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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Sincexcantakeonlypositiveintegralvalues(as−1thterm,0thterm,0.5thtermdoesnotexists),thefunctionx2isdiscontinuousandthusisnondifferentiable.
Thank you for posting this question.
You're trying to use the equality:
x2=x+x+...+x
You are trying to differentiate two functions. Let's see what we know about them.
Let f(x)=x2. 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 2 times the domain of this function is N.
Now you differentiate. Tell me how you can calculate:
g′(x0)=limx→x0x−x0g(x)−g(x0)
But you can't approach x0 on N. You can only differentiate on intervals or R.
More than this the differentiation gives you the slope of the tangent in x0, 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.
Well the you only differentiated the x's,the number of x's should also be differentiated as it is also a variable
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Since this function is non differentiable , there is no question of differentiating the no. of x's
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Well,I dint even consider that lol