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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Ashwin Upadhyay
3 years, 7 months ago
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Top NewestWhen 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.
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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 fallacy posted by @Sandeep Bhardwaj sir , I'm not able to find it . I'll give you the link if I'm able to find it .
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BTW i just found another note on this
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In fact i found a whole set of such apparent fallacies
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Wow . You've been busy :P
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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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