<s>[INST] 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 $\sqrt{2}\sqrt{3}=\sqrt{6}$ 
 [/INST] 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.  
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