It seems to be in fashion among people who know a little basic math to “prove” using sketchy math and subterfuge that zero equals one, or that one equals two, or something like that. You can even find a few of these on youtube. Most of these are just elaborate obfuscations of the same error—division by zero—and don’t have a lot of mental calories as far as I can see. But a few have something to teach us. This one, definitely, is my favorite, and I even bring it up when I teach integral calculus.
For this trick, we need standard calculus, namely the integration by parts formula. If and are any differentiable functions and and are their differentials as usual, then the formula says that
which is really just the product rule for derivatives rejangled around. Usually we apply this technique to integral involving logarithms or products of polynomials and trig functions or exponentials, but let’s see what happens when we apply it to .
Take . Then
Cancelling the like terms, . Ta-da!
Okay, looks like Dr. Cap is cheating again. Resolution after the jump.
The point here is that the indefinite integral of a function is not a function in the normal sense, it’s a family of functions. More precisely, consists of all the functions whose derivative is . So the indefinite integral is really an equivalence class of functions, where two functions are equivalent if they differ by a constant, that is, if their difference is independent of x. (Your calculus teacher, if you got one of the standard issue models, may have docked you a point each time you forgot the “+C”.) Normally the fact that indefinite integrals are really equivalence classes of functions is obvious (or at least inferrable) from context, and no confusion results. But here when we cancel like terms, we remove the context that reminds us that our equation is one where “=” only means “equals up to a constant function”.
(If you know what a quotient group is, the point here is that indefinite integral operator doesn’t really output in the additive group of functions, it outputs in the quotient group of functions modulo the subgroup of constant functions.)
Viewed in this light, our argument is not incorrect at all. What was incorrect is our interpretation of the conclusion. We have proven that the function which always returns 0 and the function which always returns 1 are the same, up to a constant. That is, the difference between 0 and 1 doesn’t depend on x. And that’s perfectly true.
I am reminded of a joke which I picked up at some conference or other which is not completely irrelevant.
Two professors of mathematics were having lunch at a diner and discussing the sorry state of basic mathematical knowledge among the general public. When one of the professors excused himself from the table in order to make a phone call, the other professor took the opportunity to discreetly call the waitress over.
“If I give you five dollars, can you just say ‘x cubed over three’ when I ask you a question later?”
“Just say ‘x cubed over three’, and don’t worry about the question. You got it?”
“x cubed over three, sure.”
“Great, here’s your five dollars. This conversation never happened.”
So the first professor returns, and they resume the conversation. Ultimately they decide to bet on whether a layperson, say, this waitress, knows any basic calculus facts, loser pays for the lunches.
So when the waitress finally came with the bill, the second professor smiles and asks, “Excuse me, miss, do you know the indefinite integral of x squared?”
The waitress looked pained, sighed, gritted her teeth, and said “x cubed over three”.
The first professor was thunderstuck, and grumbled some praise at the waitress as he paid her for their meal.
After the professors were safely out of hearing, she added “…plus a constant.”