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.
 A: 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:
$$\frac{\sqrt{a}}{\sqrt{b}}=\sqrt\frac{a}{b}$$.
A: Say $a,b\geq 0$. If $x=\sqrt{a}$ and $y= \sqrt{b}$ then $x^2=a$ and $y^2= b$ so $$\Big({x\over y}\Big)^2={x^2\over y^2} = {a\over b}\implies {x\over y} = \sqrt{a\over b}$$
and we are done.
A: By definition of $\sqrt{\phantom 3}$, and assuming $a\geq0, b>0$, we have that
$
\sqrt{\frac ab}
$
is the unique non-negative number such that
$$
\left(\sqrt{\frac ab}\right)^2=\frac ab
$$
(One must, of course, be convinced that this number is indeed unique.) Now note that
$$
\left(\frac{\sqrt a}{\sqrt b}\right)^2=\frac{\sqrt a}{\sqrt b}\cdot \frac{\sqrt a}{\sqrt b}=\frac{(\sqrt a)^2}{(\sqrt b)^2}=\frac ab
$$
so $\frac{\sqrt a}{\sqrt b}$ fulfills the defining property of $\sqrt{\frac ab}$. They must therefore be equal.
A: Assuming $a,b\in\mathbb R$, $a\geqslant 0$, and $b>0$, we have
\begin{align}
\frac{\sqrt a}{\sqrt b} &= \frac{a^{1/2}}{b^{1/2}}\\
&=\frac{e^{\log(a^{1/2})}}{e^{\log(b^{1/2})}}\\
&=\frac{e^{\frac 12\log a}}{e^{\frac12\log b}}\\
&=e^{\frac12\log a - \frac12\log b}\\
&= e^{\frac12 \log\left(\frac ab\right)}\\
&=\left(\frac ab\right)^{1/2}\\
&=\sqrt{\frac ab}.
\end{align}
A: By the definition of square-root, $x=\sqrt{y}\hspace{.15cm}$ means that $x^2=y$.  
Assuming that $a,b\in\mathbb{R}$ with $a\geq 0$ and $b>0$, then we can simply compute:
\begin{align*}
\left(\frac{\sqrt{a}}{\sqrt{b}}\right)^2&=\frac{\sqrt{a}}{\sqrt{b}}\cdot \frac{\sqrt{a}}{\sqrt{b}}
=\frac{\sqrt{a}\cdot \sqrt{a}}{\sqrt{b}\cdot\sqrt{b}}=\frac{a}{b}
\end{align*} 
Therefore, $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\hspace{.25cm}$
Note that I used the fact that $\frac{a}{b}\cdot \frac{c}{d}=\frac{ac}{bd}$
A: 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<0,b<0 $
Therefore, you need to use the module ;)
