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

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

$ 1^2 = 1$

$ 2^2 = 2 + 2$

$ 3^2 = 3 + 3 + 3$

$ \vdots$

$ x^2 = x + x + \dots + x$ ($ x$ times)

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

$ 2x = 1 + 1 + \dots + 1$ ($ x$ times)

Which simplifies to $ x=2x$, and plugging in $ x=1$ we have $ 1 = 2$, 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 $ x$, 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 $ x$ to itself depends on $ x$. In other words, $ x + x + \dots + x$ ($ x$ times) is a function of *two* variables in disguise:

$ f(u,v) = u + u + \dots + u$ ($ v$ 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:

$ x = 1 + 1 + \dots + 1$ ($ x$ times)

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

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