How can one prove that $\sqrt{ 2} \cdot \sqrt{ 3} = \sqrt{ 6}$? I have already proved that $\sqrt{2}\cdot \sqrt{2}=2$ so I hope I can now use $\sqrt{3}\cdot \sqrt{3}=3$ and the same for 6.
The exercise comes from Stillwell: Mathematics and its History. The other exercises have not been complicated so probably the solution to proving $\sqrt{2}\cdot \sqrt{2}=2$ I borrowed from Martin K was way over the top. But it cannot be adapted, I think, to this one where the roots are not of the same number.
Here is my attempt:
$r^2=2, s^2=3$.
$r\cdot r=2$ implies $r=\sqrt{2}$ and $s\cdot s=3$ 
implies $s=\sqrt{3}$
I also know that $\sqrt{6}\cdot \sqrt{6}=6$.
$r^2\cdot s^2=6$  so  $(rs)^2=6$
This implies that $(\sqrt{2}\sqrt{3})^2=\sqrt{6}\cdot \sqrt{6}$
Which I am not sure about, but hope, implies that $\sqrt{2}\sqrt{3}=\sqrt{6}$ 
 A: Let $a, b > 0$ then we need to prove that $ \sqrt{a}\sqrt{b} = \sqrt{ab} $
If we do not prove axiomatically, then we can do this:
Since we have positive numbers, their squares will also be equal.
$$ (\sqrt{a} \cdot \sqrt{b})^2 =_{?} (\sqrt{ab})^2 $$
and, $ (xy)^2 = x^2y^2 $, then $\sqrt{a}^2\cdot\sqrt{b}^2 = ab $ 
by the definition of a root, we obtain equality.  
A: Note that
$$a=\sqrt{2}\implies a^2=2$$
$$b=\sqrt{3}\implies b^2=3$$
then 
$$(ab)^2=6\implies ab=\sqrt{6}$$
A: Let $\sqrt{2}\sqrt{3} = x$.
Substitute $\sqrt{2} = \dfrac{2}{\sqrt{2}}$ and $\sqrt{3} = \dfrac{3}{\sqrt{3}}$. Then we should have: $$\left(\frac{2}{\sqrt{2}}\right)\left(\frac{3}{\sqrt{3}}\right) = \sqrt{6}.\tag1$$ Note that the $LHS$ (Left Hand Side of the equation) is equal to $\dfrac{6}{\sqrt{2}\sqrt{3}}$. It follows, then, that: $$\begin{align} \frac{6}{\sqrt{2}\sqrt{3}} &=\sqrt{6} \\ &\Downarrow \\ \frac{6}{x} &= \sqrt{6}.\end{align}$$ For you to have made your substitutions in Eq. $(1)$, it must follow that $x = \sqrt{6}$.

$$\therefore \sqrt{2}\sqrt{3} = \sqrt{6}.\tag*{$\bigcirc$}$$

