This question is motivated by my midterm exam. In this exam there was a question as follow:

Question: If the following statement is true, prove it, otherwise disprove it.

If $\mathbf{u}$ and $\mathbf{v}$ are vectors in three dimensions, then $\mathbf{u}\times\mathbf{v}=\mathbf{v}\times\mathbf{u}$. The $\times$ operation here means cross product.

For this question I actually proved that $\mathbf{u}\times\mathbf{v}=-\mathbf{v}\times\mathbf{u}$ and so the statement is false but my lecturer deducted some marks and said that my solution is not correct. She said for disproving question, you need one counter example. But still I think that I disproved it because I clearly showed that the given statement does not hold.

So my question after this long story is, what is the correct way of disproving a mathematical statement?

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    $\begingroup$ He is wrong, showing a counter example is not the only way to disprove a statement. Your proof doesn't really work either. You merely proved what you proved. That alone doesn't disprove the statement. Why couldn't both of the statements be true? $\endgroup$ – Git Gud Jun 15 '14 at 11:34
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    $\begingroup$ There can be other way(s) to disprove a claim, yet in all cases I can think of right now the only mathematical way that I can see is by means of a counterexample... $\endgroup$ – DonAntonio Jun 15 '14 at 11:55
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    $\begingroup$ In this case, you do not need to explicitly exhibit a counterexample, but you definitely need to prove that a counterexample exists. $\endgroup$ – Andrés E. Caicedo Jun 15 '14 at 23:06
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    $\begingroup$ The statement is false if in addition you prove that u x v is not always zero. Apart from this, finding a correct formula is surely more valuable than giving a counter example for an incorrect one. $\endgroup$ – gnasher729 Jun 15 '14 at 23:18

A correct solution is any solution that works. That is like saying "it's correct if it's correct" but the point is that you don't have to use a counterexample. In most cases, if you can get away by showing a short and sweet counterexample to a claim then you should certainly do that. Always opt for the short and simple demonstration.

As for your professor's comment. If he said that you proof is incorrect because you did not provide a counterexample, then that is not a correct criticism, as it seems to imply that the only way to give an answer to this question is by counterexample, which is not the case. If, however, he claimed that you did not entirely prove the claim, then that is correct, since you proved an other equality holds, not the the original one does not.

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    $\begingroup$ Disproving $\forall x: P$ is the same thing as proving $\exists x: \neg P$. While you don't have to find a counterexample to disprove $\forall x: P$, you must prove that one exists. $\endgroup$ – Hurkyl Jun 15 '14 at 12:13
  • $\begingroup$ I think the point that good counterexamples are usually shorter and thus easier to follow than other forms of disproving a claim is the most important part here. Logically, you can of course prove $\neg C$ to disprove $C$. $\endgroup$ – G. Bach Jun 15 '14 at 15:56
  • $\begingroup$ Technically, if the professor really said the exact phrase "you need one counter-example", it doesn't necessarily imply "you must construct one counter-example". So you didn't prove that the professor's comment is incorrect either. =) $\endgroup$ – user21820 Jun 16 '14 at 3:02

I agree with your professor. You should still argue that it cannot be true that both equations hold simultaneously. This might not be a $100$% clear. From $u\times v=v\times u$ and $u\times v=-v\times u$ it follows that $u\times v=v\times u=0$. So then it is sufficient to give $u,v$ such that $u\times v\not=0$.

But normally a counterexample (as explicit and simple as possible) is the correct way to disprove a mathematical statement.

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    $\begingroup$ More specifically, after showing that $u \times v = -v \times u$, it's sufficient to demonstrate at least one pair $u, v$ such that $v \times u \ne -v \times u$. $\endgroup$ – Ilmari Karonen Jun 15 '14 at 16:29

In my experience, there are three common types of proofs of falsity. These are:

  1. Finding an explicit counterexample.
  2. Showing the existence of a counterexample without exhibiting it (and in some cases with not the slightest chance of exhibiting one!)
  3. Assuming the thing you want to disprove and inferring something you know to be false. This is called reductio ad absurdum, or proof by contradiction. It may seem bizarre to assume a (possibly) false proposition, but what you are really doing is considering the truth value of propositions conditional on the proposition you want to disprove. To be more explicit, if you know "A implies B" is a true statement and you know that B is false, you also know A is false.
  • $\begingroup$ Maybe I'm wrong, but I think B being conditional on A means A is a prerequisite of B, or $\neg A\implies \neg B$, not the other way around. (Just a nitpick on wording) $\endgroup$ – David Z Jun 15 '14 at 16:09
  • $\begingroup$ I see what you mean, but I was using the word conditional exactly like in "conditional probability" (but where probabilities are 0 or 1, since this is logic). This corresponds to the proposition if A then B. As you point out, the two uses of the word might be slightly confusing: the truth value of B given the truth value of A (my use) and the truth of B being conditional on A being true are very different concepts, but the other words in the sentence makes it unambiguous which it is. $\endgroup$ – Elroch Jun 15 '14 at 18:46

There's no real answer to "does every disproof have to involve a counterexample", because the question is too vague. What exactly is a counterexample? And what exactly is the difference between disproving something and proving something? Note that proving any statement can be thought of as proving that its negation is false, so there's no hard line between proofs and disproofs.

Statement: There are finitely many prime numbers.

The proof that this is false is just the proof that there are infinitely many prime numbers, which doesn't involve any kind of counter-example. What would a counterexample to there being finitely many of something even look like?

There's no deep mathematical laws at work here, I think all that's happened is that your professor made a slightly inaccurate choice of words, and she really just wanted to push you to finish the proof. Indeed, just because you show $x=-y$ doesn't mean you can't have $x=y$. You could have $x=y=0$. You need the extra fact that there are vectors $u, v$ such that $u\times v\neq0$. It's a bit nitpicky, but still a legitimate criticism if your job is to teach people to prove things.

  • $\begingroup$ A counter example is an instantiation with a constant $a$ for a universal statement $\forall xP(x)$ such that $P(a)$ is false. Counterexamples for existential statements are not defined (at least I don't see how). To disprove something can easily be interpreted as proving the negation holds (and I can't really see any other way of interpreting this). $\endgroup$ – Git Gud Jun 15 '14 at 13:08
  • $\begingroup$ @GitGud I was actually going to put in something like "(without getting into formal logic)" after my first sentence. It slipped my mind after I started typing. I suspected there must be some meta-mathematical way of approaching it. $\endgroup$ – Jack M Jun 15 '14 at 13:20
  • $\begingroup$ The proof of infinitely many prime numbers which I have seen involves presuming there are a finite number of primes and then showing a contradiction (multiplying all integers less than it together and adding one, giving a number which cannot be the product of any of the small numbers.) $\endgroup$ – user157288 Jun 15 '14 at 22:53

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