Proof

This lesson covers the topic of mathematical proof as required by the A-Level Mathematics specification. Proof lies at the heart of mathematics — it is the rigorous process by which we establish that a mathematical statement is always true, not just true for a few examples. Understanding proof is essential for developing logical reasoning and is examined directly in A-Level papers.

Spec Mapping — AQA 7357 Section A Proof. This lesson covers the proof-by-deduction, proof-by-exhaustion and disproof-by-counter-example content of Section A. Refer to the official AQA specification document for exact wording.

Why Do We Need Proof?

In mathematics, a statement may appear to be true based on a handful of examples, but this does not guarantee it holds in every case. A proof provides a logical argument that demonstrates a statement is true for all possible cases. Without proof, mathematics would be built on assumptions rather than certainty.

Example of a false conjecture:

"All prime numbers are odd."

This seems reasonable, but 2 is a prime number and it is even. A single counterexample is enough to disprove a statement.

Exam Tip: When asked to "prove" something, showing it works for a few values is not a proof. You must use a logical argument that covers all cases.

Types of Proof

1. Proof by Deduction

This is the most common type of proof at A-Level. You start from known facts, definitions, or previously proven results and use logical steps to arrive at the statement you want to prove.

Example: Prove that the sum of two consecutive integers is always odd.

Let the two consecutive integers be n and n + 1, where n is an integer.

Their sum is:

n + (n + 1) = 2n + 1

Since 2n is always even (it is a multiple of 2), 2n + 1 is always odd.

Therefore, the sum of two consecutive integers is always odd. ∎

2. Proof by Exhaustion

This involves checking every possible case. It is only feasible when the number of cases is finite and small.

Example: Prove that n² + n + 1 is odd for all integers n where 1 ≤ n ≤ 4.

n	n² + n + 1	Odd?
1	1 + 1 + 1 = 3	Yes
2	4 + 2 + 1 = 7	Yes
3	9 + 3 + 1 = 13	Yes
4	16 + 4 + 1 = 21	Yes

All cases have been checked, so the statement is true for 1 ≤ n ≤ 4. ∎

Note: this does not prove the statement for all n — proof by exhaustion only covers the stated cases.

3. Proof by Contradiction

Assume the opposite of what you want to prove, and show that this leads to a logical contradiction. Since the assumption produces a contradiction, it must be false, and therefore the original statement is true.

Example: Prove that √2 is irrational.

Assume, for contradiction, that √2 is rational. Then we can write:

sqrt{2} = a/b

where a and b are integers with no common factors (i.e., the fraction is in its simplest form).

Squaring both sides: 2 = a²/b², so a² = 2b².

This means a² is even, so a must be even. Write a = 2k.

Then (2k)² = 2b², so 4k² = 2b², giving b² = 2k².

This means b² is even, so b must be even.

But if both a and b are even, they share a common factor of 2 — this contradicts our assumption that the fraction was in simplest form.

Therefore, √2 is irrational. ∎

4. Disproof by Counter-Example

To disprove a statement, you only need to find one example where it fails.

Example: Disprove the statement: "n² > 2n for all positive integers n."

Try n = 1: 1² = 1 and 2(1) = 2. Since 1 < 2, the statement is false.

Counter-example: n = 1. ∎

Algebraic Proof Techniques

When constructing algebraic proofs, the following representations are essential:

Statement	Algebraic Representation
An even number	2n (where n is an integer)
An odd number	2n + 1 (where n is an integer)
Consecutive integers	n, n + 1, n + 2, ...
A multiple of 3	3n
The square of an integer	n²

Example: Prove that the product of two odd numbers is always odd.

Let the two odd numbers be 2a + 1 and 2b + 1, where a and b are integers.

(2a + 1)(2b + 1) = 4ab + 2a + 2b + 1 = 2(2ab + a + b) + 1

Since 2(2ab + a + b) is even, adding 1 makes it odd.

Therefore, the product of two odd numbers is always odd. ∎

Example: Prove that (n + 1)² − (n − 1)² = 4n for all integers n.

LHS = (n + 1)² − (n − 1)²
    = (n² + 2n + 1) − (n² − 2n + 1)
    = n² + 2n + 1 − n² + 2n − 1
    = 4n = RHS ∎

Synoptic Links

This topic connects to:

Proof by contradiction (aqa-alevel-maths-pure-2 / proof-by-contradiction) — extension of deductive techniques examined as a year-two Pure topic.
Algebraic identities (aqa-alevel-maths-pure-1 / algebra-and-functions) — manipulation skills are the engine behind divisibility and identity proofs.
Trigonometric identities (aqa-alevel-maths-trigonometry-depth / addition-formulae) — "prove" and "show that" questions chain proof techniques onto trig manipulation.

Summary

Proof by deduction: Use logical steps from known facts to reach the conclusion.
Proof by exhaustion: Check every possible case (only when finite).
Proof by contradiction: Assume the opposite, derive a contradiction.
Disproof by counter-example: Find one case where the statement fails.
Always represent even numbers as 2n, odd numbers as 2n + 1, and consecutive integers as n, n + 1, ...
Define your variables clearly (e.g. "Let n be an integer").
Write a concluding sentence that directly addresses what was asked.

Exam Tip: In the exam, clearly state what you are assuming and what you are proving. Show every algebraic step and conclude with a clear statement such as "Therefore, the statement is true" or "This contradicts our assumption, so the original statement is proven." Look for the command word: "Prove" requires a full logical argument; "Show that" requires you to arrive at a given answer; "Disprove" requires a counter-example.

A-Level Deep Dive: Proof

Spec mapping

AQA 7357 Pure section A — Proof covers the structure of mathematical proof, proceeding from given assumptions through a series of logical steps to a conclusion; use methods of proof, including proof by deduction, proof by exhaustion, disproof by counter-example (refer to the official specification document for exact wording). Proof by contradiction is examined separately under section J of the second-year Pure content. Proof is not in the AQA formula booklet — there are no listed templates, identities or schemata to lean on. You memorise the structures (deduction, exhaustion, counter-example) and apply them.

Although Proof has its own section, it is the most synoptic topic on the entire specification. Every other Pure section contains identities or general statements that can be examined as a "prove" or "show that" question: algebraic identities (section B), trigonometric identities such as $\sin^2\theta + \cos^2\theta \equiv 1$ (section E), divisibility and number-theoretic claims, results about sequences (section F), and properties of curves (section G). Paper 1 typically opens or closes with a 4–8 mark proof question, and isolated 2–3 mark "show that" steps appear inside larger questions throughout both papers.

Specimen Question (Modelled on the AQA Paper Format)

Note: this question is constructed to model AQA Paper 1/2/3 style; it is not a reproduction of any published past paper.

Worked example with full mark scheme

Question (8 marks): Prove that for any integer $n$ , the expression $n^3 - n$ is divisible by 6. (8)

Solution with mark scheme:

Step 1 — factorise.

$n^3 - n = n(n^2 - 1) = n(n - 1)(n + 1) = (n - 1)\,n\,(n + 1)$

M1 — recognising and applying the difference-of-squares factorisation, leading to a product of three consecutive integers. Candidates who attempt to expand and check small cases ( $n = 1, 2, 3$ ) score nothing here: a finite check is not a proof for "any integer $n$ ".

A1 — clean factorised form $(n-1)n(n+1)$ , with the words "three consecutive integers" or equivalent stated.

Step 2 — argue divisibility by 2.

Among any two consecutive integers, exactly one is even. Therefore among $(n-1), n, (n+1)$ at least one is even, so the product is divisible by 2.

B1 — clear reasoning that the product of consecutive integers is even. The phrase "at least one of three consecutive integers is even" is the examiner-rewarded form.

Step 3 — argue divisibility by 3.

Among any three consecutive integers, exactly one is divisible by 3 (since the residues $\bmod 3$ cycle through 0, 1, 2). Therefore one of $(n-1), n, (n+1)$ is a multiple of 3, so the product is divisible by 3.

B1 — clear reasoning that the product of three consecutive integers is divisible by 3. The modular-arithmetic justification is not required at A-Level, but stating the residue cycle gains credit for AO2 reasoning.

Step 4 — combine via coprimality.

Since $\gcd(2, 3) = 1$ , divisibility by 2 and by 3 together imply divisibility by $2 \times 3 = 6$ .

M1 — explicit appeal to the fact that 2 and 3 are coprime, so their product divides the expression.

A1 — concluding statement: "Therefore $n^3 - n$ is divisible by 6 for every integer $n$ , as required."

Step 5 — closing reasoning mark.

R1 — overall logical clarity: each step follows from the previous one, the quantifier "for every integer $n$ " is honoured throughout, and the proof avoids the common error of "checking small cases".

Total: 8 marks (M2 A2 B2 R1, plus a structure mark folded into the final A1).

Specimen question modelled on the AQA 7357 Paper 1 format

Question (6 marks): Prove by deduction that the sum of the squares of any two consecutive odd integers is even but is not divisible by 4.

Mark scheme decomposition by AO:

M1 (AO1.1a) — representing two consecutive odd integers algebraically as $2n + 1$ and $2n + 3$ for some integer $n$ .
M1 (AO1.1b) — squaring and adding: $(2n+1)^2 + (2n+3)^2 = 4n^2 + 4n + 1 + 4n^2 + 12n + 9 = 8n^2 + 16n + 10$ .
A1 (AO1.1b) — correctly simplified expression $8n^2 + 16n + 10$ .
M1 (AO2.1) — factoring as $2(4n^2 + 8n + 5)$ to demonstrate divisibility by 2.
R1 (AO2.4) — observing that $4n^2 + 8n + 5$ is odd (since $4n^2 + 8n$ is even and 5 is odd), so the expression is $2 \times \text{odd}$ , hence divisible by 2 but not by 4.
A1 (AO2.2a) — concluding sentence stating both halves of the claim and tying them back to the question.

Total: 6 marks split AO1 = 3, AO2 = 3. Proof questions are AO2-heavy because the reasoning is the answer — mere algebra without justification scores at most half marks.

Synoptic links

Connects to:

Every other Pure section. Algebra (section B): proving polynomial identities by expansion or factor theorem. Trigonometry (section E): proving identities such as $\tan\theta + \cot\theta \equiv \sec\theta\csc\theta$ . Sequences (section F): proving closed forms for arithmetic and geometric sums. Calculus (sections H, I): proving derivative identities or definite-integral results. Proof is the connective tissue of the specification.
Section B — Algebra and functions: divisibility proofs and identity proofs both rely on confident factorisation. The factorisation $n^3 - n = (n-1)n(n+1)$ is purely algebraic; the proof is the divisibility argument that follows.
Section E — Trigonometry: identity proofs ("show that ...") use the same logical structure as deductive proof — start from a known identity, manipulate one side, reach the other side, conclude. The grammar is identical; only the toolkit changes.
Section F — Sequences and series: proving formulae for $\sum_{r=1}^{n} r$ or $\sum_{r=1}^{n} r^2$ at A-Level uses direct manipulation; at Further Maths, proof by induction generalises this to any $n$ -dependent statement.
Pure 2 / section J — Proof by contradiction: the fourth proof method, building on the same logical apparatus introduced here. Classic example: proving $\sqrt{2}$ is irrational. The structural skills you build now — defining variables, quantifying carefully, concluding cleanly — transfer directly.

Mark-scheme literacy

Proof questions on AQA 7357 split AO marks heavily toward AO2:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	25–35%	Setting up algebraic representations (e.g. $2n + 1$ for odd), expanding brackets, applying known identities
AO2 (reasoning / interpretation)	55–65%	Justifying each step, quantifying correctly ("for all", "there exists"), concluding with a sentence that addresses the claim
AO3 (problem-solving)	5–15%	Choosing the right proof method (deduction vs exhaustion vs counter-example), structuring a multi-step argument

Examiner-rewarded phrasing: "Let $n$ be an integer"; "for all integers $n$ "; "as required"; "this contradicts our assumption"; "we have shown that ...". Phrases that lose marks: "this works for $n = 1, 2, 3$ so it must be true" (example-checking is not proof); "obviously"; "clearly"; missing concluding sentences. A bare algebraic line with no surrounding logical commentary will not score the AO2 reasoning marks even when the algebra is correct.

A specific AQA pattern to watch: questions phrased "Prove that ..." demand a full deductive argument with quantifier; "Show that ..." typically asks you to reach a printed expression and is more procedural; "Disprove ..." means a single counter-example suffices. Read the command word first.

Grade-band model answers

3-mark question

Question: Prove that the sum of any three consecutive integers is divisible by 3.

Mid-band response (~150 words):

Let the three consecutive integers be $n$ , $n+1$ , $n+2$ .

Sum: $n + (n+1) + (n+2) = 3n + 3 = 3(n+1)$ .

This is a multiple of 3, so the sum is divisible by 3.

Examiner commentary: Full marks (3/3). The candidate defines the variable cleanly, performs the algebra correctly, factors out the 3, and states the conclusion. The argument is complete despite being brief. The only stylistic improvement an A* candidate would make is the closing phrase "for all integers $n$ , as required" — explicitly honouring the quantifier and tying back to the question stem.

Top-band response (~210 words):

Let $n$ be any integer. Then the three consecutive integers can be written as $n$ , $n+1$ and $n+2$ .

Their sum is:

$n + (n+1) + (n+2) = 3n + 3 = 3(n + 1)$

Since $n + 1$ is an integer (because $n$ is), $3(n+1)$ is by definition a multiple of 3. Therefore the sum of any three consecutive integers is divisible by 3, as required.

Examiner commentary: Full marks (3/3). Beyond procedural correctness, the candidate quantifies explicitly ("Let $n$ be any integer"), justifies why $3(n+1)$ is a multiple of 3 (because $n+1$ is an integer), and closes with "as required" to signal that the original claim has been addressed. This is the difference between a proof that scores and one that scores while looking like a proof.

6-mark question

Question: Prove by exhaustion that for every integer $n$ with $1 \le n \le 5$ , the value $n^2 + n + 1$ is odd.

Stronger response (~190 words):

I will check each case.

$n = 1$ : $1 + 1 + 1 = 3$ , which is odd. $n = 2$ : $4 + 2 + 1 = 7$ , odd. $n = 3$ : $9 + 3 + 1 = 13$ , odd. $n = 4$ : $16 + 4 + 1 = 21$ , odd. $n = 5$ : $25 + 5 + 1 = 31$ , odd.

So the statement is true.

Examiner commentary: 4/6. The candidate has performed the exhaustive check correctly but has not opened with a sentence stating the strategy ("By exhaustion, I will check each of $n = 1, 2, 3, 4, 5$ "), nor closed with a sentence tying the cases back to the claim ("In each case the value is odd, so the statement holds for all integers $1 \le n \le 5$ "). The structural marks for AO2 reasoning are lost. The arithmetic and the method choice are correct, but proof-by-exhaustion answers must explicitly frame the exhaustion to score full marks.

Top-band response (~245 words):

The claim concerns five integers: $n = 1, 2, 3, 4, 5$ . Since the domain is finite, proof by exhaustion is appropriate — I will evaluate $n^2 + n + 1$ in each case and verify the result is odd.

$n$	$n^2 + n + 1$	Odd?
1	$1 + 1 + 1 = 3$	yes
2	$4 + 2 + 1 = 7$	yes
3	$9 + 3 + 1 = 13$	yes
4	$16 + 4 + 1 = 21$	yes
5	$25 + 5 + 1 = 31$	yes

In every case the expression is odd. Since the five cases exhaust the domain $1 \le n \le 5$ , the claim holds for all integers in this range, as required.

Examiner commentary: Full marks (6/6). The candidate opens by justifying the choice of method (finite domain → exhaustion is valid), tabulates the cases for clarity, and closes with a sentence that explicitly invokes the exhaustion principle ("the five cases exhaust the domain"). The AO2 reasoning marks are secured by the framing, not just the arithmetic. Note also: the candidate does not generalise to $n^2 + n = n(n+1)$ being even and so adding 1 makes the expression odd — that would be a deductive proof, not an exhaustion proof. Method discipline matters.

9-mark question

Question: (a) Prove by deduction that $n^2 + 2$ is never divisible by 4 for any integer $n$ . (5) (b) Hence, or otherwise, disprove the claim that "for every integer $n$ , $n^2 + 2$ is even". (4)

Top-band response (~340 words):

(a) Any integer $n$ is either even or odd. I will consider the two cases separately.

Case 1: $n$ is even. Write $n = 2k$ for some integer $k$ . Then:

$n^2 + 2 = (2k)^2 + 2 = 4k^2 + 2 = 4k^2 + 2$

This equals $4k^2 + 2$ , which leaves remainder 2 on division by 4. So $n^2 + 2$ is not divisible by 4 in this case.

Case 2: $n$ is odd. Write $n = 2k + 1$ for some integer $k$ . Then:

$n^2 + 2 = (2k+1)^2 + 2 = 4k^2 + 4k + 1 + 2 = 4k^2 + 4k + 3 = 4(k^2 + k) + 3$

This leaves remainder 3 on division by 4. So $n^2 + 2$ is not divisible by 4 in this case either.

In both cases, $n^2 + 2$ is not divisible by 4. Since every integer is either even or odd, the claim holds for all integers $n$ , as required.

(b) The claim " $n^2 + 2$ is even for every integer $n$ " requires testing.

Try $n = 1$ : $1^2 + 2 = 3$ , which is odd.

This single counter-example disproves the universal claim. Therefore the statement " $n^2 + 2$ is even for every integer $n$ " is false.

Examiner commentary: Full marks (9/9). Part (a) demonstrates exemplary case analysis: variable definition, parallel structure across the two cases, explicit modular interpretation ("remainder 2", "remainder 3"), and a closing sentence that invokes the dichotomy "every integer is even or odd". Part (b) shows method discipline — the candidate recognises that one counter-example suffices to disprove a universal claim and resists the temptation to "prove" it false by a longer argument. The contrast between (a) (prove for all) and (b) (disprove via one case) is exactly the AQA examiner's intended pedagogical contrast.

A-Level-depth misconceptions

The errors that distinguish A from A* on proof questions:

Example-based "proof". Checking $n = 1, 2, 3$ and concluding "so it works for all $n$ " is the most common A-Level proof error. A finite number of confirmations does not prove a universal claim. The cure: every proof for "all integers" must use a general variable $n$ , not specific values.
Missing case for "even AND odd" (or "neither"). A case analysis must cover every integer. The two-case split "even or odd" is exhaustive over the integers, but a three-case split "positive, negative, zero" forgets that zero is even, so combining the two splits incorrectly is a common error. Be explicit about which dichotomy you are using.
Circular reasoning. "Prove that $\sin^2\theta + \cos^2\theta = 1$ " — and the candidate divides both sides by $\cos^2\theta$ to get $\tan^2\theta + 1 = \sec^2\theta$ , then squares and adds to get back to the original. This proves nothing because the manipulation assumed the identity. The cure: start from a known result and derive the target, never the reverse without clear justification.
Confusing "show that" with "prove". "Show that $(n+1)^2 - n^2 = 2n + 1$ " is a procedural identity check; "Prove that the difference of consecutive squares is odd" demands a generalised argument with a concluding sentence. The marks split differently — read the command word.
Quantifier slippage. "There exists $n$ such that ..." (existential) needs only one example; "For all $n$ ..." (universal) needs a general argument, and disproving it needs only one counter-example. Confusing these flips the proof strategy and scores zero.
Algebraic equivalence vs proof. Showing two expressions are equal algebraically (e.g. by expanding) is not always enough — if the question demands "by deduction", you must walk through logical steps with stated reasons. Pure algebra without commentary loses AO2 marks.
Forgetting the conclusion. The final sentence "therefore the claim holds for all integers $n$ , as required" is not decoration: it is what earns the AO2.4 reasoning mark on most proof questions. Many otherwise-correct proofs lose 1 mark to a missing or weak concluding sentence.

Common errors and mark-loss patterns

Three patterns repeatedly cost candidates marks on Paper 1 proof questions. They are all about presentation, not technique.

No variable definition. Jumping straight into algebra without "Let $n$ be an integer" or "let the consecutive odd integers be $2n+1$ and $2n+3$ " loses the opening method mark. The variable definition is what makes the subsequent argument general rather than a specific calculation.
Skipping the concluding sentence. Reaching the algebraic conclusion (e.g. $3(n+1)$ , factored cleanly) and stopping without "therefore the sum is a multiple of 3" loses the final reasoning mark. The proof is incomplete without the verbal tie-back to the original claim.
Mishandling the disproof command. When asked to disprove a "for all" statement, candidates often try to prove the negation in full generality, when a single well-chosen counter-example suffices. Disproof is easier than proof — but only if you recognise the command and exploit it.
Treating "show that" results as evidence. Once a "show that" question yields a printed expression, candidates sometimes use that expression as if it were a known result in a later proof — but if the "show that" working contained an error, the later proof inherits it. Always re-derive critical steps if your earlier working was uncertain.

Going further — university and academic signposting

Proof technique points directly toward several undergraduate trajectories:

Formal logic and propositional calculus (Year 1): at university, proof is formalised as inference within a system of axioms and rules. Statements like "if $P$ then $Q$ " are dissected into truth tables, contrapositives ( $\neg Q \Rightarrow \neg P$ ), and converses, with explicit attention to which directions hold.
Peano axioms and the foundations of arithmetic (Year 1/2): the integers themselves are constructed from a tiny set of axioms (zero, successor, induction). Every theorem about $\mathbb{Z}$ — including " $n^2 - n$ is even" — ultimately reduces to a chain of Peano-style deductions.
Real analysis (Year 1): the rigorous $\varepsilon$ - $\delta$ definition of a limit demands proof structures of the form "for all $\varepsilon > 0$ , there exists $\delta > 0$ such that ...". A-Level proof discipline is the entry ticket.
Gödel's incompleteness theorems (Year 3+): Gödel showed that any formal system rich enough to express arithmetic contains true statements that cannot be proved within the system. The very nature of "what counts as proof" has known limits — and recognising those limits is part of mathematical maturity.

Oxbridge interview prompt: "Prove that there are infinitely many primes. Now prove that there are infinitely many primes of the form $4k + 3$ . What's different about the proof structure for primes of the form $4k + 1$ ?"

Additional A* practice: disproof by counter-example

A-Level students often underuse counter-examples. When a question presents a universal claim ("for every integer $n$ ...") and asks you to disprove it, one counter-example suffices — and the question is usually trying to reward you for finding it quickly rather than constructing a full negation argument.

Worked example: Disprove the claim "for every positive integer $n$ , the value $n^2 - n + 41$ is prime".

Try small values: $n = 1$ gives 41 (prime); $n = 2$ gives 43 (prime); $n = 3$ gives 47 (prime); $n = 10$ gives 131 (prime). The pattern holds for many initial values, which is precisely why this claim looks true — it is the famous Euler-style prime-generating polynomial pattern.

Try $n = 41$ : $41^2 - 41 + 41 = 41^2 = 1681 = 41 \times 41$ , which is not prime.

The single counter-example $n = 41$ disproves the universal claim. Therefore the statement is false.

Why A candidates spot this immediately:* they recognise that any polynomial of the form $f(n) = n^2 + bn + c$ with integer coefficients satisfies $f(c) = c^2 + bc + c = c(c + b + 1)$ , which is composite (divisible by $c$ ) whenever $c > 1$ . So no polynomial of this form can produce only primes for every $n$ .

A subtlety: counter-examples must be valid integers in the stated domain. Disproving "for every positive integer $n$ ..." with $n = 0$ or $n = -1$ does not count. Read the domain restriction precisely, then choose your counter-example from inside it.

AQA A-Level alignment footer

This content is aligned with the AQA A-Level Mathematics (7357) specification, Paper 1 — Pure Mathematics, Section A: Proof. For the most accurate and up-to-date information, please refer to the official AQA specification document.

Visual summary

graph TD
    A["Proof question:<br/>identify command word"] --> B{"Which command?"}
    B -->|"Prove (universal)"| C["Choose method:<br/>deduction or exhaustion"]
    B -->|"Disprove"| D["Find one<br/>counter-example<br/>in the stated domain"]
    C --> E["Define variables<br/>(e.g. Let n be an integer)"]
    E --> F["Algebra + reasoning<br/>+ closing sentence<br/>tying back to claim"]
    D --> F

Proof

Proof

Why Do We Need Proof?

Types of Proof

1. Proof by Deduction

2. Proof by Exhaustion

3. Proof by Contradiction

4. Disproof by Counter-Example

Algebraic Proof Techniques

Synoptic Links

Summary

A-Level Deep Dive: Proof

Spec mapping

Specimen Question (Modelled on the AQA Paper Format)

Worked example with full mark scheme

Specimen question modelled on the AQA 7357 Paper 1 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

6-mark question

9-mark question

A-Level-depth misconceptions

Common errors and mark-loss patterns

Going further — university and academic signposting

Additional A* practice: disproof by counter-example

AQA A-Level alignment footer

Visual summary

More in Mathematics