What is the importance of proof in Pure Mathematics 4 for Edexcel International A Levels?

Proof is a fundamental concept in Pure Mathematics 4, as it provides a logical foundation for mathematical statements and theorems. In the Edexcel International A Levels curriculum, understanding proof is crucial because it helps students develop critical thinking and problem-solving skills, which are essential for advanced mathematical studies and applications.

How do I approach writing a mathematical proof in the context of Pure Mathematics 4?

When writing a mathematical proof in Pure Mathematics 4, start by clearly understanding the statement you need to prove. Identify the type of proof required, such as direct proof, proof by contradiction, or proof by induction. Structure your proof logically, beginning with known information or axioms, and use logical reasoning to arrive at the conclusion. Ensure each step is justified and clearly explained to make the proof easy to follow.

What are the common types of proof techniques covered in Pure Mathematics 4?

In Pure Mathematics 4, students typically encounter several proof techniques, including direct proof, proof by contradiction, proof by induction, and proof by exhaustion. Each technique has its own application and is used depending on the nature of the problem. Understanding when and how to apply these techniques is essential for solving complex mathematical problems in the Edexcel International A Levels curriculum.

Can you explain proof by induction with an example relevant to Pure Mathematics 4?

Proof by induction is a method used to prove statements about natural numbers. It involves two steps: the base case and the inductive step. For example, to prove that the sum of the first n natural numbers is (n(n+1))/2, start with the base case n=1, which holds true. Then, assume it holds for n=k, and prove it for n=k+1. This method is often used in Pure Mathematics 4 to prove sequences and series.

Why is proof by contradiction a powerful tool in Pure Mathematics 4?

Proof by contradiction is powerful because it allows you to prove a statement by assuming the opposite is true and showing that this assumption leads to a contradiction. This technique is particularly useful in Pure Mathematics 4 for proving statements where direct proof is challenging. It helps students understand the logical structure of mathematical arguments and reinforces the importance of logical consistency in mathematics.

Why is "Assuming what you are trying to prove" a common mistake in Proof?

Why it happens: Students take the conclusion and work both sides simultaneously, ending at 0 = 0 or 5 = 5. How to avoid it: For identities, start with ONE side (usually the more complex one) and manipulate it until it equals the other side. For 'prove that...' problems, the conclusion can only appear on the LAST line. Source: WMA14/01 examiner reports — flagged annually

Why is "Failing to name 'proof by contradiction' or 'proof by exhaustion'" a common mistake in Proof?

Why it happens: Students dive into the algebra without signalling the method. How to avoid it: Open every proof with one sentence: 'Assume for contradiction that...', or 'We check each of the finite cases...'. Mark schemes often award B1 for explicitly naming or signalling the method.

Why is "Treating a single numerical check as a proof" a common mistake in Proof?

Why it happens: Students confuse 'verifying for n = 3' with 'proving for all n'. How to avoid it: One example NEVER proves a universal statement (but one counter-example always disproves one). Algebra with a general n is required for a proof.

Why is "Omitting 'in lowest terms' in the $\sqrt{2}$ proof" a common mistake in Proof?

Why it happens: Students assume any p/q representation is fine. How to avoid it: The whole contradiction hinges on (p, q) = 1. Without that condition, finding that both p and q are even is not a contradiction. State 'in lowest terms' or 'with (p, q) = 1' clearly in the opening line. Source: WMA14/01 examiner reports — flagged annually

Why is "Claiming a counter-example without showing the failure" a common mistake in Proof?

Why it happens: Students write 'n = 11 works' but don't compute the value or explain why it disproves the claim. How to avoid it: Always: (i) state the value, (ii) compute the expression, (iii) explicitly explain why it contradicts the claim. E.g. 'n = 11: n^2 - n + 11 = 121 = 11 × 11, which is not prime — hence the claim is false.'

Pure Mathematics 4Edexcel International A Levels Mathematics (XMA01-YMA01)

Proof

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Proof Study Notes — Edexcel IAL Mathematics P4 (WMA14/01, 2018 spec — 2026 onwards)

Methods of proof in IAL Maths: deduction, exhaustion, disproof by counter-example, and contradiction. P4 examiners reward the LOGICAL STRUCTURE of a proof as much as the algebra — naming the method, stating the assumption clearly, and writing an explicit closing sentence are all mark-bearing actions.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

P4 1.1 — Construct proofs by deduction, including identities involving algebraic manipulation.
P4 1.2 — Construct proofs by exhaustion.
P4 1.3 — Disprove a statement by means of a counter-example.
P4 1.4 — Construct proofs by contradiction (e.g. irrationality of $\sqrt{2}$ , infinitude of primes).

Proof by deduction

Logical chain from known facts to the target statement. The bread and butter.

Definition. A proof by deduction proceeds from agreed starting points — axioms, definitions, or earlier-proved theorems — via a sequence of valid algebraic / logical steps to the conclusion.

Setup checklist for identity proofs.

Identify which side to start from (usually the more complex one).
State variable assumptions clearly ("Let $n$ be an integer.").
Manipulate one side until it matches the other.
Close with a one-sentence statement: "Hence ... as required."

Worked example 1. Prove that $(2n + 1)^{2} - (2n - 1)^{2} = 8n$ for all real $n$ .

LHS $= (4n^{2} + 4n + 1) - (4n^{2} - 4n + 1) = 8n =$ RHS. $\square$

Worked example 2 (divisibility). Prove that the product of two consecutive integers is even.

Let the integers be $n$ and $n + 1$ , with $n \in \mathbb{Z}$ .
Product $= n(n + 1)$ . One of $n$ , $n + 1$ is even (consecutive integers alternate parity), so the product contains a factor of $2$ .
Hence $n(n+1)$ is even. $\square$

Worked example 3 (algebraic). Prove that $n^{3} - n$ is divisible by $6$ for every positive integer $n$ .

Factorise: $n^{3} - n = n(n^{2} - 1) = (n - 1) \cdot n \cdot (n + 1)$ .
These are three CONSECUTIVE integers. Among any three consecutive integers, one is divisible by $3$ and at least one is even.
So the product is divisible by $2 \times 3 = 6$ . $\square$

Style tips.

Use $\equiv$ (identity) when proving for-all-values statements, $=$ when solving for specific values.
Number your equations only if you refer back to them.
Write a closing sentence — examiners reject "stranded" working.

Start with one side; manipulate until it matches the other.
Parametrise: even $= 2n$ , odd $= 2n \pm 1$ , consecutive $= n, n+1$ .
Factor products of consecutive integers to expose hidden divisors.
Always close with a concluding sentence.

Proof by exhaustion

Finite cases? Check them all.

Definition. A proof by exhaustion splits a statement into a finite number of cases and verifies each. Valid ONLY when the case-list is genuinely finite (e.g. integers in $[1, 5]$ , residues modulo $4$ ).

Worked example. Prove that $n^{2} + 2$ is not divisible by $5$ for integer $n$ with $1 \leq n \leq 5$ .

$n$	$n^{2}+2$	$\pmod{5}$
$1$	$3$	$3$
$2$	$6$	$1$
$3$	$11$	$1$
$4$	$18$	$3$
$5$	$27$	$2$

None is $0 \pmod 5$ . Hence the result is proved by exhaustion. $\square$

Going modular. A more elegant exhaustion: any integer $n$ is congruent to $0, 1, 2, 3$ or $4$ modulo $5$ . So $n^{2} \pmod 5 \in \{0, 1, 4, 4, 1\} = \{0, 1, 4\}$ , and $n^{2} + 2 \pmod 5 \in \{2, 3, 1\}$ . This handles ALL integers in $5$ cases — proof that $n^{2} + 2$ is never divisible by $5$ for any integer $n$ . Edexcel rarely demands modular reasoning but it's a powerful tool.

Worked example. Prove that every prime greater than $3$ is of the form $6k \pm 1$ .

Any integer is $6k, 6k+1, 6k+2, 6k+3, 6k+4$ , or $6k+5$ for some integer $k$ .
$6k$ is divisible by $6$ , so $> 3$ ⇒ not prime.
$6k + 2 = 2(3k + 1)$ divisible by $2$ ⇒ not prime if $> 3$ .
$6k + 3 = 3(2k + 1)$ divisible by $3$ ⇒ not prime if $> 3$ .
$6k + 4 = 2(3k + 2)$ ⇒ not prime if $> 3$ .
Only $6k + 1$ and $6k + 5 = 6(k+1) - 1$ remain. Hence every prime $> 3$ is of the form $6k \pm 1$ . $\square$

Finite list of cases — check each one explicitly.
Modular reasoning packs infinitely many integers into finitely many residue classes.
State the method up front: 'By exhaustion, we check each case...'

Disproof by counter-example

One specific failure refutes a universal claim.

Asymmetry of universal statements. A claim of the form "for all $n$ , $P(n)$ holds" can be:

PROVED only by a fully general argument (algebra with arbitrary $n$ ).
DISPROVED by a single explicit case where $P$ fails.

Worked example 1. A student claims: "If $n$ is a positive integer, then $n^{2} - n + 11$ is always prime."

Try $n = 11$ : $11^{2} - 11 + 11 = 121 = 11 \times 11$ , not prime.
Hence the claim is false. $\square$

Worked example 2. "For all real $x$ , $x^{2} > x$ ."

Try $x = 1/2$ : $x^{2} = 1/4 < 1/2 = x$ . Counter-example. $\square$

Worked example 3. "If $a > b$ then $a^{2} > b^{2}$ ."

Try $a = 1$ , $b = -2$ : $a > b$ but $a^{2} = 1 < 4 = b^{2}$ . Counter-example. $\square$

How to find counter-examples.

Try small values: $n = 0, 1, 2$ .
Try negative or zero (if the domain allows).
Try fractions and irrationals.
Look at "edge" cases where the algebra has a clear vulnerability.

Marking note. Always:

State the counter-example value clearly.
Compute both sides of the claim.
Write the explicit conclusion: "Hence the claim is false."

ONE counter-example suffices to disprove 'for all $n$ ' claims.
Try edge cases: $0$ , $1$ , negatives, fractions.
Show the computation — don't just assert.
Close with the explicit conclusion.

Proof by contradiction

Assume the negation. Derive nonsense. Conclude the original.

Strategy. To prove a statement $P$ :

Assume $\neg P$ (the negation of $P$ ).
Deduce consequences from $\neg P$ using valid algebra / logic.
Reach a contradiction — a statement known to be false ( $1 = 0$ , $\gcd = 1$ contradicted, finite list contains a missing element, ...).
Conclude that the assumption $\neg P$ was false, hence $P$ is true.

Worked example 1: $\sqrt{2}$ is irrational.

Assume, for contradiction, that $\sqrt{2}$ is rational. Then $\sqrt{2} = \dfrac{p}{q}$ for some integers $p$ , $q$ with $\gcd(p, q) = 1$ (lowest terms), $q \neq 0$ .

Square: $2 = p^{2}/q^{2}$ , so $p^{2} = 2q^{2}$ .
$p^{2}$ is even ⇒ $p$ is even (the square of an odd is odd).
Write $p = 2k$ . Then $4k^{2} = 2q^{2}$ , so $q^{2} = 2k^{2}$ .
$q^{2}$ is even ⇒ $q$ is even.
So $2 \mid p$ and $2 \mid q$ , contradicting $\gcd(p, q) = 1$ . $\square$

Worked example 2: Infinitely many primes (Euclid).

Assume, for contradiction, only finitely many primes exist: $p_{1}, p_{2}, \ldots, p_{k}$ .

Let $N = p_{1} p_{2} \cdots p_{k} + 1$ .
$N > 1$ , so $N$ has a prime factor $p$ .
By assumption, $p = p_{i}$ for some $i$ , so $p \mid p_{1} \cdots p_{k}$ .
$p \mid N$ and $p \mid p_{1} \cdots p_{k}$ , so $p \mid (N - p_{1} \cdots p_{k}) = 1$ .
But no prime divides $1$ . Contradiction. $\square$

Worked example 3: $\log_{2} 3$ is irrational.

Assume $\log_{2} 3 = p/q$ in lowest terms with $p, q \in \mathbb{Z}_{>0}$ .

$2^{p/q} = 3$ , so $2^{p} = 3^{q}$ .
LHS is a power of $2$ (only factor: $2$ ); RHS is a power of $3$ (only factor: $3$ ).
The only way $2^{p} = 3^{q}$ is $p = q = 0$ , contradicting $p, q > 0$ . $\square$

Structure phrases (memorise).

Opening: "Assume, for contradiction, that ..."
Middle: "Then ... Hence ..."
Contradiction: "But this contradicts ..."
Closing: "Therefore the assumption is false, so ... as required."

Always state the assumption explicitly.
For $\sqrt{2}$ : insist on 'in lowest terms' / $\gcd = 1$ .
For primes: the $+1$ in $N$ is what creates the contradiction.
Close by reversing the assumption to recover $P$ .

Choosing the right proof method

How to tell which technique a P4 question wants.

Decision tree.

'Show that' / 'Prove that' an identity ⇒ direct deduction. Expand and simplify.
'Prove for $n = 1, \ldots, k$ ' with small $k$ ⇒ exhaustion.
'Is this always true?' / 'A student claims...' ⇒ probably a counter-example.
'Prove that ... is irrational' / 'Prove that there are infinitely many...' ⇒ contradiction.
'Disprove' / 'Show that the following is false' ⇒ counter-example.

Signal words.

Phrase	Method
"Prove the identity"	Deduction
"Show that ... is always"	Deduction (general $n$ )
"Verify for $n = 1, 2, 3$ "	Exhaustion (those values)
"Disprove" / "Find a counter-example"	Counter-example
"Prove by contradiction"	Contradiction (explicit)
"Show $\sqrt{p}$ is irrational"	Contradiction (idiomatic)

Mark-scheme features to hit.

B1 for naming the method (especially contradiction / exhaustion).
B1 for stating the assumption / parametrisation clearly.
M1/A1 marks for each algebraic step.
Final A1 for the explicit concluding statement.

Identity questions ⇒ deduction.
Bounded $n$ ⇒ exhaustion.
Universal claim, suspicious ⇒ counter-example.
Irrationality / infinity ⇒ contradiction.

How it’s examined

Proof appears in WMA14/01 P4 every sitting (Jan & Jun) as Q1 or Q2 — typically $4$ - $6$ marks. Patterns: 'prove the algebraic identity' (4 marks, deduction); 'prove this divisibility statement' (5 marks, deduction with parametrisation); 'prove by contradiction that $\sqrt{2}$ / $\sqrt{3}$ / $\log_{2} 5$ is irrational' (6 marks); 'a student claims X — disprove' (3 marks, counter-example). Mark schemes are EXTRA STRICT on logical structure: a fully correct algebraic chain that lacks the concluding sentence will lose the final A. Practise writing closing statements aloud.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WMA14/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Worked examples, formulae, definitions and the mistakes examiners flag — everything you need to push from a pass to an A*.

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Step-by-step worked examples — Proof

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

1Proof by deduction — algebraic identity (4 marks, P4)
Core• Adapted from WMA14/01 January 2024 Q1• proof by deduction, identity
Question
Prove that for all real $n$ , $(2n + 1)^{2} - (2n - 1)^{2} \equiv 8n$ .
Step-by-step solution
1. Step 1
  Start with the left-hand side and expand each square.
  $(2n + 1)^{2} = 4n^{2} + 4n + 1, \qquad (2n - 1)^{2} = 4n^{2} - 4n + 1$
2. Step 2
  Subtract — the $4n^{2}$ and $+1$ cancel.
  $(4n^{2} + 4n + 1) - (4n^{2} - 4n + 1) = 8n$
3. Step 3
  LHS equals RHS for all real $n$ , so the identity is proved.
Answer
$(2n + 1)^{2} - (2n - 1)^{2} = 8n$ for all real $n$ . $\square$
Examiner tip
M1 for attempting to expand both brackets. A1 for both expansions correct. M1 for the subtraction with correct sign handling. A1 for arriving at $8n$ and writing a brief concluding sentence (or ' $\square$ '). The mark scheme always requires a closing statement — examiners refuse the final A if the candidate just leaves $8n$ on the page with no conclusion.
2Proof by deduction — divisibility (5 marks, P4)
Extended• Adapted from WMA14/01 June 2024 Q2• proof by deduction, divisibility
Question
Prove that the sum of the squares of any two consecutive odd integers is even but not divisible by $4$ .
Step-by-step solution
1. Step 1
  Let the two consecutive odd integers be $2n - 1$ and $2n + 1$ , where $n$ is an integer.
2. Step 2
  Sum of their squares:
  $(2n - 1)^{2} + (2n + 1)^{2} = (4n^{2} - 4n + 1) + (4n^{2} + 4n + 1) = 8n^{2} + 2$
3. Step 3
  Factor: $8n^{2} + 2 = 2(4n^{2} + 1)$ . The factor of $2$ shows the sum is even.
4. Step 4
  Now check divisibility by $4$ . We have $2(4n^{2} + 1)$ . Since $4n^{2}$ is even, $4n^{2} + 1$ is odd. So $2 \times (\text{odd})$ cannot have another factor of $2$ , i.e. cannot be divisible by $4$ .
5. Step 5
  Hence the sum is even but not divisible by $4$ , as required.
Answer
Sum $= 2(4n^{2} + 1)$ where $4n^{2} + 1$ is odd — even, not divisible by $4$ . $\square$
Examiner tip
B1 for clearly defining consecutive odd integers as $2n \pm 1$ (with $n$ stated as an integer). M1 for summing the squares. A1 for $8n^{2} + 2$ . M1 for the parity argument explaining why $4n^{2} + 1$ is odd. A1 for the explicit conclusion. Common failure: defining the integers as $n$ and $n + 2$ (these aren't guaranteed to both be odd unless extra reasoning is given).
3Proof by exhaustion (4 marks, P4)
Core• Adapted from WMA14/01 January 2023 Q1• proof by exhaustion
Question
Prove by exhaustion that for any integer $n$ with $1 \leq n \leq 5$ , $n^{2} + 2$ is not divisible by $5$ .
Step-by-step solution
1. Step 1
  State the method: there are only $5$ integer values to check, so check each in turn.
2. Step 2
  Tabulate $n^{2} + 2$ :
  $n = 1: 1 + 2 = 3 \\ n = 2: 4 + 2 = 6 \\ n = 3: 9 + 2 = 11 \\ n = 4: 16 + 2 = 18 \\ n = 5: 25 + 2 = 27$
3. Step 3
  Reduce each modulo $5$ : $3, 1, 1, 3, 2$ . None is $0 \pmod 5$ , so none is divisible by $5$ .
4. Step 4
  Every case has been checked, so the proof by exhaustion is complete.
Answer
$n^{2} + 2 \notin \{5, 10, 15, 20, 25\}$ for $n = 1, \ldots, 5$ . $\square$
Examiner tip
B1 for naming the method (proof by exhaustion) or for listing all $5$ cases. M1 for computing $n^{2} + 2$ for each $n$ . A1 for all values correct. A1 for the explicit conclusion that none is divisible by $5$ . Mark scheme deducts the final A if a student forgets one case (e.g. omits $n = 1$ ).
4Disproof by counter-example (3 marks, P4)
Core• Adapted from WMA14/01 June 2023 Q1• counter-example, disproof
Question
A student claims: 'If $n$ is a positive integer, then $n^{2} - n + 11$ is always prime.' Disprove this claim.
Step-by-step solution
1. Step 1
  We need a single counter-example. Try small $n$ to spot the pattern, then look for a value where the expression has an obvious factor.
2. Step 2
  Try $n = 11$ : $11^{2} - 11 + 11 = 121$ . But $121 = 11^{2}$ , which is not prime.
3. Step 3
  Hence the claim is false. The counter-example $n = 11$ disproves it.
Answer
When $n = 11$ , $n^{2} - n + 11 = 121 = 11 \times 11$ , not prime. $\square$
Examiner tip
M1 for substituting any value of $n$ . A1 for identifying $n = 11$ (or $n = 12$ , where the value is $143 = 11 \times 13$ ). A1 for the explicit statement that the result is not prime — quoting its factorisation. Common error: writing 'because $121$ is not prime' without giving the factorisation; examiners expect proof, not assertion.
5Proof by contradiction — $\sqrt{2}$ is irrational (6 marks, P4)
Challenge• Adapted from WMA14/01 January 2024 Q4• proof by contradiction, irrationality
Question
Prove by contradiction that $\sqrt{2}$ is irrational.
Step-by-step solution
1. Step 1
  Assume the negation: suppose $\sqrt{2}$ is rational. Then $\sqrt{2} = \dfrac{p}{q}$ for some integers $p$ , $q$ with no common factor (i.e. the fraction is in lowest terms), and $q \neq 0$ .
2. Step 2
  Square both sides:
  $2 = \dfrac{p^{2}}{q^{2}} \quad \Rightarrow \quad p^{2} = 2 q^{2}$
3. Step 3
  So $p^{2}$ is even, which means $p$ is even (the square of an odd integer is odd). Write $p = 2k$ for some integer $k$ .
4. Step 4
  Substitute: $p^{2} = 4 k^{2}$ , so $4 k^{2} = 2 q^{2}$ , giving $q^{2} = 2 k^{2}$ . Hence $q^{2}$ is even, so $q$ is even too.
5. Step 5
  But now both $p$ and $q$ are even, so they share a common factor of $2$ . This contradicts the assumption that $p/q$ was in lowest terms.
6. Step 6
  The contradiction shows our assumption is false. Hence $\sqrt{2}$ is irrational.
Answer
Assumption $\sqrt{2} = p/q$ in lowest terms forces both $p$ and $q$ even — contradiction. Hence $\sqrt{2}$ is irrational. $\square$
Examiner tip
B1 for stating the assumption (negation of the conclusion) clearly. B1 for the 'in lowest terms' / 'coprime' condition. M1 for squaring to get $p^{2} = 2 q^{2}$ . A1 for deducing $p$ is even. M1 for substituting $p = 2k$ and deducing $q$ is even. A1 for the contradiction and conclusion. The 'in lowest terms' premise is essential — without it the contradiction collapses. Examiners reject proofs that omit it.
6Proof by contradiction — infinitely many primes (Euclid, 6 marks, P4)
Challenge• Adapted from WMA14/01 June 2024 Q4• proof by contradiction, primes
Question
Prove by contradiction that there are infinitely many prime numbers.
Step-by-step solution
1. Step 1
  Assume the negation: there are only finitely many primes. List them all: $p_{1}, p_{2}, \ldots, p_{k}$ .
2. Step 2
  Consider the integer $N = p_{1} p_{2} \cdots p_{k} + 1$ (product of all primes, plus $1$ ). Note $N > 1$ .
3. Step 3
  Every integer $>1$ has at least one prime factor. Let $p$ be a prime that divides $N$ . Then $p$ must be one of $p_{1}, \ldots, p_{k}$ (by the assumption that this is the complete list).
4. Step 4
  But $p$ divides $p_{1} p_{2} \cdots p_{k}$ (it appears in the product) and $p$ divides $N$ . Subtracting, $p$ divides $N - p_{1} p_{2} \cdots p_{k} = 1$ .
5. Step 5
  No prime divides $1$ — primes are $\geq 2$ . Contradiction.
6. Step 6
  Hence the assumption fails: there must be infinitely many primes.
Answer
Assuming finitely many primes leads to a prime dividing $1$ — impossible. So there are infinitely many primes. $\square$
Examiner tip
B1 for stating the assumption (finitely many primes — list them). M1 for constructing $N = \prod p_{i} + 1$ . A1 for the observation that any prime divisor of $N$ must be on the list. M1 for deducing $p \mid 1$ (the $+1$ trick). A1 for the contradiction. A1 for the explicit conclusion. This is the Euclid construction — the mark scheme name-checks it but doesn't require students to. The single most common slip: not justifying why any prime factor of $N$ must lie on the assumed finite list.

Key Formulae — Proof

The formulae you need to memorise for proof on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Identity vs equation
$A(x) \equiv B(x) \;\;\text{means LHS} = \text{RHS for all valid } x; \quad A(x) = B(x) \;\;\text{is solved for specific } x$
$\equiv$
identity (always true)
$=$
equation (true for specific values)
When to use
When asked to 'prove an identity' the symbol $\equiv$ is appropriate throughout; final lines should re-state the identity. NOT in the formula booklet — convention.
Example
$\sin^{2}\theta + \cos^{2}\theta \equiv 1$ is an identity; $\sin\theta = 1$ is an equation.
Parametrising special integers
$\text{Even: } 2n; \;\; \text{Odd: } 2n + 1 \text{ or } 2n - 1; \;\; \text{Consecutive: } n, n+1, n+2$
$n$
integer parameter (state this!)
When to use
Whenever a proof involves 'even', 'odd', or 'consecutive' integers — set up the algebra by writing the integers in this parametric form, then manipulate. NOT in the formula booklet.
Example
Sum of three consecutive integers: $n + (n+1) + (n+2) = 3n + 3 = 3(n+1)$ , hence divisible by $3$ .
Proof by contradiction structure
$\text{To prove } P: \;\; \text{Assume } \neg P, \text{ derive a contradiction, conclude } P$
$P$
the statement to be proved
$\neg P$
negation of $P$
When to use
For statements where direct deduction is hard but the negation gives concrete algebraic / numerical handles (irrationality of $\sqrt{2}$ , infinitude of primes, 'no integer solutions exist'). NOT in the formula booklet — method.
Example
To prove $\sqrt{2}$ irrational, assume $\sqrt{2} = p/q$ in lowest terms and force $\gcd(p, q) > 1$ .

Key Definitions and Keywords — Proof

Definitions to memorise and the exact keywords mark schemes credit for proof answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Proof by deduction
Examiner keyword
A proof that proceeds from agreed axioms / known results via a chain of logical implications to the conclusion. Each step must be justified by algebra or a previously-quoted theorem. The default proof method in IAL Maths.
Example
Prove $(2n+1)^{2} - (2n-1)^{2} = 8n$ : expand both squares and subtract.
Proof by exhaustion
Examiner keyword
Splitting the statement into a finite number of cases and verifying each one separately. Only valid when there are finitely many cases (e.g. integers in a bounded range).
Example
Prove that $n^{2} + 2$ is not divisible by $5$ for $n = 1, \ldots, 5$ by checking each value.
Counter-example
Examiner keyword
A single specific case that makes a universal claim false. One counter-example is sufficient to DISPROVE a 'for all' statement — but no counter-examples can PROVE one.
Example
$n = 11$ is a counter-example to ' $n^{2} - n + 11$ is always prime' (gives $121 = 11^{2}$ ).
Proof by contradiction (reductio ad absurdum)
Examiner keyword
Assume the negation of the statement you wish to prove. Derive a contradiction (a statement known to be false, e.g. $1 = 0$ , or a violation of the assumptions). Conclude that the original statement must be true.
Example
Proving $\sqrt{2}$ is irrational by assuming $\sqrt{2} = p/q$ in lowest terms and forcing $\gcd(p, q) \geq 2$ .
Rational / irrational
Examiner keyword
A real number is rational if it can be written as $\dfrac{p}{q}$ for integers $p$ , $q$ with $q \neq 0$ . Otherwise it is irrational. Examples: $\tfrac{3}{4}$ rational; $\sqrt{2}$ , $\pi$ , $e$ irrational.
Prime number
Examiner keyword
An integer $p \geq 2$ whose only positive divisors are $1$ and $p$ . $1$ is NOT prime. The first few primes: $2, 3, 5, 7, 11, 13, \ldots$

Common Mistakes and Misconceptions — Proof

The traps other students keep falling into on proof questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Assuming what you are trying to prove
WMA14/01 examiner reports — flagged annually
Why it happens
Students take the conclusion and work both sides simultaneously, ending at $0 = 0$ or $5 = 5$ .
How to avoid it
For identities, start with ONE side (usually the more complex one) and manipulate it until it equals the other side. For 'prove that...' problems, the conclusion can only appear on the LAST line.
✕Failing to name 'proof by contradiction' or 'proof by exhaustion'
Why it happens
Students dive into the algebra without signalling the method.
How to avoid it
Open every proof with one sentence: 'Assume for contradiction that...', or 'We check each of the finite cases...'. Mark schemes often award B1 for explicitly naming or signalling the method.
✕Treating a single numerical check as a proof
Why it happens
Students confuse 'verifying for $n = 3$ ' with 'proving for all $n$ '.
How to avoid it
One example NEVER proves a universal statement (but one counter-example always disproves one). Algebra with a general $n$ is required for a proof.
✕Omitting 'in lowest terms' in the $\sqrt{2}$ proof
WMA14/01 examiner reports — flagged annually
Why it happens
Students assume any $p/q$ representation is fine.
How to avoid it
The whole contradiction hinges on $\gcd(p, q) = 1$ . Without that condition, finding that both $p$ and $q$ are even is not a contradiction. State 'in lowest terms' or 'with $\gcd(p, q) = 1$ ' clearly in the opening line.
✕Claiming a counter-example without showing the failure
Why it happens
Students write 'n = 11 works' but don't compute the value or explain why it disproves the claim.
How to avoid it
Always: (i) state the value, (ii) compute the expression, (iii) explicitly explain why it contradicts the claim. E.g. ' $n = 11$ : $n^{2} - n + 11 = 121 = 11 \times 11$ , which is not prime — hence the claim is false.'

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.

Proof — frequently asked questions

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

Proof

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Proof by deduction — algebraic identity (4 marks, P4)

2Proof by deduction — divisibility (5 marks, P4)

3Proof by exhaustion (4 marks, P4)

4Disproof by counter-example (3 marks, P4)

5Proof by contradiction — 2\sqrt{2}2​ is irrational (6 marks, P4)

6Proof by contradiction — infinitely many primes (Euclid, 6 marks, P4)

Identity vs equation

Parametrising special integers

Proof by contradiction structure

Proof by deduction

Proof by exhaustion

Counter-example

Proof by contradiction (reductio ad absurdum)

Rational / irrational

Prime number

✕Assuming what you are trying to prove

✕Failing to name 'proof by contradiction' or 'proof by exhaustion'

✕Treating a single numerical check as a proof

✕Omitting 'in lowest terms' in the 2\sqrt{2}2​ proof

✕Claiming a counter-example without showing the failure

Practice questions

Past paper style quiz

Video lesson

Proof — frequently asked questions

5Proof by contradiction — $\sqrt{2}$ is irrational (6 marks, P4)

✕Omitting 'in lowest terms' in the $\sqrt{2}$ proof