Proving theorems and special cases (Part 12): The sum and difference formulas for sine

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?

Usually, the answer is no. In this series of posts, we’ve seen that a conjecture could be true for the first 40 cases or even the first $10^{316}$ cases yet not always be true. We’ve also explored the computational evidence for various unsolved problems in mathematics, noting that even this very strong computational evidence, by itself, does not provide a proof for all possible cases.

However, there are plenty of examples in mathematics where it is possible to prove a theorem by first proving a special case of the theorem. For the remainder of this series, I’d like to list, in no particular order, some common theorems used in secondary mathematics which are typically proved by first proving a special case.

3. Theorem 1. $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha + \sin \beta$

Theorem 2. $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$

For angles that are not acute, these theorems can be proven using a unit circle and the following four lemmas:

Lemma 1. $\cos(x - y) = \cos x \cos y + \sin x \sin y$

Lemma 2. $\cos(x + y) = \cos x \cos y - \sin x \sin y$

Lemma 3. $\sin(\pi/2 - x) = \cos x$

Lemma 4. $\cos(\pi/2 - x) = \sin x$

Specifically, assuming Lemmas 1-4, then:

$\sin(\alpha + \beta) = \cos(\pi/2 - [\alpha + \beta])$ by Lemma 4

$= \cos([\pi/2 - \alpha] - \beta)$

$= \cos(\pi/2 - \alpha) \cos \beta + \sin(\pi/2 - \alpha) \sin \beta$ by Lemma 1

$= \sin \alpha \cos \beta + \cos \alpha \sin \beta$ by Lemmas 3 and 4.

Also,

$\sin(\alpha - \beta) = \cos(\pi/2 - [\alpha - \beta])$ by Lemma 4

$= \cos([\pi/2 - \alpha] + \beta)$

$= \cos(\pi/2 - \alpha) \cos \beta - \sin(\pi/2 - \alpha) \sin \beta$ by Lemma 2

$= \sin \alpha \cos \beta - \cos \alpha \sin \beta$ by Lemmas 3 and 4.

However, we see that what I’ve called Lemma 3, often called a cofunction identity, can be considered a special case of Theorem 2. However, this is not circular logic since the cofunction identities can be proven without appealing to Theorems 1 and 2.

Proving theorems and special cases (Part 11): The Law of Cosines

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?

2. Theorem. In $\triangle ABC$ where $a = BC$ , $b = AC$ , and $c = AB$ , we have $c^2 = a^2 + b^2 - 2 a b \cos (m \angle C)$ .

This is typically proven using the Pythagorean theorem:

Lemma. In right triangle $\triangle ABC$ , where $\angle C$ is a right angle, we have $c^2 = a^2 + b^2$ .

Though it usually isn’t thought of this way, the Pythagorean theorem is a special case of the Law of Cosines since $\cos 90^\circ = 0$ .

There are well over 100 different proofs of the Pythagorean theorem that do not presuppose the Law of Cosines. The standard proof of the Law of Cosines then uses the Pythagorean theorem. In other words, a special case of the Law of Cosines is used to prove the Law of Cosines.

Proving theorems and special cases (Part 5): Mathematical induction

Today’s post is a little bit off the main topic of this series of posts… but I wanted to give some pedagogical thoughts on yesterday’s post concerning the following proof by induction.

Theorem: If $n \ge 1$ is a positive integer, then $5^n - 1$ is a multiple of 4.

Proof. By induction on $n$ .

$n = 1$ : $5^1 - 1 = 4$ , which is clearly a multiple of 4.

$n$ : Assume that $5^n - 1$ is a multiple of 4, so that $5^n - 1 = 4q$ , where $q$ is an integer. We can also write this as $5^n = 4q + 1$ .

$n+1$ . We wish to show that $5^{n+1} - 1$ is equal to $4Q$ for some (different) integer $Q$ . To do this, notice that

$5^{n+1} - 1 = 5^1 5^n - 1$

$= 5 \times 5^n - 1$

$= 5 \times (4q + 1) - 1$ by the induction hypothesis

$= 20q + 5 - 1$

$= 20q + 4$

$= 4(5q + 1)$ .

So if we let $Q = 5q +1$ , then $5^{n+1} - 1 = 4Q$ , where $Q$ is an integer because $q$ is also an integer.

My primary observation is that even very strong math students tend to have a weak spot when it comes to simplifying exponential expressions (as opposed to polynomial expressions). For example, I find that even very good math students can struggle through the logic of this sequence of equalities:

$2^n + 2^n = 2 \times 2^n = 2^1 \times 2^n = 2^{n+1}$ .

The first step is using the main stumbling block. Students who are completely comfortable with simplifying $x + x$ as $2x$ can be perplexed by simplifying $2^n + 2^n$ as $2 \times 2^n$ . I attribute this to lack of practice with this kind of simplification in lower grade levels.

Here’s another algoebraic stumbling block that I’ve often seen: at the beginning of the $n+1$ case, some students will make the following mistake:

$5^{n+1} - 1 = 5^1 5^n - 1 = 5 (4q) = 4 (5q) = 4Q$ .

Because these students end with a multiple of 4, they fail to notice that the second equality is incorrect since

$5^1 5^n - 1 \ne 5^1 (5^n - 1)$ .

Again, I attribute this to lack of practice with simplifying exponential expressions in lower grade levels… as well as being a little bit over-excited upon seeing $5^n - 1$ and wishing to use the induction hypothesis as soon as possible.

Proving theorems and special cases (Part 4): Mathematical induction

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?

Usually, the answer is no… even checking many special cases of a conjecture does not mean that the conjecture is correct. In the previous two posts, we saw that a statement that’s true for the first 40 cases or even the first $10^{316}$ cases may not be true for all cases.

This is the reason that mathematical induction is important, as it provides a way to build from previous cases to prove that the next case is still correct, thus proving that all cases are correct.

Theorem: If $n \ge 1$ is a positive integer, then $5^n - 1$ is a multiple of 4.

Proof. By induction on $n$ .

$n = 1$ : $5^1 - 1 = 4$ , which is clearly a multiple of 4.

$n$ : Assume that $5^n - 1$ is a multiple of 4, so that $5^n - 1 = 4q$ , where $q$ is an integer. We can also write this as $5^n = 4q + 1$ .

$n+1$ . We wish to show that $5^{n+1} - 1$ is equal to $4Q$ for some (different) integer $Q$ . To do this, notice that

$5^{n+1} - 1 = 5^n 5^1 - 1$

$= 5 \times 5^n - 1$

$= 5 \times (4q + 1) - 1$ by the induction hypothesis

$= 20q + 5 - 1$

$= 20q + 4$

$= 4(5q + 1)$ .

So if we let $Q = 5q +1$ , then $5^{n+1} - 1 = 4Q$ , where $Q$ is an integer because $q$ is also an integer.

QED

In the above proof, we were able to build from the $n$ case to reach the $n +1$ case. In this sense, to answer the student’s question, it is possible to prove a theorem by first proving a special case of the theorem.

By contrast, when trying to “prove” that $n^2 - n + 41$ is prime for all integers $n$ , the proposition is true for $1 \le n \le 40$ , but it’s just a coincidence… there was no string of logic that connected these first 40 cases other than the coincidence that they all were correct.

Proving theorems and special cases (Part 3): Skewes’ number

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?

Usually, the answer is no… even checking many special cases of a conjecture does not mean that the conjecture is correct.

In the first two posts of this series, we showed conjectures could be true for the first 40 cases or even the first 900 million odd cases but fail on the next case. In today’s post, I’ll describe a conjecture that, for hundreds of years, was thought to be true for all integers $n$ and has since been shown to be true for all $n \le 10^{316}$ . However, despite being true for so many special cases, the conjecture is false.

Let’s absorb the above paragraph again. The conjecture “ $n^2 - n + 41$ is always prime” was true for the first 40 cases before failing. By contrast, the conjecture I’m about the describe is true for the first (approximately) 10,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,
000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,
000,000,000,000,000,000,000,000,000 cases before failing.

The conjecture in question concerns the number of prime numbers $\pi(n)$ that are less than or equal to $n$ . For example:

There are four prime numbers less than 10 (namely 2, 3, 5, 7), and so $\pi(10) = 4$ .

There are eight prime numbers less than 10 (namely 2, 3, 5, 7, 11, 13, 17, 19), and so $\pi(20) = 8$ .

One of the classic problems in mathematics, often called the Prime Number Theorem, is estimating the value of $\pi(n)$ when $n$ is large. It turns out that one way of approximating $\pi(n)$ is through the logarithmic integral

$\hbox{Li}(n) = \displaystyle \int_2^n \frac{dx}{\ln x}$

Indeed, the Prime Number Theorem states that

$\pi(n) \sim \hbox{Li}(n)$ ,

$\displaystyle \lim_{n \to \infty} \frac{\pi(n)}{\hbox{Li}(n)} = 1$ .

This asymptotic relationship was proven by the eminent mathematician Karl Friedrich Gauss.

With that as prelude, here’s the false conjecture. For decades, luminaries of mathematics, including Gauss and Bernhard Riemann, conjectured that

$\pi(n) < \hbox{Li}(n)$ for all integers $n$

based on the available numerical evidence at the time. However, this conjecture was proven false in the 20th century by the British mathematician John Littlewood. No only did Littlewood show that there is a value of $n$ so that $\pi(n) > \hbox{Li}(n)$ , but he also showed that the graphs of $\pi(n)$ and $\hbox{Li}(n)$ cross over each other infinitely many times!

Littlewood proved his theorem over 100 years ago in 1914. Today, even with modern computing power, we still do not precisely know the first value of $n$ for which $\pi(n) > \hbox{Li}(n)$ . This number is called Skewes’ number, in honor of the mathematician who first found (in 1955) an upper bound on this first crossing point. The bound that he found was absolutely enormous: he showed that $\pi(n) > \hbox{Li}(n)$ for some $n$ less than

$\displaystyle 10^{\displaystyle 10^{10^{34}}}$

More recent work has established that the first crossover likely occurs in the vicinity of $n \approx 1.397 \times 10^{316}$ .

Whoever first evaluates Skewes’ number exactly will surely have a nice feather in his/her cap for completing a task that mystified both Gauss and Riemann.

References:

http://mathworld.wolfram.com/PrimeNumberTheorem.html

http://mathworld.wolfram.com/SkewesNumber.html

http://mathworld.wolfram.com/PrimeCountingFunction.html

http://mathworld.wolfram.com/LogarithmicIntegral.html

Proving theorems and special cases (Part 2): The Pólya conjecture

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?

Usually, the answer is no… even checking many special cases of a conjecture does not mean that the conjecture is correct.

In yesterday’s post, we showed that the conjecture “ $n^2 - n + 41$ is a prime number for all positive integers $n$ ” is true for $1 \le n \le 40$ but fails for $n = 41$ . In today’s post, I’ll describe a conjecture that is true for plenty more special cases before becoming false.

The Pólya conjecture (see here and here for more information) stated that 50% or more of the natural numbers less than or equal to any given number have an odd number of prime factors (counting multiplicity). For example:

$2$ has one prime factor: odd

$3$ has one prime factor: odd

$4 = 2 \times 2$ has two prime factors: even

$5$ has one prime factor: odd

$6 = 2 \times 3$ has two prime factors: even

$7$ has one prime factor: odd

$8 = 2 \times 2 \times 2$ has three prime factors: odd

$9 = 3 \times 3$ has two prime factors: even

$10 = 2 \times 5$ has two prime factors: even

So of the numbers less than or equal to 10, five have an odd number of prime factors, and only four have an even number of prime factors.

The Pólya conjecture was first proven false by producing a counterexample in the vicinity of $1.845 \times 10^{361}$ . It turns out that the smallest counterexample is 906,150,257. In other words, the Pólya conjecture is true for the first 906,150,256 cases but fails on the next case.

Proving theorems and special cases (Part 1): Is n^2-n+41 always prime?

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?

Usually, the answer is no… even checking many special cases of a conjecture does not mean that the conjecture is correct.

The following example probably appears in every textbook that I’ve seen that handles mathematical induction to convince students that checking even many special cases of a conjecture is not sufficient for proving the conjecture.

Conjecture: If $n \ge 1$ is a positive integer, then $n^2 - n + 41$ is a prime number.

Is this true? Well, let’s start checking:

If $n = 1$ , then $n^2 - n + 41 = 1^2 - 1 + 41 = 41$ , which is a prime number.

If $n = 2$ , then $n^2 - n + 41 = 2^2 - 2 + 41 = 43$ , which is a prime number.

If $n = 3$ , then $n^2 - n + 41 = 3^2 - 3 + 41 = 47$ , which is a prime number.

If $n = 4$ , then $n^2 - n + 41 = 4^2 - 4 + 41 = 53$ , which is a prime number.

If $n = 5$ , then $n^2 - n + 41 = 5^2 - 5 + 41 = 61$ , which is a prime number.

If $n = 6$ , then $n^2 - n + 41 = 6^2 - 6 + 41 = 71$ , which is a prime number.

If $n = 7$ , then $n^2 - n + 41 = 7^2 - 7 + 41 = 83$ , which is a prime number.

If $n = 8$ , then $n^2 - n + 41 = 8^2 - 8 + 41 = 97$ , which is a prime number.

If $n = 9$ , then $n^2 - n + 41 = 9^2 - 9 + 41 = 113$ , which is a prime number.

If $n = 10$ , then $n^2 - n + 41 = 10^2 - 10 + 41 = 131$ , which is a prime number.

If $n = 11$ , then $n^2 - n + 41 = 11^2 - 11 + 41 = 151$ , which is a prime number.

If $n = 12$ , then $n^2 - n + 41 = 12^2 - 12 + 41 = 173$ , which is a prime number.

If $n = 13$ , then $n^2 - n + 41 = 13^2 - 13 + 41 = 197$ , which is a prime number.

If $n = 14$ , then $n^2 - n + 41 = 14^2 - 14 + 41 = 223$ , which is a prime number.

If $n = 15$ , then $n^2 - n + 41 = 15^2 - 15 + 41 = 251$ , which is a prime number.

If $n = 16$ , then $n^2 - n + 41 = 16^2 - 16 + 41 = 281$ , which is a prime number.

If $n = 17$ , then $n^2 - n + 41 = 17^2 - 17 + 41 = 313$ , which is a prime number.

If $n = 18$ , then $n^2 - n + 41 = 18^2 - 18 + 41 = 347$ , which is a prime number.

If $n = 19$ , then $n^2 - n + 41 = 19^2 - 19 + 41 = 383$ , which is a prime number.

If $n = 20$ , then $n^2 - n + 41 = 20^2 - 20 + 41 = 421$ , which is a prime number.

If $n = 21$ , then $n^2 - n + 41 = 21^2 - 21 + 41 = 461$ , which is a prime number.

If $n = 22$ , then $n^2 - n + 41 = 22^2 - 22 + 41 = 503$ , which is a prime number.

If $n = 23$ , then $n^2 - n + 41 = 23^2 - 23 + 41 = 547$ , which is a prime number.

If $n = 24$ , then $n^2 - n + 41 = 24^2 - 24 + 41 = 593$ , which is a prime number.

If $n = 25$ , then $n^2 - n + 41 = 25^2 - 25 + 41 = 641$ , which is a prime number.

If $n = 26$ , then $n^2 - n + 41 = 26^2 - 26 + 41 = 691$ , which is a prime number.

If $n = 27$ , then $n^2 - n + 41 = 27^2 - 27 + 41 = 743$ , which is a prime number.

If $n = 28$ , then $n^2 - n + 41 = 28^2 - 28 + 41 = 797$ , which is a prime number.

If $n = 29$ , then $n^2 - n + 41 = 29^2 - 29 + 41 = 853$ , which is a prime number.

If $n = 30$ , then $n^2 - n + 41 = 30^2 - 30 + 41 = 911$ , which is a prime number.

If $n = 31$ , then $n^2 - n + 41 = 31^2 - 31 + 41 = 971$ , which is a prime number.

If $n = 32$ , then $n^2 - n + 41 = 32^2 - 32 + 41 = 1033$ , which is a prime number.

If $n = 33$ , then $n^2 - n + 41 = 33^2 - 33 + 41 = 1097$ , which is a prime number.

If $n = 34$ , then $n^2 - n + 41 = 34^2 - 34 + 41 = 1163$ , which is a prime number.

If $n = 35$ , then $n^2 - n + 41 = 35^2 - 35 + 41 = 1231$ , which is a prime number.

If $n = 36$ , then $n^2 - n + 41 = 36^2 - 36 + 41 = 1301$ , which is a prime number.

If $n = 37$ , then $n^2 - n + 41 = 37^2 - 37 + 41 = 1373$ , which is a prime number.

If $n = 38$ , then $n^2 - n + 41 = 38^2 - 38 + 41 = 1447$ , which is a prime number.

If $n = 39$ , then $n^2 - n + 41 = 39^2 - 39 + 41 = 1523$ , which is a prime number.

If $n = 40$ , then $n^2 - n + 41 = 40^2 - 40 + 41 = 1601$ , which is a prime number.

Okay, a show of hands… did anyone actually carefully read and check the above 40 lines? I didn’t think so. The point is that the proposition works for $n = 1, 2, 3, \dots, 40$ . By about $n = 4$ , or so, a student (who didn’t already know the answer) actually did the above work would begin thinking, “Wow, this probably is correct for any value of $n$ .”

Of course, the catch happens at $n = 41$ :

If latex n = 41, then $n^2 - n + 41 = 41^2 - 41 + 41 = 41^2 = 41 \times 41$ ,

which is not a prime number.

All this to say, seeing a trend for the first few special cases… or the first few dozen special cases… does not necessarily mean that the trend will continue.

Engaging students: Graphing parabolas

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 again comes from my former student Tiffany Wilhoit. Her topic, from Algebra: graphing parabolas.

How did people’s conception of this topic change over time?

The parabola has been around for a long time! Menaechmus (380 BC-320 BC) was likely the first person to have found the parabola. Therefore, the parabola has been around since the ancient Greek times. However, it wasn’t until around a century later that Apollonius gave the parabola its name. Pappus (290-350) is the mathematician who discovered the focus and directrix of the parabola, and their given relation. One of the most famous mathematicians to contribute to the study of parabolas was Galileo. He determined that objects falling due to gravity fall in parabolic pathways, since gravity has a constant acceleration. Later, in the 17^th century, many mathematicians studied properties of the parabola. Gregory and Newton discovered that parabolas cause rays of light to meet at a focus. While Newton opted out of using parabolic mirrors for his first telescope, most modern reflecting telescopes use them. Mathematicians have been studying parabolas for thousands of years, and have discovered many interesting properties of the parabola.

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

A fun activity to set up for your students will include several boxes and balls, for a smaller set up, you can use solo cups and ping pong balls. Divide the class into groups, and give each group a set of boxes and balls. First, have the students set up a tower(s) with the boxes. The students will now attempt to knock the boxes down using the balls. The students can map out the parabolic curve showing the path they want to take. By changing the distance from the student throwing the ball and the boxes, the students will be able to see how the curve changes. If students have the tendency to throw the ball straight instead of in the shape of a parabola, have a member of the group stand between the thrower and the boxes. This will force the ball to be thrown over the student’s head, resulting in the parabolic curve. The students can also see what happens to the curve depending on where the student stands between the thrower and boxes. In order for the students to make a positive parabolic curve, have them throw the ball underhanded. This activity will engage the students by getting them involved and active, plus they will have some fun too! (To start off with, you can show the video from part E1, since the students are playing a real life version of Angry Birds!)

How can technology be used to effectively engage students with this topic?

A great video to show students before studying parabolas can be found on YouTube:

The video uses the popular game Angry Birds to introduce parabolic graphs. First, the video shows the bird flying a parabolic path, but the bird misses the pig. The video goes on to explain why the pig can’t be hit. It does a good job of explaining what a parabola is, why the first parabolic curve would not allow the bird to hit the pig, and how to change the curve to line up the path of the bird to the pig. This video would be interesting to the students, because a majority of the class (if not all) will know the game, and most have played the game! The video goes even further by encouraging students to look for parabolas in their lives. It even gives other examples such as arches and basketball. This will get the students thinking about parabolas outside of the classroom. (This video would be perfect to show before the students try their own version of Angry Birds discussed in part A2)

Resources:

Youtube.com/watch?v=bsYLPIXI7VQ

Parabolaonline.tripod.com/history.html

http://www-history.mcs.st-and.ac.uk/Curves/Parabola.html

Engaging students: Slope-intercept form of a line

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

This student submission again comes from my former student Kelley Nguyen. Her topic, from Algebra: slope-intercept form of a line.

How has this topic appeared in high culture (art, classical music, theatre, etc.)?

The slope-intercept form of a line is a linear function. Linear functions are dealt with in many ways in everyday life, some of which you probably don’t even notice.

One example where the slope-intercept form of a line appears in high culture is through music and arts. Suppose a band wants to book an auditorium for their upcoming concert. As most bands do, they meet with the manager of the location, book a date, and determine a payment. Let’s say it costs $1,500 to rent the building for 2 hours. In addition to this fee, the band earns 20% of each $30 ticket sold. Write an equation that determines whether the band made profit or lost money due to the number of tickets sold – the equation would be y = 0.2(30)x – 1500, where y is the amount gained or lost and x is the number of tickets sold that night. This can also help the band determine their goal on how many tickets to sell. If they want to make a profit of $2,000, they would have to sell x-many tickets to accomplish that.

In reality, most arts performances make a profit from their shows or concerts. Not only do mathematicians and scientists use slope-intercept of a line, but with this example, it shows up in many types of arts and real-world situations. Not only does the form work for calculating cost or profit, it can relate to the number of seats in a theatre, such as x rows of 30 seats and a VIP section of 20 seats. The equation to find how many seats are available in the theatre is y = 30x + 20, where x is the number of rows.

How can technology be used to effectively engage students with this topic?

A great way to engage students when learning about slope-intercept form of a line is to use Geometer’s Sketchpad. After opening a graph with an x- and y-axis, use the tools to create a line. From there, you can drag the line up or down and notice that the slope increases as you move upward and decreases as you move downward. Students can also find the equation of the line by selecting the line, clicking “Measure” in the menu bar, and selecting “Equation” in the drop-down list. This gives the students an accurate equation of the line they selected in slope-intercept form. Geometer’s Sketchpad allows students to experiment and explore directions of lines, determine whether or not it has an increasing slope, and help create a visual image for positive and negative slopes.

Also, with this program, students can play a matching game with slope-intercept equations and lines. You will instruct the student to create five random lines that move in any direction. Next, they will select all of the lines, go to “Measure” in the menu bar, and click “Equation.” From there, it’ll give them the equation of each line. Then, the student will go back and select the lines once again, go to “Edit” on the menu bar, hover over “Action Buttons,” and select “Hide/Show.” Once a box comes up, they will click the “Label” tab and type Scramble Lines in the text line. Next, the lines will scramble and stop when clicked on. Once the lines are done scrambling, the student could then match the equations with their lines. This activity gives the students the chance to look at equations and determine whether the slope is increasing and decreasing and where the line hits the y-axis.

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

With this topic, I could definitely do a project that consists of slope-intercept equations, their graphs, and word problems that involve computations. For example, growing up, some students had to earn money by doing chores around the house. Parents give allowance on daily duties that their children did.

The project will give the daily amount of allowance that each student earned. With that, say the student needed to reach a certain amount of money before purchasing the iPad Air. In part one of the project, the student will create an equation that reflects their daily allowing of $5 and the amount of money they have at the moment. In part two, the student will construct a graph that shows the rate of their earnings, supposing that they don’t skip a day of chores. In part three, the students will answer a series of questions, such as,

What will you earn after a week?
What is your total amount of money after that week?
When will you have enough money to buy that iPad Air at $540 after tax?

This would be a short project, but it’s definitely something that the students can do outside of class as a fun activity. It can also help them reach their goals of owning something they want and making a financial plan on how to accomplish that.

References

(5 Jun 2012). “The Geometer’s Sketchpad ® 5: Measuring Slopes and Equations”. YouTube. Retrieved 17 Sep 2014 from http://www.youtube.com/watch?v=nnkVdju5VyA
Lopez, Andrea. (10 Dec 2012). “Slope Game Tutorial”. YouTube. Retrieved 17 Sep 2014 from http://www.youtube.com/watch?v=hxGx-w6JEBg
(Unknown). “Solving Real World Problems”. Big Ideas Math. Retrieved 17 Sep 2014 from https://www.bigideasmath.com/protected/content/ipe_cc/grade%208/03/g8_03_04.pdf

Engaging students: Word problems involving inequalities

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

This student submission again comes from my former student Emily Bruce. Her topic, from Algebra: word problems involving inequalities.

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

Everyone learns in different ways. There are three common learning types, which are auditory, visual, and kinesthetic. The best activities and lesson plans involve all three of these learning styles. A great way to involve all of these learning styles is to use objects that students can rearrange and manipulate with their hands. When learning about inequality word problems, I would have print large numbers and symbols on pieces of paper that they could tape to a whiteboard. In groups, they would be able to rearrange their numbers and inequality symbols as they are working through a word problem, until the figure out the correct inequality. Then as a class, we could discuss their answers. This addresses the auditory, by discussing, the visual, by them seeing the inequalities as they read them, and the kinesthetic learners, by being able to manipulate it using their hands.

What are the contributions of various cultures to this topic?

The strict inequality symbols (less than and greater than) were originally seen in 1631, when used by British mathematician, Thomas Harriot. Some believe that his inspiration for these symbols came from a symbol that he saw on the arm of a Native American. The symbol he saw looked like the strict inequality symbols overlapping. The bars for the unstrict inequalities (less than or equal to and greater than or equal to) were not added until much later. It wasn’t until almost 40 years later, in 1670, that John Wallis started putting a line above the strict inequality symbols. Almost 65 years after that, in 1734, French mathematician, Pierre Bouguer, began writing a double line underneath the inequality symbols.

http://jeff560.tripod.com/relation.html

http://en.wikipedia.org/wiki/Table_of_mathematical_symbols_by_introduction_date

How can technology be used to effectively engage students with this topic?

Quizlet.com is a website that can be used as a good review for many topics. When exploring the section on inequality word problems, I found many useful and engaging things that would help students review and study the material. There were flash cards with word problems on one side and the corresponding equations on the flip side. There was also a test that they could take after studying the material, in order to examine their progress. Lastly, the website had two games that involve solving inequality word problems. This is a great way for students to study and review material. The website is not only great for inequality word problems, but topics of all kinds, in all subjects.