Rationalising the Denominator: From Surds to Rational Numbers
Why Do We Rationalise the Denominator?
The practice of rationalising denominators has both historical and practical roots. Centuries ago, before calculators, dividing by an irrational number was extremely difficult. Having a rational number in the denominator made long division and estimation much simpler. Today, the reasons are more mathematical:
- Standard Form: It is a widely accepted convention to present final answers without a surd in the denominator, making it easier to compare and combine results.
- Simplified Calculation: Even with calculators, a rationalised form like $\frac{\sqrt{2}}{2}$ is often easier to work with than $\frac{1}{\sqrt{2}}$ when adding to other fractions or substituting into further equations.
- Precision: It avoids rounding errors. For example, $\frac{1}{\sqrt{3}} \approx 0.57735$, while $\frac{\sqrt{3}}{3} \approx 0.57735$. If you round the denominator $\sqrt{3} \approx 1.732$ first, then compute $1 / 1.732$, you introduce an extra step of approximation.
The Basic Case: A Single Square Root
This is the simplest and most common scenario. When the denominator is a single surd, like $\sqrt{a}$, you multiply the fraction by $\frac{\sqrt{a}}{\sqrt{a}}$.
Example 1: Rationalise $\frac{7}{\sqrt{3}}$.
Multiply numerator and denominator by $\sqrt{3}$:
$\frac{7}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{7 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{7\sqrt{3}}{3}$.
The denominator is now the rational number 3.
Example 2: Rationalise $\frac{2\sqrt{5}}{\sqrt{10}}$.
The process is identical: $\frac{2\sqrt{5}}{\sqrt{10}} \times \frac{\sqrt{10}}{\sqrt{10}} = \frac{2\sqrt{50}}{10} = \frac{2 \times 5\sqrt{2}}{10} = \frac{10\sqrt{2}}{10} = \sqrt{2}$.
Notice how simplifying the fraction after rationalising often leads to a much cleaner result.
Handling Denominators with Two Terms: Using the Conjugate
When the denominator is a sum or difference that includes a surd, e.g., $a + \sqrt{b}$ or $\sqrt{x} - \sqrt{y}$, the "clever form of 1" is the conjugate. The conjugate of a two-term expression is formed by changing the sign between the two terms.
- Conjugate of $3 + \sqrt{2}$ is $3 - \sqrt{2}$.
- Conjugate of $\sqrt{7} - 2$ is $\sqrt{7} + 2$.
- Conjugate of $\sqrt{5} + \sqrt{3}$ is $\sqrt{5} - \sqrt{3}$.
Why does this work? It leverages the difference of squares formula: $(a-b)(a+b) = a^2 - b^2$. When you multiply a surd sum by its conjugate, the surds cancel out.
Example 3: Rationalise $\frac{4}{2+\sqrt{3}}$.
Multiply numerator and denominator by the conjugate, $2 - \sqrt{3}$:
$\frac{4}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}} = \frac{4(2-\sqrt{3})}{(2)^2 - (\sqrt{3})^2} = \frac{8 - 4\sqrt{3}}{4 - 3} = \frac{8 - 4\sqrt{3}}{1} = 8 - 4\sqrt{3}$.
Example 4: Rationalise $\frac{\sqrt{6}}{\sqrt{5}-\sqrt{2}}$.
Multiply by the conjugate $\sqrt{5}+\sqrt{2}$:
$\frac{\sqrt{6}}{\sqrt{5}-\sqrt{2}} \times \frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}} = \frac{\sqrt{6}(\sqrt{5}+\sqrt{2})}{(\sqrt{5})^2 - (\sqrt{2})^2} = \frac{\sqrt{30}+\sqrt{12}}{5 - 2} = \frac{\sqrt{30}+2\sqrt{3}}{3}$.
| Denominator Type | Multiply By (The "1") | Key Principle | Simple Example |
|---|---|---|---|
| Single Surd: $\sqrt{a}$ | $\frac{\sqrt{a}}{\sqrt{a}}$ | $\sqrt{a} \times \sqrt{a} = a$ | $\frac{3}{\sqrt{5}} \to \frac{3\sqrt{5}}{5}$ |
| Sum/Difference: $a \pm \sqrt{b}$ | $\frac{a \mp \sqrt{b}}{a \mp \sqrt{b}}$ (conjugate) | $(a-b)(a+b)=a^2-b^2$ | $\frac{1}{4-\sqrt{2}} \to \frac{4+\sqrt{2}}{14}$ |
| Sum/Difference: $\sqrt{p} \pm \sqrt{q}$ | $\frac{\sqrt{p} \mp \sqrt{q}}{\sqrt{p} \mp \sqrt{q}}$ (conjugate) | $(\sqrt{p})^2 - (\sqrt{q})^2 = p-q$ | $\frac{5}{\sqrt{3}+\sqrt{2}} \to 5(\sqrt{3}-\sqrt{2})$ |
Beyond Square Roots: Cube Roots and Higher
The principle extends to cube roots and other higher-order roots, though the technique is slightly different. The goal remains the same: multiply by a form of 1 that makes the denominator a perfect cube (or perfect nth power).
For a denominator like $\sqrt[3]{a}$, we need $\sqrt[3]{a} \times \sqrt[3]{a^2} = \sqrt[3]{a^3} = a$. So we multiply by $\frac{\sqrt[3]{a^2}}{\sqrt[3]{a^2}}$.
Example 5: Rationalise $\frac{6}{\sqrt[3]{4}}$.
Note: $\sqrt[3]{4} = \sqrt[3]{2^2}$. To make it a perfect cube, we need one more factor of $2$, i.e., $\sqrt[3]{2}$. So multiply by $\frac{\sqrt[3]{2}}{\sqrt[3]{2}}$:
$\frac{6}{\sqrt[3]{4}} \times \frac{\sqrt[3]{2}}{\sqrt[3]{2}} = \frac{6\sqrt[3]{2}}{\sqrt[3]{8}} = \frac{6\sqrt[3]{2}}{2} = 3\sqrt[3]{2}$.
Application in Solving Real-World Problems
Rationalisation is not just an abstract exercise. It appears in geometry, physics, and engineering. For instance, calculating the exact height of an equilateral triangle with side length s gives a formula $h = \frac{s\sqrt{3}}{2}$, which is a rationalised form. If you used $h = \frac{s}{\frac{2}{\sqrt{3}}}$, it would be much harder to use.
Example Scenario: You have a right-angled isosceles triangle with legs of length 1. The hypotenuse is $\sqrt{2}$. The sine of a $45^\circ$ angle is $\frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt{2}}$. In all standard math tables, this is given as $\frac{\sqrt{2}}{2}$, its rationalised form. This makes it straightforward to add $\sin 45^\circ$ to another fraction, like $\frac{1}{2}$, without converting to decimals.
Example 6 (Geometry): A circle is inscribed in a square of side 10 cm. The radius is 5 cm. The diagonal of the square is $10\sqrt{2}$ cm. The ratio of the circle's circumference to the square's diagonal is $\frac{2\pi(5)}{10\sqrt{2}} = \frac{\pi}{\sqrt{2}}$. Rationalising gives $\frac{\pi\sqrt{2}}{2}$, a much cleaner expression for further use.
Important Questions
Q1: Is it always necessary to rationalise the denominator?
While not strictly "wrong" to leave a surd in the denominator, it is a strong mathematical convention. In school exams, you are almost always required to present your answer with a rationalised denominator unless instructed otherwise. It is considered the simplified, standard form.
Q2: What if the denominator is a sum of three terms, like 1 + \sqrt{2} + \sqrt{3}?
This becomes more advanced. The technique involves using a "conjugate" that is based on more complex algebraic identities, not just the difference of squares. For elementary and high school levels, you will primarily deal with denominators that are single surds or two-term sums/differences.
Q3: Can the numerator contain surds after rationalisation?
Absolutely yes. The goal is only to make the denominator a rational number. The numerator can be any simplified expression, often containing surds. For example, rationalising $\frac{2}{\sqrt{5}}$ gives $\frac{2\sqrt{5}}{5}$, which is perfectly acceptable.
Footnote
1 Surd: An irrational number that can be expressed as the root of a rational number, especially a non-perfect square root (e.g., $\sqrt{2}, \sqrt[3]{5}$). The term often refers specifically to such roots.
2 Conjugate: In the context of rationalising denominators, the conjugate of a binomial expression $a + b$ is $a - b$. Multiplying a binomial by its conjugate results in the difference of squares, $a^2 - b^2$, which eliminates any surd terms if $b$ is a surd.
3 Rational Number: A number that can be expressed as a fraction $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$. Examples include $7, -\frac{1}{2}, 0.75$.
4 Difference of Squares: A fundamental algebraic identity: $(a - b)(a + b) = a^2 - b^2$. This is the key tool for rationalising denominators involving sums or differences.