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We can express a square number as the repeated addition of that number in this manner:

$1^2 = 1$

$2^2 = 2 + 2$

$3^2 = 3 + 3 + 3$

Generalising this, we get:

$x^2 = x + x + x...$ $x$ $times$

If we differentiate with respect to $x$ on both sides, we get:

$2x = 1 + 1 + 1...$ $x$ $times$

$2x = x$

$2 = 1$

This is obviously wrong. What's the mistake in my proof?

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    Applying continuous arguments to a discrete space is where you're wrong. – Asaf Karagila Sep 17 '18 at 14:04
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    By differentiation, all you found was the smallest natural value which will satisfy your generalization $x^2 = x$. Now the proper criteria to follow from the second last equation is to divide by $x$, however also claiming that $x\neq 0$ since division by $0$ is undefined. As you see, such a follow-up result in a contradiction, thus $x$ must be $0$. – mathnoob123 Sep 17 '18 at 14:04
  • If we agree up to $2x=x$, the only sound conclusion (see elementary algebra) is that $x=0$. – Mauro ALLEGRANZA Sep 17 '18 at 14:08
  • @mathnoob123 what do you mean by saying that the only thing we obtain from the differentiation is the smallest natural value that satisfies a certain generalisation? – Aman Vernekar Sep 17 '18 at 14:23

3 Answers3

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Because when you differentiate, you forgot to deal with the '$x$ times' expression, which is non-constant.

TheSimpliFire
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The formula only holds for natural values of $x$. Hence you can't differentiate.

Yanko
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    Clarification: It makes no sense to say "$x+x+x...$" $x$ times when $x$ is not natural. – Yanko Sep 17 '18 at 14:06
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Your expression of repeated summation is:

$$x^2=x\cdot x = \underbrace{x+x+x+\cdots}_{x\,\text{times}}$$

Underlined which $x$ you forgot to differentiate over:

$$x\cdot \underline{x} = \underbrace{x+x+x+\cdots}_{\underline{x}\,\text{times}}$$

Of course, it gets worse. Differentiation is defined for continuously varying arguments (e.g. real numbers). Repeated addition only works for natural numbers. So this cannot work in this form, until you move on from repeated summation definition.

orion
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