# Is 2=1 Possible?

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 ???

Note by Shravan Jain
7 years, 5 months ago

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

- 7 years, 5 months ago

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.

- 7 years, 3 months ago

Well the you only differentiated the x's,the number of x's should also be differentiated as it is also a variable

- 7 years, 5 months ago

Since this function is non differentiable , there is no question of differentiating the no. of x's

- 7 years, 5 months ago

Well,I dint even consider that lol

- 7 years, 4 months ago