# Simplify this division of square roots?

## $\frac{\frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}}$

Sep 1, 2016

$\sqrt{2} - 1$.

#### Explanation:

The Expression$= \frac{\frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}}$

$= \frac{\frac{\sqrt{2}}{\cancel{2}}}{\frac{2 + \sqrt{2}}{\cancel{2}}}$

$= \frac{\sqrt{2}}{2 + \sqrt{2}}$

$= \frac{\sqrt{2}}{2 + \sqrt{2}}$

=cancel(sqrt2)/(cancelsqrt2(sqrt2+1)

$= \frac{1}{\sqrt{2} + 1} \times \left(\frac{\sqrt{2} - 1}{\sqrt{2} - 1}\right)$

$= \frac{\sqrt{2} - 1}{2 - 1}$

$= \sqrt{2} - 1$.

Sep 1, 2016

$\frac{\frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}} = \sqrt{2} - 1$

#### Explanation:

We will continue under the assumption that "simplifying" requires rationalizing the denominator.

First, we can remove fractions from the numerator and denominator by multiplying both by $2$:

$\frac{\frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}} = \frac{\frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}} \cdot \frac{2}{2}$

$= \frac{\sqrt{2}}{2 + \sqrt{2}}$

Then, we rationalize the denominator by multiplying by the conjugate of the denominator, and taking advantage of the identity $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$

$\frac{\sqrt{2}}{2 + \sqrt{2}} = \frac{\sqrt{2}}{2 + \sqrt{2}} \cdot \frac{2 - \sqrt{2}}{2 - \sqrt{2}}$

$= \frac{2 \sqrt{2} - \sqrt{2} \cdot \sqrt{2}}{{2}^{2} - {\sqrt{2}}^{2}}$

$= \frac{2 \sqrt{2} - 2}{4 - 2}$

$= \frac{\cancel{2} \left(\sqrt{2} - 1\right)}{\cancel{2}}$

$= \sqrt{2} - 1$

Sep 1, 2016

$\sqrt{2} - 1$

#### Explanation:

We will make use of the fact that $\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \times d}{b \times c}$

But before we can do that, we need to add the fractions in the denominator to make one fraction.

$\frac{\frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}} \text{ = } \frac{\frac{\sqrt{2}}{2}}{\frac{2 + \sqrt{2}}{2}}$

(color(red)(sqrt2)/color(blue)(2))/(color(blue)((2+sqrt2)/color(red)(2))) " = "(color(red)(cancel2sqrt2))/ (color(blue)(cancel2(2+sqrt2)) Much better!

Now rationalise the denominator:

$\frac{\sqrt{2}}{\left(2 + \sqrt{2}\right)} \times \textcolor{\lim e}{\frac{\left(2 - \sqrt{2}\right)}{\left(2 - \sqrt{2}\right)}} = \frac{2 \sqrt{2} - {\sqrt{2}}^{2}}{{2}^{2} - {\sqrt{2}}^{2}}$

$\frac{2 \sqrt{2} - 2}{4 - 2} = \frac{\cancel{2} \left(\sqrt{2} - 1\right)}{\cancel{2}}$

=$\sqrt{2} - 1$