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?
Best Answer
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 $\sqrt{ab} = m \cdot n = \sqrt{a} \cdot \sqrt{b}$.