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-1 is not 1, so where is the mistake?
Significance of $\displaystyle\sqrt[n]{a^n} $?

$i = \sqrt{-1} = \sqrt{\frac{-1}{1}} = \sqrt{\frac{1}{-1}} = \frac{1}{i}$

hence,

$i^2 = 1$ What is wrong with the steps shown below? I start with i = sqrt(-1) but then I end up with i = sqrt(1). What is the solution out of this paradox?

Is it that by definition, i is that number whose square is -1. or is it because,

by taking the denominator '1' inside the sqrt, I am losing some information because:

$\sqrt{1} = \pm 1$

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marked as duplicate by J. M., t.b., Zev Chonoles Nov 18 '11 at 14:44

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Very clever. Yes the problem is that the square root is only well-defined up to $\pm 1$, so you're equation shows $i=\pm \dfrac{1}{i}$ which is not a paradox. –  Grumpy Parsnip Nov 18 '11 at 14:18
    
And many more of the same spirit are linked to here –  t.b. Nov 18 '11 at 14:44

1 Answer 1

First and foremost, $\sqrt{1}$ is not $1$ or $-1$. When we discuss the square root of a number $\sqrt{n}$ this denotes the positive or principal square root. Therefore, $\sqrt{1} = 1$ and $1$ alone. This is different, however, than the fact that $x^2 = 1$ has two solutions, namely $1$ and $-1$.

Secondly, square roots and complex numbers do not behave as square roots and real numbers. This is because to properly define the square root function in the complex plane, we need to take a branch cut of the function $f(z) = \sqrt{z}$.

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"don't quite behave correctly..." Well, they behave correctly; it's just that square roots of positive real numbers have some "extra" properties due to the fact that they are positive reals: the correcting factor for the argument in the product of two square roots comes out to $0$. –  Arturo Magidin Nov 18 '11 at 14:22
    
@Arturo: That's fair. Thank you for the comment. –  JavaMan Nov 18 '11 at 14:24

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