It seems that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ is only defined for $a,b\geq 0$, because it doesn't work for $a,b<0$.

I can see that it doesn't work. I would like to know why it doesn't work. Is there a less circular reason than "by definition"?

This comes up in Khan Academy's i as the principal root of -1 and wikipedia's square root faulty proof:

$$ \begin{align*} -1&=ii\\ &=\sqrt{-1}\sqrt{-1}\\ &=\sqrt{(-1)(-1)}\\ &=\sqrt{1}\\ &=1 \end{align*} $$

They say $ii=\sqrt{-1}\sqrt{-1}$ is OK, and the faulty step is $\sqrt{-1}\sqrt{-1 }=\sqrt{(-1)(-1)}$, because $\sqrt{ab}=\sqrt{a}\sqrt{b}$ is only defined for $a,b\geq 0$.

marked as duplicate by dxiv algebra-precalculus Jul 3 at 6:10

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    "by definition" is not a circular explanation, it is the only possible explanation. What might be circular is an answer to why this is the definition that people use. – vadim123 Jul 3 at 6:01
  • Actually, the simplest way to clear it out in my mind is to agree that $\sqrt{-1}$ is never an "ok" notation, because it leads to meaningless confusions. Once we are given the usual abstract symbol $i$, there is no reason to consider it anymore. – Suzet Jul 3 at 6:01
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    A better question is: why is $\sqrt{ab} = \sqrt{a}\sqrt{b}$ for positive $a$, $b$? – Frank Vel Jul 3 at 6:08
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    It seems to be a common thing among students to assume that operations are "distributive". For example $\sqrt{x^2+y^2}=x+y$, $\log(x+y) = \log(x) + \log(y)$, etc. – steven gregory Jul 3 at 6:09
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    You are assuming that the rule $(ab)^r=a^rb^r$ extends to the complex numbers. It does not, that's it. – Yves Daoust Jul 3 at 6:47

Nice question. Think of it this way.

Let $A=\sqrt{ab}$

Let $B=(\sqrt{a})(\sqrt{b})$


$B^2=((\sqrt{a} )^2)((\sqrt{b} )^2)=ab$

Which means-


Which further means - $\pm{A}=\pm{B}$

Now you understood what happens? $A$ is not always equal to $B$.

Now I'll prove why it's not equal only if $a,b\lt0$

Condition: $a$ and $b$ are negative.

Let $\sqrt{a}=ix$ and $\sqrt{b}=iy$

Where $i=\sqrt{-1}$


(note: $ab$ is positive so there roots would simply be $x$ and $y$)


Thus, when $a,b\lt0$ then $A=-B$

I hope I answered your doubts.

Since $(\sqrt{a}\sqrt{b})^2=ab$, $\sqrt{a}\sqrt{b}=\pm\sqrt{ab}$. The only way to get rid of the $\pm$ sign for a more specific conclusion is by comparing the two sides of the equation. For $a,\,b\ge 0$, square roots are non-negative do we're fine. With negative numbers, the two $i$ factors on the LHS ruin it. If you want a fancier explanation than that, no inverse of $z^2$ on $\mathbb{C}$ is continuous, as can be seen by considering its phase.

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