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For example: Why is $\sqrt{\dfrac{1}{2}} = \dfrac{1}{2}\sqrt{2}$ ?

How do you solve (simple) roots of fractions? I'm having trouble because I need this for my upcoming trig test but I haven't done algebra in a long, long time..

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$\frac{\sqrt 2}{2} = \frac{\sqrt 2}{\sqrt 4} = \sqrt \frac{2}{4} = \sqrt \frac{1}{2}$ using the rules (identities) $\frac{\sqrt a}{\sqrt b} = \sqrt \frac{a}{b}$ and $x = \pm \sqrt{x^2}$ – Arthur Sep 15 '12 at 12:12
You can choose a best answer by clicking the check mark next to the answer. It helps a lot! – David Sep 16 '12 at 13:35
Please pick a best answer. – Mew Sep 18 '12 at 7:19

There are three things to remember here:

  1. $$\displaystyle\large\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$$
  2. $$\displaystyle\large\frac{a}{b} = \frac{1}{b}\times a$$
  3. It's convention to simplify fractions with a radical denominator by multiplying the fraction by the denominator.

$$\displaystyle\large\frac{a}{\sqrt{b}} = \frac{a}{\sqrt{b}}\times\frac{\sqrt{b}}{\sqrt{b}}=\frac{a\sqrt{b}}{b}$$

So it's clear what happens as follows:

$$\begin{align*}\displaystyle\large\sqrt{\frac{1}{2}}= \\ & \displaystyle\large\frac{\sqrt{1}}{\sqrt{2}}\\ & = \displaystyle\large\frac{1}{\sqrt{2}}\\ & = \displaystyle\large\frac{1}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}\\ & = \displaystyle\large\frac{\sqrt{2}}{2}\displaystyle\large\\ & = \displaystyle\large\frac{1}{2}\times\sqrt{2}\end{align*}$$

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Start with $\sqrt{\dfrac{1}{2}}$
This is the same as ${\dfrac{\sqrt{1}}{\sqrt{2}}} = {\dfrac{{1}}{\sqrt{2}}}$
Now multiply both both the numerator and denominator of ${\dfrac{{1}}{\sqrt{2}}}$ by $\sqrt{2}$ to get:
${\dfrac{{\sqrt{2}}}{\sqrt{2}\sqrt{2}}} = {\dfrac{{\sqrt{2}}}{\sqrt{4}}} = {\dfrac{{\sqrt{2}}}{2}}$

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In general, $\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}$ if $a\ge0$ and $b>0$. But notice that, by multiplying both the numerator and denominator of the fraction by $\sqrt{b}$, we get $$\dfrac{1}{\sqrt{b}} = \dfrac{\sqrt{b}}{\sqrt{b} \cdot \sqrt{b}} = \dfrac{1}{b}\sqrt{b}$$ and so $$\sqrt{\dfrac{a}{b}} = \dfrac{1}{b} \sqrt{ab}$$

Put $a=1$ and $b=2$ into the above to get your answer.

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I added a condition on $a,b$ to avoid people turning this into one of those "standard proofs" of $1=2$. – Hagen von Eitzen Sep 15 '12 at 12:25
@HagenvonEitzen: Sure, thanks for that. – Clive Newstead Sep 15 '12 at 12:28

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