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.

My Favorite One Liners: Part 2

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 doing a large computation, I’ll often leave plenty of blank space on the board to fill it later. For example, when proving by mathematical induction that

$1 + 3 + 5 + \dots + (2n-1) = n^2$ ,

the inductive step looks something like

$1 + 3 + 5 \dots + (2k-1) + (2[k+1]-1) =$

$~$

$= (k+1)^2$

So I explained that, to complete the proof by induction, all we had to do was convert the top line into the bottom line.

As my class swallowed hard as they thought about how to perform this task, I told them, “Yes, this looks really intimidating. Indeed, to quote the great philosopher, ‘You might think that I’m insane. But I’ve got a blank space, baby… so let’s write what remains.’ “

And, just in case you’ve been buried under a rock, here’s the source material for the one-liner (which, at the time of this writing, is the fifth-most watched video on YouTube):

Predicate Logic and Popular Culture (Part 123): Willie Nelson

Let $M(t)$ be the proposition “You were on my mind at time $t$ .” Translate the logical statement

$\forall t < 0 (M(t))$ .

Naturally, this matches the classic song by Willie Nelson (though Elvis did record it before him).

Context: This semester, I taught discrete mathematics for the first time. Part of the discrete mathematics course includes an introduction to predicate and propositional logic for our math majors. As you can probably guess from their names, students tend to think these concepts are dry and uninteresting even though they’re very important for their development as math majors.

In an effort to making these topics more appealing, I spent a few days mining the depths of popular culture in a (likely futile) attempt to make these ideas more interesting to my students. In this series, I’d like to share what I found. Naturally, the sources that I found have varying levels of complexity, which is appropriate for students who are first learning prepositional and predicate logic.

When I actually presented these in class, I either presented the logical statement and had my class guess the statement in actual English, or I gave my students the famous quote and them translate it into predicate logic. However, for the purposes of this series, I’ll just present the statement in predicate logic first.

Predicate Logic and Popular Culture (Part 122): Queen

Let $p$ be the proposition “I cross a million rivers,” let $q$ be the proposition “I rode a million miles,” and let $r$ be the proposition “I still am where I started.” Translate the logical statement

$(p \land q) \Rightarrow r$ .

This matches a line from this classic by Queen.

Predicate Logic and Popular Culture (Part 121): OneRepublic

Let $F(x)$ be the proposition “ $x$ is a right friend,” let $P(y)$ be the proposition “ $y$ is a right place,” let $I(x,y)$ be the proposition “ $x$ is located at place $y$ ,” and let $H(x,y)$ be the proposition “They have $x$ at place $y$ ,” and let $p$ be the proposition “We’re going down.” Translate the logical statement

$\forall x \forall y(F(x) \land P(y) \land I(x,y) \Rightarrow H(x,y)) \land p$ .

This matches the chorus of this song by OneRepublic.

Predicate Logic and Popular Culture (Part 120): Crossfade

Let $C(t)$ be the proposition “At time $t$ , I meant to be so cold.” Translate the logical statement

$\forall t < 0 \lnot C(t)$ .

This matches the echo of this song by Crossfade.

Predicate Logic and Popular Culture (Part 119): Billy Joel

Let $p$ be the proposition “I’m gonna try for an uptown girl,” let $B(x)$ the proposition “ $x$ has hot blood,” let $q$ be the proposition “She’s looking for a downtown man,” and let $r$ be the proposition “I’m a downtown man.” Also, define the function $f(x)$ to be how long $x$ has lived in a white bread world. Translate the logical statement

$p \land \forall x (B(x) \Rightarrow (f(x) \le f(\hbox{she})) \land q \land r$ .