I have a question where we are given a statement and asked to prove whether or not the statement is true. If the statement is true, then we must prove it; otherwise we must provide a counterexample to prove the statement incorrect.

If the statement is false, and the student provides a counter-example; should the student also provide an explanation of why the counter-example disproves the statement?


2 Answers 2


I believe this depends on how clearly the counterexample is stated.

Consider this claim: $f:\mathbb{R}\to\mathbb{R}$ is continuous $\implies f$ is differentiable.

Imagine a student, let's call him Karl, who says

Take any function of the form $\sum_{n=0} ^\infty a^n \cos(b^n \pi x)$, where $0<a<1$, $b$ is a positive odd integer, and $ab > 1+\frac{3}{2} \pi$.

Would you give him credit for this counterexample? Even presuming that the difficulty level of your course and the talents of your students are such that you did, indeed, expect them to come up with this result, I don't feel like this answer is sufficient. Obviously, Karl must have understood something about the exercise to be able to come up with this answer (never mind the great ingenuity required) but his written argument fails to demonstrate that knowledge.

Contrast that with a disproof of this claim: For any function $f:A\to B$ and any sets $S,T\subseteq A$, $f[S]\subseteq f[T]\implies S\subseteq T$.

This is the kind of statement you might consider in an introductory analysis course. I'd expect a student to find an easy counterexample, say by defining $A=\{\heartsuit,\spadesuit\}$ and $B=\{\star\}$ and setting $f(\heartsuit)=f(\spadesuit)=\star$ and $S=\{\heartsuit\}$ and $T=\{\spadesuit\}$.

Would I expect the student to have to show all the details? Do they really need to explicitly say, "Observe that $S,T\subseteq A$ and $f[S]=f[T]$ and yet $S\not\subseteq T$"? Well, that depends on your standards. If you're really training them to write clearly and explicitly (concision be darned), then you might require them to show these details. If you're just checking to see that they can generate such examples, then merely stating the example would suffice.

Main answer: It depends on your standards for such a written argument, which in turn should depend on what you expect the student to be able to understand and demonstrate.

  • 6
    $\begingroup$ Sure, let's call him Karl. $\endgroup$ Sep 25, 2014 at 1:26
  • $\begingroup$ Maybe give extra credit to someone who described how to create a counter example, perhaps in the above case by using S={<heart>} and T={<something besides a heart>} $\endgroup$
    – rbp
    Sep 29, 2014 at 15:13

Certainly the student should be aware of the expectations beforehand. To this end, I think there are at least two approaches. One is to specify on individual questions whether counterexamples should be explained, and the other is to establish (preferably from the outset, i.e., setting "norms" at the beginning of a course) the expectation that reasoning/justification for all answers will be provided.

More generally, I think explaining counterexamples is important. Let me give just two reasons.

1) Consider the following problem:

Is every whole number divisible by 6 and 8 also divisible by 6 $\times$ 8 = 48? If so, prove your answer; if not, please provide a counterexample. Answer: 24.

But what does such a student understand? Perhaps she just remembered 24 is divisible by both and realized it is not divisible by 48. Or perhaps she realizes that the issue is that the two divisors in this proposed rule are not relatively prime, and could correct it by using, e.g., 3 and 16 instead. Or, even more seriously, perhaps she has the deeper understanding, but accidentally writes 32. In this last case, the understanding is the deepest, but it would be tough to justify giving the response any credit.

Asking that the counterexample be justified ameliorates this potential problem. Personally, I would expect more than just an explanation of the particular counterexample, e.g., more than:

No, and 24 is a counterexample: It is divisible by 6 because 24 = 4 $\times$ 6, and it is also divisible by 8 because 24 = 3 $\times$ 8. But it is not divisible by 48, because 24/48 is not a whole number.

I would prefer to see an answer that mentions being "relatively prime" or something equivalent, and perhaps even one that gives an example of two numbers that would yield a divisibility rule for 48.

2) More subtle and perhaps not related to the specific question you have in mind (if there is one): Note that (speaking slightly messily) counterexamples can settle for all questions, but not there exists questions. The previous example is the assertion that all whole numbers (etc). If this turns out to be false, then it can be settled with a counterexample. If it turns out to be true, then it will need a further proof or justification. On the other hand, a there exists statement that is true might be settled with a single example. If it turns out to be false, then it will need a further proof or justification.

This latter point about proving/disproving by (counter)example in the context of for all vs. there exists assertions may seem a bit pedantic, but it is a place in which many students who are just beginning to understand ideas around "proof" can struggle quite a bit.

  • $\begingroup$ I don't think there exists statements are really that different because you can rewrite them as for all statements, i.e. $\forall x\ P(x) \iff \neg\exists x\ \neg P(x)$. (Of course maybe students don't know this, in which case to their minds there is a difference...) $\endgroup$
    – David Z
    Sep 25, 2014 at 3:53
  • 1
    $\begingroup$ @DavidZ Yes, this is the underlying "reason" for the phenomenon mentioned in 2 above (i.e., that an example can disprove a for all statement or prove a there exists statement). And yes, one of the primary difficulties is precisely the one to which you allude parenthetically: For many students who are just getting into proof-based mathematics (or other mathematics with more formal reasoning) the equivalence you mention is neither known nor obvious. $\endgroup$ Sep 25, 2014 at 18:08

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