What are exponents and how are they used in Cambridge IGCSE Mathematics?

Exponents, also known as powers or indices, represent repeated multiplication of a number by itself. In Cambridge IGCSE Mathematics, exponents are used to simplify expressions, solve equations, and understand exponential growth and decay. For example, 3^4 means 3 multiplied by itself 4 times, which equals 81.

How do you simplify expressions with surds in IGCSE Mathematics?

Simplifying expressions with surds involves expressing the surd in its simplest form. This is done by factoring the number under the square root into its prime factors and simplifying. For example, √50 can be simplified to 5√2 because 50 equals 25 times 2, and the square root of 25 is 5.

What are the rules for multiplying and dividing exponents in IGCSE Mathematics?

In IGCSE Mathematics, the rules for multiplying and dividing exponents are: when multiplying like bases, add the exponents (a^m * a^n = a^(m+n)); when dividing like bases, subtract the exponents (a^m / a^n = a^(m-n)). These rules help simplify complex expressions and solve equations involving exponents.

How do you rationalize the denominator of a surd in Cambridge IGCSE Mathematics?

Rationalizing the denominator involves removing the surd from the denominator of a fraction. This is done by multiplying both the numerator and the denominator by the conjugate of the denominator. For example, to rationalize 1/√3, multiply by √3/√3 to get √3/3.

What is the significance of understanding exponents and surds in the IGCSE Mathematics curriculum?

Understanding exponents and surds is crucial in the IGCSE Mathematics curriculum as they form the foundation for more advanced topics like algebra, calculus, and scientific notation. Mastery of these concepts enables students to solve complex mathematical problems, understand growth patterns, and perform calculations involving irrational numbers.

Why is "Adding the bases of powers with the same index" a common mistake in Exponents and Surds?

Why it happens: Students mis-apply the index laws: 2^3 + 5^3 ≠ 7^3. How to avoid it: Index laws apply to **products**, not sums. Compute each power separately, then add.

Why is "Treating $8^{2/3}$ as $\dfrac{8^{2}}{3}$" a common mistake in Exponents and Surds?

Why it happens: Students confuse fractional indices with division by the denominator. How to avoid it: a^m/n = √[n]a^m = (√[n]a)^m. Take the root, then the power. Source: 0580/42 — fractional powers

Why is "Stopping at $\sqrt{18}$ instead of $3\sqrt{2}$" a common mistake in Exponents and Surds?

Why it happens: Students miss that 18 has a perfect-square factor. How to avoid it: Always factor out the **largest** perfect square: 18 = 9 × 2.

Why is "Leaving a surd in the denominator" a common mistake in Exponents and Surds?

Why it happens: Students think 1/√(2) is fully simplified. How to avoid it: Rationalise unless the question explicitly says you may leave it. Source: 0580/42 Oct/Nov 2023 — examiner report Q15

Why is "Treating $a^{-1}$ as $-a$ instead of $\dfrac{1}{a}$" a common mistake in Exponents and Surds?

Why it happens: Confusing the negative sign with subtraction. How to avoid it: a^-n = 1/a^n — flip, don't negate.

NumberCambridge IGCSE Maths 0580 Extended

Exponents and Surds

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Exponents and Surds — Cambridge IGCSE 0580 Maths Extended (2026)

Indices (positive, negative, zero, fractional) and surds (square roots in their exact form). The two together unlock most algebraic manipulation on the Extended paper.

At a glance

$a^{m} \times a^{n} = a^{m+n}$ . $a^{m} \div a^{n} = a^{m-n}$ . $(a^{m})^{n} = a^{mn}$ .
$a^{0} = 1$ (for $a \ne 0$ ). $a^{-n} = \dfrac{1}{a^{n}}$ .
Fractional index: $a^{1/n} = \sqrt[n]{a}$ . So $a^{p/q} = \sqrt[q]{a^{p}}$ .
Surds: $\sqrt{a} \sqrt{b} = \sqrt{ab}$ . $\sqrt{a} + \sqrt{b}$ does NOT simplify.
Crucial: $\sqrt{a + b} \ne \sqrt{a} + \sqrt{b}$ .
Simplify a surd by pulling out the largest perfect-square factor: $\sqrt{50} = 5\sqrt{2}$ .
Rationalising: multiply by the surd (single) or its CONJUGATE (binomial denominator).
$a^{m/n}$ on a calculator: $a$ → $\hat{}$ → $($ → $m$ → $\div$ → $n$ → $)$ .

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E1.3 — Calculate squares, square roots, cubes and cube roots of numbers, and other powers and roots.
E1.9 — Manipulate surds, including simplifying and rationalising the denominator.
Use index laws including positive, negative, zero and fractional exponents.

The index laws

Six rules cover every index manipulation Cambridge will throw at you.

Every index question reduces to one of these rules. Memorise the lot.

Rule	Statement
Multiplication	$a^{m} \times a^{n} = a^{m+n}$
Division	$a^{m} \div a^{n} = a^{m-n}$
Power of a power	$(a^{m})^{n} = a^{mn}$
Zero exponent	$a^{0} = 1$ (for $a \ne 0$ )
Negative exponent	$a^{-n} = \dfrac{1}{a^{n}}$
Fractional exponent	$a^{1/n} = \sqrt[n]{a}$ , $a^{p/q} = (\sqrt[q]{a})^{p} = \sqrt[q]{a^{p}}$

Worked walkthroughs.

$2^{5} \times 2^{3} = 2^{8} = 256$ .
$\dfrac{3^{7}}{3^{4}} = 3^{3} = 27$ .
$(5^{2})^{4} = 5^{8}$ .
$7^{0} = 1$ .
$4^{-2} = \dfrac{1}{4^{2}} = \dfrac{1}{16}$ .
$9^{1/2} = \sqrt{9} = 3$ .
$8^{2/3} = (\sqrt[3]{8})^{2} = 2^{2} = 4$ .
$16^{-3/4} = \dfrac{1}{16^{3/4}} = \dfrac{1}{(\sqrt[4]{16})^{3}} = \dfrac{1}{2^{3}} = \dfrac{1}{8}$ .

Combined indices. Apply rules step by step. $\dfrac{2^{5} \times 2^{-2}}{2^{3}} = 2^{5 + (-2) - 3} = 2^{0} = 1.$

Different bases. Index laws apply only to the SAME base. $2^{3} \times 3^{2}$ does NOT simplify to $6^{5}$ — it stays as $8 \times 9 = 72$ .

Same base, multiplied: ADD the indices.
Same base, divided: SUBTRACT the indices.
Power of a power: MULTIPLY the indices.
$a^{0} = 1$ , $a^{-n} = 1/a^{n}$ , $a^{1/n} = \sqrt[n]{a}$ .
Different bases: NO shortcut — compute separately.

Fractional indices and roots

$a^{1/n}$ is the $n$ -th root. $a^{p/q}$ is a power AND a root rolled into one.

$a^{1/n} = \sqrt[n]{a}$ — the $n$ -th root.

$9^{1/2} = 3$ (square root).
$27^{1/3} = 3$ (cube root).
$32^{1/5} = 2$ (fifth root).

$a^{p/q} = \sqrt[q]{a^{p}} = (\sqrt[q]{a})^{p}$ — these two forms are equivalent. Pick the one that's easiest to compute by hand. Usually taking the root FIRST keeps the numbers small.

$8^{2/3}$ : take cube root of $8$ first ( $= 2$ ), then square ( $= 4$ ). MUCH easier than $\sqrt[3]{64} = 4$ .
$4^{3/2}$ : take square root of $4$ first ( $= 2$ ), then cube ( $= 8$ ).

Negative fractional indices combine the negative-index and fractional-index rules:

$a^{-p/q} = \dfrac{1}{a^{p/q}}$ .
e.g. $16^{-1/2} = \dfrac{1}{\sqrt{16}} = \dfrac{1}{4}$ .

Calculator entry. $8^{2/3}$ → press 8 → ^ → ( 2 ÷ 3 ). The brackets are essential — without them, the calculator interprets it as $8^{2} / 3 \approx 21.33$ , which is wrong.

$a^{1/n} = \sqrt[n]{a}$ .
$a^{p/q} =$ root first, power second (easiest).
Negative fractional: $a^{-p/q} = 1 / a^{p/q}$ .
Calculator: wrap $p/q$ in brackets.

Surds — simplifying

A surd is an irrational root of a non-perfect number. Simplify by pulling out the largest perfect-square factor.

A surd is a root (usually a square root) of a number that doesn't simplify to a rational. $\sqrt{2}$ , $\sqrt{3}$ , $\sqrt{50}$ are surds; $\sqrt{4} = 2$ is not.

The three rules.

$\sqrt{a} \sqrt{b} = \sqrt{a b}$ (only for $a, b \ge 0$ ).
$\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}$ (for $a \ge 0$ , $b > 0$ ).
$\sqrt{a} + \sqrt{b}$ does NOT simplify in general.

Simplifying. Pull out the largest perfect-square factor:

Split the number so one factor is a perfect square, then take its root out front.

$\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25}\sqrt{2} = 5\sqrt{2}.$ $\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}.$ $\sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5}.$

If you don't spot the largest perfect square, peel off any perfect square and repeat: $\sqrt{72} = \sqrt{4 \times 18} = 2\sqrt{18} = 2\sqrt{9 \times 2} = 2 \times 3 \sqrt{2} = 6\sqrt{2}.$ Same answer either way.

Combining surds. Once everything is simplified, surds with the SAME radicand combine like algebraic terms: $\sqrt{50} + \sqrt{18} = 5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}.$

Multiplying brackets. Use the distributive law as you would with $a + b$ . $(\sqrt{2} + 3)(\sqrt{2} - 1) = \sqrt{2}\cdot\sqrt{2} - \sqrt{2} + 3\sqrt{2} - 3 = 2 + 2\sqrt{2} - 3 = -1 + 2\sqrt{2}.$

$\sqrt{a}\sqrt{b}=\sqrt{ab}$ but $\sqrt{a}+\sqrt{b} \ne \sqrt{a+b}$ .
Simplify by pulling out the largest perfect-square factor.
Like surds combine: $5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}$ .
Brackets containing surds expand by the distributive law.

Rationalising the denominator

Cambridge expects the denominator to be rational. Multiply top and bottom by the right thing to clear the surd.

Final answers in Cambridge style have NO surds in the denominator.

Single surd in the denominator. Multiply top and bottom by the same surd. $\dfrac{6}{\sqrt{3}} = \dfrac{6}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{6\sqrt{3}}{3} = 2\sqrt{3}.$

Multiplying by √3 over √3 is multiplying by 1 — it clears the surd without changing the value.

Binomial denominator with a surd. Multiply top and bottom by the conjugate — same expression with the surd's sign flipped. The conjugate trick uses the difference-of-two-squares identity to clear the surd: $\dfrac{1}{2 + \sqrt{3}} \times \dfrac{2 - \sqrt{3}}{2 - \sqrt{3}} = \dfrac{2 - \sqrt{3}}{(2)^{2} - (\sqrt{3})^{2}} = \dfrac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}.$

The conjugate of $a + b\sqrt{c}$ is $a - b\sqrt{c}$ . The conjugate of $\sqrt{p} - \sqrt{q}$ is $\sqrt{p} + \sqrt{q}$ .

Worked. Rationalise $\dfrac{1}{\sqrt{5} - \sqrt{2}}$ . $\dfrac{1}{\sqrt{5} - \sqrt{2}} \times \dfrac{\sqrt{5} + \sqrt{2}}{\sqrt{5} + \sqrt{2}} = \dfrac{\sqrt{5} + \sqrt{2}}{5 - 2} = \dfrac{\sqrt{5} + \sqrt{2}}{3}.$

Single surd: multiply top and bottom by that surd.
Binomial denominator: multiply by the CONJUGATE.
Conjugate flips the SIGN of the surd term.
Difference of two squares clears the surd in the denominator.

How it’s examined

Index-law and surd questions appear on every paper. Paper 2 typically has 1-2 mark questions: simplify $a^{m} \times a^{n}$ , evaluate $8^{2/3}$ , simplify $\sqrt{50}$ . Paper 4 escalates to multi-step manipulations, often inside a coordinate-geometry or trig question, where exact surd answers are required. Examiner reports flag fractional-index errors (forgetting the brackets on the calculator) and writing $\sqrt{a + b}$ as $\sqrt{a} + \sqrt{b}$ as the most-recurring failures.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E1.3, E1.9); 0580/22 May/Jun 2024 — Q14 (fractional index); 0580/42 Oct/Nov 2024 — Q12 (rationalise the denominator); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

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Step-by-step worked examples — Exponents and Surds

Step-by-step solutions to past-paper-style questions on exponents and surds, written exactly the way a tutor would explain them at the board.

1Apply the index laws to simplify
Extended• Adapted from 0580/22 May/Jun 2024 Q14• index laws
Question
Simplify (a) $a^{5} \times a^{-2}$ , (b) $\dfrac{x^{8}}{x^{3}}$ , (c) $(2y^{3})^{4}$ .
Step-by-step solution
1. Step 1
  (a) Add indices: $a^{5 + (-2)} = a^{3}$ .
2. Step 2
  (b) Subtract: $x^{8 - 3} = x^{5}$ .
3. Step 3
  (c) Apply the power to each factor: $2^{4} \times y^{12} = 16y^{12}$ .
Answer
(a) $a^{3}$ (b) $x^{5}$ (c) $16y^{12}$
2Evaluate a fractional index
Extended• fractional index
Question
Evaluate (a) $25^{1/2}$ , (b) $8^{2/3}$ , (c) $16^{-3/4}$ .
Step-by-step solution
1. Step 1
  (a) $25^{1/2} = \sqrt{25} = 5$ .
2. Step 2
  (b) $8^{2/3} = (8^{1/3})^{2} = 2^{2} = 4$ .
3. Step 3
  (c) $16^{-3/4} = \dfrac{1}{16^{3/4}} = \dfrac{1}{(16^{1/4})^{3}} = \dfrac{1}{2^{3}} = \dfrac{1}{8}$ .
Answer
(a) $5$ (b) $4$ (c) $\dfrac{1}{8}$
Examiner tip
Always take the root before applying the power — easier numbers, fewer slips.
3Simplify a surd
Extended• surds
Question
Simplify $\sqrt{72}$ .
Step-by-step solution
1. Step 1
  Find the largest perfect-square factor of $72$ : $36$ .
  $72 = 36 \times 2$
2. Step 2
  Use $\sqrt{ab} = \sqrt{a}\sqrt{b}$ .
  $\sqrt{72} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2}$
Answer
$6\sqrt{2}$
4Rationalise the denominator
Extended• Adapted from 0580/42 Oct/Nov 2023 Q15• rationalising
Question
Rationalise $\dfrac{6}{\sqrt{3}}$ and simplify.
Step-by-step solution
1. Step 1
  Multiply numerator and denominator by $\sqrt{3}$ .
  $\dfrac{6}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{6\sqrt{3}}{3}$
2. Step 2
  Simplify the fraction.
  $= 2\sqrt{3}$
Answer
$2\sqrt{3}$
Examiner tip
Always simplify the resulting fraction. $\dfrac{6\sqrt{3}}{3}$ left unsimplified is partial credit.
5Add and subtract surds
Extended• surd arithmetic
Question
Simplify $3\sqrt{20} - \sqrt{45}$ .
Step-by-step solution
1. Step 1
  $\sqrt{20} = 2\sqrt{5}$ , so $3\sqrt{20} = 6\sqrt{5}$ .
2. Step 2
  $\sqrt{45} = 3\sqrt{5}$ .
3. Step 3
  Subtract: $6\sqrt{5} - 3\sqrt{5} = 3\sqrt{5}$ .
Answer
$3\sqrt{5}$
6Evaluate zero and negative powers
Core• zero power, negative power
Question
Without a calculator, evaluate (a) $7^{0}$ , (b) $5^{-2}$ , (c) $\left(\dfrac{2}{3}\right)^{-1}$ .
Step-by-step solution
1. Step 1
  (a) Any non-zero base to the power $0$ is $1$ . So $7^{0} = 1$ .
2. Step 2
  (b) $5^{-2} = \dfrac{1}{5^{2}} = \dfrac{1}{25}$ .
3. Step 3
  (c) The reciprocal: $\left(\dfrac{2}{3}\right)^{-1} = \dfrac{3}{2}$ .
Answer
(a) $1$ (b) $\dfrac{1}{25}$ (c) $\dfrac{3}{2}$
Examiner tip
Mark schemes do not accept $5^{-2} = -25$ or $-10$ . Always remember: negative power means reciprocal, not negative value.
7Apply the index laws in a chain of operations
Core• index laws, chain
Question
Simplify $\dfrac{a^{6} \times a^{2}}{a^{3}}$ .
Step-by-step solution
1. Step 1
  Apply the multiplication law in the numerator: $a^{6} \times a^{2} = a^{6 + 2} = a^{8}$ .
2. Step 2
  Apply the division law: $\dfrac{a^{8}}{a^{3}} = a^{8 - 3} = a^{5}$ .
Answer
$a^{5}$
Examiner tip
The mark scheme awards method marks for showing each index law step. Skipping to the answer $a^{5}$ is risky if an arithmetic slip happens.
8Convert between surd and fractional-index form
Extended• Adapted from 0580/22 Oct/Nov 2023 Q8• fractional index, surd
Question
Write (a) $\sqrt[3]{x^{2}}$ in index form, (b) $y^{5/2}$ in surd form.
Step-by-step solution
1. Step 1
  (a) Use $a^{m/n} = \sqrt[n]{a^{m}}$ in reverse: $\sqrt[3]{x^{2}} = x^{2/3}$ .
2. Step 2
  (b) Rewrite the index as a root and power: $y^{5/2} = \sqrt{y^{5}}$ , which simplifies to $y^{2}\sqrt{y}$ .
Answer
(a) $x^{2/3}$ (b) $y^{2}\sqrt{y}$ (or $\sqrt{y^{5}}$ )
Examiner tip
The 2023 mark scheme accepts either $\sqrt{y^{5}}$ or the fully simplified $y^{2}\sqrt{y}$ . Always check whether 'simplest form' is required — if so, factor out the largest perfect-square power.
9Rationalise a binomial denominator (conjugate)
Extended• Adapted from 0580/42 May/Jun 2024 Q15• rationalising, conjugate, surds
Question
Rationalise $\dfrac{4}{2 + \sqrt{3}}$ and simplify.
Step-by-step solution
1. Step 1
  Multiply numerator and denominator by the conjugate $2 - \sqrt{3}$ .
  $\dfrac{4}{2 + \sqrt{3}} \times \dfrac{2 - \sqrt{3}}{2 - \sqrt{3}} = \dfrac{4(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})}$
2. Step 2
  Apply the difference-of-squares identity to the denominator: $(2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1$ .
3. Step 3
  Expand the numerator: $4(2 - \sqrt{3}) = 8 - 4\sqrt{3}$ .
  $\dfrac{8 - 4\sqrt{3}}{1} = 8 - 4\sqrt{3}$
Answer
$8 - 4\sqrt{3}$
Examiner tip
The 2024 examiner report notes that candidates frequently use $2 + \sqrt{3}$ (same sign) as the conjugate instead of $2 - \sqrt{3}$ . The conjugate must reverse the middle sign so the irrational cross-term cancels.
10Solve an equation involving surds
Extended• Adapted from 0580/42 Oct/Nov 2022 Q18• surd equation, algebra
Question
Solve $\sqrt{x + 5} = 7$ for $x$ . Then check your answer.
Step-by-step solution
1. Step 1
  Square both sides to remove the square root.
  $(\sqrt{x + 5})^{2} = 7^{2}$
2. Step 2
  Simplify.
  $x + 5 = 49$
3. Step 3
  Subtract $5$ : $x = 44$ .
4. Step 4
  Check: $\sqrt{44 + 5} = \sqrt{49} = 7$ . ✓
Answer
$x = 44$
Examiner tip
Always check the answer in the original equation — squaring can introduce extraneous solutions. The 2022 mark scheme awards a verification mark for this step.
11Solve an equation involving a fractional index (Paper 4 stretch)
Challenge• Adapted from 0580/42 Oct/Nov 2024 Q23• fractional index, equation, stretch
Question
Solve $x^{3/2} = 27$ for $x$ , where $x > 0$ .
Step-by-step solution
1. Step 1
  Raise both sides to the reciprocal power $\dfrac{2}{3}$ to isolate $x$ .
  $\left(x^{3/2}\right)^{2/3} = 27^{2/3}$
2. Step 2
  The left side simplifies to $x^{1} = x$ .
3. Step 3
  Evaluate $27^{2/3} = \left(27^{1/3}\right)^{2} = 3^{2} = 9$ .
4. Step 4
  So $x = 9$ . Check: $9^{3/2} = (9^{1/2})^{3} = 3^{3} = 27$ . ✓
Answer
$x = 9$
Examiner tip
The 2024 examiner report flags that candidates often square both sides (getting $x^{3} = 729$ ) and then take the cube root — both work, but require more steps. Using the reciprocal power is the cleanest method.
12Expand and simplify $(\sqrt{a} + \sqrt{b})^{2}$
Challenge• Adapted from 0580/42 May/Jun 2023 Q21• surd expansion, algebra
Question
Expand and simplify $(\sqrt{7} + \sqrt{3})^{2}$ , leaving your answer in the form $a + b\sqrt{c}$ where $a, b, c$ are integers and $c$ is as small as possible.
Step-by-step solution
1. Step 1
  Use $(p + q)^{2} = p^{2} + 2pq + q^{2}$ with $p = \sqrt{7}$ and $q = \sqrt{3}$ .
2. Step 2
  Compute each term: $p^{2} = 7$ , $q^{2} = 3$ , $2pq = 2\sqrt{7 \times 3} = 2\sqrt{21}$ .
3. Step 3
  Add: $7 + 2\sqrt{21} + 3 = 10 + 2\sqrt{21}$ .
  $(\sqrt{7} + \sqrt{3})^{2} = 10 + 2\sqrt{21}$
Answer
$10 + 2\sqrt{21}$
Examiner tip
The 2023 mark scheme awards method marks for explicitly using the $(p+q)^{2}$ identity. Candidates who write $\sqrt{7} \times \sqrt{3} = \sqrt{10}$ (adding inside the surd) make a common slip — surd multiplication multiplies the radicands.
13Show that $\sqrt{ab} = \sqrt{a}\sqrt{b}$ for $a, b \geq 0$
Challenge• Adapted from 0580/42 Oct/Nov 2023 Q24• show that, surd identity, proof
Question
Let $a$ and $b$ be non-negative real numbers. Show that $\sqrt{a}\sqrt{b} = \sqrt{ab}$ by squaring both sides.
Step-by-step solution
1. Step 1
  Both $\sqrt{a}\sqrt{b}$ and $\sqrt{ab}$ are non-negative since $a, b \geq 0$ . If their squares are equal, then they themselves are equal.
2. Step 2
  Square the left side.
  $(\sqrt{a}\sqrt{b})^{2} = (\sqrt{a})^{2} (\sqrt{b})^{2} = a \cdot b = ab$
3. Step 3
  Square the right side.
  $(\sqrt{ab})^{2} = ab$
4. Step 4
  Both squared values equal $ab$ . Since both original expressions are non-negative, $\sqrt{a}\sqrt{b} = \sqrt{ab}$ . QED.
Answer
$(\sqrt{a}\sqrt{b})^{2} = ab = (\sqrt{ab})^{2}$ , and as both sides are non-negative, $\sqrt{a}\sqrt{b} = \sqrt{ab}$ .
Examiner tip
On 'show that' surd identity questions, examiners expect explicit mention of the non-negativity condition; without it, the proof is incomplete because $x^{2} = y^{2}$ alone only gives $x = \pm y$ . The 2023 mark scheme awards a separate mark for stating this.

Key Formulae — Exponents and Surds

The formulae you need to memorise for exponents and surds on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Index laws
$a^{m} \cdot a^{n} = a^{m+n},\ \ \dfrac{a^{m}}{a^{n}} = a^{m-n},\ \ (a^{m})^{n} = a^{mn}$
$a$
non-zero base
$m, n$
any integers (or rationals for fractional powers)
When to use
All algebraic simplifications involving powers.
Zero, negative and fractional powers
$a^{0} = 1,\ \ a^{-n} = \dfrac{1}{a^{n}},\ \ a^{m/n} = \sqrt[n]{a^{m}}$
$a$
non-zero base
When to use
Whenever the index is zero, negative or a fraction.
Surd manipulation rules
$\sqrt{ab} = \sqrt{a}\sqrt{b},\quad \sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}$
$a, b$
non-negative reals (with $b \neq 0$ in the second)
When to use
Simplifying surds and dividing surds.
Rationalising a single surd denominator
$\dfrac{c}{\sqrt{a}} = \dfrac{c\sqrt{a}}{a}$
$a$
positive integer that is not a perfect square
When to use
Always when the question requires a rational denominator.

Key Definitions and Keywords — Exponents and Surds

Definitions to memorise and the exact keywords mark schemes credit for exponents and surds answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Base and index
Examiner keyword
In $a^{n}$ , $a$ is the base and $n$ is the index (or exponent or power).
Surd
Examiner keyword
A square (or other) root that cannot be simplified to a rational number, such as $\sqrt{5}$ .
Rationalise
Examiner keyword
Manipulate an expression so that no surd appears in the denominator.
Perfect square
A non-negative integer of the form $n^{2}$ .

Common Mistakes and Misconceptions — Exponents and Surds

The traps other students keep falling into on exponents and surds questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Adding the bases of powers with the same index
Why it happens
Students mis-apply the index laws: $2^{3} + 5^{3} \neq 7^{3}$ .
How to avoid it
Index laws apply to products, not sums. Compute each power separately, then add.
✕Treating $8^{2/3}$ as $\dfrac{8^{2}}{3}$
0580/42 — fractional powers
Why it happens
Students confuse fractional indices with division by the denominator.
How to avoid it
$a^{m/n} = \sqrt[n]{a^{m}} = (\sqrt[n]{a})^{m}$ . Take the root, then the power.
✕Stopping at $\sqrt{18}$ instead of $3\sqrt{2}$
Why it happens
Students miss that $18$ has a perfect-square factor.
How to avoid it
Always factor out the largest perfect square: $18 = 9 \times 2$ .
✕Leaving a surd in the denominator
0580/42 Oct/Nov 2023 — examiner report Q15
Why it happens
Students think $\dfrac{1}{\sqrt{2}}$ is fully simplified.
How to avoid it
Rationalise unless the question explicitly says you may leave it.
✕Treating $a^{-1}$ as $-a$ instead of $\dfrac{1}{a}$
Why it happens
Confusing the negative sign with subtraction.
How to avoid it
$a^{-n} = \dfrac{1}{a^{n}}$ — flip, don't negate.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Exponents and Surds — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Exponents and Surds

Short Study Notes in the form of Slides

Short Study Notes — Exponents and Surds

Detailed Notes

What you’ll learn

How it’s examined

1Apply the index laws to simplify

2Evaluate a fractional index

3Simplify a surd

4Rationalise the denominator

5Add and subtract surds

6Evaluate zero and negative powers

7Apply the index laws in a chain of operations

8Convert between surd and fractional-index form

9Rationalise a binomial denominator (conjugate)

10Solve an equation involving surds

11Solve an equation involving a fractional index (Paper 4 stretch)

12Expand and simplify (a+b)2(\sqrt{a} + \sqrt{b})^{2}(a​+b​)2

13Show that ab=ab\sqrt{ab} = \sqrt{a}\sqrt{b}ab​=a​b​ for a,b≥0a, b \geq 0a,b≥0

Index laws

Zero, negative and fractional powers

Surd manipulation rules

Rationalising a single surd denominator

Base and index

Surd

Rationalise

Perfect square

✕Adding the bases of powers with the same index

✕Treating 82/38^{2/3}82/3 as 823\dfrac{8^{2}}{3}382​

✕Stopping at 18\sqrt{18}18​ instead of 323\sqrt{2}32​

✕Leaving a surd in the denominator

✕Treating a−1a^{-1}a−1 as −a-a−a instead of 1a\dfrac{1}{a}a1​

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Exponents and Surds — frequently asked questions

12Expand and simplify $(\sqrt{a} + \sqrt{b})^{2}$

13Show that $\sqrt{ab} = \sqrt{a}\sqrt{b}$ for $a, b \geq 0$

✕Treating $8^{2/3}$ as $\dfrac{8^{2}}{3}$

✕Stopping at $\sqrt{18}$ instead of $3\sqrt{2}$

✕Treating $a^{-1}$ as $-a$ instead of $\dfrac{1}{a}$