Rationalisation of Denominators

Mathematicians (and exam papers) like fractions whose denominators are rational. A denominator like $\sqrt{2}$ or $\sqrt{2}+1$ is technically fine but awkward , you cannot easily compare $\dfrac{1}{\sqrt{2}+1}$ with $\dfrac{1}{\sqrt{3}-1}$, and adding such fractions is a chore. Rationalisation is the technique that scrubs surds out of the denominator using one short trick: multiply by a clever form of $1$.

Definitions

The conjugate of a binomial surd $\sqrt{a} + \sqrt{b}$ is $\sqrt{a} - \sqrt{b}$. The conjugate of $a + \sqrt{b}$ is $a - \sqrt{b}$. The conjugate of $\sqrt{a} - \sqrt{b}$ is $\sqrt{a} + \sqrt{b}$. In every case, multiplying a binomial surd by its conjugate gives a difference of squares with no surd left: $(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b, \qquad (a + \sqrt{b})(a - \sqrt{b}) = a^2 - b.$

To rationalise a fraction is to rewrite it as an equivalent fraction whose denominator contains no surd.

Concept and technique

Single-term surd in the denominator. Multiply numerator and denominator by the same surd: $\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}.$ The result is exact: $\dfrac{1}{\sqrt{2}}$ and $\dfrac{\sqrt{2}}{2}$ are the same number. We have only multiplied by $1$.

Binomial surd in the denominator. Multiply by the conjugate: $\frac{1}{\sqrt{2} + 1} = \frac{1}{\sqrt{2} + 1} \cdot \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \frac{\sqrt{2} - 1}{(\sqrt{2})^2 - 1^2} = \frac{\sqrt{2} - 1}{1} = \sqrt{2} - 1.$

That same trick works for $\dfrac{1}{\sqrt{3} - \sqrt{2}}$: multiply by $\dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}}$ to get $\dfrac{\sqrt{3} + \sqrt{2}}{3 - 2} = \sqrt{3} + \sqrt{2}$.

General form. For a fraction $\dfrac{N}{a + b\sqrt{c}}$ (with $N$ any expression and $a, b, c$ rationals, $c \ge 0$), multiply top and bottom by $a - b\sqrt{c}$. The new denominator is $a^2 - b^2 c$, which is rational. The new numerator might still contain surds, but that is exactly the goal: shift them upstairs.

Why we do it. Three big reasons.

1. Comparison. $\dfrac{1}{\sqrt{2}+1}$ rationalised becomes $\sqrt{2} - 1 \approx 0.414$. Now it is obvious it is less than $1$.
2. Addition. Adding $\dfrac{1}{\sqrt{2}}$ and $\dfrac{1}{\sqrt{3}}$ is messy. Rationalise to $\dfrac{\sqrt{2}}{2}$ and $\dfrac{\sqrt{3}}{3}$, then a common denominator of $6$ does it.
3. Standard form. A "neat" answer in board exams is one with no surd in the denominator.

The trick is the difference of squares. Every rationalisation in this lesson is just $(a+b)(a-b) = a^2 - b^2$ applied to surds. If you remember that identity, the technique is automatic.

For cube roots and higher, conjugates are more elaborate. To rationalise $\dfrac{1}{\sqrt[3]{a}}$, multiply by $\dfrac{\sqrt[3]{a^2}}{\sqrt[3]{a^2}}$ to get $\dfrac{\sqrt[3]{a^2}}{a}$. We focus on square roots here.

Worked examples

Example 1. Rationalise the denominator of $\dfrac{1}{\sqrt{7}}$.

Multiply by $\dfrac{\sqrt{7}}{\sqrt{7}}$: $\dfrac{1}{\sqrt{7}} = \dfrac{\sqrt{7}}{7}$.

Example 2. Rationalise $\dfrac{1}{2 + \sqrt{3}}$.

Multiply by $\dfrac{2 - \sqrt{3}}{2 - \sqrt{3}}$: numerator $= 2 - \sqrt{3}$, denominator $= 4 - 3 = 1$. So the answer is $2 - \sqrt{3}$.

Example 3. Rationalise $\dfrac{5}{\sqrt{5} - \sqrt{2}}$.

Multiply by $\dfrac{\sqrt{5} + \sqrt{2}}{\sqrt{5} + \sqrt{2}}$: numerator $= 5(\sqrt{5} + \sqrt{2})$, denominator $= 5 - 2 = 3$. Answer: $\dfrac{5(\sqrt{5} + \sqrt{2})}{3}$.

Example 4. Simplify $\dfrac{1}{3 + \sqrt{2}} + \dfrac{1}{3 - \sqrt{2}}$.

Combine over a common denominator: $\dfrac{(3 - \sqrt{2}) + (3 + \sqrt{2})}{(3+\sqrt{2})(3-\sqrt{2})} = \dfrac{6}{9 - 2} = \dfrac{6}{7}$. Notice how the conjugate pair in the denominator did the rationalisation automatically.

Example 5. If $\dfrac{1}{\sqrt{5} - 2} = a + b\sqrt{5}$, find $a$ and $b$.

Rationalise: $\dfrac{1}{\sqrt{5} - 2} \cdot \dfrac{\sqrt{5} + 2}{\sqrt{5} + 2} = \dfrac{\sqrt{5} + 2}{5 - 4} = \sqrt{5} + 2$. So $a = 2, b = 1$.

Try it yourself

1. Rationalise $\dfrac{1}{\sqrt{11}}$.
2. Rationalise $\dfrac{1}{\sqrt{7} - \sqrt{6}}$.
3. Rationalise $\dfrac{1}{\sqrt{5} + \sqrt{2}}$.
4. Simplify $\dfrac{4}{2 + \sqrt{3}}$.
5. Rationalise $\dfrac{6}{\sqrt{3} - \sqrt{2}}$.
6. Show that $\dfrac{1}{\sqrt{2} + \sqrt{3}} + \dfrac{1}{\sqrt{2} - \sqrt{3}}$ is rational.
7. If $\dfrac{7 + \sqrt{5}}{7 - \sqrt{5}} - \dfrac{7 - \sqrt{5}}{7 + \sqrt{5}} = a + b\sqrt{5}$, find $a$ and $b$.
8. Rationalise $\dfrac{1}{\sqrt{2} - 1} - \dfrac{1}{\sqrt{2} + 1}$.
9. Simplify $\dfrac{3}{\sqrt{6} - \sqrt{3}}$.
10. If $x = \dfrac{1}{\sqrt{3} - \sqrt{2}}$, find $x^2$ in rationalised form.

Pitfalls / Insight

• Conjugates flip only the sign between the two surd terms. The conjugate of $\sqrt{3} + 2$ is $\sqrt{3} - 2$, not $-\sqrt{3} + 2$. (Both work mathematically here, but the standard convention is the first.)
• Don't change the value. You multiply by $\dfrac{c}{c}$ where $c$ is non-zero. If $c = 0$, your "trick" was illegal because you cannot multiply by $0/0$.
• Simplify after rationalising. $\dfrac{\sqrt{12}}{6}$ should be written $\dfrac{\sqrt{12}}{6} = \dfrac{2\sqrt{3}}{6} = \dfrac{\sqrt{3}}{3}$.

Insight. When you encounter $a + b\sqrt{c}$ anywhere, imagine its conjugate $a - b\sqrt{c}$ next to it. Together they kill the surd via $a^2 - b^2 c$. This is the only idea in the lesson. Everything else is bookkeeping.