Why/How does this sqrt term work? The inverse of a fraction in a sqrt Why does this term work?
$$
\frac{1}{\sqrt{\frac{g}{l}}} = \sqrt{\frac{l}{g}}
$$
 A: $\sqrt{\frac{1}{a}}=\frac{1}{\sqrt{a}}$ and $\frac{1}{a/b}=b/a$
A: $$\dfrac{1}{\sqrt{\dfrac{g}{l}}} = \dfrac{1}{\left(\dfrac{g}{l}\right)^{\frac{1}{2}}} =   \dfrac{\left(\dfrac{g}{l}\right)^{0}}{\left(\dfrac{g}{l}\right)^{\frac{1}{2}}} = \left(\dfrac{g}{l}\right)^{0-\frac{1}{2}} =\left(\dfrac{g}{l}\right)^{-\frac{1}{2}} = \left(\left(\dfrac{g}{l}\right)^{-1}\right)^{\frac{1}{2}} = \left(\dfrac{l}{g}\right)^{\frac{1}{2}} =  \sqrt{\dfrac{l}{g}}$$
A: There are two things happening here. One is that $$\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$$ for all nonnegative real numbers where $b \neq 0$. Second is that $$\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a}{b} \cdot \frac{d}{c}$$ for all real numbers where $b,c,d \neq 0$. For your problem, let $a=b=1$, and $c=\sqrt{g},d = \sqrt{l}$. The equality follows.
A: If both $g,l\ne 0$, you apply roots property $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ and reals ratios property $\frac{1}{\left(\frac{c}{d}\right)}=\frac{d}{c}$, and get:
$$
\frac{1}{\sqrt{\frac{g}{l}}}=
\frac{1}{\left(\frac{\sqrt{g}}{\sqrt{l}}\right)}=
\frac{\sqrt{l}}{\sqrt{g}}=
\sqrt{\frac{l}{g}}.
$$
A: With $l \cdot g \neq 0$
$$
\frac{1}{\sqrt{\frac{g}{l}}} = \frac{1\times\sqrt{\frac{l}{g}}}{\sqrt{\frac{g}{l}}\times\sqrt{\frac{l}{g}}}= \frac{\sqrt{\frac{l}{g}}}{\sqrt{\frac{g}{l}\times\frac{l}{g}}} = \frac{\sqrt{\frac{l}{g}}}{1} = \sqrt{\frac{l}{g}}
$$
