5

Possible Duplicate:
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 $2$.

L. F.
  • 1,940
Bazinga
  • 1,949

2 Answers2

20

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

Pedro
  • 122,002
  • 5
    Besides, it's also not clear what it means to add a number to itself a non-integer number of times. That interpretation of multiplication breaks down when you move beyond integers to fractions. – Neal Jun 29 '12 at 04:06
  • @Neal True. I'm thinking about it from the differentiation perspective. – Pedro Jun 29 '12 at 04:20
  • 4
    I'm sure that Neal is also thinking 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 context of calculus on the real line. – Jonas Meyer Jun 29 '12 at 04:27
8

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?

Michael Hardy
  • 1
  • The arrangement of math equations in the question is very ugly. Help to edit it, such as $\underbrace{x+x+\cdots+x}_{x\textrm{ times}}$ instead of the one in the question, and displaymath is not necessary. His $dy/dx$ can be arranged with inline math. – Yai0Phah Jun 29 '12 at 04:21
  • 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. – ThisIsNotAnId Jul 27 '12 at 03:36