4

Why does this term work?

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

graydad
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ColdStormy
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5 Answers5

19

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.

graydad
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  • Welp, now I got it. Thanks a lot :) – ColdStormy Mar 25 '15 at 15:23
  • For your second your say it is true when $b$ and $c$ are different from $0$, but what about $d$? – Alice Ryhl Mar 25 '15 at 16:56
  • @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. – graydad Mar 25 '15 at 16:57
  • Well, it's certainly not true when $d=0$, so it's a good idea to leave it in. – Alice Ryhl Mar 25 '15 at 16:58
  • @KristofferRyhl There is nothing wrong with having $d=0$ in the quantity $\frac{a}{b} \cdot \frac{d}{c}$. – graydad Mar 25 '15 at 16:59
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    No, that equation is well defined for $d=0$, but the other is not. $$\text{undefined}\ne\frac ab\cdot\frac 0c$$ – Alice Ryhl Mar 25 '15 at 17:01
  • Again, it is probably better to include $d\neq 0$ to avoid any uncertainty. If I knew $\frac{a}{b}\cdot {d}{c} = 0$ I'd be inclined to define $\frac{\frac{a}{b}}{\frac{c}{d}} \equiv 0$ – graydad Mar 25 '15 at 17:04
  • It does not hold for all real numbers with $b\neq 0$. Consider $(a,b)=(1,-1)$. Change that to 'for all non-negative numbers where $b\neq 0$'. – user26486 Mar 29 '15 at 18:20
9

$\sqrt{\frac{1}{a}}=\frac{1}{\sqrt{a}}$ and $\frac{1}{a/b}=b/a$

vudu vucu
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    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. – Erick Wong Mar 25 '15 at 18:01
4

$$\dfrac{1}{\sqrt{\dfrac{g}{l}}} = \dfrac{1}{\left(\dfrac{g}{l}\right)^{\frac{1}{2}}} = \dfrac{\left(\dfrac{g}{l}\right)^{0}}{\left(\dfrac{g}{l}\right)^{\frac{1}{2}}} = \left(\dfrac{g}{l}\right)^{0-\frac{1}{2}} =\left(\dfrac{g}{l}\right)^{-\frac{1}{2}} = \left(\left(\dfrac{g}{l}\right)^{-1}\right)^{\frac{1}{2}} = \left(\dfrac{l}{g}\right)^{\frac{1}{2}} = \sqrt{\dfrac{l}{g}}$$

3d0
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0

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

MattAllegro
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0

With $l \cdot g \neq 0$

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

phuclv
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