Roots of fractions

Question

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..

Answer 1

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:
${\dfrac{{\sqrt{2}}}{\sqrt{2}\sqrt{2}}} = {\dfrac{{\sqrt{2}}}{\sqrt{4}}} = {\dfrac{{\sqrt{2}}}{2}}$

Answer 2

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.

Answer 3

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:

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