Dividing by possibly zero value when proving an identity?

In my trigonometry course, we're currently proving identities. I'm wonder if I can divide by something that could be zero while proving it. For example, $\sin{x}$, is it still a valid proof if I divide by it?

edit: Sorry, what I really meant to ask was if it's alright to divide by a variable when DISproving an identity. I would assume it's alright if the disproof stands, right?

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strictly, you should create a separate case for what happens when $sinx = 0$, although this is often ignored. –  Ronald Apr 24 '12 at 7:56
You have to assume that it differs from zero and consider that individual case separately. –  Thomas E. Apr 24 '12 at 7:56
Alright, so every time I divide by something that could be zero, I have a new case to prove? –  mowwwalker Apr 24 '12 at 7:57
yes. two cases, one where it's not zero (you can safely divide) and one where it is zero (it should simplify). –  Ronald Apr 24 '12 at 7:58
This is why division by zero should be considered a separate case: $$a = b$$ $$a^{2} = ab$$ $$a^{2} - b^{2} = ab - b^{2}$$ $$(a - b)(a + b) = b(a - b)$$ $$\dfrac{(a - b)(a + b)}{a - b} = \dfrac{b(a - b)}{a - b}$$ $$a + b = b$$ $$b + b = b$$ $$2b = b$$ $$2 = 1$$ –  Neil Apr 24 '12 at 8:00

3 Answers

Good job noticing this issue. It is not valid to divide by a quantity that could be zero. Fortunately, there are two easy ways to deal with the situation.

1. Split the proof into two cases: one where the quantity is zero, and one where it is not zero. The first case is usually pretty easy, and division is valid in the second case

2. Instead of dividing, try factoring instead. e.g. to solve the equation $$x \sin x = \sin x,$$ rather than dividing by $\sin x$, instead do $$x \sin x - \sin x = 0,$$ $$(x-1) \sin x = 0$$ and now you can invoke the usual theorem that says this is equivalent to "$\sin x = 0$ or $x-1 = 0$" and continue from there. Of course, this isn't too much different from option #1.

In some cases, there are other things you can do. For example, you might be able to prove an identity under the assumption that a certain quantity is nonzero, and then use some other means to prove it in the remaining case.

As a contrived example, suppose I had to assume $x \neq 0$ to prove $$2 \sin x \cos x = \sin 2x.$$ I could finish the proof by observing both sides are continuous: $$2 \sin 0 \cos 0 = \lim_{x \to 0} 2 \sin x \cos x = \lim_{x \to 0} \sin 2x = \sin (2 \cdot 0).$$ This is a good trick for a variety of situations. (of course, in this particular case, it would be easier to just plug in $x=0$ to verify the identity holds) (penartur's answer has other examples)

Also, I should point out that this isn't just an abstract concern, or a pedantic point to counter silly "proofs" that 2=1. Sometimes, those extra cases really matter.

For example, in multi-variable calculus, a lot of people have trouble with Lagrange multipliers, since the answer you want quite frequently comes from the "assume the quantity is zero" case, rather than from "divide by the quantity because I'm going to assume it's nonzero" case.

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Alright, and about disproving an identity? –  mowwwalker Apr 24 '12 at 8:07
Disproving an identity is usually just looking for a counterexample, right? You can assume whatever you want in your scratch work, as long as at the end you verify your counterexample actually is a counterexample. Verification is important; sometimes you might do your work and discover that $x = 0$ looks like a counterexample, but it turns out not to be. Then when you reanalyze your work, you find out that you assumed half-way through that $x \neq 0$. –  Hurkyl Apr 24 '12 at 8:22
e.g. For another contrived example, to disprove the identity $x^2 = x$, you might simplify by dividing through by $x$ to get $x=1$, and then say "$0 \neq 1$, so $x=0$ is a counterexample to the identity!". Alas, $x=0$ turns out to be one of the few cases where the equation actually holds! –  Hurkyl Apr 24 '12 at 8:24
I can't think of any simple examples off hand, but I have seen things like this: some equation $f(x) = 0$ that holds for every nonzero $x$ but fails for $x=0$ that gets turned into a valid identity $x f(x) = 0$. –  Hurkyl Apr 24 '12 at 8:31

@Hurkyl is right about dividing by zero in the common case, but I'd clarify something for your specific case.

You've said that you're dealing with identities, not equations (the identity is like $\sin{2x} = 2\sin{x}\cos{x}$). In that sense, dividing by something which could be zero (but only at some points: such as $\sin{x}$, but not $\min(0, |x|-1)$) is legitimate, because:

• On the both sides of the equation you have functions with no discontinuities (except for the countable set of points where the limits from both sides are equal to $\infty$, and the value in the point itself is not defined);

• When dividing by e.g. $\sin{x}$, you achieve the continuous (in the sense above) function;

• Given two continuous (in the sense above) functions which are equal almost everywhere (e.g. everywhere except for the countable set of points), one can deduce by continuity that the functions are equal everywhere where they're defined, so it is safe to say two functions are equal / identical.

Thus you can prove e.g. that $\tan^2{x} = \dfrac{\sin^2{x}}{\cos^2{x}} = \dfrac{1-\cos^2{x}}{\cos^2{x}} = \dfrac{1}{\cos^2{x}}-1$ without having to bother about the points where $\cos{x}$ is zero.

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+1: nice clarification –  Ronald Apr 24 '12 at 8:21

Well, in the real field a division by zero is always forbidden! When in doubt, please check the case in which the denominator is zero separately.

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