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This question already has an answer here:

Suppose $a,b\in \mathbb{Z}$.
Is it true $\sqrt{a}\sqrt{b}=\sqrt{ab}$.
If so, then $\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)}=\sqrt{1}=1$

But we know $\sqrt{-1}=i$ and so $i^2=-1.$
Finally we get $i^2=-1=1.$ Which is not true.
What is the logic behind it?

Thank you in advance.

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marked as duplicate by mrf, Travis, André Nicolas, Claude Leibovici, Jonas Meyer Sep 30 '14 at 4:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Not what you typed, but here are questions that ask what you meant: math.stackexchange.com/questions/438/… and math.stackexchange.com/questions/49169/… $\endgroup$ – Henry Swanson Sep 25 '14 at 7:30
  • $\begingroup$ $(-1)^{\frac{1}{2}+{\frac{1}{2}}}=1$? Really? $\endgroup$ – Did Sep 25 '14 at 7:43
  • $\begingroup$ your hypotesis is true only for $a, b \in \mathbb{N}$, since square root is a multivalued function and you have to choose a single value $\endgroup$ – mau Sep 25 '14 at 8:42
  • $\begingroup$ Any radical is a choice between n possible options. $\sqrt1$ can be $+1$ as well as $-1$, if we are talking about complex roots. $\endgroup$ – Lucian Sep 25 '14 at 8:59
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The statement is true for positive, real $a, b$, but in general it does not hold for complex numbers. This can be seen as a consequence of the fact that we don't have a natural choice of inverse for $f(z) = z^2$ on $\mathbb{C}$. For certain choices of inverses (i.e., choices of square root functions), the above statement holds for complex numbers $a, b$ whose arguments satisfy certain inequalities, but when first learning about complex numbers, it's perhaps safest to treat the identity as though it only holds for positive reals.

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