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If we have $$\sqrt[2]{x}^{2} = \sqrt[2]{x^2} = |x|$$ so : $$\sqrt[2]{-1}^{2} = \sqrt[2]{-1^2} = |-1| = 1$$ so why we say $i^2 = -1$?

David
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    Well I say that $i^2$ equals $-1$ because it equals $-1$. I don't say $i^2$ equal $1$ because it doesn't equal $1$. – Angina Seng Nov 29 '17 at 19:28
  • This question is massively a duplicate, but I can't find the One True Answer - https://math.stackexchange.com/questions/1096917/can-someone-prove-why-sqrtab-sqrta-sqrtb-is-only-valid-when-a-and-b-ar?rq=1 is an example. – Patrick Stevens Nov 29 '17 at 19:30
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    Logical reason? It's the definition of $i$... – ziggurism Nov 29 '17 at 19:31

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$$\sqrt{x^2}\neq (\sqrt{x})^2$$

Botond
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