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.