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.

green_speech_bubble

 

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

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