I was tasked with proving a trig identity -- the particular identity doesn't much matter, as I'm interested far more in whether this is a valid method of proof, generally -- and a professor recommended this strategy: equate both sides and, via dot and cross products and rules of vector algebra, produce a trivially true statement. The conclusion, then, is that the original identity is correct.

This strategy is a bit new to me. I've usually been told to start on one side of the equation and generate the other. But, in thinking about this more, and specifically in trying to provide a counterexample, it seems that this strategy is valid provided that we can reverse each of the steps we took to get down to some trivial statement. In various proofs involving inequalities, for example, which require reversing signs, this is likely to fail.

Is my intuition here correct? I would be very interested in hearing of possible counterexamples to this or conditions where a proof like this is appropriate, as well as whether such a proof is frowned upon -- proofs like this do seem rather uncommon, and I believe this may be the first time I've seen such a statement proved like this.

  • 1
    $\begingroup$ Yes, your intuition is absolutely correct. You must be able to reverse each step. In fact, the forward direction is entirely unimportant. Producing reverse proofs is a handy technique for cracking a problem, but the working produced cannot always be turned into a proof. $\endgroup$ – Theo Bendit Sep 27 '18 at 0:50

First let us look at an example where this kind of arguments fail to work.

Let $V=<3,4,5>$ and $W=<-4,3,0>$ and $Z=<0,-5,4>.$ Prove that $W=Z$

Let $W=Z$

Now find the dot product of both sides with $V$.

We find $V.W=0$ and $V.Z=0$ Therefore, $V.W=V.Z$ which is true in this case.

We started with a false assumption and via dot product we came to a true statement.

Thus this method does not work in general.

On the other hand if every step is reversible, then the method will work.

So the key is that every operation on both sides must be reversible so we can go backward and deduct the original statement from the terminal statement.


No, this is not a valid method. Let your trig identity be A and your 'trivial statement' be B. Then with your method you show 'A implies B', in general this does not equate to 'B implies A'(*) - which is what I assume you believe to be true. A non-mathematical (just logical) analogy would be 'I just shot myself in the head and now I'm dead'. Assume 'I shot myself in the head' (A), the trivial statement following this would be 'I am dead' (B). However, if I am dead (B), I did not necessarily shoot myself in the head (A).

A similar, ubiquitous method is proof by contradiction. (I assume you've come across this before) But if 'not A' implies 'not B', then this is equivalent to 'B implies A'(*)

  • 2
    $\begingroup$ Or, a classic, more directly mathematical example: assume $-1 = 1$. Then $(-1)^2 = 1^2$ and so $1 = 1$. Therefore $-1 = 1$? $\endgroup$ – Theo Bendit Sep 27 '18 at 0:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.