A Game for Bored Math Teachers

Let's say you're an instructor in a low-level math course; maybe something like college algebra, a liberal-arts math course, or something similar. It's possible that you consider this to be beneath you and you're bored in class. Here's a little game you can play with yourself that will spice things up a bit:

When you ask a question to the class, and someone answers incorrectly, see if you can conversationally edit the question in a way that the student would have been right. That is, respond by starting with, "Well, that would be right if the problem said ____", and fill in the end of that sentence in appropriate way. 

I'm very fond of this technique. It actually accomplishes several things:

• Makes things a bit more challenging for the instructor, keeping them on their toes
• Cushions the "no you're wrong" response to the student (a bit like the "shit sandwich" feedback protocol)
• Force you to diagnose & clarify the misdirected mental pathway for you and the student (and in fact usually the student has misperceived some pattern that's just adjacent to the given problem).

Try it and see how it feels. To be clear: I don't do this because I'm bored in class, but nevertheless I've found it to be a compelling and clarifying technique.

Radicals and Absolute Values

Here's a fact that I've never seen expressed clearly, or in this way, in any of the several college-level algebra books from which I've taught. Say we're working in the domain of real numbers, and have a radical of some index over a variable to a power. In broad strokes, we can divide the power by the radical index -- however, in some cases, distressingly, you'd need an absolute value to express the result. The question is, exactly when do you need that absolute value?

$$\sqrt[n]{a^m} = a^{m \over n} \text{ or } |a^{m \over n}|? $$

Say we're in this situation, with m and n whole numbers, and m is evenly divisible by n. The primary issue is that when the initial power m is even, it makes any product nonnegative, wiping out any negatives that the base a might represent. So when the reduced power m/n is odd, then it would fraudulently claim to possibly produce negatives, which our initial expression cannot do -- and so require the absolute value as a correction. 

Let's give more detail by inspecting all the permutations of even/odd possibilities between the starting and ending powers in the expression:

1. Odd m, odd m/n: The odd starting power m can produce values of any sign, and so can the odd reduced power m/n. So all is fine here, and we don't need the absolute value.
2. Even m, even m/n: The even power m wipes out any negatives, and the reduced even power m/n does the same thing. So again they're aligned, and no absolute value is needed.
3. Even m, odd m/n: This is the case alluded to above -- the even start power m wipes out negatives, but the odd ending power m/n would deceive us into thinking negatives could be a possible product. This is the situation in which we need the absolute value as a correction.
4. Odd m, even m/n: This case is impossible. If m/n is even, then any multiple (e.g., by n to produce m) is also even.
   

So it's only that third case in which the power switches from even to odd where we need the absolute value bars for full fidelity. Interestingly, since the fourth case can't happen, we could express the protocol briefly as follows:

When the powers switch parity, then we need absolute value bars.