E1.18 - Rationalising denominators | Cambridge IGCSE Mathematics

An exact fraction can be written with a surd in its denominator or in an equivalent form with a rational denominator. You will make that change by multiplying by one, using a single surd or a conjugate pair, then simplify the result. No decimal approximation is needed.

A single-surd denominator

Can moving a square root out of a denominator leave a fraction's value unchanged? For instance, $6/\sqrt3$ looks quite different from $2\sqrt3$ , but multiplying either by $\sqrt3$ gives $6$ .

You already know that multiplying a fraction's numerator and denominator by the same non-zero number preserves its value: you are multiplying by one. Here choose a number that makes the new denominator rational.

Rationalising a denominator

To rationalise the denominator is to write an equivalent fraction with a rational denominator. For a single square-root denominator, multiplying by that square root works because $\sqrt b\,\sqrt b=b$ , for $b>0$ .

For example, $\sqrt3/\sqrt3=1$ , so it changes the appearance but not the value. A denominator must be non-zero before and after the change. There is no need to rationalise $\sqrt9=3$ : that denominator is already rational.

Worked example 1: a single square root

Question. Rationalise the denominator and simplify $6/\sqrt3$ .

Plan. Multiply both numerator and denominator by $\sqrt3$ , then simplify the rational denominator.

\begin{aligned} \frac6{\sqrt3} &=\frac{6\sqrt3}{\sqrt3\,\sqrt3} && \text{multiply by one}\\ &=\frac{6\sqrt3}{3} && \text{square the root}\\ &=2\sqrt3 && \text{divide the coefficient by three} \end{aligned}

Answer. $2\sqrt3$ .

Check. $2\sqrt3\times\sqrt3=2(3)=6$ , so multiplying the proposed answer by the original denominator returns the numerator.

Add a coefficient of $2$ to the denominator. Only that factor changes; multiplying by $\sqrt3$ still removes the radical.

Worked example 2: a coefficient beside the root

Question. Rationalise the denominator and simplify $6/(2\sqrt3)$ .

Plan. Keep the coefficient while turning the pair of square roots into $3$ .

\begin{aligned} \frac6{2\sqrt3} &=\frac{6\sqrt3}{2\sqrt3\,\sqrt3} && \text{multiply top and bottom}\\ &=\frac{6\sqrt3}{6} && \text{denominator is }2(3)\\ &=\sqrt3 && \text{cancel the common factor six} \end{aligned}

Answer. $\sqrt3$ .

Check. $\sqrt3(2\sqrt3)=6$ . Also, doubling a non-zero denominator halves the fraction, matching half of the first answer.

In general, for rational $p,q$ , with $q\ne0$ and $b>0$ ,

\frac{p}{q\sqrt b}=\frac{p\sqrt b}{qb}.

When $b$ is rational, the new denominator is rational. Simplify common factors afterwards.

Watch out

Multiplying only the bottom changes the value. The line $6/\sqrt3=6/3$ is wrong: the top must also acquire the factor $\sqrt3$ .

Conjugates and a difference of squares

Multiplying $1/(3+\sqrt5)$ by $\sqrt5/\sqrt5$ would leave a term $3\sqrt5$ in the denominator. With two terms, use the cancellation you know from expanding two brackets.

The conjugate of $3+\sqrt5$ is $3-\sqrt5$ : keep both terms, but change the sign between them. Their middle terms cancel:

(3+\sqrt5)(3-\sqrt5)=9-3\sqrt5+3\sqrt5-5=4.

Conjugates and a difference of squares

(u+v)(u-v)=u^2-v^2.

For $u=a$ and $v=\sqrt b$ , where $a,b$ are rational, $b\ge0$ and $a^2\ne b$ , the product $a^2-b$ is a non-zero rational number. Thus multiplying by the conjugate over itself rationalises $a+\sqrt b$ or $a-\sqrt b$ .

Not given in the List of formulas: learn it.

The condition $a^2\ne b$ ensures that both the original denominator and the conjugate used as a multiplier are non-zero. A conjugate is a convenient choice, rather than the only possible multiplier: a non-zero rational multiple of it can work too.

Worked example 3: a plus sign in the denominator

Question. Rationalise the denominator of $\dfrac1{3+\sqrt5}$ .

Plan. Multiply by the conjugate over itself, then use a difference of squares.

\begin{aligned} \frac1{3+\sqrt5} &=\frac{3-\sqrt5}{(3+\sqrt5)(3-\sqrt5)} && \text{multiply by one}\\ &=\frac{3-\sqrt5}{9-5} && \text{middle terms cancel}\\ &=\frac{3-\sqrt5}{4} && \text{simplify the denominator} \end{aligned}

Answer. $(3-\sqrt5)/4$ .

Check. Multiplying by $3+\sqrt5$ gives $(9-5)/4=1$ . Since $3>\sqrt5$ , both the original fraction and this answer are positive.

Change only the denominator's plus sign to a minus sign. Its conjugate changes sign too, but the difference-of-squares denominator is unchanged.

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E1.18 - Rationalising denominators

A single-surd denominator

Conjugates and a difference of squares

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