TITLE: Proving $\sqrt{ab} = \sqrt a\sqrt b$
QUESTION [12 upvotes]: I am currently in high school and we are studying radicals. I had asked my math teacher why $\sqrt{ab}=\sqrt{a}\sqrt{b}$ (for all a,b>0) and he tries to prove it by arguing that $a^{1/2}*b^{1/2}=(ab)^{1/2}$ (an exponent law). However, I find this proof problematic since $x^{1/2}$ is simply defined as $\sqrt{x}$, so the reasoning is circular. 
My view is that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ because once we square both sides we get $ab=ab$. Since we're obviously referring to the positive root, and the function $f(x) = \sqrt{x}$ is injective, it necessarily follows that the original expressions $\sqrt{ab}$ and $\sqrt{a}\sqrt{b}$ are equivalent because for injective functions it is not possible to map distinct elements in the domain to the same element in the range ($ab$). Hence they are equivalent expressions. 
My question therefore is, is my proof valid and/or rigorous (I find it convincing but maybe it's wrong; I just want to be clear) and secondly was my teacher's proof correct?

REPLY [8 votes]: Could we proceed along the following lines, as long as we restrict ourselves to the positive reals?  We observe $\sqrt{x}$ to be the unique positive real number $r$ such that $r \cdot r = x$.
Let $m = \sqrt{a}, n = \sqrt{b}$.  Then $m \cdot m = a, n \cdot n = b$, and
$$
\begin{align}
a \cdot b & = (m \cdot m) \cdot b \\
          & = m \cdot (m \cdot b) \\
          & = m \cdot (m \cdot (n \cdot n)) \\
          & = m \cdot ((m \cdot n) \cdot n) \\
          & = m \cdot ((n \cdot m) \cdot n) \\
          & = m \cdot (n \cdot (m \cdot n)) \\
          & = (m \cdot n) \cdot (m \cdot n)
\end{align}
$$
where we rely on the associativity and commutativity of multiplication.  Hence $\sqrt{ab} = m \cdot n = \sqrt{a} \cdot \sqrt{b}$.