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We can see that

1^2 =1 ; 2^2 =2+2 ; 3^2=3+3+3 ; . . . x^2=x+x+x+..... (x times)

differentiation on both sides gives

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

2x=x

What's happening hear.How is this possible.

Assume X be as integer and non-integer ,both cases.

user239656
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2 Answers2

1

This is because, if $x$ is not a positive integer what does, "$x$ times" mean? Like what is, $\sqrt{2}$ times? What does that even mean?

Nicolas Bourbaki
  • 5,064
  • But what if X is an integer.Why it is wrong in that case also. – user239656 Feb 15 '15 at 21:11
  • @user239656 When you try to write function as $x+x+...+x$ ($x$ times), then your function is defined only at integers. Because of that, it is impossible to differentiate this function. – Wojowu Feb 15 '15 at 21:22
0

You're trying to use the linearity property of derivatives: ie, that $\frac{d}{dx}(\sum_{i=1}^n f_i(x)) = \sum_{i=1}^n\frac{d}{dx}f_i(x)$, but how do you write $\underbrace{x + x + \cdots + x}_{x \text{ times}}$ in the form $\sum_{i=1}^n f_i(x)$?

Answer: You can't, so you're proof is invalid.

oxeimon
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  • I don't know why this was downvoted. In general most incorrect proofs are incorrect because some step in the proof was invalid. There is rarely an intuitive reason why the proof is wrong. – oxeimon Feb 16 '15 at 18:09