My Favorite One-Liners: Part 98

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.

I’ll use today’s quip just after introducing the methodology of mathematical induction to my students:

Induction is so easy that even the army uses it.

My Favorite One-Liners: Part 89

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.

Here’s a problem that might arise in my discrete mathematics class:

Find the negation of p \Rightarrow q.

This requires a couple of reasonably complex steps. First, we use the fact that p \Rightarrow q is logically equivalent to $\lnot p \lor q$:

\lnot(p \Rightarrow q) \equiv \lnot (\lnot p \lor q).

Next, we have to apply DeMorgan’s Law to find the negation:

\lnot (p \Rightarrow q) \equiv \lnot(\lnot p \lor q) \equiv \lnot(\lnot p) \land \lnot q

Finally, we arrive at the final step: simplifying \lnot(\lnot p). At this point, I tell my class, it’s a bit of joke, especially after the previous, more complicated steps. “Not not p,” of course, is the same as p. So this step is a bit of a joke. Which steps up the following cringe-worthy pun:

In fact, you might even call this a not-not joke.

After the groans settle down, we finish the derivation:

\lnot(p \Rightarrow q) \equiv \lnot(\lnot p \lor q) \equiv \lnot(\lnot p) \land \lnot q \equiv p \land \lnot q.

My Favorite One-Liners: Part 44

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.

Today’s quip is something that I’ll use to emphasize that the meaning of the word “or” is a little different in mathematics than in ordinary speech. For example, in mathematics, we could solve a quadratic equation for x:

x^2 + 2x - 8 = 0

(x+4)(x-2) = 0

x + 4 = 0 \qquad \hbox{OR} \qquad x - 2 = 0

x = -4 \qquad \hbox{OR} \qquad x = 2

In this example, the word “or” means “one or the other or maybe both.” It could be that both statements are true, as in the next example:

x^2 + 2x +1 = 0

(x+1)(x+1) = 0

x + 1 = 0 \qquad \hbox{OR} \qquad x + 1= 0

x = -1 \qquad \hbox{OR} \qquad x = -1

However, in plain speech, the word “or” typically means “one or the other, but not both.” Here the quip I’ll use to illustrate this:

At the end of “The Bachelor,” the guy has to choose one girl or the other. He can’t choose both.

My Favorite One-Liners: Part 38

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.

When I was a student, I heard the story (probably apocryphal) about the mathematician who wrote up a mathematical paper that was hundreds of pages long and gave it to the departmental administrative assistant to type. (This story took place many years ago before the advent of office computers, and so typewriters were the standard for professional communication.) The mathematician had written “iff” as the standard abbreviation for “if and only if” since typewriters did not have a button for the \Leftrightarrow symbol.

Well, so the story goes, the administrative assistant saw all of these “iff”s, muttered to herself about how mathematicians don’t know how to spell, and replaced every “iff” in the paper with “if”.

And so the mathematician had to carefully pore through this huge paper, carefully checking if the word “if” should be “if” or “iff”.

I have no idea if this story is true or not, but it makes a great story to tell students.

My Favorite One-Liners: Part 34

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.

Suppose that my students need to prove a theorem like “Let n be an integer. Then n is odd if and only if n^2 is odd.” I’ll ask my students, “What is the structure of this proof?”

The key is the phrase “if and only if”. So this theorem requires two proofs:

  • Assume that n is odd, and show that n^2 is odd.
  • Assume that n^2 is odd, and show that n is odd.

I call this a blue-light special: Two for the price of one. Then we get down to the business of proving both directions of the theorem.

I’ll also use the phrase “blue-light special” to refer to the conclusion of the conjugate root theorem: if a polynomial f with real coefficients has a complex root z, then \overline{z} is also a root. It’s a blue-light special: two for the price of one.

 

My Favorite One-Liners: Part 28

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. Today’s quip is one that I’ll use when simple techniques get used in a complicated way.

Consider the solution of the linear recurrence relation

Q_n = Q_{n-1} + 2 Q_{n-2},

where F_0 = 1 and F_1 = 1. With no modesty, I call this one the Quintanilla sequence when I teach my students — the forgotten little brother of the Fibonacci sequence.

To find the solution of this linear recurrence relation, the standard technique — which is a pretty long procedure — is to first solve the characteristic equation, from Q_n - Q_{n-1} - 2 Q_{n-2} = 0, we obtain the characteristic equation

r^2 - r - 2 = 0

This can be solved by any standard technique at a student’s disposal. If necessary, the quadratic equation can be used. However, for this one, the left-hand side simply factors:

(r-2)(r+1) = 0

r=2 \qquad \hbox{or} \qquad r = -1

(Indeed, I “developed” the Quintanilla equation on purpose, for pedagogical reasons, because its characteristic equation has two fairly simple roots — unlike the characteristic equation for the Fibonacci sequence.)

From these two roots, we can write down the general solution for the linear recurrence relation:

Q_n = \alpha_1 \times 2^n + \alpha_2 \times (-1)^n,

where \alpha_1 and \alpha_2 are constants to be determined. To find these constants, we plug in n =0:

Q_0 = \alpha_1 \times 2^0 + \alpha_2 \times (-1)^0.

To find these constants, we plug in n =0:

Q_0 = \alpha_1 \times 2^0 + \alpha_2 \times (-1)^0.

We then plug in n =1:

Q_1 = \alpha_1 \times 2^1 + \alpha_2 \times (-1)^1.

Using the initial conditions gives

1 = \alpha_1 + \alpha_2

1 = 2 \alpha_1 - \alpha_2

This is a system of two equations in two unknowns, which can then be solved using any standard technique at the student’s disposal. Students should quickly find that \alpha_1 = 2/3 and \alpha_2 = 1/3, so that

Q_n = \displaystyle \frac{2}{3} \times 2^n + \frac{1}{3} \times (-1)^n = \frac{2^{n+1} + (-1)^n}{3},

which is the final answer.

Although this is a long procedure, the key steps are actually first taught in Algebra I: solving a quadratic equation and solving a system of two linear equations in two unknowns. So here’s my one-liner to describe this procedure:

This is just an algebra problem on steroids.

Yes, it’s only high school algebra, but used in a creative way that isn’t ordinarily taught when students first learn algebra.

I’ll use this “on steroids” line in any class when a simple technique is used in an unusual — and usually laborious — way to solve a new problem at the post-secondary level.

 

 

My Favorite One-Liners: Part 27

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.

Here’s an anecdote that I’ll share when teaching students about factorials:

1! = 1

2! = 1 \times 2 = 2

3! = 1 \times 2 \times 3 = 6

4! = 1 \times 2 \times 3 \times 4 = 24

5! = 1 \times 2 \times 3 \times 4 \times 5 = 120

The obvious observation is that the factorials get big very, very quickly.

Here’s my anecdote:

Many years ago, I was writing lesson plans while the TV show “Wheel of Fortune” was on in the background. And the contestant solved the puzzle at the end, and Pat Sajak declared, “You have just won $40,320 in cash in prizes.

So I immediately thought to myself, “Ah, 8 factorial.”

Then I thought, ugh [while slapping myself in the forehead, grimacing, and shaking my head, pretending that I can’t believe that that was the first thought that immediately came to mind].

[Finishing the story:] Not surprisingly, I was still single when this happened.

My Favorite One-Liners: Part 26

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.

Here’s a problem that could appear early in a probability class:

Let P(A) = 0.2, P(B) = 0.4, and P(A \cup B) = 0.5. Find P(A \mid B).

The standard technique for solving this problem involves first finding P(A \cap B) using the Addition Rule:

P(A \cup B) = P(A) + P(B) - P(A \cap B)

0.5 = 0.2 + 0.4 - P(A \cap B)

P(A \cap B) = 0.1

From here, the Multiplication Rule can be used (or, equivalently, the definition of a conditional probability):

P(B \cap A) = P(B) \cdot P(A \mid B)

0.1 = 0.4 P(A \mid B)

0.25 = P(A \mid B)

So far, so good.

Now let me add a small twist to the original problem that creates a small difficulty when solving:

Let P(A) = 0.2, P(B) = 0.4, and P(A \cup B) = 0.5. Find P(A \cap B \mid A \cup B).

Proceeding as before, we obtain

P( [A \cap B] \cup [A \cup B] ) = P(A \cup B) \cdot P(A \cap B \mid A \cup B)

The value of $P(A \cup B)$ is obvious. But how do we evaluate the left side?

If I’m teaching an advanced probability class, I might expect them to use DeMorgan’s Laws. However, it’s a whole lot easier to reason out the left hand side: I’m looking for the probability that both A and B happen or else at least one of A and B happen. Well, that’s clearly redundant: if both A and B happen, then certainly at least one of A and B happen.

Here’s my one-liner, which I say, if possible, using only one breath of air:

Clearly, this is redundant. It’s like saying Dr. Q is my professor and he’s a total stud. It’s redundant. It’s obvious. There’s no need to actually say it.

After the laughter settles from this bit of braggadocio, the A \cup B can be safely dropped from the left side:

P( A \cap B) = P(A \cup B) \cdot P(A \cap B \mid A \cup B)

0.1 = 0.5 \cdot P(A \cap B \mid A \cup B)

0.2 = P(A \cap B \mid A \cup B)

However, I need to emphasize that dropping the term on the left side is a special feature of this particular problem since one set was a subset of the other, and that students shouldn’t expect to always be able to do this when computing conditional probabilities.

My Favorite One-Liners: Part 16

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.

One of the basic notions of functions that’s taught in Precalculus and in Discrete Mathematics is the notion of an inverse function: if f: A \to B is a one-to-one and onto function, then there is an inverse function f^{-1}: B \to A so that

f^{-1}(f(a)) = a for all a \in A and

f(f^{-1}(b)) = b for all b \in B.

If A = B = \mathbb{R}, this is commonly taught in high school as a function that satisfies the horizontal line test.

In other words, if the function f is applied to a, the result is f(a). When the inverse function is applied to that, the answer is the original number a. Therefore, I’ll tell my class, “By applying the function f^{-1}, we uh-uh-uh-uh-uh-uh-uh-undo it.”

If I have a few country music fans in the class, this always generates a bit of a laugh.

See also the amazing duet with Carrie Underwood and Steven Tyler at the 2011 ACM awards:

My Favorite One-Liners: Part 12

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.

Often in mathematics, one proof is quite similar to another proof. For example, in Precalculus or Discrete Mathematics, students encounter the theorem

\sum_{k=1}^n (a_k + b_k) = \sum_{k=1}^n a_k + \sum_{k=1}^n b_k.

The formal proof requires mathematical induction, but the “good enough” proof is usually convincing enough for most students, as it’s just the repeated use of the commutative and associative properties to rearrange the terms in the sum:

\sum_{k=1}^n (a_k + b_k)= (a_1 + b_1) + (a_2 + b_2) + \dots + (a_n + b_n)

= (a_1 + a_2 + \dots + a_n) + (b_1 + b_2 + \dots + b_n)

= \sum_{k=1}^n a_k + \sum_{k=1}^n b_k.

Next, I’ll often present the new but closely related theorem

\sum_{k=1}^n (a_k - b_k) = \sum_{k=1}^n a_k -\sum_{k=1}^n b_k.

The proof of this would take roughly the same amount of time as the first proof, but there’s often little pedagogical value in doing all the steps over again in class. So here’s the line I’ll use: “At this point, I invoke the second-most powerful word in mathematics…” and then let them guess what this mysterious word is.

After a few seconds, I tell them the answer: “Similar.” The proof of the second theorem exactly parallels the proof of the first except for some sign changes. So I’ll tell them that mathematicians often use this word in mathematical proofs when it’s dead obvious that the proof can be virtually copied-and-pasted from a previous proof.

Eventually, students will catch on to my deliberate choice of words and ask, “What the most powerful word in mathematics?” As any mathematician knows, the most powerful word in mathematics is “Trivial”… the proof is so easy that it’s not necessary to write the proof down. But I warn my students that they’re not allowed to use this word when answering exam questions.

The third most powerful phrase in mathematics is “It is left for the student,” thus saving the professor from writing down the proof in class and encouraging students to figure out the details on their own.