The following is invalid, since the operation is not defined when $a, b < 0$: $\sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)} = \sqrt{(-1)^2} = \sqrt{1} = 1$. This is not correct, because $ii = -1$. This shows that $\sqrt{a}\sqrt{b} = \sqrt{ab}$ is invalid when $a, b< 0$.

However, say we have $\sqrt{-5}$. In order to simplify this, we do the following: $\sqrt{-5} = \sqrt{(-1)(5)} = \sqrt{-1}\sqrt{5} = i\sqrt{5}$. Why is this a valid manipulation given the previous statement that $\sqrt{a}\sqrt{b} = \sqrt{ab}$ is invalid when $a, b< 0$?

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    $\begingroup$ It is invalid when both $a$ and $b$ are $\lt0$, not when one of them is $\lt0$ and the other $\gt0$. $\endgroup$ – Workaholic Jan 19 '16 at 0:36
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    $\begingroup$ Keep in mind there are always 2 square roots and which we choose to define as the square root by convention is arbitrary. To take -1*-1 = 1 we are deliberately trying to deliberately destroy and hid information. Basically $a = \sqrt{a^2}$ isn't true and we shouldn't ever say it. $\endgroup$ – fleablood Jan 19 '16 at 0:38
  • $\begingroup$ workaholic and jamie seem to be claiming opposite opinions. I'm inclined to accep Workaholic's $\endgroup$ – fleablood Jan 19 '16 at 0:40
  • $\begingroup$ @JoelReyesNoche No, I said that it is invalid when $ab < 0$, and not invalid when $a$ or $b < 0$. Having $-1$ for both $a$ and $b$ would make $ab < 0$, making it invalid. $\endgroup$ – Jamie Sanborn Jan 20 '16 at 2:29
  • $\begingroup$ @JamieSanborn, If $a=-1$ and $b=-1$, then $ab=1>0$. $\endgroup$ – Joel Reyes Noche Jan 20 '16 at 4:32

$a = \sqrt{a^2}$ is simply not acceptable and we shouldn't fall for it.

There are always two square roots (if one is $v$ the other is $-1*v$) and it is convention that if $x \in \mathbb R; x > 0$ then $\sqrt{x}$ is defined to be the positive one. $x \in \mathbb R; x < 0$ then $\sqrt{i} = i\sqrt{|x|}$. In not sure what the convention is for $x \in \mathbb C$. It may be that we choose the root with a positive real part. Or maybe it is we choose the root with the positive imaginary part. Or maybe there is no convention.

Anyway $\sqrt{a b} = \pm \sqrt{a} \sqrt{b}$ is acceptable if you choose the appropriate conventions for the parity of a and b.

  • $\begingroup$ As an additional note, $|a| = \sqrt{a^2}$ in $\mathbb{R}$, not $a = \sqrt{a^2}$. $\endgroup$ – Clarinetist Jan 19 '16 at 0:58
  • $\begingroup$ That's a good point. $\endgroup$ – fleablood Jan 19 '16 at 1:38

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