Proving $\sqrt{ab} = \sqrt a\sqrt b$

Solution 1:

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