My Favorite One-Liners: Part 20

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.

Perhaps the world’s most famous infinite series is

S = \displaystyle \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots

as this is the subject of Zeno’s paradox. When I teach infinite series in class, I often engage the students by reminding students about Zeno’s paradox and then show them this clip from the 1994 movie I.Q.

This clip is almost always a big hit with my students. After showing this clip, I’ll conclude, “When I was single, this was part of my repertoire of pick-up lines… but it never worked.”

Even after showing this clip, some students resist the idea that an infinite series can have a finite answer. For such students, I use a physical demonstration: I walk half-way across the classroom, then a quarter, and so on… until I walk head-first into a walk at full walking speed. The resulting loud thud usually confirms for students that an infinite sum can indeed have a finite answer.

For further reading, see my series on arithmetic and geometric series.

 

 

My Favorite One-Liners: Part 19

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. This is a quip that I’ll use when a theoretical calculation can be easily confirmed with a calculator. Today’s post is less of a one-liner than a story.

When I teach Algebra II or Precalculus (or train my  future high school teachers to teach these subjects), we eventually land on the Rational Root Test and Descartes’ Rule of Signs as an aid for finding the roots of cubic equations or higher. Before I get too deep into this subject, however, I like to give a 10-15 minute pseudohistory about the discovery of how polynomial equations can be solved. Historians of mathematics will certain take issue with some of this “history.” However, the main purpose of the story is not complete accuracy but engaging students with the history of mathematics. I think the story I tell engages students while remaining reasonably accurate… and I always refer students to various resources if they want to get the real history.

To begin, I write down the easiest two equations to solve (in all cases, a \ne 0:

ax + b = 0 \qquad and \qquad ax^2 + bx + c = 0

These are pretty easy to solve, with solutions well known to students:

x = -\displaystyle \frac{b}{a} \qquad and \qquad x = \displaystyle \frac{-b \pm \sqrt{b^2-4ac}}{2a}

In other words, there are formulas that you can just stick in the coefficients and get the answer out without thinking too hard. Sure, there are alternate ways of solving for x that could be easier, like factoring, but the worst-case scenario is just plugging into the formula.

These formulas were known to Babylonian mathematicians around 2000 B.C. (When I teach this in class, I write the date, and all other dates and discoverers, next to the equations for dramatic pedagogical effect.) Though not written in these modern terms, basically every ancient culture on the globe that did mathematics had some version of these formulas: for example, the ancient Egyptians, Greeks, Chinese, and Mayans.

Naturally, this leads to a simple question: is there a formula for the cubic:

ax^3 + bx^2 + cx + d = 0

Is there some formula that we can just plug a, b, c, and d to just get the answer?  The answer is, Yes, there is a formula. But it’s nasty. The formula was not discovered until 1535 A.D., and it was discovered by a man named Tartaglia. During the 1500s, the study of mathematics was less about the dispassionate pursuit of truth and more about exercising machismo. One mathematician would challenge another: “Here’s my cubic equation; I bet you can’t solve it. Nyah-nyah-nyah-nyah-nyah.” Then the second mathematician would solve it and challenge the first: “Here’s my cubic equation; I bet you can’t solve it. Nyah-nyah-nyah-nyah-nyah.” And so on. Well, Tartaglia came up with a formula that would solve every cubic equation. By plugging in a, b, c, and d, you get the answer out.

Tartaglia’s discovery was arguably the first triumph of the European Renaissance. The solution of the cubic was perhaps the first thing known to European mathematicians in the Middle Ages that was unknown to the ancient Greeks.

In 1535, Tartaglia was a relatively unknown mathematician, and so he told a more famous mathematician, Cardano, about his formula. Cardano told Tartaglia, why yes, that is very interesting, and then published the formula under his own name, taking credit without mention of Tartaglia. To this day, the formula is called Cardano’s formula.

So there is a formula. But it would take an entire chalkboard to write down the formula. That’s why we typically don’t make students learn this formula in high school; it’s out there, but it’s simply too complicated to expect students to memorize and use.

This leads to the next natural question: what about quartic equations?

ax^4 + bx^3 + cx^2 + dx + e = 0

The solution of the quartic was discovered less than five years later by an Italian mathematician named Ferrari. Ferrari found out that there is a formula that you can just plug in a, b, c, d, and e, turn the crank, and get the answers out. Writing out this formula would take two chalkboards. So there is a formula, but it’s also very, very complicated.

Of course, Ferrari had some famous descendants in the automotive industry.

So now we move onto my favorite equation, the quintic. (If you don’t understand why it’s my favorite, think about my last name.)

ax^5 + bx^4 + cx^3 + dx^2 + ex + f = 0

After solving the cubic and quartic in rapid succession, surely there should also be a formula for the quintic. So they tried, and they tried, and they tried, and they got nowhere fast. Finally, the problem was solved nearly 300 years later, in 1832 (for the sake telling a good story, I don’t mention Abel) by a French kid named Evariste Galois. Galois showed that there is no formula. That takes some real moxie. There is no formula. No matter how hard you try, you will not find a formula that can work for every quintic. Sure, there are some quintics that can be solved, like x^5 = 0. But there is no formula that will work for every single quintic.

Galois made this discovery when he was 19 years old… in other words, approximately the same age as my students. In fact, we know when wrote down his discovery, because it happened the night before he died. You see, he was living in France in 1832. What was going on in France in 1832? I ask my class, have they seen Les Miserables?

France was torn upside-down in 1832 in the aftermath of the French Revolution, and young Galois got into a heated argument with someone over politics; Galois was a republican, while the other guy was a royalist. More importantly, both men were competing for the hand of the same young woman. So they decided to settle their differences like honorable Frenchmen, with a duel. So Galois wrote up his mathematical notes one night, and the next day, he fought the duel, he lost the duel, and he died.

Which bring me to the conclusion of this story: we have complete and total proof that tremendous mathematical genius does not prevent somebody from being a complete idiot.

For the present, there are formulas for cubic and quartic equations, but they’re long and impractical. And for quintic equations and higher, there is no formula. So that’s why we teach these indirect methods like the Rational Root Test and Descartes’ Rule of Signs, as they give tools to use to guess at the roots of higher-order polynomials without using something like the quadratic formula.

green lineReal references:

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

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

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

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

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

http://library.wolfram.com/examples/quintic/

http://library.wolfram.com/examples/quintic/timeline.html

My Favorite One-Liners: Part 18

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. This is a quip that I’ll use when a theoretical calculation can be easily confirmed with a calculator.

Sometimes I teach my students how people converted decimal expansions into fractions before there was a button on a calculator to do this for them. For example, to convert  x = 0.\overline{432} = 0.432432432\dots into a fraction, the first step (from the Bag of Tricks) is to multiply by 1000: How do we change this into a decimal? Let’s call this number x.

1000x = 432.432432\dots

x = 0.432432\dots

Notice that the decimal parts of both x and 1000x are the same. Subtracting, the decimal parts cancel, leaving

999x = 432

or

x = \displaystyle \frac{432}{999} = \displaystyle \frac{16}{37}

In my experience, most students — even senior math majors who have taken a few theorem-proof classes and hence are no dummies — are a little stunned when they see this procedure for the first time.

To make this more real and believable to them, I then tell them my one-liner: “I can see that no one believes me. OK, let’s try something that you will believe. Pop out your calculators. Then punch in 16 divided by 37.”

Indeed, my experience many students really do need this technological confirmation to be psychologically sure that it really did work. Then I’ll tease them that, by pulling out their calculators, I’m trying to speak my students’ language.

TI1637

See also my fuller post on this topic as well as the index for the entire series.

 

My Favorite One-Liners: Part 17

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.

Sometimes it’s pretty easy for students to push through a proof from beginning to end. For example, in my experience, math majors have little trouble with each step of the proof of the following theorem.

Theorem. If z, w \in \mathbb{C}, then \overline{z+w} = \overline{z} + \overline{w}.

Proof. Let z = a + bi, where a, b \in \mathbb{R}, and let w = c + di, where c, d \in \mathbb{R}. Then

\overline{z + w} = \overline{(a + bi) + (c + di)}

= \overline{(a+c) + (b+d) i}

= (a+c) - (b+d) i

= (a - bi) + (c - di)

= \overline{z} + \overline{w}

\square

For other theorems, it’s not so easy for students to start with the left-hand side and end with the right-hand side. For example:

Theorem. If z, w \in \mathbb{C}, then \overline{z \cdot w} = \overline{z} \cdot \overline{w}.

Proof. Let z = a + bi, where a, b \in \mathbb{R}, and let w = c + di, where c, d \in \mathbb{R}. Then

\overline{z \cdot w} = \overline{(a + bi) (c + di)}

= \overline{ac + adi + bci + bdi^2}

= \overline{ac - bd + (ad + bc)i}

= ac - bd - (ad + bc)i

= ac - bd - adi - bci.

A sharp math major can then provide the next few steps of the proof from here; however, it’s not uncommon for a student new to proofs to get stuck at this point. Inevitably, somebody asks if we can do the same thing to the right-hand side to get the same thing. I’ll say, “Sure, let’s try it”:

\overline{z} \cdot \overline{w} = \overline{(a + bi)} \cdot \overline{(c + di)}

= (a-bi)(c-di)

= ac -adi - bci + bdi^2

= ac - bd - adi - bci.

\square

I call working with both the left and right sides to end up at the same spot the Diamond Rio approach to proofs: “I’ll start walking your way; you start walking mine; we meet in the middle ‘neath that old Georgia pine.” Not surprisingly, labeling this with a catchy country song helps the idea stick in my students’ heads.

Though not the most elegant presentation, this is logically correct because the steps for the right-hand side can be reversed and appended to the steps for the left-hand side:

Proof (more elegant). Let z = a + bi, where a, b \in \mathbb{R}, and let w = c + di, where c, d \in \mathbb{R}. Then

\overline{z \cdot w} = \overline{(a + bi) (c + di)}

= \overline{ac + adi + bci + bdi^2}

= \overline{ac - bd + (ad + bc)i}

= ac - bd - (ad + bc)i

= ac - bd - adi - bci

= ac -adi - bci + bdi^2

= (a-bi)(c-di)

= \overline{(a + bi)} \cdot \overline{(c + di)}

\overline{z} \cdot \overline{w}.

\square

 For further reading, here’s my series on complex numbers.

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 15

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. Let me describe a one-liner that I’ll use when I want my class to figure out a pattern, thus developing a theorem by inductive logic rather than deductive logic.

Today’s one-liner is easily stated: “Gosh, I’ve seen that somewhere before.”

For example, In my statistics class, here’s the very first illustration that I show to demonstrate how to compute a standard deviation:

Find the standard deviation of the following data set: 1, 4, 6, 7, 8, 10.

The first step is finding the average:

\overline{x} = \displaystyle \frac{1+4+6+7+8+10}{6} = 6.

We then find the deviations from average by subtracting 6 from all of the original data values:

Deviations from average = -5, -2, 0, 1, 2, 4

With these numbers, we can compute the standard deviation:

s = \displaystyle \sqrt {\frac{ (-5)^2 + (-2)^2 + 0^2 + 1^2 + 2^2 + 4^2}{5} } = \sqrt{10}.

After asking if there are any questions of clarification about the nuts and bolts of this calculation, I’ll proceed to the next example:

Find the standard deviation of the following data set: 5, 8, 10, 11, 12, 14.

The first step is finding the average:

\overline{x} = \displaystyle \frac{5+8+10+11+12+14}{6} = 10.

We then find the deviations from average by subtracting 10 from all of the original data values and then constructing the square root as before:

s = \displaystyle \sqrt {\frac{ (-5)^2 + (-2)^2 + 0^2 + 1^2 + 2^2 + 4^2}{5} } = \sqrt{10}.

Then, in a loud obvious voice, I’ll declare, “Gosh, I’ve seen that somewhere before” and then wait a few seconds for the answer. Students can obviously see that the two answers are the same — which gets them thinking about why that happened.

Obviously, the two answers are the same. The real conceptual question that I want my students to figure out is why the two answers are same. Eventually, someone will come up with the correct answer — the second data set was made by adding 4 to all the values in the first data set, which may change the average but does not change how spread out the numbers are… so the standard deviation should be unchanged.

I love the “Gosh, I’ve seen that somewhere before” line after a couple of carefully chosen examples, as it cues my class that they really need to think a little harder than the dull and mechanical operations toward a deeper conceptual understanding of what’s really happening.

My Favorite One-Liners: Part 14

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. This quip is similar to the “bag of tricks” one-liner, and I’ll use this one if the “bag of tricks” line is starting to get a little dry.

Sometimes in math, there’s a step in a derivation that, to the novice, appears to make absolutely no sense. For example, to find the antiderivative of \sec x, the first step is far from obvious:

\displaystyle \int \sec x \, dx = \displaystyle \int \sec x \frac{\sec x + \tan x}{\sec x + \tan x} \, dx

While that’s certainly correct, it’s from from obvious to a student that this such a “simplification” is actually helpful.

To give a simpler example, to convert

x = 0.\overline{432} = 0.432432432\dots

into a decimal, the first step is to multiply x by 1000:

1000x = 432.432432\dots

Students often give skeptical, quizzical, and/or frustrated looks about this non-intuitive next step… they’re thinking, “How did you know to do that?” To lighten the mood, I’ll explain with a big smile that I’m clairvoyant… when I got my Ph.D., I walked across the stage, got my diploma, someone waved a magic wand at me, and poof! I became clairvoyant.

Clairvoyance is wonderful; I highly recommend it.

The joke, of course, is that the only reason that I multiplied by 1000 is that someone figured out that multiplying by 1000 at this juncture would actually be helpful. Subtracting x from 1000x, the decimal parts cancel, leaving

999x = 432

or

x = \displaystyle \frac{432}{999} = \displaystyle \frac{16}{37}.

In my experience, most students — even senior math majors who have taken a few theorem-proof classes and hence are no dummies — are a little stunned when they see this procedure for the first time. I learned this procedure when I was very young; however, in modern times, this procedure appears to be a dying art. I’m guessing that this algorithm is a dying art because of the ease and convenience of modern calculators. As always, I hold my students blameless for the things that they were simply not taught at a younger age, and part of my job is repairing these odd holes in their mathematical backgrounds so that they’ll have their best chance at becoming excellent high school math teachers.

For further reading, here’s my series on rational numbers and decimal expansions.

My Favorite One-Liners: Part 13

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 story that I’ll tell my students when, for the first time in a semester, I’m about to use a previous theorem to make a major step in proving a theorem. For example, I may have just finished the proof of

\hbox{Var}(X+Y) = \hbox{Var}(X) + \hbox{Var}(Y),

where X and Y are independent random variables, and I’m about to prove that

\hbox{Var}(X-Y) = \hbox{Var}(X) + \hbox{Var}(Y).

While this can be done by starting from scratch and using the definition of variance, the easiest thing to do is to write

\hbox{Var}(X-Y) = \hbox{Var}(X+[-Y]) = \hbox{Var}(X) + \hbox{Var}(-Y),

thus using the result of the first theorem to prove the next theorem.

And so I have a little story that I tell students about this principle. I think I was 13 when I first heard this one, and obviously it’s stuck with me over the years.

At MIT, there’s a two-part entrance exam to determine who will be the engineers and who will be the mathematicians. For the first part of the exam, students are led one at a time into a kitchen. There’s an empty pot on the floor, a sink, and a stove. The assignment is to boil water. Everyone does exactly the same thing: they fill the pot with water, place it on the stove, and then turn the stove on. Everyone passes.

For the second part of the exam, students are led one at a time again into the kitchen. This time, there’s a pot full of water sitting on the stove. The assignment, once again, is to boil water. Nearly everyone simply turns on the stove. These students are led off to become engineers. The mathematicians are ones who take the pot off the stove, dump the water into the sink, and place the empty pot on the floor… thereby reducing to the original problem, which had already been solved.

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 11

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.

Every once in a while, I’ll cover a theorem in class that looks utterly surprising to students at first glance. For example, in trigonometry, I might state that

\sin^{-1} \left( \sin \pi \right) \ne \pi,

so that the inverse function doesn’t quite behave like it’s supposed to (because of the restricted domain used to define inverse sine.)

Before explaining why \sin^{-1} \left( \sin \pi \right) isn’t equal to \pi, I’ll get the discussion started by saying, “Don’t believe me? Just watch.”… a tip of the cap to this recent hit song (at the time of this writing, the third-most watched video on YouTube).

While on this topic, I have to tip my cap to Kelli Hauser, a sixth-grade teacher in my city who made the following motivational video for students about to take their end-of-year high-stakes test (called, here in Texas, the STAAR exam).

One more parody concerning a recent spacecraft that visited Pluto:

For further reading, here’s my series on inverse functions.