$$x^2=x + x + \cdots + x$$ Where x is repeated on the right hand side x times.
Take the derivative of both sides: $$2x=1 + 1 + \cdots + 1$$ $$2x = x$$ $$2 = 1$$
Where did I go wrong?
Asked
Active
Viewed 70 times
0
-
The right hand side sum's number of terms also depends on $x$ so you cannot just differentiate term by term like this! – SEWillB May 16 '17 at 20:25
-
What does $x+x+x+\cdots+x$ mean when $x=\sqrt{2}$? – Thomas Andrews May 16 '17 at 20:25
-
The higher-rated question, I guess, is: https://math.stackexchange.com/questions/1096/where-is-the-flaw-in-this-proof-that-1-2-derivative-of-repeated-addition?rq=1 – Thomas Andrews May 16 '17 at 20:27
-
First off, the original equation is $x^2=x\cdot x$ so $x=0$. So then an equivalent way of expressing the fallacy that @SEWillB pointed out is that $(x\cdot x)'=2x\neq x$. This first fallacy is that $x$ is a variable, not a constant so it can't be treated as such. But that's not even the only fallacy, you've also got the fact that if $2x=x$ (and $x=0$) then division by $x$ is undefined. It's an easy mistake to make since it's not immediately apparent that you can't always divide through by $x$ but that last step should have been solved as $2x=x\Rightarrow 2x-x=0 \Rightarrow x=0$ – Jam May 16 '17 at 21:38