How do you proof $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$?

It is not always true. To see the clear picture, we solve some clear examples.
Remember, $\sqrt{\frac{a}{b}}$ means "find $a$ divided by $b$ then evaluate the square root".
While $\frac{\sqrt{a}}{\sqrt{b}}$ means "find the square roots of $a$ and $b$ first then divide the result".
To proof our case, we will develop four cases, 
Case 1: Let $a,b\in\mathbb{R}^+$ then 
$\sqrt{\frac{64}{16}}=\sqrt{4}=2$ and $\frac{\sqrt{64}}{\sqrt{16}}=\frac{8}{4}=2$.
These shows that $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ is true. 

Case 2: Let $a\in\mathbb{R}^-$ and $b\in\mathbb{R}^+$ then $\sqrt{\frac{-64}{16}}=\sqrt{-4}=2i$ and $\frac{\sqrt{-64}}{\sqrt{16}}=\frac{8i}{4}=2i$.

Case 3: $a\in\mathbb{R}^+$ and $b\in\mathbb{R}^-$ then $\sqrt{\frac{64}{-16}}=\sqrt{-4}=2i$ and $\frac{\sqrt{64}}{\sqrt{-16}}=\frac{8}{4i}=-2i$ therefore, it has failed to hold here, $\sqrt{\frac{a}{b}}\neq\frac{\sqrt{a}}{\sqrt{b}}$?

Case 4: $a\in\mathbb{R}^-$ and $b\in\mathbb{R}^-$ then $\sqrt{\frac{-64}{-16}}=\sqrt{4}=2$ and $\frac{\sqrt{-64}}{\sqrt{-16}}=\frac{8i}{4i}=2$. Therefore, $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ is true.


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