A proof of 2=1 using derivative

Question

We know that $2^2=2+2$

$3^2=3+3+3$

Similarly

$x^2=x+x+...$ ( upto $x$ times)

Now I want to differentiate both side with respect to $x$

This gives $2x=1+1+1+.....$ ( upto $x$ times)

$2x=x$

Cancelling $x$ I have

$2=1$

Which should be false. Where am I going wrong? Please tell. The proof has to be wrong somewhere.

Deriving is a continuity argument, however, your equality works only at one point! I.e. As functions your equality does not hold — Felix, Nov 23 '19 at 13:17

score -1 · Answer 1 · answered Nov 23 '19 at 13:16

-1

You assume that $x$ is a natural number, but derivatives don't exist for functions on the set of natural numbers since any neighborhood of a natural number contains infinitely many non-natural numbers.

answered Nov 23 '19 at 13:16

Nicolas

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A proof of 2=1 using derivative

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