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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1$\sqrt{-a}=i\sqrt{a}$, for $a\geq 0$. – Theoretical Economist Dec 09 '16 at 01:53
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5Possible duplicate of Why is $\sqrt{-x}*\sqrt{-x}=-x?$. See also Why $\sqrt{-1 \times {-1}} \neq \sqrt{-1}^2$? – Winther Dec 09 '16 at 02:26
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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.
WW1
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spaceisdarkgreen
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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 $\sqrt {-a} \cdot \sqrt {-b}= i\sqrt a\cdot i\sqrt b=-\sqrt {ab}$
Teh Rod
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