Engaging students: Solving one-step and two-step inequalities

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 first student submission comes from my former student Jesse Faltys (who, by the way, was the instigator for me starting this blog in the first place). Her topic: how to engage students when teaching one-step and two-step inequalities.

A. Applications – How could you as a teacher create an activity or project that involves your topic?

Index Card Game: Make two sets of cards. The first should consist of different inequalities. The second should consist of the matching graph. Put your students in pairs and distribute both sets of cards. The students will then practice solving their inequalities and determine which graph illustrates which inequality.
Inequality Friends: Distribute index cards with simple inequalities to a handful of your students (four or five different inequalities) and to the rest of the students pass of cards that only contain numbers. Have your students rotate around the room and determine if their numbers and inequalities are compatible or not. If they know that their number belongs with that inequality then the students should become “members” and form a group. Once all the students have formed their groups, they should present to the class how they solved their inequality and why all their numbers are “members” of that group.

Both applications allow for a quick assessment by the teacher. Having the students initially work in pairs to explore the inequality and its matching graph allows for discover on their own. While ending the class with a group activity allows the teacher to make individual assessments on each student.

B. Curriculum: How does this topic extend what your students should have learned in previous courses?

In a previous course, students learned to solve one- and two-step linear equations. The process for solving one-step equality is similar to the process of solving a one-step inequality. Properties of Inequalities are used to isolate the variable on one side of the inequality. These properties are listed below. The students should have knowledge of these from the previous course; therefore not overwhelmed with new rules.

Properties of Inequality

1. When you add or subtract the same number from each side of an inequality, the inequality remains true. (Same as previous knowledge with solving one-step equations)

2. When you multiply or divide each side of an inequality by a positive number, the inequality remains true. (Same as previous knowledge with solving one-step equations)

3. When you multiply or divide each side of an inequality by a negative number, the direction of the inequality symbol must be reversed for the inequality to remain true. (THIS IS DIFFERENT)

There is one obvious difference when working with inequalities and multiply/dividing by a negative number there is a change in the inequality symbol. By pointing out to the student, that they are using what they already know with just one adjustment to the rules could help ease their mind on a new subject matter.

C. Culture – How has this topic appeared in pop culture?

Amusement Parks – If you have ever been to an amusement park, you are familiar with the height requirements on many of the rides. The provide chart below shows the rides at Disney that require 35 inches or taller to be able to ride. What rides will you ride?

(Height of Student $\ge$ Height restriction)

Blizzard Beach	Summit Plummet	48″
Magic Kingdom	Barnstormer at Goofy’s Wiseacres Farm	35″
Animal Kingdom	Primeval Whirl	48″
Blizzard Beach	Downhill Double Dipper	48″
DisneyQuest	Mighty Ducks Pinball Slam	48″
Typhoon Lagoon	Bay Slide	52″
Animal Kingdom	Kali River Rapids	38″
DisneyQuest	Buzz Lightyear’s AstroBlaster	51″
DisneyQuest	Cyberspace Mountain	51″
Epcot	Test Track	40″
Epcot	Soarin’	40″
Hollywood Studios	Star Tours: The Adventures Continue	40″
Magic Kingdom	Space Mountain	44″
Magic Kingdom	Stitch’s Great Escape	40″
Typhoon Lagoon	Humunga Kowabunga	48″
Animal Kingdom	Expedition Everest	44″
Blizzard Beach	Cross Country Creek	48″
Epcot	Mission Space	44″
Hollywood Studios	The Twilight Zone Tower of Terror	40″
Hollywood Studios	Rock ‘n’ Roller Coaster Starring Aerosmith	48″
Magic Kingdom	Splash Mountain	40″
Magic Kingdom	Big Thunder Mountain Railroad	40″
Animal Kingdom	Dinosaur	40″
Epcot	Wonders of Life / Body Wars	40″
Blizzard Beach	Summit Plummet	48″
Magic Kingdom	Barnstormer at Goofy’s Wiseacres Farm	35″
Animal Kingdom	Primeval Whirl	48″
Blizzard Beach	Downhill Double Dipper	48″
DisneyQuest	Mighty Ducks Pinball Slam	48″
Typhoon Lagoon	Bay Slide	52″

Sports – Zdeno Chara is the tallest person who has ever played in the NHL. He is 206 cm tall and is allowed to use a stick that is longer than the NHL’s maximum allowable length. The official rulebook of the NHL state limits for the equipment players can use. One of these rules states that no hockey stick can exceed160 cm. (Hockey stick $\le$ 160 cm) The world’s largest hockey stick and puck are in Duncan, British Columbia. The stick is over 62 m in length and weighs almost 28,000 kg. Is your equipment legal?

Weather – Every time the news is on our culture references inequalities by the range in the temperature throughout the day. For example, the most extreme change in temperature in Canada took place in January 1962 in Pincher Creek, Alberta. A warm, dry wind, known as a chinook, raised the temperature from -19 °C to 22 °C in one hour. Represent the temperature during this hour using a double inequality. (-19 < the temperature < 22) What Inequality is today from the weather in 1962?

Geometric magic trick

This is a magic trick that my math teacher taught me when I was about 13 or 14. I’ve found that it’s a big hit when performed for grade-school children.

Magician: Tell me a number between 3 and 10.

Child: (gives a number, call it $x$ )

Magician: On a piece of paper, draw a shape with $x$ corners.

Child: (draws a figure; an example for $x=6$ is shown)

Important Note: For this trick to work, the original shape has to be convex… something shaped like an L or M won’t work. Also, I chose a maximum of 10 mostly for ease of drawing and counting (and, for later, calculating).

Magician: Tell me another number between 3 and 10.

Child: (gives a number, call it $y$ )

Magician: Now draw that many dots inside of your shape.

Child: (starts drawing $y$ dots inside the figure; an example for $y = 7$ ) While the child does this, the Magician calculates $2y + x - 2$ , writes the answer on a piece of paper, and turns the answer face down.

Magician: Now connect the dots with lines until you get all triangles. Just be sure that no two lines cross each other.

Child: (connects the dots until the shape is divided into triangles; an example is shown)

Magician: Now count the number of triangles.

Child: (counts the triangles)

Magician: Was your answer… (and turns the answer over)?

The reason this magic trick works so well is that it’s so counter-intuitive. No matter what convex $x-$ gon is drawn, no matter where the $y$ points are located, and no matter how lines are drawn to create triangles, there will always be $2y + x - 2$ triangles. For the example above, $2y+x-2 = 2\times 7 + 6 - 2 = 18$ , and there are indeed $18$ triangles in the figure.

Why does this magic trick work? I offer a thought bubble if you’d like to think about it before scrolling down to see the answer.

This trick works by counting the measures of all the angles in two different ways.

Method #1: If there are $T$ triangles created, then the sum of the measures of the angles in each triangle is $180$ degrees. So the sum of the measures of all of the angles must be $180 T$ degrees.

Method #2: The sum of the measures of the angles around each interior point is $360$ degrees. Since there are $y$ interior points, the sum of these angles is $360y$ degrees.

The measures of the remaining angles add up to the sum of the measures of the interior angles of a convex polygon with $x$ sides. So the sum of these measures is $180(x-2)$ degrees.

In other words, it must be the case that

$180T = 360y + 180(x-2)$ , or $T = 2y + x - 2$ .

Statistical significance

When teaching my Applied Statistics class, I’ll often use the following xkcd comic to reinforce the meaning of statistical significance.

The idea that’s being communicated is that, when performing an hypothesis test, the observed significance level $P$ is the probability that the null hypothesis is correct due to dumb luck as opposed to a real effect (the alternative hypothesis). So if the significance level is really about $0.05$ and the experiment is repeated about 20 times, it wouldn’t be surprising for one of those experiments to falsely reject the null hypothesis.

In practice, statisticians use the Bonferroni correction when performing multiple simultaneous tests to avoid the erroneous conclusion displayed in the comic.

Source: http://www.xkcd.com/882/

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.

A cute mathematical challenge

Show each number from 0 to 25 as a combination of four $4$ s. You can do any operation (multiply, divide, exponent, etc.) you want in any combination, but you must use all four $4$ s and you cannot use any other numbers.

For example: $16 = 4+4+4+4$ (using all four $4$ s and addition)

Fourier transform

Source: http://www.smbc-comics.com/index.php?db=comics&id=2874#comic

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 $9$ s. 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:

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.
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.)
Subtract the smaller of the two numbers from the bigger, and write down the difference. Use a calculator if you wish.
Pick any nonzero digit in the difference, and scratch it out.
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.

What 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.