# Product of imaginary numbers. [duplicate]

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.

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 $\sqrt {-a} \cdot \sqrt {-b}= i\sqrt a\cdot i\sqrt b=-\sqrt {ab}$
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.