# Rationalise Denominator: 1/(1 + √2 + √3) by using: (1 - √2 + √3) as conjugate?

Feb 24, 2018

Explained below :-

#### Explanation:

k = 1/(1 + {√2 + √3})

rArr k = 1/({√3+1}+√2)

rArrk = ({√3+1}-√2)/[({√3+1}+√2).({√3+1}-√2)]

rArrk = ({√3+1}-√2)/[(√3+1)^2-(√2)^2]

rArrk = ({√3+1}-√2)/[3+1+2√3-2]

:.k = ({√3+1}-√2)/[2(1+√3)]

Mar 6, 2018

$\frac{1}{1 + \sqrt{2} + \sqrt{3}} = \frac{2 + \sqrt{2} - \sqrt{6}}{4}$

#### Explanation:

The difference of squares identity can be written:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

This is the key to eliminating square roots from the denominator.

Note that $\left(1 - \sqrt{2} + \sqrt{3}\right)$ is only a partial conjugate for $\left(1 + \sqrt{2} + \sqrt{3}\right)$. Multiplying these two expressions will eliminate terms in $\sqrt{2}$ but leave terms in $\sqrt{3}$.

If we want to rationalise the denominator, we will also need to multiply by some expression of the form $a + b \sqrt{3}$ (actually I will use $\left(\sqrt{3} - 1\right)$)

$\frac{1}{1 + \sqrt{2} + \sqrt{3}}$

$= \frac{\left(1 + \sqrt{3}\right) - \sqrt{2}}{\left(\left(1 + \sqrt{3}\right) - \sqrt{2}\right) \left(\left(1 + \sqrt{3}\right) + \sqrt{2}\right)}$

$= \frac{1 + \sqrt{3} - \sqrt{2}}{{\left(1 + \sqrt{3}\right)}^{2} - {\left(\sqrt{2}\right)}^{2}}$

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

$= \frac{1 + \sqrt{3} - \sqrt{2}}{2 \left(\sqrt{3} + 1\right)}$

$= \frac{\left(\sqrt{3} - 1\right) \left(1 + \sqrt{3} - \sqrt{2}\right)}{2 \left(\sqrt{3} - 1\right) \left(\sqrt{3} + 1\right)}$

$= \frac{\left(\sqrt{3} + 3 - \sqrt{6}\right) - \left(1 + \sqrt{3} - \sqrt{2}\right)}{2 \left(3 - 1\right)}$

$= \frac{2 + \sqrt{2} - \sqrt{6}}{4}$

Note that having got to this result, the numerator is a proper conjugate for $1 + \sqrt{2} + \sqrt{3}$ in that:

$\left(1 + \sqrt{2} + \sqrt{3}\right) \left(2 + \sqrt{2} - \sqrt{6}\right) = 4$