1. Is 2i less than 3i? (Index) | Mean Green Math

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

w

Connecting to %s

%d bloggers like this: