What is the correct way of disproving a mathematical statement?
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?
$\endgroup$ 44 Answers
$\begingroup$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.
$\endgroup$ 3 $\begingroup$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.
$\endgroup$ 1 $\begingroup$In my experience, there are three common types of proofs of falsity. These are:
- Finding an explicit counterexample.
- Showing the existence of a counterexample without exhibiting it (and in some cases with not the slightest chance of exhibiting one!)
- 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.
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.
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