Square root property proof Can anyone provide a link to a proof of the following square root property $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$. Could not find it online anywhere.
In the following proof, we will use the following definition of the (principal) square root of a non-negative number:


For a given real number $p\ge 0$, we say that a real number $q$ is its (principal) square root ($q=\sqrt{p}$), iff $q\ge 0$ and $q^2=p$.


It is possible to prove that such a number $q$ exists (not easily - the proof uses some fundamental properties of real numbers), and is unique (fairly easy), so the above definition allows us to view the square root as a function of non-negative real numbers.



Now, to the proof. Let $a,b\ge 0, b\ne 0$ - real numbers, and let $x=\sqrt{a}, y=\sqrt{b}$. This means that $x,y\ge 0$ and $x^2=a, y^2=b$. Now this implies that $y\ne 0, \frac{x}{y}\ge 0$, and also:

$$\left(\frac{x}{y}\right)^2=\frac{x^2}{y^2}=\frac{a}{b}$$

Therefore, $\frac{x}{y}=\sqrt\frac{a}{b}$. Remembering what $x$ and $y$ were in the first place, we conclude:

$$\frac{\sqrt{a}}{\sqrt{b}}=\sqrt\frac{a}{b}$$.
I think that 
$$\sqrt\frac{a}{b}=\frac{\sqrt |a|}{\sqrt |b|} $$
Because $$\frac{a}{b}\ge0 $$
But $a\ge0, b\ge0 $
This is only a special case, because we may lose the solution
$\frac{a}{b} $ then $ a&lt;0,b&lt;0 $
Therefore, you need to use the module ;)