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.

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

  1. 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 “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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