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?


marked as duplicate by J. M. is a poor mathematician, GEdgar, user2468, Ross Millikan, M Turgeon Aug 22 '12 at 17:13

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  • $\begingroup$ Square roots behave slightly differently on complex numbers. This has been covered on the site before. $\endgroup$ – Asaf Karagila Aug 22 '12 at 16:42
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    $\begingroup$ The issue is that $\sqrt{ab}\neq \sqrt{a}\sqrt{b}$. This question has been asked here before in numerous variations. $\endgroup$ – Alex Becker Aug 22 '12 at 16:42
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    $\begingroup$ $\sqrt{ab} = \sqrt{a}\sqrt{b}$ only holds when $a, b \ge 0.$ Hence $\sqrt{\frac{1}{-1}} \neq \frac{\sqrt{1}}{\sqrt{-1}}.$ $\endgroup$ – user2468 Aug 22 '12 at 16:56

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.


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