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Possible Duplicate:
-1 is not 1, so where is the mistake?
$i^2$ why is it $-1$ when you can show it is $1$?

Can someone please point out what I'm doing wrong here?

$ \frac{1}{-1} = \frac{-1}{1} \implies \sqrt{\frac{1}{-1}} = \sqrt{\frac{-1}{1}} $ which should imply that $ \frac{\sqrt{1}}{\sqrt{-1}} = \frac{\sqrt{-1}}{\sqrt{1}} $

But the last equality is not true. Is there something really obvious that I'm overlooking here?

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    $\sqrt{ab} = \sqrt{a}\sqrt{b}$ only holds when $a, b \ge 0.$ Hence $\sqrt{\frac{1}{-1}} \neq \frac{\sqrt{1}}{\sqrt{-1}}.$2012-08-22

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Assume that $f \colon A \to B$ is a given function. From $x=y$ you can get $f(x)=f(y)$, but only when $x$ and $y$ belong to $A$. In your example, $f=\sqrt{\cdot}$, $A=[0,+\infty)$ and $B=\mathbb{R}$. But $x=-1/1 = -1 \notin A$, and you can't conclude.

Of course, you may say that $A =\mathbb{C}$, but then $f$ becomes multi-valued.