0

We know $x.x=x^2$(Consider $x\ne 0$)

$x.x$ is adding $x$ $x$ times. So we have

$x+x+\dots+x$($x$ times $x$ is added)=$x^2$......(1)

Differentiating both sides of (1) we get,

$1+1+\dots+1$(x times)=$2.x$

$\Rightarrow x=2x$

$1=2$ Proved.

Find out the mistake in this proof.

Abhra Abir Kundu
  • 7,910
  • 18
  • 36
  • 3
    How does something occur $x$ times if $x\notin\mathbb{N}$? – Ben West Feb 15 '13 at 07:01
  • 1
    Also [How to disprove this fallacy that derivatives of $x^2$ and $x+x+x+⋯$ (x times) are not same](http://math.stackexchange.com/questions/164444/how-to-disprove-this-fallacy-that-derivatives-of-x2-and-xxx-cdots-x-tex) – MJD Feb 15 '13 at 07:03
  • You forgot the chain rule; you need to account for the derivative of the operation "... (x times)" (which, of course, doesn't exist since it's restricted to $x$ being an integer) –  Feb 15 '13 at 07:04
  • there is one evident mistake: x times is valid when x is integer. when it is integer differentiation is not valid. –  Feb 15 '13 at 07:06
  • The same reasoning shows that differentiating $x^2=\underbrace{1+\cdots+1}_{x^2\text{ times}}$ we get $0$. – Stefan Hansen Feb 15 '13 at 07:13

0 Answers0