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:

enter image description here

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


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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ $\sqrt{-ab}=\sqrt{-a}\sqrt b$ does not hold. $\endgroup$ – Yves Daoust Jul 7 '18 at 15:51
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ $\sqrt{-\frac 12} = \sqrt{-\frac 24} = \sqrt{(\frac 12)^2*-2} = \frac 12\sqrt{-2}$. $\endgroup$ – fleablood Jul 7 '18 at 19:04

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.

  • $\begingroup$ Thank you so much for this awesome reply, and so many comments. $\endgroup$ – Silent Jul 8 '18 at 5:43

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).


Not the answer you're looking for? Browse other questions tagged or ask your own question.