Rationalising a Simple Surd

Statement

When a fraction has a single square root in the denominator, we can rationalise it by multiplying top and bottom by the same square root:

\[ \frac{1}{\sqrt{a}} = \frac{\sqrt{a}}{\sqrt{a}\cdot\sqrt{a}} = \frac{\sqrt{a}}{a}, \quad (a>0). \]

Why it’s true

• Multiplying numerator and denominator by \(\sqrt{a}\) eliminates the surd from the denominator.
• Since \(\sqrt{a}\times \sqrt{a} = a\), the denominator becomes rational.

Recipe (how to use it)

1. Identify a denominator of the form \(\sqrt{a}\).
2. Multiply numerator and denominator by \(\sqrt{a}\).
3. Simplify the result as \(\tfrac{\sqrt{a}}{a}\).

Spotting it

This occurs in simple fractions like \(\tfrac{1}{\sqrt{2}}\), \(\tfrac{1}{\sqrt{5}}\), or more generally \(\tfrac{1}{\sqrt{a}}\).

Common pairings

• Surds simplification in GCSE algebra.
• Exact trigonometric values (e.g. \(\sin 45^\circ = \tfrac{1}{\sqrt{2}} = \tfrac{\sqrt{2}}{2}\)).
• Probability and geometry questions requiring simplified denominators.

Mini examples

1. \(\tfrac{1}{\sqrt{2}} = \tfrac{\sqrt{2}}{2}\).
2. \(\tfrac{1}{\sqrt{5}} = \tfrac{\sqrt{5}}{5}\).

Pitfalls

• Forgetting to multiply both numerator and denominator by the surd.
• Leaving the denominator irrational when the question explicitly asks for rational form.
• Not simplifying the final fraction fully.

Exam strategy

• Always show the rationalisation step explicitly.
• Check if the surd simplifies before rationalising (e.g. \(\sqrt{12} = 2\sqrt{3}\)).
• In trigonometry, remember to give exact forms (like \(\sqrt{2}/2\)) instead of decimals.

Summary

Rationalising a simple surd is a straightforward process: multiply top and bottom by the same surd to remove it from the denominator. The result is cleaner, exact, and consistent with exam expectations.