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A book stated that $\sqrt{-a}•\sqrt{-b} = -\sqrt{ab}$ where, a and b are positive real numbers. I know the proof of the above equation but why isn't this $\sqrt{-a}•\sqrt{-b} = \sqrt{ab}$. It also stated that this equality $\sqrt{a}•\sqrt{b} = \sqrt{ab}$ is true only when either of a and b are positive or zero but is false when a and b are negative, but why? Thank you for your help.

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So it's weird but you gotta evaluate underneath the radical first. In PEDMAS we have exponents before multiplications and square roots are exponents so $\sqrt {-a} \cdot \sqrt {-b}= i\sqrt a\cdot i\sqrt b=-\sqrt {ab}$

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If we define $\sqrt{-a} = i\sqrt{a}$ for positive $a,$ then $$\sqrt{-a}\sqrt{-b} = i\sqrt{a}i\sqrt{b} = -\sqrt{a}\sqrt{b} = -\sqrt{ab}$$ where we multiplied the i's together to get -1.

It is not true that $\sqrt{a}\sqrt{b} = \sqrt{ab}$ for negative a and b, because that would imply $\sqrt{-a}\sqrt{-b} = \sqrt{ab}$ for positive a and b, which we just showed was false.

I should add that there's a bit of ambiguity in how you define the square root of a negative number (remember, there's always two roots), but under the convention I gave, that's what you get.

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