Once my teacher gave a proof of 2=1 which is as follow:

\( { 1 }^{ 2 }\quad =\quad 1\\ { 2 }^{ 2 }\quad =\quad 2\quad +\quad 2\\ { 3 }^{ 2 }\quad =\quad 3\quad +\quad 3\quad +\quad 3\\ .\\ .\\ .\\ .\\ .\\ { x }^{ 2 }\quad =\quad x\quad +\quad x\quad +\quad x\quad +\quad ........\quad upto\quad x\quad terms.\\ \\ Now\quad differentiating\quad both\quad sides\quad w.r.t\quad x,\\ 2x\quad =\quad 1\quad +\quad 1\quad +\quad 1\quad +\quad ........\quad upto\quad x\quad terms\\ ie,\quad 2x\quad =\quad x\\ \quad \quad \boxed { { 2\quad =\quad 1 } } \)

Now we were challenged to find the flaw in this proof ( Which we won )

Can you find the flaw in this proof ???

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TopNewest\( Since\quad x\quad can\quad take\quad only\quad positive\quad integral\quad values\quad \\ (as\quad { -1 }^{ th }\quad term,\quad { 0 }^{ th }\quad term,\quad { 0.5 }^{ th }\quad term\quad does\quad not\quad exists),\quad \\ the\quad function\quad { x }^{ 2 }\quad is\quad discontinuous\quad \\ and\quad thus\quad is\quad non\quad differentiable. \) – Shravan Jain · 3 years ago

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Thank you for posting this question.

You're trying to use the equality:

\({x}^{2} = x + x + ... + x\)

You are trying to differentiate two functions. Let's see what we know about them.

Let \(f(x)={x}^{2}\). The domain for this function is clearly \(R\). And yeah when we differentiate we get \(f'(x)=2x\).

Let \(g(x)= x + x + ... + x\) (for the equality to be true x repeats in the sum exactly x times), then since a number can't repeat itself \(\sqrt { 2 } \) times the domain of this function is \(N\).

Now you differentiate. Tell me how you can calculate:

\(g'\left( { x }_{ 0 } \right) =\lim _{ x\rightarrow { x }_{ 0 } }{ \frac { g\left( x \right) -g\left( { x }_{ 0 } \right) }{ x-{ x }_{ 0 } } } \)

But you can't approach \({ x } _ { 0 }\) on \(N\). You can only differentiate on intervals or \(R\).

More than this the differentiation gives you the slope of the tangent in \({ x } _ { 0 }\), but you can't have a slope to a function that's made of dots.

Even if we took \(f\) defined on \(N\) we still wouldn't have been able to differentiate it either.

So no, 2 isn't 1. – Adrian Neacșu · 2 years, 11 months ago

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Well the you only differentiated the x's,the number of x's should also be differentiated as it is also a variable – Murlidhar Sharma · 3 years ago

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– Shravan Jain · 3 years ago

Since this function is non differentiable , there is no question of differentiating the no. of x'sLog in to reply

– Murlidhar Sharma · 3 years ago

Well,I dint even consider that lolLog in to reply