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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.

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  • $\begingroup$ Could you give a bit more context? In particular, are $a$ and $b$ real or complex? $\endgroup$ May 6, 2018 at 14:46
  • $\begingroup$ @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. $\endgroup$ May 6, 2018 at 14:50

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The core problem is that there are two square roots for any non-zero numbers. In the case of positive real numbers, they are easy to distinguish. One square root is positive and the other is negative. In the case of negative real numbers, the two square roots are pure imaginary conjugates of each other. If there is only one of these involved, then there is no loss of generality in choosing the one with positive imaginary part. However, if there are two such square roots, then they could be chosen independently and different. This is the case in the Hall and Knight example

$$\sqrt{-a}\times\sqrt{-b}=\sqrt{a}\cdot\sqrt{-1}\times\sqrt{b}\cdot\sqrt{-1}$$

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.

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