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Why does this term work?

$$ \frac{1}{\sqrt{\frac{g}{l}}} = \sqrt{\frac{l}{g}} $$

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  • $\begingroup$ Multiply numerator and denominator (on the left) by $\sqrt l$. $\endgroup$ Mar 25, 2015 at 17:05
  • $\begingroup$ Is there a reason for g and l? $\endgroup$ Mar 25, 2015 at 18:19
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    $\begingroup$ @EvanCarroll probably for gravity and length. It looks like a quantity you frequently encounter in periodic motion. $\endgroup$
    – graydad
    Mar 25, 2015 at 19:50
  • $\begingroup$ @graydad yeah, if the variables represent gravity and length, then this expression represents a frequency, since it has the dimensions of [1/time] $\endgroup$
    – Dancrumb
    Mar 25, 2015 at 20:51

5 Answers 5

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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.

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  • $\begingroup$ Welp, now I got it. Thanks a lot :) $\endgroup$
    – ColdStormy
    Mar 25, 2015 at 15:23
  • $\begingroup$ For your second your say it is true when $b$ and $c$ are different from $0$, but what about $d$? $\endgroup$
    – Alice Ryhl
    Mar 25, 2015 at 16:56
  • $\begingroup$ @KristofferRyhl I debated including that in there. Certainly if it were just $\frac{c}{d}$ I would want to say that. I'll include it to be safe. $\endgroup$
    – graydad
    Mar 25, 2015 at 16:57
  • $\begingroup$ Well, it's certainly not true when $d=0$, so it's a good idea to leave it in. $\endgroup$
    – Alice Ryhl
    Mar 25, 2015 at 16:58
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    $\begingroup$ No, that equation is well defined for $d=0$, but the other is not. $$\text{undefined}\ne\frac ab\cdot\frac 0c$$ $\endgroup$
    – Alice Ryhl
    Mar 25, 2015 at 17:01
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$\sqrt{\frac{1}{a}}=\frac{1}{\sqrt{a}}$ and $\frac{1}{a/b}=b/a$

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    $\begingroup$ In one regard this is superior to many of the other answers, in that it doesn't rely on $\sqrt{l}$ or $\sqrt{g}$ being well-defined. $\endgroup$
    – Erick Wong
    Mar 25, 2015 at 18:01
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$$\dfrac{1}{\sqrt{\dfrac{g}{l}}} = \dfrac{1}{\left(\dfrac{g}{l}\right)^{\frac{1}{2}}} = \dfrac{\left(\dfrac{g}{l}\right)^{0}}{\left(\dfrac{g}{l}\right)^{\frac{1}{2}}} = \left(\dfrac{g}{l}\right)^{0-\frac{1}{2}} =\left(\dfrac{g}{l}\right)^{-\frac{1}{2}} = \left(\left(\dfrac{g}{l}\right)^{-1}\right)^{\frac{1}{2}} = \left(\dfrac{l}{g}\right)^{\frac{1}{2}} = \sqrt{\dfrac{l}{g}}$$

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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: $$ \frac{1}{\sqrt{\frac{g}{l}}}= \frac{1}{\left(\frac{\sqrt{g}}{\sqrt{l}}\right)}= \frac{\sqrt{l}}{\sqrt{g}}= \sqrt{\frac{l}{g}}. $$

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With $l \cdot g \neq 0$

$$ \frac{1}{\sqrt{\frac{g}{l}}} = \frac{1\times\sqrt{\frac{l}{g}}}{\sqrt{\frac{g}{l}}\times\sqrt{\frac{l}{g}}}= \frac{\sqrt{\frac{l}{g}}}{\sqrt{\frac{g}{l}\times\frac{l}{g}}} = \frac{\sqrt{\frac{l}{g}}}{1} = \sqrt{\frac{l}{g}} $$

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