# How do you rationalize the denominator of 1/(sqrt(2)+sqrt(3)+sqrt(5))?

##### 3 Answers
May 8, 2015

This is a very good question!

If it were simply $\frac{1}{\sqrt{a} + \sqrt{b}}$, we would use the conjugate,

and we would multiply by $\frac{\sqrt{a} - \sqrt{b}}{\sqrt{a} - \sqrt{b}}$.

Let's try something like that and see if it works. (This is what we do with problems of kinds we have not seen before. Try something and see if it works.)

$\frac{1}{\sqrt{2} + \sqrt{3} + \sqrt{5}} = \frac{1}{\sqrt{2} + \left(\sqrt{3} + \sqrt{5}\right)}$

$= \frac{1}{\left[\sqrt{2} + \left(\sqrt{3} + \sqrt{5}\right)\right]} \frac{\left[\sqrt{2} - \left(\sqrt{3} + \sqrt{5}\right)\right]}{\left[\sqrt{2} - \left(\sqrt{3} + \sqrt{5}\right)\right]}$

$= \frac{\left[\sqrt{2} - \left(\sqrt{3} + \sqrt{5}\right)\right]}{\left[\sqrt{2} + \left(\sqrt{3} + \sqrt{5}\right)\right] \left[\sqrt{2} - \left(\sqrt{3} + \sqrt{5}\right)\right]}$

$= \frac{\left[\sqrt{2} - \left(\sqrt{3} + \sqrt{5}\right)\right]}{2 - {\left(\sqrt{3} + \sqrt{5}\right)}^{2}}$

$= \frac{\left[\sqrt{2} - \left(\sqrt{3} + \sqrt{5}\right)\right]}{2 - \left(3 + 2 \sqrt{15} + 5\right)}$

$= \frac{\left[\sqrt{2} - \left(\sqrt{3} + \sqrt{5}\right)\right]}{- 6 - 2 \sqrt{15}}$

Did that help? (Yes, it did. We now have a more familiar looking problem.

$= \frac{\left[\sqrt{2} - \left(\sqrt{3} + \sqrt{5}\right)\right] \left[- 6 + 2 \sqrt{15}\right]}{\left[- 6 - 2 \sqrt{15}\right] \left[- 6 + 2 \sqrt{15}\right]}$

$= \frac{\left[\sqrt{2} - \left(\sqrt{3} + \sqrt{5}\right)\right] \left[- 6 + 2 \sqrt{15}\right]}{36 - 4 \left(15\right)}$

$= \frac{\left[\sqrt{2} - \left(\sqrt{3} + \sqrt{5}\right)\right] \left[- 6 + 2 \sqrt{15}\right]}{-} 24$

$= \frac{\left[\sqrt{2} - \left(\sqrt{3} + \sqrt{5}\right)\right] \left[3 - \sqrt{15}\right]}{12}$

Multiply the numerator if you like, to get:

$= \frac{\left[\sqrt{2} - \sqrt{3} - \sqrt{5}\right] \left[3 - \sqrt{15}\right]}{12}$

$= \frac{\left[\sqrt{2} - \sqrt{3} - \sqrt{5}\right] \left[3 - \sqrt{15}\right]}{12}$

$= \frac{3 \sqrt{2} - 3 \sqrt{3} - 3 \sqrt{5} - \sqrt{30} + \sqrt{45} + \sqrt{75}}{12}$

$= \frac{3 \sqrt{2} - 3 \sqrt{3} - 3 \sqrt{5} - \sqrt{30} + 3 \sqrt{5} + 5 \sqrt{3}}{12}$

$= \frac{3 \sqrt{2} + 2 \sqrt{3} - \sqrt{30}}{12}$

May 8, 2015

I've decided to add my version to Jim's as a demonstration or perhaps a warning about the variety of forms the outcome could take:
$\frac{1}{\sqrt{2} + \sqrt{3} + \sqrt{5}}$

Consider a simpler problem:
If we were asked to rationalize the denominator of
$\frac{1}{x + \sqrt{5}}$

we would simply multiply both the numerator and denominator by the conjugate of the denominator

$\frac{1}{x + \sqrt{5}} \times \frac{x - \sqrt{5}}{x - \sqrt{5}}$

$= \frac{x - \sqrt{5}}{{x}^{2} - 5}$

In this case $x = \left(\sqrt{2} + \sqrt{3}\right)$
and the result would be

=(sqrt(2)+sqrt(3)-sqrt(5))/((sqrt(2)+sqrt(3))^2-5

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

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

In order to complete the rationalization of the denominator we would need to multiply both the numerator and denominator by $\sqrt{6} = \sqrt{2 \cdot 3}$

$\frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{2 \sqrt{2 \cdot 3}} \times \frac{\sqrt{6}}{\sqrt{2 \cdot 3}}$

$= \frac{\sqrt{12} + \sqrt{18} - \sqrt{30}}{2 \left(2 \cdot 3\right)}$

$= \frac{2 \sqrt{3} + 3 \sqrt{2} - \sqrt{2 \cdot 3 \cdot 5}}{12}$

I avoided posting this because I am bothered by the result seems to imply that the sequence of the denominator terms is reflected in the result. This should not be true but I have not (yet) been able to come up with a final result where the terms are interchangeable.

Mar 23, 2017

A more general denominator

#### Explanation:

If we employ the more general denominator of
$\sqrt{a} + \sqrt{b} + \sqrt{c}$

then we arrive at a numerator of

$\left(a - b - c\right) \sqrt{a} + \left(b - a - c\right) \sqrt{b} + \left(c - a - b\right) \sqrt{c} + 2 \sqrt{a b c}$

and a more general (rational) denominator of

${a}^{2} + {b}^{2} + {c}^{2} - 2 \left(a b + a c + b c\right)$

The same work as we see for 2, 3, and 5 yields this result.

In the example given, c = a + b, which eliminates the term that would have contained $\sqrt{5}$. Furthermore any time c = a + b, the more general solution simplifies to a numerator of

$b \sqrt{a} + a \sqrt{b} - \sqrt{a b \left(a + b\right)}$
and a denominator of 2ab.

In the specific instance here we have
$\frac{3 \sqrt{2} + 2 \sqrt{3} - \sqrt{30}}{12}$