Another $1=2$ proof So a friend shows me this :
$x^4= x^2+x^2+ \cdots +x^2 $ ( i.e.  $x^2$ added $x^2$ times)
Now take the derivative of both side;
$4x^3 = 2x + 2x + \cdots + 2x $;
So $4x^3 = 2x^3  \cdots $(1)
And so dividing by $x^3$ gives  $2=1  \cdots $(2).
I know we can't divide by 0 so that makes (2) false, but to show that (1) is false too ?

*

*First of all, the statement 



$x^2$ added $x^2$ times

makes sense only if $x^2$ is a positive integer. Else if $x^2$ is not a positive integer, then the statement is meaningless.


*Moreover, from $(1)$, we have that $4x^3=2x^3 \Rightarrow 4x^3-2x^3=0 \Rightarrow 2x^3=0 \Rightarrow x^3=0$


And hence division by $x^3$ in the next step is meaningless.
