Rationalising the Denominator

Eliminating surds from the bottom of fractions

📚 Mathematics 🎯 Difficulty: ⭐ ⏱️ Reading time: 20 minutes 📋 Edexcel & AQA

📚 Introduction

When a surd appears in the denominator of a fraction, it’s often desirable to “rationalise” it — that is, to eliminate the surd from the denominator. Rationalising helps simplify expressions and prepares them for further algebraic manipulation.

🎯 Learning Objectives

Understand why rationalising the denominator is useful
Apply rationalising techniques to single and binomial surd denominators
Recognise when rationalisation is necessary

🤔 Why Rationalise?

Historically, having surds in the denominator was considered “untidy,” especially before calculators. More importantly, rationalising makes it easier to compare or combine expressions, particularly in algebra and calculus.

🛠️ Methods

Case 1: Single Surd Denominator

To rationalise $\frac{1}{\sqrt{3}}$, multiply both numerator and denominator by $\sqrt{3}$:

\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}

Case 2: Binomial Surd Denominator

To rationalise a denominator like $\frac{1}{2 + \sqrt{3}}$, multiply top and bottom by the conjugate of the denominator.

\frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{2 - \sqrt{3}}{4 - 3} = \frac{2 - \sqrt{3}}{1} = \boxed{2 - \sqrt{3}}

📝 Worked Examples

Example 1

Question: Rationalise $\frac{5}{\sqrt{2}}$.

Solution:

\frac{5}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{5\sqrt{2}}{2}

Example 2

Question: Rationalise $\frac{3}{1 - \sqrt{2}}$.

Solution:

\frac{3}{1 - \sqrt{2}} \times \frac{1 + \sqrt{2}}{1 + \sqrt{2}} = \frac{3(1 + \sqrt{2})}{(1 - \sqrt{2})(1 + \sqrt{2})} = \frac{3(1 + \sqrt{2})}{1 - 2} $$ $$= \frac{3(1 + \sqrt{2})}{-1} = \boxed{-3 - 3\sqrt{2}}

🎯 Practice Question 1

(A) Rationalise: $\frac{7}{\sqrt{5}}$

(B) Rationalise: $\frac{2}{\sqrt{6} + 1}$

(C) Rationalise: $\frac{4}{3 - \sqrt{2}}$

(D) Rationalise: $\frac{5}{\sqrt{3} - \sqrt{2}}$

🔍 Show Solution

Solutions:

(A) $\frac{7}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{7\sqrt{5}}{5}$

(B) $\frac{2}{\sqrt{6} + 1} \times \frac{\sqrt{6} - 1}{\sqrt{6} - 1} = \frac{2(\sqrt{6} - 1)}{6 - 1} = \frac{2\sqrt{6} - 2}{5}$

(C) $\frac{4}{3 - \sqrt{2}} \times \frac{3 + \sqrt{2}}{3 + \sqrt{2}} = \frac{4(3 + \sqrt{2})}{9 - 2} = \frac{12 + 4\sqrt{2}}{7}$

(D) $\frac{5}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = \frac{5(\sqrt{3} + \sqrt{2})}{3 - 2} = 5\sqrt{3} + 5\sqrt{2}$

🧾 Summary

Rationalising the denominator means removing surds from the denominator of a fraction
For single surds, multiply top and bottom by the surd
For binomial denominators, use the conjugate to simplify
Helps simplify expressions and is useful in higher-level maths