Surds and rationalisation

Type $\dfrac{1}{\sqrt3}$ into a division and you hit a wall: you cannot divide neatly by an irrational number. Yet $\dfrac{\sqrt3}{3}$ — the very same value — divides just fine. The difference is entirely in where the root sits. Numbers like $\sqrt2$ , $\sqrt3$ and $\sqrt[3]{5}$ are called surds: irrational roots that refuse to simplify away. This article, part of the Pure Arithmetic chapter of ICSE Class 9 Mathematics, builds the two skills that make surds usable — combining them correctly under addition, multiplication and division, and rationalising a denominator so no root remains underneath a fraction. The key trick, multiplying by a cleverly chosen conjugate, turns a messy irrational denominator into a clean whole number, and it rests on nothing more advanced than the identity $(x+y)(x-y)=x^2-y^2$ you already know from algebra.

The core insight

A surd denominator is cleared by multiplying top and bottom by the surd itself (single-term case) or by its conjugate (two-term case), using $(x+y)(x-y)=x^2-y^2$ to turn the denominator into a whole number.

The core idea

A surd is an irrational root of a rational number that cannot be reduced to remove the root: $\sqrt2$ , $\sqrt3$ , $\sqrt[3]{5}$ are surds, but $\sqrt4=2$ is not, since it simplifies to a rational number. Surds combine under three simple laws, valid for $a,b \geq 0$ :

\sqrt a \times \sqrt b = \sqrt{ab}, \qquad \frac{\sqrt a}{\sqrt b} = \sqrt{\frac{a}{b}}, \qquad k\sqrt a \pm m\sqrt a = (k \pm m)\sqrt a

where: $a$ and $b$ are the numbers under the root (the radicands), and $k$ , $m$ are the whole-number coefficients multiplying a shared surd $\sqrt a$ . The third law only applies when the radical part is identical on both sides — $\sqrt2$ and $\sqrt2$ can combine, but $\sqrt2$ and $\sqrt3$ cannot be added directly any more than $2x$ and $3y$ can. This is exactly why a surd should first be reduced to its lowest form, pulling out any perfect-square factor: $\sqrt{18} = \sqrt{9\times2} = 3\sqrt2$ , so that hidden common radical parts become visible before you try to add or subtract.

Rationalisation is the process of rewriting a fraction so that its denominator contains no surd at all, by multiplying by a form of $1$ chosen to cancel the root.

Single-surd denominator. Multiply numerator and denominator by that same surd:

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

Binomial surd denominator. Multiply by the conjugate — the same two terms with the middle sign flipped — using $(x+y)(x-y)=x^2-y^2$ :

\frac{1}{\sqrt a+\sqrt b}\times\frac{\sqrt a-\sqrt b}{\sqrt a-\sqrt b} = \frac{\sqrt a - \sqrt b}{a-b}

where: $a - b$ in the final denominator is always a rational number once $a$ and $b$ are themselves rational, because the middle cross-terms $\sqrt{ab}$ cancel exactly when you expand $(\sqrt a+\sqrt b)(\sqrt a-\sqrt b)$ . Multiplication and division of surds combine cleanly only when the roots share the same order: $\sqrt a\times\sqrt b=\sqrt{ab}$ works for two square roots, but mixing a square root with a cube root, such as $\sqrt2\times\sqrt[3]2$ , does not simplify by this rule and lies beyond the Class 9 syllabus — every surd in this article's rationalisation work is a square root. The figure below shows why the conjugate is the only multiplier that works for a two-term surd denominator: multiplying by the same expression again leaves a root behind, while the conjugate's cross-terms are equal and opposite and vanish.

Figure

Comparing multiplying a binomial surd by itself versus by its conjugate

Derivation ARationalising a binomial denominator

Rationalise $\dfrac{5}{3+\sqrt2}$ and simplify it — the goal is to turn this into an equal fraction whose denominator has no surd, using only the conjugate trick and the difference-of-squares identity, with every intermediate value checkable by direct substitution back into the original expression.

Step 1. Identify the conjugate of the denominator. The denominator is $3+\sqrt2$ , so its conjugate is $3-\sqrt2$ — the same two terms, with the middle sign flipped from plus to minus. Note that $3$ itself is unaffected; only the sign in front of the surd changes.

Step 2. Multiply numerator and denominator by that conjugate. This is a legal move because $\dfrac{3-\sqrt2}{3-\sqrt2}=1$ , so the value of the fraction does not change even though its appearance does:

\frac{5}{3+\sqrt2}\times\frac{3-\sqrt2}{3-\sqrt2} = \frac{5(3-\sqrt2)}{(3+\sqrt2)(3-\sqrt2)}

Step 3. Expand the denominator using $(x+y)(x-y)=x^2-y^2$ with $x=3$ , $y=\sqrt2$ . Squaring removes the root entirely, since $(\sqrt2)^2=2$ :

(3+\sqrt2)(3-\sqrt2) = 3^2-(\sqrt2)^2 = 9-2 = 7

Step 4. Expand the numerator by distributing the $5$ across the bracket:

5(3-\sqrt2) = 15-5\sqrt2

Step 5. Combine numerator and denominator into the final rationalised form:

\frac{5}{3+\sqrt2} = \frac{15-5\sqrt2}{7}

The denominator is now the rational number $7$ ; no surd remains beneath the fraction line, so the expression is fully rationalised. As a check, notice the general pattern behind Step 3: rationalising $\dfrac{c}{a+\sqrt b}$ with the conjugate $a-\sqrt b$ always produces denominator $a^2-b$ , a whole number whenever $a$ and $b$ are themselves whole numbers — here $3^2-2=7$ , matching exactly.

Worked example

Simplify $\sqrt{50}+\sqrt{18}-\sqrt8$ , then rationalise $\dfrac{1}{\sqrt2}$ and combine the two results.

Step 1. Reduce each surd by pulling out its largest perfect-square factor: $\sqrt{50}=\sqrt{25\times2}=5\sqrt2$ , $\sqrt{18}=\sqrt{9\times2}=3\sqrt2$ , $\sqrt8=\sqrt{4\times2}=2\sqrt2$ .

Step 2. All three terms now share the radical part $\sqrt2$ , so combine coefficients directly: $5\sqrt2+3\sqrt2-2\sqrt2 = (5+3-2)\sqrt2 = 6\sqrt2$ .

Step 3. Rationalise $\dfrac{1}{\sqrt2}$ by multiplying top and bottom by $\sqrt2$ : $\dfrac{1}{\sqrt2}\times\dfrac{\sqrt2}{\sqrt2} = \dfrac{\sqrt2}{2}$ .

Step 4. Add the two simplified results, since both now carry the radical part $\sqrt2$ : $6\sqrt2 + \dfrac{\sqrt2}{2} = \left(6+\dfrac12\right)\sqrt2 = \dfrac{13}{2}\sqrt2$ . So $\sqrt{50}+\sqrt{18}-\sqrt8 = 6\sqrt2$ , $\dfrac{1}{\sqrt2}=\dfrac{\sqrt2}{2}$ , and their sum is $\dfrac{13}{2}\sqrt2$ .

Competitive stretch

Rationalise $\dfrac{1}{\sqrt5-\sqrt3}$ and check it against the value of $\sqrt5+\sqrt3$ . The conjugate of $\sqrt5-\sqrt3$ is $\sqrt5+\sqrt3$ , so multiply both parts of the fraction by it: $\dfrac{1}{\sqrt5-\sqrt3}\times\dfrac{\sqrt5+\sqrt3}{\sqrt5+\sqrt3} = \dfrac{\sqrt5+\sqrt3}{(\sqrt5)^2-(\sqrt3)^2} = \dfrac{\sqrt5+\sqrt3}{5-3} = \dfrac{\sqrt5+\sqrt3}{2}$ . This kind of denominator, built from two different surds rather than a whole number and a surd, rationalises by exactly the same conjugate rule — only the identification of $a$ and $b$ in $a^2-b^2$ changes, here $a=\sqrt5$ and $b=\sqrt3$ , so $a^2-b^2=5-3=2$ .

Common traps

Adding surds with different radical parts directly, e.g. wrongly writing $\sqrt2+\sqrt3=\sqrt5$ — surds only combine like unlike terms in algebra; the radical part must match first.
Rationalising $\dfrac{1}{\sqrt a+\sqrt b}$ by multiplying by $\sqrt a+\sqrt b$ again instead of the conjugate $\sqrt a-\sqrt b$ — this leaves a surd in the denominator instead of removing it, as the figure above shows.
Multiplying only the denominator by the conjugate and forgetting the numerator — this changes the value of the fraction; both must be multiplied by the same form of $1$ .
Treating $\sqrt{a+b}$ as equal to $\sqrt a+\sqrt b$ — roots do not distribute over addition. Check with numbers: $\sqrt{9+16}=\sqrt{25}=5$ , but $\sqrt9+\sqrt{16}=3+4=7$ , and $5 \neq 7$ .
Leaving the final answer only partly simplified — after rationalising, always check whether the numerator and denominator share a common factor that cancels, and whether any surd term inside the numerator itself can be reduced to a smaller surd.

Ask the tutor: If a rationalisation problem with three terms in the denominator, like $\dfrac{1}{2+\sqrt3-\sqrt5}$ , trips you up, ask your AI tutor for a step-by-step walkthrough at your own pace.

Why can I not just add root 2 and root 3 the way I add whole numbers?

Because a surd's radical part behaves like a variable name, not a plain number — $\sqrt2$ and $\sqrt3$ are as different from each other as $x$ and $y$ are in algebra. You can only combine surds directly, using $k\sqrt a \pm m\sqrt a = (k\pm m)\sqrt a$ , when they share the exact same radicand $a$ . If two surds look different, always simplify them first by factoring out perfect squares — sometimes a hidden common radical part appears, as with $\sqrt{50}$ and $\sqrt{18}$ both reducing to multiples of $\sqrt2$ .

Why do we multiply by the conjugate instead of just squaring the denominator?

Squaring the denominator alone would change the value of the whole fraction, since you would only be altering the bottom and not the top. Multiplying by $\dfrac{\sqrt a-\sqrt b}{\sqrt a-\sqrt b}$ is really multiplying by $1$ , so the value of the fraction is preserved, while the identity $(x+y)(x-y)=x^2-y^2$ guarantees the two square-root cross-terms cancel exactly, leaving a whole number underneath. The conjugate is simply the one factor guaranteed to trigger that cancellation.

Is a fraction like 15 minus 5 root 2, all over 7, considered fully simplified even though it still has a root in it?

Yes — the ICSE convention for "fully rationalised" only requires that no surd remains in the denominator; a surd is perfectly acceptable in the numerator. Always do one more check though: see whether the numerator and denominator share a common factor that cancels, and whether any surd term inside the numerator itself can be reduced to a smaller surd. In $\dfrac{15-5\sqrt2}{7}$ , neither $15$ nor $5\sqrt2$ shares a factor with $7$ , so this form is final.

Surds and rationalisation

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads