# My Favorite One-Liners: Part 54

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.

The complex plane is typically used to visually represent complex numbers. (There’s also the Riemann sphere, but I won’t go into that here.) The complex plane looks just like an ordinary Cartesian plane, except the “$x-$axis” becomes the real axis and the “$y-$axis” becomes the imaginary axis. It makes sense that this visualization has two dimensions since there are two independent components of complex numbers. For real numbers, only a one-dimensional visualization is needed: the number line that (hopefully) has been hammered into my students’ brains ever since elementary school.

While I’m on the topic, it’s unfortunate that “complex numbers” are called complex, as this often has the connotation of difficult. However, that’s not why our ancestors chose the word complex was chosen. Even today, there is a second meaning of the word: a group of associated buildings in close proximity to each other is often called an “apartment complex” or an “office complex.” This is the real meaning of “complex numbers,” since the real and imaginary parts are joined to make a new number.

When I teach my students about complex number, I tell the following true story of when my daughter was just a baby, and I was extremely sleep-deprived and extremely desperate for ways to get her to sleep at night.

I tried counting monotonously, moving my finger to the right on a number line with each number:

$1, 2, 3, 4, ...$

That didn’t work, so I tried counting monotonously again, but this time moving my finger to the left on a number line with each number:

$-1, -2, -3, -4, ...$

That didn’t work either, so I tried counting monotonously once more, this time moving my finger up the imaginary axis:

$i, 2i, 3i, 4i...$

For the record, that didn’t work either. But it gave a great story to tell my students.

# My Favorite One-Liners: Part 49

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 post is certainly not a one-liner but instead is my pseudohistory for how the roots of polynomials were found.

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.

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

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.

Thus giving 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.

Real references:

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 48

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 techniques commonly taught in Algebra II or Precalculus is the Rational Root Test, which is a way of making a list of candidates of rational numbers that might (emphasis, might) be roots of the polynomial. This is a commonly taught method for finding the roots of polynomials whose degree is higher than 3. (Other techniques that are typically taught to students are Descartes’ Rule of Signs and (less commonly) the Upper and Lower Bound Rules.) For example, for the polynomial $f(x) = 2x^3 + 5 x^2 - 2x - 15$.

• The factors of the constant term are $\pm 1, \pm 3, \pm 5$ and $\pm 15$, and so the numerator of any rational root must be one of these numbers.
• The factors of the leading coefficient are $\pm 1$ and $\pm 2$, and so the denominator of any rational root must be one of these numbers.
• In conclusion, if there’s a rational root, then it’s $\pm 1, \pm 3, \pm 5, \pm 15, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{5}{2}$ and $\pm \frac{15}{2}$. In other words, we have a list of 16 possible rational roots. Not all of them will be roots, of course, since the cubic polynomial only has at most three distinct roots. Also, there’s no guarantee that any of them will be roots. The only way to find out if any of them work is by testing them, usually using synthetic division.

So, after a practice problem or two, I’ll ask my students,

What guarantee do you have that at least one of the possible rational roots will actually work?

After letting them think for a few seconds, I give them the answer:

In other words, there is no guarantee that any of the possible rational roots will actually work, except that the instructor (or author of the textbook) has rigged things so that it happens.

# My Favorite One-Liners: Part 42

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.

The function $f(x) = a^x$ typically exhibits exponential growth (if $a > 1$) or exponential decay (if $a < 1$). The one exception is if $a = 1$, when the function is merely a constant. Which often leads to my favorite blooper from Star Trek. The crew is trying to find a stowaway, and they get the bright idea of turning off all the sound on the ship and then turning up the sound so that the stowaway’s heartbeat can be heard. After all, Captain Kirk boasts, the Enterprise has the ability to amplify sound by 1 to the fourth power.

# My Favorite One-Liners: Part 32

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 story is a continuation of yesterday’s post. I call today’s one-liner “Method #1… Method #2.”

Every once in a while, I want my students to figure out that there’s a clever way to do a problem that will save them a lot of time, and they need to think of it.

For example, in Algebra II, Precalculus, or Probability, I might introduce the binomial coefficients to my students, show them the formula for computing them and how they’re related to combinatorics and to Pascal’s triangle, and then ask them to compute $\displaystyle {100 \choose 3}$. We write down

$\displaystyle {100 \choose 3} = \displaystyle \frac{100!}{3!(100-3)!} = \displaystyle \frac{100!}{3! \times 97!}$

So this fraction needs to be simplified. So I’ll dramatically announce:

Method #1: Multiply out the top and the bottom.

This produces the desired groans from my students. If possible, then I list other available but undesirable ways of solving the problem.

Method #2: Figure out the 100th row of Pascal’s triangle.

Method #3: List out all of the ways of getting 3 successes in 100 trials.

All of this gets the point across: there’s got to be an easier way to do this. So, finally, I’ll get to what I really want my students to do:

Method #4: Write $100! = 100 \times 99 \times 98 \times 97!$, and cancel.

The point of this bit of showman’s patter is to get my students to think about what they should do next as opposed to blindly embarking in a laborious calculation.

As another example, consider the following problem from Algebra II/Precalculus: “Show that $x-1$ is a factor of $f(x)=x^{78} - 4 x^{37} + 2 x^{15} + 1$.”

As I’m writing down the problem on the board, someone will usually call out nervously, “Are you sure you mean $x^{78}$?” Yes, I’m sure.

“So,” I announce, “how are we going to solve the problem?”

Method #1: Use synthetic division.

Then I’ll make a point of what it would take to write down the procedure of synthetic division for this polynomial of degree 78.

Method #2: (As my students anticipate the real way of doing the problem) Use long division.

Understanding laughter ensures. Eventually, I tell my students — or, sometimes, my students will tell me:

Method #3: Calculate $f(1)$.

# 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 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.

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

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.

Real references:

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 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 10

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.

While I can’t take credit for today’s one-liner, I’m more than happy to share it.

A colleague was explaining his expectations for simplifying expressions such as

$\displaystyle \frac{\displaystyle ~~~\frac{2x}{x^2+1}~~~}{\displaystyle ~~~\frac{x}{x^2-1}~~~}$

Of course, this isn’t yet simplified, but his students were balking about doing the required work. So, on the spur of the moment, he laid down a simple rule:

Not simplifying a fraction in a fraction is an infraction.

Utterly brilliant.

# My Favorite One-Liners: Part 8

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.

At many layers of the mathematics curriculum, students learn about that various functions can essentially commute with each other. In other words, the order in which the operations is performed doesn’t affect the final answer. Here’s a partial list off the top of my head:

1. Arithmetic/Algebra: $a \cdot (b + c) = a \cdot b + a \cdot c$. This of course is commonly called the distributive property (and not the commutative property), but the essential idea is that the same answer is obtained whether the multiplications are performed first or if the addition is performed first.
2. Algebra: If $a,b > 0$, then $\sqrt{ab} = \sqrt{a} \sqrt{b}$.
3. Algebra: If $a,b > 0$ and $x$ is any real number, then $(ab)^x = a^x b^x$.
4. Precalculus: $\displaystyle \sum_{i=1}^n (a_i+b_i) = \displaystyle \sum_{i=1}^n a_i + \sum_{i=1}^n b_i$.
5. Precalculus: $\displaystyle \sum_{i=1}^n c a_i = c \displaystyle \sum_{i=1}^n a_i$.
6. Calculus: If $f$ is continuous at an interior point $c$, then $\displaystyle \lim_{x \to c} f(x) = f(c)$.
7. Calculus: If $f$ and $g$ are differentiable, then $(f+g)' = f' + g'$.
8. Calculus: If $f$ is differentiable and $c$ is a constant, then $(cf)' = cf'$.
9. Calculus: If $f$ and $g$ are integrable, then $\int (f+g) = \int f + \int g$.
10. Calculus: If $f$ is integrable and $c$ is a constant, then $\int cf = c \int f$.
11. Calculus: If $f: \mathbb{R}^2 \to \mathbb{R}$ is integrable, $\iint f(x,y) dx dy = \iint f(x,y) dy dx$.
12. Calculus: For most differentiable function $f: \mathbb{R}^2 \to \mathbb{R}$ that arise in practice, $\displaystyle \frac{\partial^2 f}{\partial x \partial y} = \displaystyle \frac{\partial^2 f}{\partial y \partial x}$.
13. Probability: If $X$ and $Y$ are random variables, then $E(X+Y) = E(X) + E(Y)$.
14. Probability: If $X$ is a random variable and $c$ is a constant, then $E(cX) = c E(X)$.
15. Probability: If $X$ and $Y$ are independent random variables, then $E(XY) = E(X) E(Y)$.
16. Probability: If $X$ and $Y$ are independent random variables, then $\hbox{Var}(X+Y) = \hbox{Var}(X) + \hbox{Var}(Y)$.
17. Set theory: If $A$, $B$, and $C$ are sets, then $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.
18. Set theory: If $A$, $B$, and $C$ are sets, then $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.

However, there are plenty of instances when two functions do not commute. Most of these, of course, are common mistakes that students make when they first encounter these concepts. Here’s a partial list off the top of my head. (For all of these, the inequality sign means that the two sides do not have to be equal… though there may be special cases when equality happens to happen.)

1. Algebra: $(a+b)^x \ne a^x + b^x$ if $x \ne 1$. Important special cases are $x = 2$, $x = 1/2$, and $x = -1$.
2. Algebra/Precalculus: $\log_b(x+y) = \log_b x + \log_b y$. I call this the third classic blunder.
3. Precalculus: $(f \circ g)(x) \ne (g \circ f)(x)$.
4. Precalculus: $\sin(x+y) \ne \sin x + \sin y$, $\cos(x+y) \ne \cos x + \cos y$, etc.
5. Precalculus: $\displaystyle \sum_{i=1}^n (a_i b_i) \ne \displaystyle \left(\sum_{i=1}^n a_i \right) \left( \sum_{i=1}^n b_i \right)$.
6. Calculus: $(fg)' \ne f' \cdot g'$.
7. Calculus $\left( \displaystyle \frac{f}{g} \right)' \ne \displaystyle \frac{f'}{g'}$
8. Calculus: $\int fg \ne \left( \int f \right) \left( \int g \right)$.
9. Probability: If $X$ and $Y$ are dependent random variables, then $E(XY) \ne E(X) E(Y)$.
10. Probability: If $X$ and $Y$ are dependent random variables, then $\hbox{Var}(X+Y) \ne \hbox{Var}(X) + \hbox{Var}(Y)$.

All this to say, it’s a big deal when two functions commute, because this doesn’t happen all the time.

I wish I could remember the speaker’s name, but I heard the following one-liner at a state mathematics conference many years ago, and I’ve used it to great effect in my classes ever since. Whenever I present a property where two functions commute, I’ll say, “In other words, the order of operations does not matter. This is a big deal, because, in real life, the order of operations usually is important. For example, this morning, you probably got dressed and then went outside. The order was important.”