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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.

Md Kutubuddin Sardar
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  • Not what you typed, but here are questions that ask what you meant: http://math.stackexchange.com/questions/438/1-is-not-1-so-where-is-the-mistake and http://math.stackexchange.com/questions/49169/i2-why-is-it-1-when-you-can-show-it-is-1 – Henry Swanson Sep 25 '14 at 07:30
  • $(-1)^{\frac{1}{2}+{\frac{1}{2}}}=1$? Really? – Did Sep 25 '14 at 07:43
  • 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 – mau Sep 25 '14 at 08:42
  • Any radical is a choice between n possible options. $\sqrt1$ can be $+1$ as well as $-1$, if we are talking about complex roots. – Lucian Sep 25 '14 at 08:59

1 Answers1

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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.

Travis Willse
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