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Why does $\sqrt{-ab}=\sqrt{-a}\sqrt b$ hold, where $a,b\in\Bbb R$?

I know that splitting sqrts is not permissible for complex numbers generally, eg, $\sqrt{-1\cdot-1}\ne\sqrt{-1}\sqrt{-1}$.

Also, is $\sqrt{-a}$ unique or we can assign two values, one additive inverse of other?

Edit/Context: My question was based on the last paragraph of this example from Dummit and Foote, Abstract Algebra:

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Here, even though $D=f^2D'$ be negative both sides, they have taken square root and concluded $\sqrt D=f\sqrt D$.

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closed as unclear what you're asking by Did, José Carlos Santos, Leucippus, Xander Henderson, Chris Custer Jul 8 '18 at 1:33

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    $\begingroup$ What you have written is wrong even for reals. What if $a = b = 1$? There is no $\sqrt{-1}$. It's not $i$. That and $-i$ both have square $-1$ and neither is properly called "the" square root. $\endgroup$ – Ethan Bolker Jul 7 '18 at 15:21
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    $\begingroup$ For me $\sqrt{a}=\{x\in \mathbb{C}: x^2=a\}$ for any complex number $a$, so what you wrote doesn't make any sense $\endgroup$ – Jakobian Jul 7 '18 at 15:22
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    $\begingroup$ $\sqrt{-ab}=\sqrt{-a}\sqrt b$ does not hold. $\endgroup$ – Yves Daoust Jul 7 '18 at 15:51
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    $\begingroup$ If $b > 0$ and if $a \in \mathbb R$ then $\sqrt{-ab} = \sqrt{-a}\sqrt{b}$ which may or may not be imaginary depending on whether $a$ is negative or non-negative. But for that to hold $b$ must be non-negative real. $\endgroup$ – fleablood Jul 7 '18 at 19:02
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    $\begingroup$ $\sqrt{-\frac 12} = \sqrt{-\frac 24} = \sqrt{(\frac 12)^2*-2} = \frac 12\sqrt{-2}$. $\endgroup$ – fleablood Jul 7 '18 at 19:04
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Why does $\sqrt{-ab}=\sqrt{-a}\sqrt{b}$, where $a,b∈\mathbb R$?

It doesn't. It holds if $a,b \ge 0$ and then then $\sqrt{-ab} = i\sqrt{ab}=i\sqrt{a}\sqrt{b}= \sqrt{-a}\sqrt{b}$. However if $a$ and $b$ are negative or complex more care needs to be given.

Note: if one of them is non-negative real, Say $b \ge 0$ then we can say that $\sqrt{-ab} = \sqrt{-a}\sqrt{b}$ whether $-a$ is positive of negative. However it must be that $b$ is non-negative.

Or if $-a$ is positive. Then $-a = u\ge 0$ and $\sqrt{-ab} = \sqrt{ub}=\sqrt{u}\sqrt{b}=\sqrt{-a}{b}$.

But $\sqrt{ab} = \sqrt{a}\sqrt{b}$ is only valid if one or the other is non-negative real. Otherwise.... well, care must be taken.

My question was based on the last paragraph of this example from Dummit and Foote, Abstract Algebra:

The last paragraph is to note that any rational $D$ can be uniquely expressed as $f^2*D'$ where $f$ is rational and $D'$ is a "square-free" integer. (I will admit the definition is a bit clunky.)

In other words if $D \in Q$ then $D = f^2D'$ and $\sqrt{D} =\sqrt{f^2D'} = f\sqrt{D'}$. This is acceptable because $f^2 > 0$.

Pf: $D = \pm \frac ab; a,b \in \mathbb Z; a\ge 0; b> 0; \gcd(a,b) = 1$.

Let $a = \prod p_i^{m_i}$ be the unique prime factorization of $a$ and let $b = \prod q_i^{n_i}$ be the unique prime factorization of $b$.

Now each $m_i$ and each $n_i$ is either odd or even. Let $m_i' = \frac {m_i}2$ if $m_i$ is even and let $m_i' = \frac {m_i - 1}2$ if $m_i$ is odd. (In other words let $m_i' = \lfloor \frac {m_i}2 \rfloor$. ) Likewise define $n_i'$ in the same ways so that $n_i' = \frac {n_i}2$ if $n_i$ is even and let $n_i' = \frac {n_i - 1}2$ if $n_i$ is odd.

Then $a = \prod p_i^{m_i} = \prod p_i^{2m_i'}\prod_{m_i\text{ is odd}} p_i = (\prod p_i^{m_i'})^2 *\prod_{m_i\text{ is odd}} p_i$. Let $e = \prod p_i^{m_i'}$ and $D_1 = \prod_{m_i\text{ is odd}} p_i$. Notice that $D_1$ is a prime free integer.

Likewise $b = \prod q_i^{n_i} = \prod q_i^{2n_i'}\prod_{n_i\text{ is odd}} q_i = (\prod q_i^{n_i'})^2 *\prod_{n_i\text{ is odd}} q_i$. Let $g = \prod q_i^{n_i'}$ and $D_2 = \prod_{n_i\text{ is odd}} q_i$.

So $D = \pm\frac ab = \frac {e^2D_1}{f^2D_2} = \pm\frac {e^2D_1D_2}{g^2D_2^2} = (\frac {e}{gD_2})^2(\pm D_1D_2)$.

Let $f=\frac {e}{gD_2}$ is a uniquely determined positive rational number and $D' = \pm D_1D_2$ is a square-free integer also uniquely determined.

As $D$ was not a square of a rational, it is not possible that $D' = 1$. But it is possible that $D' = -1$. But $D'$ is squarefree, an integer and possibly positive and possibly negative. But $f^2$ is a square (and therefore positive) of a rational number.

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  • $\begingroup$ Thank you so much for this awesome reply, and so many comments. $\endgroup$ – Silent Jul 8 '18 at 5:43
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You just gave a counter-example to your own question. That is if you pick $a=1$ and $b=-1$ the equation turns out to be $\sqrt{-1\cdot -1} = \sqrt{-1}\sqrt{-1}$.

The reason for this (as you guessed) is that $\sqrt{-1}$, or any square above complex numbers is not uniquely determined. For example $\sqrt{-1}=i$ in one hand, while in the other hand $\sqrt{-1}=-i$, unlike real numbers there is no "natural" way to distinguish between $i$ and $-i$ (because there is no such thing "a positive complex number").

We can define the square root of a complex number $a$ to be any number which satisfies the equation $x^2-a=0$. In this setting it is easy to see that if $b$ is one solution, then it's negative, $-b$ is another solution (and there are no more solutions).

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