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does 2=-2?

I was thinking recently and got stuck on a strange problem:

\((-2)^{2}=4=(-2)^{0.5\times4}=((-2)^{0.5})^{4}=4=\sqrt{2}^{4}\)

Take the fourth route if both sides:

\((-2)^{0.5}=\sqrt{2}\)

\(\sqrt{-2}=\sqrt{2}\)

\(-2=2\)

I don't know what the problem is: have I made a mistake? Or is their a way of explaining this that I am not aware of? Please leave a comment if you have any ideas!

Note by Katie Marsden
1 year, 12 months ago

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\(4=x^{4}\) has \(4\) solutions for \(x\). They are like roots for polynomials. These roots are \(\pm\sqrt{2}\) and \(\pm\sqrt{-2}\).

btw I edited the Latex of your note. Julian Poon · 1 year, 12 months ago

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@Julian Poon Thank! I still won't pretend I undestand how that rule accounts for the other possible answers to that (i.e. Wouldn't that mean there were 4 answers to root-2, one of which being root 2?). But thanks that's really interesting-and thanks for the editing I'm really bad at that kind of thing! Katie Marsden · 1 year, 12 months ago

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@Katie Marsden I'll give a simpler example: \(x^{2} = 4\) would have 2 solutions for \(x\), namely \(-2\) and \(2\). Even though \((-2)^{2}=4\) and \((2)^{2}=4\), \(-2\neq 2\) Julian Poon · 1 year, 12 months ago

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