Practice Maths

Rationalising Denominators

Key Terms

A fraction with a surd in the denominator (e.g. 3/√5) is not in simplest form — we rationalise by eliminating the surd from the denominator.
Simple surd denominator
multiply top and bottom by the same surd — e.g. 3/√5 × √5/√5 = 3√5/5.
Binomial surd denominator
multiply by the conjugate. The conjugate of (a + √b) is (a − √b).
Why the conjugate works: (a + √b)(a − √b) = a² − b — this is the difference of two squares, giving a rational result.
Always simplify the result after rationalising — cancel common factors.
Denominator typeTechnique & Result
Single surd: 1/√aMultiply by √a/√a → √a/a
Coeff. surd: p/(√a)Multiply by √a/√a → p√a/a
Binomial: 1/(a + √b)Multiply by (a − √b)/(a − √b) → (a − √b)/(a² − b)
Binomial: 1/(√a + √b)Multiply by (√a − √b)/(√a − √b) → (√a − √b)/(a − b)
DOTS identity(p + q)(p − q) = p² − q²
Hot Tip The conjugate flips the sign between the two terms — NOT both signs. The conjugate of (3 + √5) is (3 − √5), and the conjugate of (√7 − 2) is (√7 + 2).

Worked Example 1 — Simple Surd Denominator

Question: Rationalise the denominator of √7/√3.

Step 1: Multiply numerator and denominator by √3: √7/√3 × √3/√3

Step 2: Numerator: √7 × √3 = √21. Denominator: √3 × √3 = 3.

Answer: √21/3

Worked Example 2 — Binomial Surd Denominator

Question: Rationalise the denominator of 5/(3 + √2).

Step 1: The conjugate of (3 + √2) is (3 − √2). Multiply top and bottom by (3 − √2).

Step 2: Numerator: 5(3 − √2) = 15 − 5√2.

Step 3: Denominator: (3 + √2)(3 − √2) = 3² − (√2)² = 9 − 2 = 7.

Answer: (15 − 5√2)/7

Why Rationalise?

Historically, before calculators existed, dividing by a surd was computationally painful. Suppose you want 1/√2. To get a decimal, you would need to divide 1 by 1.41421… — a nightmare by hand. But √2/2 = 1.41421…/2 is easy: just halve a number you already know. Rationalising moves the surd from denominator to numerator, where it is far easier to work with.

Today, the reason is more about mathematical convention and exact form. A fraction is considered fully simplified when the denominator is rational. Just as we simplify 4/6 to 2/3, we simplify 3/√5 to 3√5/5. Both represent the same number, but the second form is standard.

There is also a deeper algebraic reason: rationalised forms are easier to add, compare, and manipulate. If you have 1/√3 + 1/√5, it is much harder to add than √3/3 + √5/5, where you can find a common denominator of 15 directly.

Simple Rationalisation: Multiplying by √a/√a

For a denominator of √a, multiply top and bottom by √a. This works because √a × √a = a (a rational number). You are multiplying by √a/√a = 1, so the value does not change — only the form does.

Example: 5/√3 × √3/√3 = 5√3/3. Note the denominator is now 3 (rational), and the surd has moved to the numerator. Always check: can the fraction be simplified further? Here, 5 and 3 share no common factors, so 5√3/3 is fully simplified.

Pitfall: if the denominator is 2√7, multiply by √7/√7, giving 2√7 × ... hmm, wait. Only the surd part needs rationalising: multiply numerator and denominator by √7, giving denominator 2 × 7 = 14. Then simplify if possible.

Conjugate Pairs: Making Surds Disappear

When the denominator is a binomial like (3 + √5), simply multiplying by √5/√5 would not help — it would leave the 3 in the denominator irrational in a different sense. Instead, we use the conjugate.

The conjugate of (a + √b) is (a − √b), and vice versa. The key property is the difference of two squares identity: (a + √b)(a − √b) = a² − (√b)² = a² − b. This is always rational, because a² and b are both rational numbers. The surd disappears completely from the denominator.

Similarly, the conjugate of (√a + √b) is (√a − √b), giving (√a)² − (√b)² = a − b when multiplied together.

Rationalising with Conjugates: Step-by-Step

To rationalise 7/(2 + √3):

  1. Identify the conjugate: (2 − √3).
  2. Multiply numerator and denominator by (2 − √3).
  3. Numerator: 7(2 − √3) = 14 − 7√3.
  4. Denominator: (2 + √3)(2 − √3) = 4 − 3 = 1.
  5. Result: (14 − 7√3)/1 = 14 − 7√3.

Notice the denominator simplified to 1 here, making the result particularly clean. This will not always happen, but the denominator will always become rational.

When Rationalising Changes the Look But Not the Value

It is important to understand that rationalising is an algebraic manipulation that does not change the mathematical value of an expression — it only changes how it looks. The number 1/√2 and √2/2 are identical; they sit at exactly the same point on the number line (approximately 0.7071).

This is why we say we multiply by a form of 1: √a/√a = 1 (for a ≠ 0). Any number multiplied by 1 is unchanged. Conjugates work the same way: (a − √b)/(a − √b) = 1. We are always multiplying by something that equals 1.

In exam questions, if you rationalise and your answer looks completely different from the provided answer, first check whether they are actually equal by finding decimal approximations of both. Equivalent surd expressions can look very different in form.

Exam Tip: The conjugate of (3 + √5) is (3 − √5). It is NOT (3 − 5) = −2, and it is NOT (−3 − √5). Only the sign between the two terms flips. Students frequently negate the wrong term or change both signs. If in doubt, check: does the product give a rational number?
Exam Tip: After rationalising, always check whether the result simplifies further. For example, (6 − 2√3)/4 should be simplified to (3 − √3)/2 by dividing numerator and denominator by 2. Leaving an unsimplified fraction will cost marks in a show-that or exact-value question.

Mastery Practice

See Answers ➔

  1. Rationalise each denominator and simplify. Fluency

    1. (a) 1/√3
    2. (b) 5/√5
    3. (c) 2/√7
    4. (d) √3/√2
    5. (e) 6/√6
    6. (f) 4/(3√2)

  2. State the conjugate of each expression. Fluency

    1. (a) 2 + √5
    2. (b) √3 − 1
    3. (c) √7 + √2
    4. (d) 4 − √11
    5. (e) 3√2 + 5

  3. Expand each product using the difference of two squares identity. Fluency

    1. (a) (2 + √3)(2 − √3)
    2. (b) (√5 + 1)(√5 − 1)
    3. (c) (√7 − √2)(√7 + √2)
    4. (d) (3 + 2√5)(3 − 2√5)
    5. (e) (4√3 − 1)(4√3 + 1)

  4. Rationalise each denominator and simplify. Fluency

    1. (a) 1/(1 + √2)
    2. (b) 3/(4 − √5)
    3. (c) √2/(√2 + 1)
    4. (d) 6/(√3 + √2)
    5. (e) (√5 + 2)/(√5 − 2)

  5. Rationalise each denominator and express in simplest form. Fluency

    1. (a) 10/√5
    2. (b) √18/(2√3)
    3. (c) (3 + √2)/(3 − √2)
    4. (d) (√6 − √2)/√2

  6. Rationalising in geometry. Understanding

    Rectangle design. A rectangle has width √3 cm and area 12 cm². A student writes the length as 12/√3 cm.
    1. (a) Rationalise 12/√3 to express the length without a surd in the denominator.
    2. (b) Calculate the perimeter of the rectangle, giving your answer in simplest surd form.
    3. (c) Verify your answer for the length by multiplying length × width and confirming you get 12 cm².

  7. Equivalent expressions. Understanding

    Two students' answers. Two students rationalise 2/(√6 − √2). Alicia writes (√6 + √2)/2 and Ben writes (√3 + 1)/(√2). Determine whether either or both are correct.
    1. (a) Rationalise 2/(√6 − √2) by multiplying by the conjugate.
    2. (b) Simplify your answer. Is it equal to Alicia's answer? Show full working.
    3. (c) Is Ben's answer equivalent? Rationalise Ben's expression and compare.

  8. Solving equations with surd denominators. Understanding

    Solve and simplify. Rationalise, then solve for x.
    1. (a) x = 1/√2 + √2. Rationalise 1/√2 first, then simplify.
    2. (b) 3/(2 + √3) = x − √3. Find x by rationalising the left side then rearranging.

  9. Show that the following expressions are equal. Problem Solving

    Algebraic identity. These problems require you to rationalise and simplify both sides to show they are equal. Show all steps clearly.
    1. (a) Show that 1/(√n + √(n+1)) = √(n+1) − √n for any positive integer n.
    2. (b) Hence evaluate 1/(√1 + √2) + 1/(√2 + √3) + 1/(√3 + √4) + … + 1/(√8 + √9). (Hint: use the result from part (a) and look for a telescoping pattern.)

  10. Resistance in electrical circuits. Problem Solving

    Physics application. In a circuit, two resistors in parallel have a combined resistance R given by 1/R = 1/R⊂1; + 1/R⊂2;. A physicist has R⊂1; = √3 ohms and R⊂2; = (1 + √3) ohms.
    1. (a) Write 1/R⊂1; and rationalise 1/R⊂2; = 1/(1 + √3) by multiplying by the conjugate.
    2. (b) Find 1/R = 1/R⊂1; + 1/R⊂2; by adding the simplified fractions (find a common denominator).
    3. (c) Hence find R by taking the reciprocal of your answer, rationalising if necessary.
    4. (d) Give R as a decimal correct to 3 significant figures and interpret this value.