High School Students Finding New Proofs of Old Theorems (Part 1): Dividing a line segment with straightedge and compass

This is one of my all-time favorite stories to share with students: how a couple of ninth graders in 1995 played with Geometer’s Sketchpad and stumbled upon a brand-new way of using only a straightedge and compass to divide a line segment into any number of equal-sized parts. This article was published in 1997 and made quite a media sensation at the time.

1997_litchfieldd._euclidfibonaccisketchpad Download

Happy Pythagoras Day!

Let me take a one-day break from my current series of posts to wish everyone a Happy Pythagoras Day! Today is 7/24/25 (or 24/7/25 in other parts of the world), and $7^2 = 24^2 + 25^2$ .

Bonus points if you can figure out (without Googling) when the next Pythagoras Days will be. Hint: It’s going to be a while.

Higher derivatives in ordinary speech

Just about every calculus student is taught that the first derivative is useful for finding the slope of a curve and finding velocity from position, and that the second derivative is useful for finding the concavity of a curve and finding acceleration from position.

I recently came across a couple of quotes that, taken literally, are statements about third and fifth derivatives.

Per Wikipedia, President Nixon announced in 1972 that the rate of increase of inflation was decreasing. Taken literally, this claims that “the second derivative of inflation is negative, and so the third derivative of purchasing power [since inflation is the derivative of purchasing power] is negative.” As dryly stated in the Notices of the American Mathematical Society, “[t]his was the first time a sitting president used the third derivative to advance his case for reelection”; the article then ponders the implications of the abuse of mathematics.

More recently, the popular blog Math With Bad Drawings had some fun analyzing a clause that appeared in a 2013 op-ed piece: “As the rate of acceleration of innovation increases…” Taken literally, the words rate, innovation and increases all refer to a first derivative (innovation would be the rate at which technology changes), while the word acceleration refers to a second derivative. Therefore, taken literally and not rhetorically (which was clearly the authors’ intent), this brief clause is a claim that the fifth derivative of technology is positive.

Musical Scales

Saturday Night Live: Washington’s Dream

Happy Fourth of July.

I’m a sucker for G-rated ways of using humor to engage students with concepts in the mathematical curriculum. I never thought that Saturday Night Live would provide a wonderful source of material for this effort.

Horrible False Analogy

I had forgotten the precise assumptions on uniform convergence that guarantees that an infinite series can be differentiated term by term, so that one can safely conclude

$\displaystyle \frac{d}{dx} \sum_{n=1}^\infty f_n(x) = \sum_{n=1}^\infty f_n'(x)$ .

This was part of my studies in real analysis as a student, so I remembered there was a theorem but I had forgotten the details.

So, like just about everyone else on the planet, I went to Google to refresh my memory even though I knew that searching for mathematical results on Google can be iffy at best.

And I was not disappointed. Behold this laughably horrible false analogy (and even worse graphic) that I found on chegg.com:

Suppose Arti has to plan a birthday party and has lots of work to do like arranging stuff for decorations, planning venue for the party, arranging catering for the party, etc. All these tasks can not be done in one go and so need to be planned. Once the order of the tasks is decided, they are executed step by step so that all the arrangements are made in time and the party is a success.

Similarly, in Mathematics when a long expression needs to be differentiated or integrated, the calculation becomes cumbersome if the expression is considered as a whole but if it is broken down into small expressions, both differentiation and the integration become easy.

Pedagogically, I’m all for using whatever technique an instructor might deem necessary to to “sell” abstract mathematical concepts to students. Nevertheless, I’m pretty sure that this particular party-planning analogy has no potency for students who have progressed far enough to rigorously study infinite series.

Predicate Logic and Popular Culture (Part 277): Kellie Pickler

Let $T$ be the set of all times, and let $G(t)$ measure how good day $t$ is. Translate the logical statement

$\exists t_1 < 0 \exists t_2 < 0 \forall t \in T ( (t \ne t_1 \land t \ne t_2) \Longrightarrow (G(t) < G(t_1) \land G(t) < G(t_2)),$

where time $0$ is today.

This matches the chorus of “Best Days of Your Life” by Kellie Pickler, co-written by and featuring Taylor Swift.

Context: Part of a discrete mathematics course includes an introduction to predicate and propositional logic for our math majors. As you can probably guess from their names, students tend to think these concepts are dry and uninteresting even though they’re very important for their development as math majors.

In an effort to making these topics more appealing, I spent a few days mining the depths of popular culture in a (likely futile) attempt to make these ideas more interesting to my students. In this series, I’d like to share what I found. Naturally, the sources that I found have varying levels of complexity, which is appropriate for students who are first learning prepositional and predicate logic.

When I actually presented these in class, I either presented the logical statement and had my class guess the statement in actual English, or I gave my students the famous quote and them translate it into predicate logic. However, for the purposes of this series, I’ll just present the statement in predicate logic first.

Predicate Logic and Popular Culture (Part 276): Heart

Let $T$ be the set of all times, and let $G(t)$ be the statement “I got by on my own at time $t$ .” Translate the logical statement

$\forall t \in T ( ((t<0) \longrightarrow G(t) ) \land (t \ge 0) \longrightarrow \sim G(t)),$

where time $0$ is today.

This matches the opening line of the fabulous power ballad “Alone” by Heart.

And while I’ve got this song in mind, here’s the breakout performance by a young unknown Carrie Underwood on American Idol.

Roman numerals

Solving Problems Submitted to MAA Journals (Part 7i)

The following problem appeared in Volume 131, Issue 9 (2024) of The American Mathematical Monthly.

Let $X$ and $Y$ be independent normally distributed random variables, each with its own mean and variance. Show that the variance of $X$ conditioned on the event $X>Y$ is smaller than the variance of $X$ alone.

In previous posts, we reduced the problem to showing that if $f(x) = 1 + \sqrt{2\pi} x e^{x^2/2} \Phi(x)$ , then $f(x)$ is always positive, where

$\Phi(z) = \displaystyle \frac{1}{\sqrt{2\pi}} \int_{-\infty}^z e^{-z^2/2} \, dz$

is the cumulative distribution function of the standard normal distribution. If we can prove this, then the original problem will be true.

Motivated by the graph of $f(x)$ , I thought of a two-step method for showing $f$ must be positive: show that $f$ is an increasing function, and show that $\displaystyle \lim_{x \to -\infty} f(x) = 0$ . If I could prove both of these claims, then that would prove that $f$ must always be positive.

I was able to show the second step by demonstrating that, if $x<0$ ,

$\displaystyle f(x) = |x| e^{x^2/2} \int_{-\infty}^x \frac{1}{t^2} e^{-t^2/2} \, dt$ .

As discussed in the last post, the limit $\displaystyle \lim_{x \to -\infty} f(x) = 0$ follows from this equality. However, I just couldn’t figure out the first step.

So I kept trying.

And trying.

Until it finally hit me: I’m working too hard! The goal is to show that $f(x)$ is positive. Clearly, clearly, the right-hand side of the last equation is positive! So that’s the entire proof for $x<0$ … there was no need to prove that $f$ is increasing!

For $x \ge 0$ , it’s even easier. If $x$ is non-negative, then

$f(x) = 1 + \sqrt{2\pi} x e^{x^2/2} \Phi(x) \ge 1 + \sqrt{2\pi} \cdot 0 \cdot 1 \cdot \frac{1}{2} = 1 > 0$ .

So, in either case, $f(x)$ must be positive. Following the logical thread in the previous posts, this demonstrates that $\hbox{Var}(Z_1 \mid Z_1 > a+bZ_2) < 1$ , so that $\hbox{Var}(X \mid X <Y) < \hbox{Var}(X)$ , thus concluding the solution.

And I was really annoyed at myself that I stumbled over the last step for so long, when the solution was literally right in front of me.