October 2017 – BAD MATHEMATICS

Harry scowled at a picture of a French girl in a bikini. Fred nudged Harry, man-to-man. “Like that, Harry?” he asked.

“Like what?”

“The girl there.”

“That’s not a girl. That’s a piece of paper.”

“Looks like a girl to me.” Fred Rosewater leered.

“Then you’re easily fooled,” said Harry. It’s done with ink on a piece of paper. That girl isn’t lying there on the counter. She’s thousands of miles away, doesn’t even know we’re alive. If this was a real girl, all I’d have to do for a living would be to stay at home and cut out pictures of big fish.”

Kurt Vonnegut, God Bless you, Mr. Rosewater

It is fundamental to be able to distinguish appearance from reality. That it is very easy to confuse the two is famously illustrated by Magritte’s The Treachery of Images (La Trahison des Images):

The danger of such confusion is all the greater in mathematics. Mathematical images, graphs and the like, have intuitive appeal, but these images are mere illustrations of deep and easily muddied ideas. The danger of focussing upon the image, with the ideas relegated to the shadows, is a fundamental reason why the current emphasis on calculators and graphical software is so misguided and so insidious.

Which brings us, once again, to Mathematical Methods. Question 5 on Section Two of the second 2015 Methods exam is concerned with the function $V:[0,5]\rightarrow\Bbb R$ , where

$\phantom{\quad} V(t) = de^{\frac{t}3} + (10-d)e^{\frac{-2t}3}\,.$

Here, $d \in (0,10)$ is a constant, with $d=2$ initially; students are asked to find the minimum (which occurs at $t = \log_e8$ ), and to graph $V$ . All this is par for the course: a reasonable calculus problem thoroughly trivialised by CAS calculators. Predictably, things get worse.

In part (c)(i) of the problem students are asked to find “the set of possible values of $d$ ” for which the minimum of $V$ occurs at $t=0$ . (Part (c)(ii) similarly, and thus boringly and pointlessly, asks for which $d$ the minimum occurs at $t=5$ ). Arguably, the set of possible values of $d$ is $(0,10)$ , which of course is not what was intended; the qualification “possible” is just annoying verbiage, in which the examiners excel.

So, on to considering what the students were expected to have done for (c)(ii), a 2-mark question, equating to three minutes. The Examiners’ Report pointedly remarks that “[a]dequate working must be shown for questions worth more than one mark.” What, then, constituted “adequate working” for 5(c)(i)? The Examiners’ solution consists of first setting $V'(0)=0$ and solving to give $d=20/3$ , and then … well, nothing. Without further comment, the examiners magically conclude that the answer to (c)(i) is $20/3 \leqslant d< 10$ .

Only in the Carrollian world of Methods could the examiners’ doodles be regarded as a summary of or a signpost to any adequate solution. In truth, the examiners have offered no more than a mathematical invocation, barely relevant to the question at hand: why should $V$ having a stationary point at $t=0$ for $d=20/3$ have any any bearing on $V$ for other values of $d$ ? The reader is invited to attempt a proper and substantially complete solution, and to measure how long it takes. Best of luck completing it within three minutes, and feel free to indicate how you went in the comments.

It is evident that the vast majority of students couldn’t make heads or tails of the question, which says more for them than the examiners. Apparently about half the students solved $V'(0)=0$ and included $d = 20/3$ in some form in their answer, earning them one mark. Very few students got further; 4% of students received full marks on the question (and similarly on (c)(ii)).

What did the examiners actually hope for? It is pretty clear that what students were expected to do, and the most that students could conceivably do in the allotted time, was: solve $V'(0)=0$ (i.e. press SOLVE on the machine); then, look at the graphs (on the machine) for two or three values of $d$ ; then, simply presume that the graphs of $V$ for all $d$ are sufficiently predictable to “conclude” that $20/3$ is the largest value of $d$ for which the (unique) turning point of $V$ lies in $[0,5]$ . If it is not immediately obvious that any such approach is mathematical nonsense, the reader is invited to answer (c)(i) for the function $W:[0,5]\rightarrow\Bbb R$ where $W(t) = (6-d)t^2 + (d-2)t$ .

Once upon a time, Victorian Year 12 students were taught mathematics, were taught to prove things. Now, they’re taught to push buttons and to gaze admiringly at pictures of big fish.

The Marriage Theorem is a beautiful piece of mathematics, proved in the 1930s by mathematician Philip Hall. Suppose we have a number of men and the same number of women. Each man is happy to marry some (but perhaps not all) of the women, and similarly for each woman. The question is, can we pair up all the men and women so that everyone is happily married?

Obviously this will be impossible if too many people are too fussy. We’ll definitely require, for example, each woman to be happy to marry at least one man. Similarly, if we take any pair of women then there’s no hope if those two women are both just keen on the one and same man. More generally, we can take any collection W the women, and then we can consider the collection M of men who are acceptable to at least one of those women. The marriage condition states that, no matter the collection W, the corresponding collection M is at least as large as W.

If the marriage condition is not satisfied then there’s definitely no hope of happily marrying everyone off. (If the condition fails for some W then there simply aren’t enough acceptable men for all the women in W.) The Marriage Theorem is the surprising result that the marriage condition is all we need to check; if the marriage condition is satisfied then everyone can be happily married.

That’s all well and good. It’s a beautiful theorem, and you can check out a very nice proof at (no pun intended) cut-the-knot. This, however, is a blog about mathematical crap. So, where’s the crap? For that, we head off to Sydney’s University of New South Wales.

It appears that a lecturer at UNSW who has been teaching the Marriage Theorem has requested that students not refer to the theorem by that name, because of the “homophobic implications”; use of the term in student work was apparently marked as “offensive”. How do we know this? Because one of the affected students went on Sky News to tell the story.

And there’s your crap.

But, at least we have a new theorem:

The “Marriage Theorem” Theorem

a) Any mathematician who whines to her students about the title “Marriage Theorem” is a trouble-making clown with way too much time on her hands.

b) Any student who whines about the mathematician in (a) to a poisonously unprincipled pseudonews network is a troublemaking clown with way too much time on his hands.

Proofs: Trivial.