# False Proof: 1 = 2 (with Calculus)

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.

## 3 thoughts on “False Proof: 1 = 2 (with Calculus)”

1. Fredrik Meyer

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

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2. Allan Boone

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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• j2kun

So that is an issue here, but with a slick enough definition of what it means to add something to itself “ $x$ times,” one can extend the function to the real line. For instance, take $x + x + \dots + x$ ( $\left \lfloor x \right \rfloor$ addends) $+ (x - \left \lfloor x \right \rfloor ) x$

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