Given that \(6\times6 = 6+6+6+6+6+6\), or equivalently

\(x\cdot x = x + x + x +x +x+x...\), where \(x= 6\).

Differentiate both sides with respect to \(x\), we get

\( \dfrac{d}{dx} (x^2) = \dfrac d{dx} x + \dfrac d{dx} x + \dfrac d{dx} x + \dfrac d{dx} x + \dfrac d{dx} x + \dfrac d{dx} x... \)

\(2x = 1 + 1 + 1 + 1 + 1 + 1 ...= 1\cdot x \)

\(2x = x \)

\(2 = 1\).
why?

## Comments

Sort by:

TopNewestYou can't differentiate an equation until it's an identity. Also, both sides must be defined in the neighbourhood of the point at which you're differentiating. As these statements don't hold for the above equation, the proof is flawed. – Deeparaj Bhat · 6 months, 1 week ago

Log in to reply

2x = 1 +1+1+1+1+1......(2x times) 2x =2x Not 2x=x So 2 not same with 1 – Ivan Tanuwijaya · 5 months, 3 weeks ago

Log in to reply

A link to solution

Brilliant Logo

Log in to reply