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Where is the flaw in this argument of a proof that 1=2? (Derivative of repeated addition)

$\begin{align*} x^2 &= \underbrace{x + x + x + \dots + x}_{x \text{ times}}, \\ \therefore \frac{\mathrm{d}}{\mathrm{d}x} (x^2) &= \frac{\mathrm{d}}{\mathrm{d}x} (\underbrace{x + x + x + \dots + x}_{x \text{ times}}) \\ &= \underbrace{1 + 1 + 1 + \dots + 1}_{x \text{ times}} \\ &= x. \end{align*}$

But we know that $ \frac{\mathrm{d}}{\mathrm{d}x} (x^2) = 2x. $

So what is the problem?

My take is that we cannot differentiate both sides because $\underbrace{{x+x+x+\cdots+x}}_{x \text{ times}}$ is not fixed and thus $1$ is not equal to $2$.

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    See [here](http://mathoverflow.net/questions/38856/jokes-in-the-sense-of-littlewood-examples/39950#39950) too.2012-06-29

2 Answers 2

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Simply because "$x \text{ times}$" is also a "function" of $x$. One mistake is not considering that in the derivation.

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    I'm sure that Neal is also thi$n$king about it from a differentiation perspective. You can't take the derivative of a function with no definition, or that is defined only at the integers, in the conte$x$t of calculus on the real line.2012-06-29
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You say "$x\text{ times}$". The number of "times" you add it up---the number of terms in the sum---keeps changing as $x$ changes. An what if $x=1.6701$? How do you add up $x$ $1.6701$ times?

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    Concerning defining repeated addition for the reals, was your question for the OP's author's consideration? I think the problem is not with defining such addition, but with using the definition for $x^2$ that has been used in the OP.2012-07-27