# How do you rationalize the denominator 1/ (1+ sqrt3 - sqrt5)?

Apr 3, 2015

$\frac{1}{1 + \sqrt{3} - \sqrt{5}}$

The rationalization of this denominator is a two-set process.
First rationalize the $\sqrt{5}$
and then, after some simplification
rationalize the $\sqrt{3}$

Let $a = \left(1 + \sqrt{3}\right)$
So our initial stage is to rationalize the denominator of
$\frac{1}{a - \sqrt{5}}$

As usual to do this we multiply the numerator and the denominator by the conjugate of the denominator:
$\frac{1}{a - \sqrt{5}} \cdot \frac{a + \sqrt{5}}{a + \sqrt{5}}$

$= \frac{a + \sqrt{5}}{{a}^{2} - 5}$

Substituting $\left(1 + \sqrt{3}\right)$ back in to this expression in place of $a$.
we get
$= \frac{1 + \sqrt{3} + \sqrt{5}}{{\left(1 + \sqrt{3}\right)}^{2} - 5}$

$= \frac{1 + \sqrt{3} + \sqrt{5}}{1 + 2 \sqrt{3} + 3 - 5}$

$= \frac{1 + \sqrt{3} + \sqrt{5}}{2 \sqrt{3} - 1}$

Rationalizing this denominator (using the conjugate of $2 \sqrt{3} - 1$
results in
$\frac{\left(1 + \sqrt{3} + \sqrt{5}\right)}{\left(2 \sqrt{3} - 1\right)} \cdot \frac{\left(2 \sqrt{3} + 1\right)}{\left(2 \sqrt{3} + 1\right)}$

$= \frac{7 + 3 \sqrt{3} + \sqrt{5} + 2 \sqrt{15}}{11}$