**Problem**: Show 1 = 2 (with calculus)

“**Solution”**: Consider the following:

( times)

And since this is true for all values of , we may take the derivative of both sides, and the equality remains true. In other words:

( times)

Which simplifies to , and plugging in we have , as desired.

**Explanation**: Though there are some considerations about the continuity of adding something to itself a variable number of times, the true error is as follows. If we are taking the derivative of a function with respect to , then we need to take into account *all* parts of that function which involve the variable. In this case, we ignored that the number of times we add to itself depends on . In other words, ( times) is a function of *two* variables in disguise:

( times)

And our mistake was to only take the derivative with respect to the first variable, and ignore the second variable. Unsurprisingly, we made miracles happen after that.

Addendum: Continuing with this logic, we could go on to say:

( times)

But certainly the right hand side is *not* constant with respect to , even though each term is.

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Clever! (although there is a typo under “following:”. You mean 3^2, but write 3^3.

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Also, the phrase x^2 = x + … x (x times) [I wish I could type that in sigma notation in this post] is only defined on the natural numbers. And you can’t really even define a derivative on a discrete metric space like the natural numbers.

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So that is an issue here, but with a slick enough definition of what it means to add something to itself “ times,” one can extend the function to the real line. For instance, take ( addends)

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