# Question on elementary Algebra: Taking square roots of products

I was reading Higher Algebra, by H. S. Hall and S. R. Knight when I stumbled across something that was confusing. On page 75, the book states:

"Since $$(-a)\times(-b)=ab,$$ by taking the square root, we have $$\sqrt{-a}\times\sqrt{-b}=\pm \sqrt{ab}$$. Thus in forming the product of $$\sqrt{-a}$$ and $$\sqrt{-b}$$, it would appear that either of the signs $$+$$ or $$-$$ might be placed before $$\sqrt{ab}$$. This is not the case, for $$\sqrt{-a}\times\sqrt{-b}=\sqrt{a}\cdot\sqrt{-1}\times\sqrt{b}\cdot\sqrt{-1}=\sqrt{ab}(\sqrt{-1})^2=-\sqrt{ab}$$" (Hall and Knight 75).

On the other hand, I thought by taking the square root in the first step, one would get: $$\pm\sqrt{(-a)\cdot(-b)} = \pm\sqrt{ab}$$ which comes out to: $$\pm\sqrt{ab}=\pm\sqrt{ab}$$. This is clearly different from what was done in the book, so which one is correct?

This stackexchange question on a similar topic actually has some stuff that resolves the confusion: Laws of Exponents if base(s) negative.

• Could you give a bit more context? In particular, are $a$ and $b$ real or complex? May 6, 2018 at 14:46
• @BillO'Haran That is part of my confusion. The topic he is discussing this is in imaginary quantities, but he never states whether they are imaginary or real. I'm looking for a bit more clarification. May 6, 2018 at 14:50

$$\sqrt{-a}\times\sqrt{-b}=\sqrt{a}\cdot\sqrt{-1}\times\sqrt{b}\cdot\sqrt{-1}$$
but the second $$\sqrt{-1}$$ could be the negative of the first and this case is not considered by Hall and Knight. This is assuming, as I think in this example, that $$a$$ and $$b$$ are positive real numbers.
In the general case $$\sqrt{a\cdot b} = \pm\sqrt{a}\cdot\sqrt{b}$$ because each square root has two values. Thus the left side has two values and the right side $$\pm$$ is needed because of the combined four choices of the two square roots there.
The crux here is to decide what $$\,\sqrt{-1}\cdot\sqrt{-1}\,$$ is equal to. I think that the value is $$\,\pm1\,$$ because the two square roots are independent. On the other hand, $$\,i\cdot i=-1\,$$ by definition. Thus, $$\,\sqrt{-1} = \pm i\,$$ because there are two square roots for any non-zero number. However, it is very common and convenient to make a special case exception so that $$\,\sqrt{-1} := i\,$$ by definition.