# My Favorite One-Liners: Part 17

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Sometimes it’s pretty easy for students to push through a proof from beginning to end. For example, in my experience, math majors have little trouble with each step of the proof of the following theorem.

Theorem. If $z, w \in \mathbb{C}$, then $\overline{z+w} = \overline{z} + \overline{w}$.

Proof. Let $z = a + bi$, where $a, b \in \mathbb{R}$, and let $w = c + di$, where $c, d \in \mathbb{R}$. Then

$\overline{z + w} = \overline{(a + bi) + (c + di)}$

$= \overline{(a+c) + (b+d) i}$

$= (a+c) - (b+d) i$

$= (a - bi) + (c - di)$

$= \overline{z} + \overline{w}$

$\square$

For other theorems, it’s not so easy for students to start with the left-hand side and end with the right-hand side. For example:

Theorem. If $z, w \in \mathbb{C}$, then $\overline{z \cdot w} = \overline{z} \cdot \overline{w}$.

Proof. Let $z = a + bi$, where $a, b \in \mathbb{R}$, and let $w = c + di$, where $c, d \in \mathbb{R}$. Then

$\overline{z \cdot w} = \overline{(a + bi) (c + di)}$

$= \overline{ac + adi + bci + bdi^2}$

$= \overline{ac - bd + (ad + bc)i}$

$= ac - bd - (ad + bc)i$

$= ac - bd - adi - bci$.

A sharp math major can then provide the next few steps of the proof from here; however, it’s not uncommon for a student new to proofs to get stuck at this point. Inevitably, somebody asks if we can do the same thing to the right-hand side to get the same thing. I’ll say, “Sure, let’s try it”:

$\overline{z} \cdot \overline{w} = \overline{(a + bi)} \cdot \overline{(c + di)}$

$= (a-bi)(c-di)$

$= ac -adi - bci + bdi^2$

$= ac - bd - adi - bci$.

$\square$

I call working with both the left and right sides to end up at the same spot the Diamond Rio approach to proofs: “I’ll start walking your way; you start walking mine; we meet in the middle ‘neath that old Georgia pine.” Not surprisingly, labeling this with a catchy country song helps the idea stick in my students’ heads.

Though not the most elegant presentation, this is logically correct because the steps for the right-hand side can be reversed and appended to the steps for the left-hand side:

Proof (more elegant). Let $z = a + bi$, where $a, b \in \mathbb{R}$, and let $w = c + di$, where $c, d \in \mathbb{R}$. Then

$\overline{z \cdot w} = \overline{(a + bi) (c + di)}$

$= \overline{ac + adi + bci + bdi^2}$

$= \overline{ac - bd + (ad + bc)i}$

$= ac - bd - (ad + bc)i$

$= ac - bd - adi - bci$

$= ac -adi - bci + bdi^2$

$= (a-bi)(c-di)$

$= \overline{(a + bi)} \cdot \overline{(c + di)}$

$\overline{z} \cdot \overline{w}$.

$\square$

For further reading, here’s my series on complex numbers.

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