First, the course context: The course "Concepts of Mathematics" (whence this blog's title!) is a transition from computation-based mathematics to proof-writing and abstract problem-solving. During the academic year, this is taught to university freshman and sophomores (mostly) in one semester. We cover sets, logic, and proof techniques (direct, contradiction, induction), and then apply these techniques to chosen topics: relations, functions, cardinality, number theory, combinatorics, and probability (depending on the instructor's choices and time). During the summer, this is also taught to high-school students (simultaneously with university students) in an advanced placement program.
Next, the problem's context: This was an assigned, written homework problem during the unit on relations. We had just spend many days studying logical proof techniques and were beginning to apply them to newly-learned topics. We had just seen relations in lecture, with a definition, several examples, and a discussion and formal definitions of the standard properties: reflexivity, symmetry, and transitivity. This homework assignment was their first opportunity writing formal proofs about these properties. Here is the original problem statement:
1. Define the relation \(T\) on \(\mathbb{R}\) by saying, for any \(x,y\in\mathbb{R}\), \[ (x,y)\in T\;\Leftrightarrow\; \left(\;\frac{y}{x}\in\mathbb{R}\;\wedge\; \frac{y}{x}\geq 0\right) \] (a) Find \(x\in\mathbb{R}\) such that \((x,x)\notin T\). Does this mean \(T\) is not reflexive? Why or why not?There were three other parts to the question, but part (a) is the one I'd like to discuss. Think about the problem for a minute, if you wish, or check out my solution below.
(b) \(\ldots\)
Consider \(x=0\). Notice that \((0,0)\notin T\) because \(\frac{0}{0}\) is undefined, so \(\frac{0}{0}\notin\mathbb{R}\).This makes sense, right? The property of being a reflexive relation is one of universal quantification: either for all elements \(x\) in the set \(T\), something happens; or else, that something fails somewhere. What could be more clear-cut? This is a "Yes/No" question. Of course, this requires some thinking and "playing around" to come up with an answer, but I predicted this would be a straightforward homework problem, the "easing in" to the assignment's material, as it were. Indeed, it was the first problem on the assignment (for what that's worth).
Since we have exhibited a counterexample to the property of reflexivity (that is, we proved that \(\exists x\in\mathbb{R}{\large.}\;(x,x)\notin T\)), we conclude that \(T\) is not reflexive.
I was wrong.
I wasn't vastly wrong, but I was decidedly incorrect in my assumptions that (a) students would formalize their "playing around" in the correct way, and that (b) they would recognize this as a "Yes/No" question. I was, and still am, far more surprised that they shattered assumption (b).
Several students got this question correct, with no issues. (FYI, there were 45 students in this class. About 30 were from the advanced high-school program, and about 15 were from the university.) And a small number of students were way off (misinterpreting the problem, failing to turn it in, etc.). But the majority of students made the same mistake that simultaneously showed (a) they understood the problem statement and thought about it carefully (and correctly, even!), but (b) failed to present those ideas formally and, more importantly, correctly.
It wasn't really the mistakes that caught my attention, but their universality. I expected some students to forget the \(\frac{y}{x}\in\mathbb{R}\) condition, perhaps, or to neglect to factor in negative real numbers. Neither of these occurred. Instead, many, many students said something like this:
Only when \(x=0\), the fraction \(\frac{x}{x}=\frac{0}{0}\) is undefined. But elsewhere, \(\frac{x}{x}=1\). Thus, \(T\) is reflexive everywhere except the origin.What? Did you expect this? I certainly did not. Clearly, all the ideas are correct. Indeed, \((0,0)\notin T\) but \(\forall x\in\mathbb{R}-\{0\}{\large .}\; (x,x)\in T\). But the point is this: a single counterexample to the "for all" claim means that \(T\) is not reflexive.
This idea had been drilled over and over. I saw that the students had trouble with quantifiers when we studied them, so I chose to spend extra class time then (even during the already time-crunched summer term). I assumed that issue with the idea of "generality vs. specificity" had been learned. But it reared its head again here. The students all saw that some specific example occurred, but they didn't think that meant anything in general.
At the time, I was baffled. The first few homework assignments I saw this on, I wrote some comments about how the negation of "\(\forall\)" is "\(\exists\)". Eventually, I wrote shorter and shorter comments, and decided to make a blanket statement in lecture the next day. And I did. I pointed out that negation again. I told them that a relation is either reflexive or it isn't. It can't be "kinda" reflexive. It's not a sliding scale. This is the Law of the Excluded Middle all over again. Reread the definition. Etc etc etc. I asked them how they collectively and independently (I assume!) came up with this notion of relative reflexivity. And I got … nothing in response. They shrugged their shoulders and I reiterated my prior ideas more concisely and moved on. Thereafter, I was careful to indicate the "Yes/No" nature of such questions, and I believe the ideas sunk in over time. But for that assignment, I was concerned.
But the oddest thing is that, in retrospect, I feel like them. Or at least, I appreciate their response and believe that they were expressing some other aspects of learning mathematics. For one, they were indicating how difficult it is to learn abstract mathematics. They can't "see" a binary relation, let alone visualize its properties. This goes against most things students learn in pre-calc and calc. (Keep in mind, 2/3 of my students were high-school juniors/seniors.) There, you graph a function and assess it pictorially (for the most part). The "vertical line test" is a visual thing, not a formal, quantified statement that you negate. And so, when you see a definition like \[ \text{A relation }R\text{ defined on a set }S\text{ is reflexive iff }\forall x\in S{\large .}\; (x,x)\in R\] it's difficult to "see" what it means; in particular, it's difficult to "see" what its opposite (read: logical negation) means!
For that reason, I somewhat understand why the students mostly failed to properly write up (formally) that "this relation is not reflexive". Mind you, they lost points because, yes, their answers were incorrect. It represents a lack of full understanding that the negation of generality is specificity. But in another sense, I am retrospectively proud of these students. I see that they realized the relation \(T\) came "oh so close" to being reflexive. "There's only this one counterexample, man … Only 0 doesn't work!" In that sense, "kinda reflexive" is an appropriate assessment of this relation. (Thankfully, no one used that much of an informal phrasing!) Accordingly, I like to think that their investment in figuring out the problem and presenting their work in written homework elicited an emotional response from them. In assessing \(T\)'s reflexivity, they saw it came close and presented their ideas as such; they wrote up their thoughts, not just "the right answer".
For that, I am proud. One of my major goals in teaching this course has always been to dispel certain myths about what mathematics "is" and what mathematicians "do" and how mathematics is all about "True or False, Right or Wrong". With this homework problem, my students showed me that they could appreciate those feelings, and not just because I told them to. They all surprised me with this development. And I love it. I intend to assign this problem again in the future, just to see what happens …
Now, that said, I'm considering adjusting the problem statement to elicit different responses. For example, I might eliminate the "kinda non-reflexive" answers by asking this:
"… Does this mean \(T\) is reflexive or not reflexive? Explain your answer."I could be even more explicit by saying: "Explain your answer by appealing to a definition from lecture/the book."
I also wonder what would happen if I doctored the example to be "more non-reflexive". I hope you know what I mean by that by now :-) . For instance, I might define \(T\) as this: \[ (x,y)\in T\;\Leftrightarrow\; \left(\;\frac{y}{(x-2)(x-3)}\in\mathbb{R}\;\wedge\; \frac{y}{(x-2)(x-3)}\geq 0\right) \] Then, students have to think about \(x=2\) and \(x=3\), and even perhaps \(2 < x < 3 \), and even … (Do you see what happens?) There might be more to their answer than just, "It's reflexive everywhere but …". Maybe by making it "less reflexive", they'll shed those intuitions about the property being a sliding scale and realize it's "Yes/No" aspect. Where does that intuition cross to the other side? How "much" does the relation have to be reflexive to get a subjective answer from them?
I'm very much looking forward to experimenting with this question and learning more about student learning!
