Engaging students: Using Pascal’s triangle

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Daniel Herfeldt. His topic, from Precalculus: using Pascal’s triangle.

A great activity for Pascal’s triangle would be to first have the students find a pattern of odds and evens. The first thing that you would do is to print out blank Pascal’s triangle. You would give each student a paper for them to fill out. They would have to first fill out the triangle themselves. This would give them practice on which numbers to add as well as further see a pattern of what the next one would potentially look like. After they finish, they would have to color in all of the odd numbers a certain color, and followed by coloring all of the even ones a different color. From here, they will see that once you color it is, the even numbers will make an upside down triangle. Next to the biggest triangles, you will see smaller triangles. An example is shown below. When the students have finished, you will show them why it is like that. Then explain what the name of the colored triangle is, which is called the Sierpinski Triangle.

Pascal’s Triangle is used all over mathematics. It is mainly recognized as how to find the coefficients of binomials, as well as a lot of other uses for binomials. What students and many other people do not know, is that this triangle can be used for much more. For example, you are able to use Pascal’s triangle to find the Fibonacci sequence. Although it may be a little harder to find than the coefficients of binomials, it is still possible. If you add up the numbers in a diagonal pattern from right to left, you will be able to find the Fibonacci sequence. Below will be a picture of how this is implemented. Another way that this will help in future courses is that it allows you to find squares of a number easily. If you look at the 3^rd diagonal row, adding two consecutive numbers from left to right will give the square of a number. A picture of this will also be posted below. Another way that this is implemented in future courses is statistics and probability. This triangle can be used to find the probability of many different things. This is only a few ways that the triangle can be used in future courses, considering that there are plenty of other ways it can be used. In all, this is a very important topic for someone that is pursuing mathematics.

Fibonacci sequence:

Triangular numbers:

This video would be a great way to either start a lesson on Pascal’s Triangle or to review the lesson before a test. The video shows different ways that you can implement the triangle to solve different things in mathematics. If this was the video to start the lesson, I would have each student take out a notebook and writing utensil while watching the video. Throughout the video the students would have to find at least three different ways a person may use Pascal’s triangle that they found particularly interesting. This should lead to most of the ways to be picked by at least one student. After they share their answers, explain further why these work. This could make students more intrigued with the subject. If the video was for a review of the topic, I would also have the students have out a writing utensil and a notebook. For this instance, I would have each individual write down what they had forgotten about Pascal’s triangle. From here the teacher will review the points that were most forgotten, serving as a review.

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.”