Operations on real numbers

Add two rational numbers and you always get another rational number — but mix in a square root, and the rules change. Operations on real numbers looks at what happens when addition, subtraction, multiplication, and division act on rational and irrational numbers together, and it hands you a toolkit — the surd laws — for simplifying expressions built from square roots. The star move of this chapter is rationalising the denominator: rewriting a fraction like $\dfrac{1}{3+2\sqrt2}$ so that the surd disappears from the bottom and lands, tamed, in the numerator instead. This one trick — multiply by a cleverly chosen partner factor — turns messy expressions into ones you can add, compare, or evaluate to a decimal in seconds. Everything here builds directly on the real number line and irrational numbers you have already met in this unit, and it feeds straight into simplifying algebraic surds in later chapters.

The core insight

Real numbers combine under addition, subtraction, multiplication, and division exactly like any other numbers, and surd expressions become friendlier once you rationalise their denominators using the difference-of-squares pattern $(x+y)(x-y) = x^2 - y^2$ .

The core idea

Closure under operations. The real numbers are closed under addition, subtraction, multiplication, and division (except by zero) — combine any two real numbers this way and the result is still a real number. A rational combined with an irrational usually stays irrational: $2 + \sqrt3$ has no simpler rational form, and $5 \times \sqrt2$ stays irrational too. But two irrationals can combine to give either a rational or an irrational result — $\sqrt2 \times \sqrt2 = 2$ is rational, while $\sqrt2 + \sqrt3$ stays irrational, since there is no way to merge unlike radical parts into a single term.

Laws for surds. For $a, b > 0$ , these identities let you simplify square-root expressions the same way you simplify algebraic ones:

\sqrt{ab} = \sqrt a \cdot \sqrt b, \qquad \sqrt{\frac{a}{b}} = \frac{\sqrt a}{\sqrt b}, \qquad (\sqrt a + \sqrt b)(\sqrt a - \sqrt b) = a - b

where: $a$ and $b$ are the numbers under the root signs. The third identity is just $(x+y)(x-y) = x^2-y^2$ with $x = \sqrt a$ , $y = \sqrt b$ — it is the engine behind rationalisation, since squaring a surd removes the root.

Combining like surds. Surds with the same radical part add and subtract like algebraic like terms, treating $\sqrt x$ as though it were a variable: $3\sqrt2 + 5\sqrt2 = 8\sqrt2$ , but $3\sqrt2 + 5\sqrt3$ will not simplify further, since $\sqrt2$ and $\sqrt3$ are different radical parts. It often helps to break a surd into a perfect-square factor first, for example $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt2$ , so that it can combine with other multiples of $\sqrt2$ .

Rationalising the denominator means multiplying a fraction by a form of 1 so that its denominator becomes rational. For $\dfrac{1}{a+b\sqrt x}$ , multiply by $\dfrac{a-b\sqrt x}{a-b\sqrt x}$ ; for $\dfrac{1}{\sqrt x + \sqrt y}$ , multiply by $\dfrac{\sqrt x - \sqrt y}{\sqrt x - \sqrt y}$ . In both cases the denominator collapses to a difference of squares — $a^2 - b^2x$ or $x - y$ — which is rational, so the surd survives only in the numerator, where it is easy to work with.

Why bother at all? A fraction like $\dfrac{1}{3+2\sqrt2}$ is perfectly correct as it stands, but it is awkward: to compare it with another number, add it to a similar fraction, or plug in a decimal approximation for $\sqrt2$ , you would have to divide by an irrational number every time. Once rationalised, the surd sits alone in the numerator, multiplication and comparison become ordinary arithmetic, and estimating a decimal value takes one substitution instead of long division by an irrational quantity. This is also why textbooks and exams insist an answer is "not fully simplified" until any surd has been cleared from the denominator — it is the standard final form for this kind of expression, the same way a fraction is expected to be reduced to lowest terms.

Figure

A fraction with a surd in the denominator transforming into an equivalent fraction with a rational denominator after multiplying by the conjugate factor

Derivation AWhy the difference-of-squares trick clears the denominator

The goal is to show, with nothing beyond the identity $(x+y)(x-y) = x^2-y^2$ , why multiplying $\dfrac{1}{a+b\sqrt x}$ by $\dfrac{a-b\sqrt x}{a-b\sqrt x}$ always leaves a rational number sitting in the denominator, no matter what natural number $x$ is chosen or what integers $a, b$ are used, as long as $a + b\sqrt x \ne 0$ .

Step 1. Write the multiplication out fully. $\dfrac{1}{a+b\sqrt x} \times \dfrac{a-b\sqrt x}{a-b\sqrt x} = \dfrac{a-b\sqrt x}{(a+b\sqrt x)(a-b\sqrt x)}$ . Multiplying by a fraction equal to 1 changes the form of the expression but not its value.

Step 2. Expand the denominator using $(x+y)(x-y) = x^2 - y^2$ with $x = a$ playing the role of the first term and $y = b\sqrt x$ the second: $(a+b\sqrt x)(a-b\sqrt x) = a^2 - (b\sqrt x)^2$ .

Step 3. Simplify $(b\sqrt x)^2 = b^2 \cdot (\sqrt x)^2 = b^2 x$ , since squaring a square root of $x$ returns $x$ itself. So the denominator becomes $a^2 - b^2x$ , a combination of integers with no square root left in it — it is rational.

Step 4. Read off the result: $\dfrac{1}{a+b\sqrt x} = \dfrac{a-b\sqrt x}{a^2 - b^2x}$ . The surd has moved entirely into the numerator, and the fraction can now be evaluated numerically or combined with other rational terms without a root sign underfoot.

Worked example

Rationalise the denominator of $\dfrac{5}{\sqrt7 + \sqrt2}$ , then evaluate it to three decimal places using $\sqrt7 \approx 2.646$ and $\sqrt2 \approx 1.414$ .

Step 1. Identify the form: this is $\dfrac{1}{\sqrt x + \sqrt y}$ with an extra factor of 5, $x = 7$ , $y = 2$ . The rationalising factor is $\sqrt7 - \sqrt2$ .

Step 2. Multiply numerator and denominator by that factor: $\dfrac{5}{\sqrt7+\sqrt2} \times \dfrac{\sqrt7-\sqrt2}{\sqrt7-\sqrt2} = \dfrac{5(\sqrt7-\sqrt2)}{(\sqrt7)^2-(\sqrt2)^2} = \dfrac{5(\sqrt7-\sqrt2)}{7-2}$ .

Step 3. Simplify the denominator: $7 - 2 = 5$ , so the fives cancel, leaving $\dfrac{5(\sqrt7-\sqrt2)}{5} = \sqrt7 - \sqrt2$ .

Step 4. Substitute the approximate values: $\sqrt7 - \sqrt2 \approx 2.646 - 1.414 = 1.232$ . So $\dfrac{5}{\sqrt7+\sqrt2} \approx \textbf{1.232}$ .

Competitive stretch

The same fraction also shows why rationalising helps with combining surds. Simplify $\sqrt{45} + \sqrt{20} - \sqrt5$ : break each into a perfect-square factor, $\sqrt{45} = 3\sqrt5$ and $\sqrt{20} = 2\sqrt5$ , then combine coefficients like algebra, $(3+2-1)\sqrt5 = \textbf{4}\sqrt5$ . Rationalised or simplified, every surd expression in this chapter reduces to a single clean term once matching radical parts are grouped.

Common traps

Multiplying only the denominator by the rationalising factor and forgetting the numerator too — this changes the value of the expression instead of just rewriting it.
Using the wrong sign when forming the rationalising factor: for $a+b\sqrt x$ the correct factor is $a-b\sqrt x$ , matching the difference-of-squares pattern; the wrong sign leaves a surd in the denominator.
Squaring a surd term carelessly, e.g. writing $(2\sqrt2)^2$ as $2\sqrt2^2$ or $2^2$ instead of the correct $4 \times 2 = 8$ .
Treating $\sqrt a + \sqrt b$ as equal to $\sqrt{a+b}$ — square roots do not distribute over addition or subtraction.

Ask the tutor: If a rationalisation problem is not simplifying the way you expect, ask the AI tutor to walk through your working line by line.

Why does multiplying by a fraction equal to 1 not change the value of the expression?

Any fraction with the same numerator and denominator, like $\dfrac{a-b\sqrt x}{a-b\sqrt x}$ , is exactly equal to 1, since a nonzero number divided by itself is always 1. Multiplying anything by 1 leaves its value unchanged — only its form changes. That is the whole trick behind rationalising: you are not solving an equation or approximating, you are rewriting the same number in a more useful shape.

How do I know whether to use a + b√x or √x + √y as the rationalising pattern?

Look at the denominator itself. If it has one rational term and one surd term, like $3 + 2\sqrt2$ , use the $a+b\sqrt x$ pattern and flip the middle sign to $a-b\sqrt x$ . If it has two surd terms added together, like $\sqrt7+\sqrt2$ , use the $\sqrt x+\sqrt y$ pattern and flip to $\sqrt x-\sqrt y$ . Either way, the rule is the same: keep the terms, just flip the sign between them.

Why bother rationalising at all if a calculator can just evaluate the surd directly?

In Class 9 you are expected to work with exact values, not decimal approximations, and many later steps — adding fractions, comparing sizes, proving identities — become far easier once denominators are rational. Rationalising also matches how answers are checked and marked on paper: an unrationalised denominator is usually treated as an incomplete simplification, even if the decimal value is correct. As a bonus, once a denominator is rational, substituting a decimal approximation for the leftover surd — as in the worked example above — takes one multiplication instead of a division by an irrational quantity.

Operations on real numbers

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads