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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