# The Product Rule of Square Roots with Negative Numbers

In the statement $\forall a, b \geq0, \sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$, why is it necessary to restrict $a$ and $b$ to being $\geq 0$? It seems that one should be able to say, for example, $(-3)^{1/2} \cdot (-3)^{1/2} = (-3 \cdot -3)^{1/2} = 9^{1/2} = 3$, so where is the flaw in this statement, since it just seems to be using laws of exponents.

Edit: I don't think people are understanding what I'm asking. To reemphasize, my question is: why do we say $\forall a, b \geq 0, \sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$, instead of just $\forall a, b \in \mathbb{R}, \sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$

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What do you think is $(-3)^{1/2}$ ? – zuggg Jul 7 '13 at 5:00
$a,b$ are $\ge0$ – lab bhattacharjee Jul 7 '13 at 5:00
You've already given a counter-example. If $\sqrt{-3}^2 = -3$, then $\sqrt{-3}\sqrt{-3} \neq \sqrt{9}$. – Thomas Andrews Jul 7 '13 at 5:24

Let us freely use these supposed laws of exponents.

You used one to show that $(-3)^{1/2}(-3)^{1/2}=9^{1/2}$. Presumably this is $3$.

But another familiar law of exponents is $a^xa^y=a^{x+y}$. Use this with $a=-3$, $x=y=1/2$. We get $(-3)^{1/2}(-3)^{1/2}=(-3)^1=-3$.

Remark: Ordinary laws can give inconsistent results, giving lawyers opportunities to get rich. Mathematical laws have to meet a higher standard.

Response to edit: Restricting attention to real $a$ and $b$ will not work, since there is no way to assign a real number to the expression $(-a)^{1/2}$ without violating some laws of the algebra of real numbers. So to give meaning to the expression, we have to introduce complex numbers. But then as shown above, and in other answers, we can by application of the "laws" reach the highly undesirable conclusion that $3=-3$.

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$\sqrt a$ is only defined for $a\ge 0$. This is the reason.

And if you work with complex numbers? Well, then the property could be false.

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Try $(-3)^{1/2}(-3)^{1/2}=(-3)^{1/2+1/2}$ along with Emanuele Paolini's answer.

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