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I found this proof of $2 = 1$ online and as usual tried to debug it.

Consider the following true statement:

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

If we differentiate both sides, we get:

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

Or

$$2 = 1$$

Now, the mistake was that $x^2 = x + x + x + ... + x$ was not true $\forall x \in \mathbb{R}$ and therefore we cannot differentiate both sides and except them to be equal.

However, I learnt of discrete calculus and tried to take the discrete derivative of both sides and got:

$$2x + 1 = x$$ which is also not true for all $x \in \mathbb{N}$.

So, my questions are: Why doesn't the discrete derivative give correct results?

EDIT: I have already mentioned that the statement is not true for all reals and only for natural numbers. I think this makes some of the comments here redundant.

Truth-seek
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  • I think this has already been answered here. Thing is, you can't rewrite $x^2$ as $x + \dots + x$ $x$ times. Since $x^2$ is a function of $x$ that just doesn't make sense – Andrew Li May 12 '18 at 04:15
  • In short, based on the duplicate, the problem is with the definition of $x^2$, to be able to differentiate (which means taking a limit) you need to have a definition for $x^{1.9}$ and $x^{2.1}$ (as examples). – Jared May 12 '18 at 04:18
  • @truth-seek:Does it make a sense? $$x=\sqrt{2}$$ so $$(\sqrt{2})^\sqrt{2} = \sqrt{2} + \sqrt{2}+ \sqrt{2} + ... + \sqrt{2} (\sqrt{2}) times $$? – Khosrotash May 12 '18 at 05:31
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    I think this answer to the duplicate question has you covered. – Micah May 12 '18 at 13:26
  • @Micah Thnx a lot. Thats just awesome. Thnx. – Truth-seek May 12 '18 at 14:00
  • @Micah yes, the wording confused me. – imranfat May 12 '18 at 21:50

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