Recently, I was doodling in my rough notebook and found something strange:

Let **x** belong to N (natural no.)

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

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

(Differentiating on both sides)

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

\(\Rightarrow 2x = x \)

Where was I wrong ??

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## Comments

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TopNewestWhen you are defining x^2=x+x+..+x(x times), you are tacitly defining the domain of f(x) to be N, the set of natural numbers. However, f(x) cannot be differentiable in the domain of N. This is because, if a function f defined from a domain D (a subset of R, the set of reals) to R (the set of reals) has to be differentiable at a real point c, then a necessary criterion is that c has to be a limit point of D and c has to be an element of D itself. In other words, c has to be a member of D such that every arbitrarily small neighbourhood of c has an element of D other than c. But no natural number is a cluster point of the set N. Here lies the fallacy.

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Actually you cannot differentiate the function

f(x)= \(x^{2}\) when you have selected the Domain as Natural numbers . After all , the Natural numbers as a domain will comprise just discrete points , so it'll not be differentiable.Log in to reply

Firstly I differentiated it generally but in that case i couldn't write the no. x times as x could be fraction or 0. Thank you now i have understood.

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Anytime . Btw there was a question on this

fallacyposted by @Sandeep Bhardwaj sir , I'm not able to find it . I'll give you the link if I'm able to find it .Log in to reply

note on this

BTW i just found anotherLog in to reply

set of such apparent

In fact i found a wholefallaciesLog in to reply

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Very good. He is fooling.

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When you are differentiating on both sides, you assumed that in x times the x is constant. Also you can't talk abt differention, of non continuous function

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I couldn't understand your first reason.

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