Category: Engagement

Engaging students in a different discipline

I have no expertise about how to teach any other subject besides mathematics. But this article from the May/June 2013 issue of the Stanford alumni magazine made a lot of sense to me about how to teach history to middle- and high-school students. The basic principle appears to be the same that governs my classes: figure out a way to make students want to come to class each day. A sample quote:

I easily could have told them in one minute that the Dust Bowl was the result of overgrazing and over-farming and World War I overproduction, combined with droughts that had been plaguing that area forever, but they wouldn’t remember it.” By reading these challenging documents and discovering history for themselves, he says, “not only will they remember the content, they’ll develop skills for life.

For history, the widespread implementation of this teaching philosophy has apparently been hindered by the lack of adequate teaching materials, which is also addressed in this article.

Dimensions

As described by the March 2013 issue of the American Mathematical Monthly, the (free!) two-hour movie Dimensions  is “an impressive computer-generated video of almost 2 hours that describes geometry in two, three and four dimensions. The video assumes an elementary geometry background possessed by most viewers and leads up to an interesting geometric structure, the Hopf fibration of the unit sphere in four-dimensional space.”

The website of this project can be found at http://www.dimensions-math.org.

Here’s the 4-minute trailer for the movie:

This full two-hour movie was uploaded to YouTube in several chapters. The full YouTube playlist is given here.

The links to the 9 separate chapters are below.

 

More on divisibility

Based on my students’ reactions, I gave my best math joke in years as I went over the proofs for checking that an integer was a multiple of 3 or a multiple of 9. I started by proving a lemma that 9 is always a factor of 10^k - 1. I asked my students how I’d write out 10^k - 1, and they correctly answered 99{\dots}9, a numeral with k consecutive 9s. So I said, “Who let the dogs out? Me. See: k nines.”

Some of my students laughed so hard that they cried.

There are actually at least three ways of proving this lemma. I love lemmas like these, as they offer a way of, in the words of my former professor Arnold Ross, to think deeply about simple things.

(1) By subtracting, 10^k - 1 = 99{\dots}9 = 9 \times 11{\dots}1, which is clearly a multiple of 9.

(2) We can use the rule

a^k - b^k = (a-b) \left(a^{k-1} + a^{k-2} b + \dots + a b^{k-2} + b^{k-1} \right)

The conclusion follows by letting a = 10 and b =1.

From my experience, my senior math majors all learned the rule for factoring the difference of two squares, but very few learned the rule for factoring the difference of two cubes, while almost none of them learned the general factorization rule above. As always, it’s not my students’ fault that they weren’t taught these things when they were younger.

I also supplement this proof with a challenge to connect Proof #2 with Proof #1… why does 11{\dots}1 = \left(a^{k-1} + a^{k-2} b + \dots + a b^{k-2} + b^{k-1} \right)?

(3) We can use mathematical induction.

If k = 0, then 10^k - 1 = 0, which is a multiple of 9.

We now assume that 10^k - 1 is a multiple of 9.

To show that 10^{k+1}-1 is a multiple of 9, we observe that

10^{k+1}-1 = \left(10^{k+1} - 10^k \right) + \left(10^k - 1\right) = 10^k (10-1) + \left(10^k - 1\right),

and both terms on the right-hand side are multiples of 9. (I also challenge my students to connect the right-hand side with the original expression 99{\dots}9.)

\hbox{QED}

Divisibility tricks

Based on personal experience, about half of our senior math majors never saw the basic divisibility rules (like adding the digits to check if a number is a multiple of 3 or 9) when they were children. I guess it’s also possible that some of them just forgot the rules, but I find that hard to believe since they’re so simple and math majors are likely to remember these kinds of tricks from grade school. Some of my math majors actually got visibly upset when I taught these rules in my Math 4050 class; they had been part of gifted and talented programs as children and would have really enjoyed learning these tricks when they were younger.

Of course, it’s not my students’ fault that they weren’t taught these tricks, and a major purpose of Math 4050 is addressing deficiencies in my students’ backgrounds so that they will be better prepared to become secondary math teachers in the future.

My guess that the divisibility rules aren’t widely taught any more because of the rise of calculators. When pre-algebra students are taught to factor large integers, it’s no longer necessary for them to pre-check if 3 is a factor to avoid unnecessary long division since the calculator makes it easy to do the division directly. Still, I think that grade-school students are missing out if they never learn these simple mathematical tricks… if for no other reason than to use these tricks to make factoring less dull and more engaging.

A mathematical magic trick

In case anyone’s wondering, here’s a magic trick that I did my class for future secondary math teachers while dressed as Carnac the Magnificent. I asked my students to pull out a piece of paper, a pen or pencil, and (if they wished) a calculator. Here were the instructions I gave them:

  1. Write down just about any number you want. Just make sure that the same digit repeated (not something like 88,888). You may want to choose something that can be typed into a calculator.
  2. Scramble the digits of your number, and write down the new number. Just be sure that any repeated digits appear the same number of times. (For example, if your first number was 1,232, your second number could be 2,231 or 1,322.)
  3. Subtract the smaller of the two numbers from the bigger, and write down the difference. Use a calculator if you wish.
  4. Pick any nonzero digit in the difference, and scratch it out.
  5. Add up the remaining digits (that weren’t scratched out).

I asked my students one at a time what they got after Step 5, and I responded, as the magician, with the number that they had scratched out. One student said 34, and I answered 2. Another said 24, and I answered 3. After doing this a couple more times, one student simply stated, “My mind is blown.”

This is actually a simple trick to perform, and the mathematics behind the trick is fairly straightforward to understand. Based on personal experience, this is a great trick to show children as young as 2nd or 3rd grade who have figured out multiple-digit subtraction and single-digit multiplication.

I offer the following thought bubble if you’d like to think about it before looking ahead to find the secret to this magic trick.

green_speech_bubbleWhat the magician does: the magician finds the next multiple of 9 greater than the volunteer’s number, and answers with the difference. For example, if the volunteer answers 25, the magician figures out that the next multiple of 9 after 25 is 27. So 27-25 = 2 was the digit that was scratched out.

This trick works because of two important mathematical facts.

(1) The difference D between the original number and the scrambled number is always a multiple of 9. For example, suppose the volunteer chooses 3417, and suppose the scrambled number is 7431. Then the difference is

7431 - 3417 = (7000 + 400 + 30 + 1) - (3000 + 400 + 10 + 7)

= (7000 - 7) + (400 - 400) + (30 - 3000) + (1 - 10)

= 7 \times (1000-1) + 4 \times (100-100) + 3 \times (10-1000) + 1 \times (1-10)

= 7 \times (999) + 1 \times (0) + 4 \times (-990) + 3 \times (-9)

Each of the numbers in parentheses is a multiple of 9, and so the difference D must also be a multiple of 9.

A more algebraic proof of (1) is set apart in the block quote below; feel free to skip it if the above numerical example is convincing enough.

More formally, suppose that the original number is a_n a_{n-1} \dots a_1a_0 in base-10 notation, and suppose the scrambled number is a_{\sigma(n)} a_{\sigma(n-1)} \dots a_{\sigma(1)} a_{\sigma(0)}, where \sigma is a permutation of the numbers \{0, 1, \dots, n\}. Without loss of generality, suppose that the original number is larger. Then the difference D is equal to

D = a_n a_{n-1} \dots a_1a_0 - a_{\sigma(n)} a_{\sigma(n-1)} \dots a_{\sigma(1)} a_{\sigma(0)}

D = \displaystyle \sum_{i=0}^n a_i 10^i - \sum_{i=0}^n a_{\sigma(i)} 10^i

D = \displaystyle \sum_{i=0}^n a_{\sigma(i)} 10^{\sigma(i)} - \sum_{i=0}^n a_{\sigma(i)} 10^i

D = \displaystyle \sum_{i=0}^n a_{\sigma(i)} \left(10^{\sigma(i)} - 10^i \right)

The transition from the second to the third line work because the terms of the first sum are merely rearranged by the permutation \sigma.

To show that D is a multiple of 9, it suffices to show that each term 10^{\sigma(i)} - 10^i is a multiple of 9.

  • If \sigma(i) > i, then 10^{\sigma(i)} - 10^i = 10^i \left( 10^{\sigma(i) - i} - 1 \right), and the term in parentheses is guaranteed to be a multiple of 9.
  • If \sigma(i) < i, then 10^{\sigma(i)} - 10^i = 10^{\sigma(i)} \left( 1-10^{i-\sigma(i)} \right) = -10^{\sigma(i)} \left( 10^{i-\sigma(i)} - 1 \right), and the term in parentheses is guaranteed to be a (negative) multiple of 9.
  • If \sigma(i) = i, then 10^{\sigma(i)} - 10^i = 0, a multiple of 9.

\hbox{QED}

Because the difference D is a multiple of 9, we use the important fact (2) that a number is a multiple of 9 exactly when the sum of its digits is a multiple of 9. Therefore, when the volunteer offers the sum of all but one of the digits of D, the missing digit is found by determining the nonzero number that has to be added to get the next multiple of 9. (Notice that the trick specifies that the volunteer scratch out a nonzero digit. Otherwise, there would be an ambiguity if the volunteer answered with a multiple of 9; the missing digit could be either 0 or 9.)

As I mentioned earlier, I showed this trick (and the proof of why it works) to a class of senior math majors who are about to become secondary math teachers. I think it’s a terrific and engaging way of deepening their content knowledge (in this case, base-10 arithmetic and the rule of checking that a number is a multiple of 9.)

As thanks for reading this far, here’s a photo of me dressed as Carnac as I performed the magic trick. Sadly, most of the senior math majors of 2013 were in diapers when Johnny Carson signed off the Tonight Show in 1992, so they didn’t immediately get the cultural reference.

542597_10200255338162345_209138628_n

Measuring terminal velocity

Using a simultaneously falling softball as a stopwatch, the terminal velocity of a whiffle ball can be obtained to surprisingly high accuracy with only common household equipment. In the January 2013 issue of College Mathematics Monthly, we describe an classroom activity that engages students in this apparently daunting task that nevertheless is tractable, using a simple model and mathematical techniques at their disposal.

Hollywood Hates Math

Dan Meyer spliced together scenes from various movies where knowledge of mathematics is denigrated. Since a big part of my job is instilling confidence in my students that they can indeed succeed in my classes, it’s a little depressing to see that I have a big opponent in popular culture.

This video has the occasional PG language and innuendo, while I prefer to keep my classes rated G to every extent possible. Some time ago, Dan was kind enough to post the original movie sources for this clip, and someday I might edit down this clip to something that I would be comfortable showing in class.