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Why $\sqrt{a}$ * $\sqrt{b}$ = $\sqrt{ab}$ ?

It's easy to verify that it works in some cases, like in $\sqrt{4}$ * $\sqrt{9}$ = $\sqrt{36}$. But why does the general rule work?

João Rimu
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  • hint: definition of sqrt; note multiplication is commutative – John Dvorak Sep 30 '13 at 20:31
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    It works only if you consider exclusively non-negative numbers. Then it works because multiplication is commutative, $$(\sqrt{a}\sqrt{b})^2 = \sqrt{a}\sqrt{b}\sqrt{a}\sqrt{b} = \sqrt{a}^2\sqrt{b}^2 = ab = \sqrt{ab}^2$$ – Daniel Fischer Sep 30 '13 at 20:32

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Squaring gives $(\sqrt a*\sqrt b)^2=\sqrt a*\sqrt b*\sqrt a*\sqrt b=(\sqrt a)^2(\sqrt b)^2=ab$. Also the product $\sqrt a*\sqrt b$ of two non-negative real numbers is non-negative. By definition this makes $\sqrt a*\sqrt b$ into the square root of $ab$. (Both $a$ and $b$ must be non-negative real numbers for these expressions to be well defined; using the principal square root of complex numbers to define $\sqrt z$, the equation fails, in particular for $a=b=-1$.)

Marc van Leeuwen
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