My Favorite One-Liners: Part 43

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. q Q

Years ago, my first class of students decided to call me “Dr. Q” instead of “Dr. Quintanilla,” and the name has stuck ever since. And I’ll occasionally use this to my advantage when choosing names of variables. For example, here’s a typical proof by induction involving divisibility.

Theorem: If $n \ge 1$ is a positive integer, then $5^n - 1$ is a multiple of 4.

Proof. By induction on $n$.

$n = 1$: $5^1 - 1 = 4$, which is clearly a multiple of 4.

$n$: Assume that $5^n - 1$ is a multiple of 4.

At this point in the calculation, I ask how I can write this statement as an equation. Eventually, somebody will volunteer that if $5^n-1$ is a multiple of 4, then $5^n-1$ is equal to 4 times something. At which point, I’ll volunteer:

Yes, so let’s name that something with a variable. Naturally, we should choose something important, something regal, something majestic… so let’s choose the letter $q$. (Groans and laughter.) It’s good to be the king.

So the proof continues:

$n$: Assume that $5^n - 1 = 4q$, where $q$ is an integer.

$n+1$. We wish to show that $5^{n+1} - 1$ is also a multiple of 4.

At this point, I’ll ask my class how we should write this. Naturally, I give them no choice in the matter:

We wish to show that $5^{n+1} - 1 = 4Q$, where $Q$ is some (possibly different) integer.

Then we continue the proof:

$5^{n+1} - 1 = 5^n 5^1 - 1$

$= 5 \times 5^n - 1$

$= 5 \times (4q + 1) - 1$ by the induction hypothesis

$= 20q + 5 - 1$

$= 20q + 4$

$= 4(5q + 1)$.

So if we let $Q = 5q +1$, then $5^{n+1} - 1 = 4Q$, where $Q$ is an integer because $q$ is also an integer.

QED

On the flip side of braggadocio, the formula for the binomial distribution is

$P(X = k) = \displaystyle {n \choose k} p^k q^{n-k}$,

where $X$ is the number of successes in $n$ independent and identically distributed trials, where $p$ represents the probability of success on any one trial, and (to my shame) $q$ is the probability of failure.