Rationalising the denominator

A frac­tion such as $\displaystyle \frac{1}{\sqrt{2}}$ is hard to work with: divid­ing $1$ by $1.41421\ldots$ by hand is no fun. But mul­ti­ply the top and bot­tom by $\sqrt{2}$ and it becomes $\displaystyle \frac{\sqrt{2}}{2}$, which is sim­ply half of $1.41421\ldots$ Remov­ing the surd from the denom­i­na­tor like this is called ratio­nal­is­ing the denom­i­na­tor. It is the main use of ratio­nal­is­ing fac­tors, and the topic of this les­son.

The method#

1. Find a ratio­nal­is­ing fac­tor of the denom­i­na­tor. For $\sqrt{a}$ it is $\sqrt{a}$; for a bino­mial surd it is the con­ju­gate.
2. Mul­ti­ply both the numer­a­tor and the denom­i­na­tor by it. This does not change the value of the frac­tion, because you are mul­ti­ply­ing by $\displaystyle \frac{\text{R.F.}}{\text{R.F.}} = 1$.
3. Sim­plify. The denom­i­na­tor is now ratio­nal.

A sin­gle surd in the denom­i­na­tor#

Exam­ple 1#

Ratio­nalise the denom­i­na­tors of (a) $\displaystyle \frac{1}{\sqrt{2}}$ (b) $\displaystyle \frac{6}{\sqrt{3}}$ (c) $\displaystyle \frac{1}{\sqrt[3]{2}}$.

(a) $\displaystyle \frac{1}{\sqrt{2}} = \frac{1 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{\sqrt{2}}{2}$

(b) $\displaystyle \frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}$

(c) The ratio­nal­is­ing fac­tor of $\sqrt[3]{2}$ is $\sqrt[3]{4}$, since $\sqrt[3]{2} \times \sqrt[3]{4} = \sqrt[3]{8} = 2$. So $\displaystyle \frac{1}{\sqrt[3]{2}} = \frac{\sqrt[3]{4}}{2}$.

A bino­mial surd in the denom­i­na­tor#

Mul­ti­ply by the con­ju­gate, and use $(a + b)(a - b) = a^{2} - b^{2}$ in the denom­i­na­tor.

Exam­ple 2#

Ratio­nalise the denom­i­na­tor of $\displaystyle \frac{1}{3 + \sqrt{2}}$.

$$\displaystyle \begin{aligned} \frac{1}{3 + \sqrt{2}} &= \frac{1(3 - \sqrt{2})}{(3 + \sqrt{2})(3 - \sqrt{2})} \\ &= \frac{3 - \sqrt{2}}{9 - 2} \\ &= \frac{3 - \sqrt{2}}{7} \end{aligned}$$

Exam­ple 3#

Ratio­nalise the denom­i­na­tor of $\displaystyle \frac{1}{\sqrt{8} - \sqrt{7}}$.

$$\displaystyle \frac{1}{\sqrt{8} - \sqrt{7}} = \frac{1(\sqrt{8} + \sqrt{7})}{(\sqrt{8} - \sqrt{7})(\sqrt{8} + \sqrt{7})} = \frac{\sqrt{8} + \sqrt{7}}{8 - 7} = \sqrt{8} + \sqrt{7}$$

The denom­i­na­tor came out as $1$, so the frac­tion dis­ap­pears alto­gether. In sim­plest form the answer is $2\sqrt{2} + \sqrt{7}$.

Com­par­ing ratio­nal and surd parts#

When two expres­sions of the form $p + q\sqrt{3}$ are equal, with $p$ and $q$ ratio­nal, their ratio­nal parts are equal and their surd parts are equal. This lets you find unknown num­bers.

Exam­ple 4#

If $\displaystyle \frac{5 + \sqrt{3}}{7 - 4\sqrt{3}} = 47a + \sqrt{3}b$, find $a$ and $b$.

Ratio­nalise the left side, using the con­ju­gate $7 + 4\sqrt{3}$:

$$\displaystyle \begin{aligned} \frac{5 + \sqrt{3}}{7 - 4\sqrt{3}} &= \frac{(5 + \sqrt{3})(7 + 4\sqrt{3})}{(7 - 4\sqrt{3})(7 + 4\sqrt{3})} \\ &= \frac{35 + 20\sqrt{3} + 7\sqrt{3} + 12}{49 - 48} \\ &= 47 + 27\sqrt{3} \end{aligned}$$

So $47 + 27\sqrt{3} = 47a + \sqrt{3}b$. Com­par­ing ratio­nal parts, $47a = 47$, so $a = 1$. Com­par­ing surd parts, $b = 27$.

Adding frac­tions with surd denom­i­na­tors#

To add or sub­tract such frac­tions, ratio­nalise each one first, then col­lect like terms.

Exam­ple 5#

Sim­plify $\displaystyle \frac{4 + \sqrt{5}}{4 - \sqrt{5}} + \frac{4 - \sqrt{5}}{4 + \sqrt{5}}$.

Mul­ti­ply the first frac­tion by $\displaystyle \frac{4 + \sqrt{5}}{4 + \sqrt{5}}$ and the sec­ond by $\displaystyle \frac{4 - \sqrt{5}}{4 - \sqrt{5}}$. Both denom­i­na­tors become $(4)^{2} - (\sqrt{5})^{2} = 16 - 5 = 11$.

$$\displaystyle \begin{aligned} &= \frac{(4 + \sqrt{5})^{2}}{11} + \frac{(4 - \sqrt{5})^{2}}{11} \\ &= \frac{16 + 5 + 8\sqrt{5}}{11} + \frac{16 + 5 - 8\sqrt{5}}{11} \\ &= \frac{21 + 8\sqrt{5} + 21 - 8\sqrt{5}}{11} \\ &= \frac{42}{11} \end{aligned}$$

Exam­ple 6: every­thing can­cels#

Sim­plify $\displaystyle \frac{1}{2 + \sqrt{3}} + \frac{2}{\sqrt{5} - \sqrt{3}} + \frac{1}{2 - \sqrt{5}}$.

Ratio­nalise each frac­tion sep­a­rately.

• $\displaystyle \frac{1}{2 + \sqrt{3}} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}$
• $\displaystyle \frac{2}{\sqrt{5} - \sqrt{3}} = \frac{2(\sqrt{5} + \sqrt{3})}{5 - 3} = \sqrt{5} + \sqrt{3}$
• $\displaystyle \frac{1}{2 - \sqrt{5}} = \frac{2 + \sqrt{5}}{4 - 5} = -(2 + \sqrt{5}) = -2 - \sqrt{5}$

Now add:

$$2 - \sqrt{3} + \sqrt{5} + \sqrt{3} - 2 - \sqrt{5} = 0$$

Watch the third frac­tion: its denom­i­na­tor became $4 - 5 = -1$, which turned every sign in the numer­a­tor round.

Exam­ple 7#

Sim­plify $\displaystyle \frac{2}{\sqrt{5} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{2}} - \frac{3}{\sqrt{5} + \sqrt{2}}$.

• $\displaystyle \frac{2}{\sqrt{5} + \sqrt{3}} = \frac{2(\sqrt{5} - \sqrt{3})}{5 - 3} = \sqrt{5} - \sqrt{3}$
• $\displaystyle \frac{1}{\sqrt{3} + \sqrt{2}} = \frac{\sqrt{3} - \sqrt{2}}{3 - 2} = \sqrt{3} - \sqrt{2}$
• $\displaystyle \frac{3}{\sqrt{5} + \sqrt{2}} = \frac{3(\sqrt{5} - \sqrt{2})}{5 - 2} = \sqrt{5} - \sqrt{2}$

So the expres­sion is

$$(\sqrt{5} - \sqrt{3}) + (\sqrt{3} - \sqrt{2}) - (\sqrt{5} - \sqrt{2}) = \sqrt{5} - \sqrt{3} + \sqrt{3} - \sqrt{2} - \sqrt{5} + \sqrt{2} = 0$$

Prac­tice#

1. Ratio­nalise the denom­i­na­tor of $\displaystyle \frac{5}{\sqrt{5}}$.
2. Ratio­nalise the denom­i­na­tor of $\displaystyle \frac{3}{2\sqrt{6}}$.
3. Ratio­nalise the denom­i­na­tor of $\displaystyle \frac{1}{\sqrt[3]{3}}$.
4. Ratio­nalise the denom­i­na­tor of $\displaystyle \frac{1}{2 + \sqrt{5}}$.
5. Ratio­nalise the denom­i­na­tor of $\displaystyle \frac{4}{\sqrt{7} - \sqrt{3}}$.
6. Ratio­nalise the denom­i­na­tor of $\displaystyle \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$.
7. Ratio­nalise the denom­i­na­tor of $\displaystyle \frac{6}{2\sqrt{3} - \sqrt{6}}$.
8. If $\displaystyle \frac{3 - \sqrt{2}}{3 + \sqrt{2}} = a + b\sqrt{2}$, find $a$ and $b$.
9. Sim­plify $\displaystyle \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}} + \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}$.
10. Sim­plify $\displaystyle \frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + 2}$.

Answers#