# To prove that a mathematical statement is false is it enough to find a counterexample?

I am trying to show if the following statements are true or false.

1. Is it true that $|a + b| = |a| + |b|$ for general vectors $a$ and $b$?

2. If $a \cdot b = a \cdot c$ for equally-sized non-zero vectors $a$, $b$, $c$, does it follow that $b = c$?

For the first one I found a counterexample that shows that the statement is false.

If vector $a=\langle 1,4,5\rangle$ and $b=\langle 2,2,2\rangle$ then $|a|+|b|=\sqrt{42}$+$2\cdot\sqrt{3}=9.945$, and then, $|a+b|=\sqrt{1^2+4^2+5^2+2^2+2^2+2^2}=7.348$,

then we can conclude that $9.945 \ne7.348$ and the statement is false. Also by the triangle identity $|a + b| \le |a| + |b|$

For the second statement I also found an counterexample that proves that is false. If vector $a=\langle 1,0,0\rangle$, $b=\langle 0,1,0\rangle$ and $c=\langle 0,0,1\rangle$, we will obtain the following dot product:

$a \cdot b = 0$

$a \cdot c = 0$

then $a \cdot b = a \cdot c = 0$ and $b \ne c$

My question is:

To prove the two statements is it enough to find a counterexample and say if it is true or false. Or should I try to provide a more mathematical proof like induction or contradiction?

• It suffices to find a counterexample to show that a statement is not true. Feb 17, 2017 at 18:26
• Yes, this is enough. The only thing that is unclear is your statement "Also by the triangle identity $|a + b| = |a| \le |b|$". First, did you mean $|a + b| - |a| \le |b|$? Second, what do you mean to prove by this? Feb 17, 2017 at 18:29
• Okay, I see it was a typo. But more importantly, you should realize that citing the triangle inequality is not enough to answer the question, because it doesn't in itself contradict statement (1). The triangle inequality should suggest to you that you can probably find $a,b$ such that $|a+b| < |a|+|b|$, which would contradict statement (1), but you don't need to mention the inequality. Showing that counterexample (as you did) is the real proof. (I just noticed that @zipirovich effectively made the same comment about this in their answer below.) Feb 17, 2017 at 19:27
• In response to Einsteins theory of relativity, a book titled A Hundred Authors Against Einstien had been written. When asked about the book, Einstein retorted by saying “Why 100 authors? If I were wrong, then one would have been enough!” Feb 17, 2017 at 22:45
• @palsch: I think you mean "example that refutes an assertion or claim". Feb 19, 2017 at 12:47

When considering a statement that claims that something is always true or true for all values of whatever its "objects" or "inputs" are: yes, to show that it's false, providing a counterexample is sufficient, because such a counterexample would demonstrate that the statement it not true for all possible values. On the other hand, to show that such a statement is true, an example wouldn't be sufficient, but it has to be proven in some general way (unless there's a finite and small enough number of possibilities so that we can actually check all of them one after another).

So logically speaking, for these two specific examples, you're right — each one can be demonstrated to be false with an appropriate counterexample. And both your counterexamples do work, but make sure that the math supporting your claim is right: in the first example you computed $|a+b|$ incorrectly.

By the way, the reference to the triangle inequality is a good touch, but it doesn't prove anything. Rather, it's a very strong hint that suggests that there have got to be examples when the inequality rather than equality holds.

• So, is it not correct to mention the triangle inequality in this case? Feb 17, 2017 at 19:00
• Here's my opinion on problem-solving in general, proof questions in particular. First there's the process of looking for a solution or a proof, thinking about it -- and in this example, the triangle inequality is an important hint suggesting that the statement is false. But then we write up a final narrative of a solution or proof, without any sidetracking notes and such -- and for this problem the triangle inequality wouldn't be there. Feb 17, 2017 at 19:04
• In short, saying that this statement is false by the triangle inequality would not qualify as a proof. The triangle inequality per se does not claim that vectors with $|a+b|<|a|+|b|$ exist. It claims that $|a+b|\le|a|+|b|$ for all $a$ and $b$. Note that logically even if "$|a+b|=|a|+|b|$" were true for all $a$ and $b$, that wouldn't contradict the triangle inequality. Feb 17, 2017 at 19:07
• I wouldn't even call the triangle inequality a strong hint. I am sure part of the purpose in giving this assignment was to convince the students that the triangle inequality cannot be strengthened into an equality, by having them find counterexamples themselves. Feb 17, 2017 at 22:34
• @PaulSinclair: I partially agree with you, and partially don't. I agree that that's part of the purpose of the exercise. But what I meant was that from a student's perspective reasoning could go like this: I need to prove or disprove "$|a+b|=|a|+|b|$" -> they taught us the triangle inequality saying that $|a+b|\le|a|+|b|$ -> why didn't they make it equals? -> either they are not telling the truth, or in fact it's not always equals and thus had to be less or equal -> so I guess I should be looking for a counterexample. Feb 17, 2017 at 23:37

If a statement ${\cal S}$ is of the form "all $x\in A$ have the property $P$" then a single $x_0\in A$ not having the property $P$ proves that the statement ${\cal S}$ is wrong.

But not all statements are of this form. For example the statement ${\cal S}\!:\>$"$\pi$ is rational" cannot be disproved by some "easy" counterexample, but only by means of hard work.

Yes, any counterexample will do. It's often instructive to look for the simplest counterexample. For example, take the one-dimensional vector space $\Bbb R$ and the vectors $a=1$ and $b=-1$ in the case of the first statement.

If the statement is true, then you give a mathematical proof. Since you can't find all feasible inputs to prove that the statement is true, even computers can't do this for some statements.

If the statement is false, then you give one counter example. Since the statement says that it is true for all the feasible inputs, you just have to find one feasible input which doesn't satisfy the statement.

As for your solutions, everything's seem correct with two mistakes. $a+b$ in the first example is <3,6,7> and triangle inequality is $|a+b| \leq |a|+|b|$.

• I did a mistake. I edited my question. Thank you for your help. Feb 17, 2017 at 18:41
• Glad to be of help @user290335 Feb 18, 2017 at 10:10
• Someone downvoted this. Can I know what to improve in this answer? Feb 18, 2017 at 10:10

As other answers have explained, if you have a claim that something is true for all possible inputs, then a single counterexample disproves the claim. Period.

Perhaps, though, you may have some lingering doubt about how "generic" the counterexample is. Sure, maybe we can prove that $\lvert a+b \rvert = \lvert a \rvert + \lvert b \rvert$ is false for $a=(1,4,5)$ and $b=(2,2,2)$. However, for all we know, this might be the only counterexample. We already know the claim is false (beyond any doubt), but maybe it's only "technically" false. Maybe it's true "in the typical case", and we just happened to find the only exception.

For example, suppose I claim "$x^4 + y^4 \ne z^4$ for all integers $x$, $y$, and $z$". You say, "That's not true: $0^4 + 1^4 = 1^4$". You'd be right: Indeed my claim is incorrect. But if you're willing to let me move the goal posts a little, I can easily salvage the claim by excluding "trivial" counterexamples like $0^4 + 1^4 = 1^4$. The claim remains true "in the typical case" - "technically" false but true "in spirit".

Most of the time you won't see people explicitly address how "generic" their counterexample is. They'll just give one counterexample and be done. And implicitly, they're saying that this one counterexample is convincing enough that the claim is "generally" false. Because:

When a statement is false, it's "generally" false... generally.

This belief is completely non-rigorous and not formally justified, but that's okay - there was no formal meaning for "typical case", "generally true/false", and "trivial" to begin with.

How could we make a more convincing argument that a statement is not only false, but "generally" false?

Depending on the situation, we might try to show that:

• There are infinitely many counterexamples (not just one)
• There are infinitely many counterexamples that aren't just obvious variations of each other (e.g. not just constant multiples of each other)
• The set of counterexamples has infinite volume
• A random input (drawn from a chosen probability distribution) is a counterexample with probability $1$

... the list goes on. In the most fortunate case, we can find a definite answer by describing exactly which inputs satisfy the claim and which ones don't. For example, it turns out that "$\lvert a+b \rvert = \lvert a \rvert + \lvert b \rvert$" is true if and only if $a$ and $b$ are positive scalar multiples of each other, or one or both of them is $0$. This is a rigorous statement which is stronger than merely disproving "$\lvert a+b \rvert = \lvert a \rvert + \lvert b \rvert$ for all $a$ and $b$", and can be informally interpreted as saying that "$\lvert a+b \rvert = \lvert a \rvert + \lvert b \rvert$" is "usually" false.

These are the ways you could go beyond proving that a claim is false, to prove that it's "very" false.

But enough talk about false statements being "generally false" or "only technically false". It's still false. All such talk involves moving the goal posts, replacing the original statement with a different one.

For whatever actual, specific claim you might be considering, one counterexample is enough to disprove it.