Proof by Deduction

Proof by deduction — also known as direct proof — is the most fundamental and widely used method of proof in A-Level Mathematics. It involves starting from known facts, definitions, axioms, or previously established results, and applying logical reasoning step by step until you reach the conclusion you wish to establish. Every step in a deductive proof must follow logically from the previous one.

The AQA A-Level Mathematics specification (7357) requires students to "construct and present mathematical arguments through appropriate use of diagrams; sketching graphs; logical deduction; precise statements involving correct use of symbols and connecting language." Proof by deduction is the backbone of this requirement.

The Structure of a Deductive Proof

A deductive proof typically follows this pattern:

State clearly what you are trying to prove.
Introduce variables and define them precisely.
Perform a sequence of logical algebraic steps, each justified by known results.
Arrive at the required conclusion.
Write a clear concluding statement.

Exam Tip: Always define your variables. If you write "Let n be an integer," the examiner knows you understand what you are working with. Never assume the reader knows what your variables represent.

Algebraic Representations for Proof

Before you can construct a deductive proof about properties of integers, you must be able to represent those properties algebraically. The following representations are essential:

Property	Algebraic Form
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, ...
Consecutive even numbers	2n, 2n + 2, 2n + 4, ...
Consecutive odd numbers	2n + 1, 2n + 3, 2n + 5, ...
A multiple of k	kn, where n is an integer
A number that leaves remainder r when divided by k	kn + r
A perfect square	n², where n is an integer

Important: When proving results involving two different integers, use different letters. For example, represent two even numbers as 2a and 2b, not 2n and 2n — the latter would force them to be the same number.

Proving Results About Even and Odd Numbers

Example 1: Prove that the sum of two even numbers is always even.

Let the two even numbers be 2a and 2b, where a and b are integers.

2a + 2b = 2(a + b)

Since a + b is an integer (the integers are closed under addition), 2(a + b) is a multiple of 2, and therefore even.

Therefore, the sum of two even numbers is always even. ∎

Example 2: Prove that the sum of two odd numbers is always even.

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

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

Since a + b + 1 is an integer, 2(a + b + 1) is a multiple of 2, and therefore even.

Therefore, the sum of two odd numbers is always even. ∎

Example 3: 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 2ab + a + b is an integer, the product has the form 2k + 1 for integer k, which is odd.

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

Proving Results About Divisibility

Example 4: Prove that for any integer n, n³ − n is divisible by 6.

n³ − n = n(n² − 1) = n(n − 1)(n + 1) = (n − 1)n(n + 1)

This is the product of three consecutive integers. Among any three consecutive integers:

At least one is divisible by 2 (since every other integer is even).
At least one is divisible by 3 (since every third integer is a multiple of 3).

Therefore, the product is divisible by both 2 and 3. Since 2 and 3 are coprime, the product is divisible by 2 × 3 = 6.

Therefore, n³ − n is divisible by 6 for all integers n. ∎

Example 5: Prove that for any integer n, n² + n is always even.

n² + n = n(n + 1)

This is the product of two consecutive integers. Of any two consecutive integers, exactly one is even. Therefore the product contains a factor of 2 and is even.

Therefore, n² + n is always even. ∎

Proving Algebraic Identities

Example 6: Prove that (a + b)² − (a − b)² ≡ 4ab.

LHS = (a + b)² − (a − b)²
    = (a² + 2ab + b²) − (a² − 2ab + b²)
    = a² + 2ab + b² − a² + 2ab − b²
    = 4ab
    = RHS ∎

Example 7: Prove that (n + 3)² − (n + 1)² = 4(n + 2).

LHS = (n + 3)² − (n + 1)²
    = (n² + 6n + 9) − (n² + 2n + 1)
    = 4n + 8
    = 4(n + 2)
    = RHS ∎

Exam Tip: When proving an identity, always start from one side (usually the LHS) and manipulate it until you reach the other side. Never work from both sides simultaneously towards a middle expression — this is not logically valid as a proof.

Proving Results About Squares

Example 8: Prove that if n is even, then n² is even.

Let n = 2k, where k is an integer.

n² = (2k)² = 4k² = 2(2k²)

Since 2k² is an integer, n² = 2(2k²) is even.

Therefore, if n is even, then n² is even. ∎

Example 9: Prove that if n is odd, then n² is odd.

Let n = 2k + 1, where k is an integer.

n² = (2k + 1)² = 4k² + 4k + 1 = 2(2k² + 2k) + 1

Since 2k² + 2k is an integer, n² has the form 2m + 1, which is odd.

Therefore, if n is odd, then n² is odd. ∎

Common Pitfalls in Deductive Proof

Using the same variable for different quantities. Writing "Let the two even numbers be 2n and 2n" forces them to be the same number.
Assuming what you are trying to prove. This is circular reasoning. You must start from known facts and derive the conclusion.
Checking specific cases instead of proving generally. Showing that a result works for n = 1, 2, 3 is not a proof — it is verification of particular cases.
Missing the concluding statement. Always write a sentence at the end that directly states what has been proved.
Not defining variables. Always state "Let n be an integer" or similar before using a variable.

Summary

Proof by deduction (direct proof) uses logical steps from known facts to reach a conclusion.
Represent even numbers as 2n, odd numbers as 2n + 1, and use different letters for different quantities.
For divisibility proofs, factorise the expression and identify factors.
For identity proofs, work from one side to the other.
Always define your variables, show every algebraic step, and write a clear concluding statement.
Three consecutive integers always include a multiple of 2 and a multiple of 3, so their product is divisible by 6.

Exam Tip: In AQA exam papers, proof by deduction questions often ask you to "Prove that..." followed by an algebraic statement. Read the question carefully — if it says "for all integers n," you must give a general proof, not check specific values. Structure your proof clearly: define variables, perform algebraic manipulation, and conclude with a sentence that refers back to the original statement.

A-Level Deep Dive: Proof by Deduction

Spec mapping

AQA 7357 specification, Paper 1 — Pure Mathematics, Section A: Proof, sub-strands A1 and A2 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 (refer to the official specification document for exact wording). Proof by deduction is the gateway technique for the whole specification — it underwrites every "show that" and "prove that" instruction across Papers 1, 2 and 3. It is examined directly in Section A (Proof) but appears synoptically in Section B (Algebra and functions, where identities such as $(a + b)^2 = a^2 + 2ab + b^2$ require deductive justification), Section E (Trigonometry, where identities such as $\sin^2\theta + \cos^2\theta = 1$ are derived deductively) and Section J (Sequences and series, where closed forms are proved from recurrence relations). The AQA formula booklet provides standard identities, but the deductive chain connecting them must be produced by the candidate.

Worked example with full mark scheme

Question (8 marks):

(a) Prove by deduction that the sum of any two odd integers is even. (4)

(b) Prove by deduction that for any integer $n$ , $(2n+1)^2 - (2n-1)^2 = 8n$ . (4)

Solution with mark scheme:

(a) Step 1 — define what "odd" means algebraically.

Let the two odd integers be $2a + 1$ and $2b + 1$ , where $a, b \in \mathbb{Z}$ .

B1 — correct algebraic representation of any odd integer. Common error: writing the two integers as $2n + 1$ and $2n + 1$ (same letter), which would only prove the result for equal odd integers. Using two distinct letters is essential.

Step 2 — form the sum.

$(2a + 1) + (2b + 1) = 2a + 2b + 2$

M1 — correct addition with terms collected.

Step 3 — factor out 2.

$2a + 2b + 2 = 2(a + b + 1)$

M1 — extracting the factor of 2 from the sum.

Step 4 — conclude.

Since $a + b + 1 \in \mathbb{Z}$ (the integers are closed under addition), the expression $2(a + b + 1)$ is an even integer by definition.

A1 — explicit conclusion citing closure of $\mathbb{Z}$ under addition and the definition of "even".

(b) Step 1 — expand both squares.

$(2n + 1)^2 = 4n^2 + 4n + 1$ $(2n - 1)^2 = 4n^2 - 4n + 1$

M1 — correct expansion of both brackets. The cross term $\pm 4n$ is the load-bearing piece; sign errors here collapse the whole proof.

Step 2 — subtract.

$(2n + 1)^2 - (2n - 1)^2 = (4n^2 + 4n + 1) - (4n^2 - 4n + 1)$

M1 — correct subtraction with brackets retained around the second expansion. Forgetting the bracket and writing $4n^2 + 4n + 1 - 4n^2 - 4n + 1$ is the canonical sign-error trap.

Step 3 — simplify.

$= 4n^2 + 4n + 1 - 4n^2 + 4n - 1 = 8n$

A1 — correct simplification.

Step 4 — conclude.

Hence $(2n + 1)^2 - (2n - 1)^2 = 8n$ for all integers $n$ , as required.

A1 — explicit "as required" closure naming the universal quantifier.

Total: 8 marks (B1 M2 A1 + M2 A2).

Specimen question modelled on the AQA 7357 Paper 1 format

Question (6 marks): Prove by deduction that for all real $x$ , $x^2 - 6x + 10 > 0$ . (6)

Mark scheme decomposition by AO:

M1 (AO2.1) — recognising that completing the square is the appropriate deductive route: rewrite $x^2 - 6x + 10$ as $(x - 3)^2 + k$ for some constant $k$ .
M1 (AO1.1a) — completing the square: $x^2 - 6x + 10 = (x - 3)^2 - 9 + 10 = (x - 3)^2 + 1$ .
A1 (AO1.1b) — correct completed-square form $(x - 3)^2 + 1$ .
M1 (AO2.2a) — citing the property $(x - 3)^2 \geq 0$ for all real $x$ (square of a real number is non-negative).
A1 (AO2.4) — combining: $(x - 3)^2 + 1 \geq 0 + 1 = 1 > 0$ , hence $x^2 - 6x + 10 > 0$ .
A1 (AO2.5) — explicit closing statement with the universal quantifier "for all real $x$ " and "as required".

Total: 6 marks split AO1 = 2, AO2 = 4. This is an AO2-heavy question — AQA uses proof items primarily to test reasoning, with procedural fluency (completing the square) earning only the foundation marks.

Synoptic links

Connects to:

Section B — Algebra and functions: completing the square, factorising, and manipulating polynomial identities are the algebraic engine of most deductive proofs. Without confident algebra, deduction stalls at Step 2.
Number theory (foundational): proofs about odd/even integers, divisibility, and modular structure ( $n \equiv 0, 1, 2 \pmod{3}$ ) recur throughout the proof section. The result " $n^3 - n$ is divisible by 6" sits at this intersection.
Section E — Trigonometry: proving identities such as $\sin(A + B) = \sin A \cos B + \cos A \sin B$ from the unit-circle definition, or $\tan^2\theta + 1 = \sec^2\theta$ from $\sin^2 + \cos^2 = 1$ , is pure deduction dressed in trigonometric clothing.
Section J — Sequences and series: deriving the closed form $\sum_{r=1}^{n} r = \tfrac{n(n+1)}{2}$ by pairing terms is a classic deductive argument (often attributed to Gauss). Proof by deduction precedes proof by induction in the AQA scheme of work.
Section A2 — Disproof by counter-example and proof by contradiction: deduction is the constructive counterpart to these alternative methods. Knowing when each technique is appropriate is itself an AO2 reasoning skill.

Mark-scheme literacy

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

AO	Typical share	Earned by
AO1 (knowledge / procedure)	20–35%	Algebraic manipulation, expansion, factorisation, applying standard identities
AO2 (reasoning / interpretation)	55–70%	Selecting the deductive route, justifying each logical step, citing closure properties, writing the closing statement
AO3 (problem-solving)	5–15%	Choosing between deduction, contradiction, exhaustion or counter-example for unfamiliar prompts

Examiner-rewarded phrasing: "let $n$ be an arbitrary integer"; "since $\mathbb{Z}$ is closed under addition"; "for all real $x$ "; "as required" or "QED" at the close. Phrases that lose marks: starting "assume the result is true" (that is contradiction or induction, not deduction); using a single numerical example as the "proof"; ending without a closing statement; using the same letter for two arbitrary objects (e.g. $2n + 1$ for both odd integers in part (a) above).

A specific AQA pattern to watch: questions phrased "prove that for all integers $n$ " demand a general algebraic argument from the start. Substituting $n = 1, 2, 3$ to "check the pattern" is verification, not proof, and earns zero marks for the deductive structure.

Grade-band model answers

3-mark question

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

Grade C response (~150 words):

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

Their sum is $n + (n + 1) + (n + 2) = 3n + 3 = 3(n + 1)$ .

Since $n + 1$ is an integer, $3(n + 1)$ is divisible by 3.

Examiner commentary: Full marks (3/3). The candidate represents three arbitrary consecutive integers correctly, sums and factorises, and concludes with explicit divisibility. The argument is brief but every step is justified. Many candidates lose marks here by using $n - 1$ , $n$ , $n + 1$ (still correct) but failing to write the closing statement that $n + 1 \in \mathbb{Z}$ — the closure-of-integers citation is the AO2 reasoning mark.

Grade A response (~180 words):*

Let $n \in \mathbb{Z}$ . Then three consecutive integers may be written as $n$ , $n + 1$ , $n + 2$ .

Their sum is:

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

Since $\mathbb{Z}$ is closed under addition, $n + 1 \in \mathbb{Z}$ . Therefore $3(n + 1)$ is an integer multiple of 3, and hence the sum is divisible by 3 for any choice of starting integer $n$ , as required.

Examiner commentary: Full marks (3/3). The "let $n \in \mathbb{Z}$ " opening fixes the universal quantifier explicitly. Closure of $\mathbb{Z}$ is named, not assumed. The closing "for any choice of starting integer $n$ , as required" returns to the universal claim of the question. This is the kind of self-monitoring phrasing that signals deductive maturity and matters more on longer proofs.

6-mark question

Question: Prove by deduction that for all integers $n$ , $n^2 + n$ is even. Then deduce that $n(n + 1)(n + 2)$ is divisible by 2 for all integers $n$ .

Grade B response (~240 words):

$n^2 + n = n(n + 1)$ . Either $n$ is even or $n$ is odd.

If $n$ is even, then $n(n + 1)$ has an even factor and is even. If $n$ is odd, then $n + 1$ is even, so $n(n + 1)$ has an even factor and is even. Either way, $n^2 + n$ is even.

For $n(n + 1)(n + 2)$ : this contains $n(n + 1)$ which we just proved is even, multiplied by $(n + 2)$ . So the whole product is even.

Examiner commentary: Method correct, secured 4/6. The case-split is clean and the closure on $n^2 + n$ is explicit. The deduction for $n(n + 1)(n + 2)$ is logically sound but the candidate doesn't explicitly cite "an even integer times any integer is even" — that's the missing AO2 line. They also don't write a closing statement. An A* candidate would name the closure rule used and end with "as required". 4/6.

Grade A response (~290 words):*

Consider $n^2 + n = n(n + 1)$ . We argue by case analysis on the parity of $n$ .

Case 1: $n$ is even. Then $n = 2k$ for some $k \in \mathbb{Z}$ , so $n(n + 1) = 2k(n + 1)$ , which is divisible by 2.

Case 2: $n$ is odd. Then $n + 1 = 2m$ for some $m \in \mathbb{Z}$ (since the successor of an odd integer is even), so $n(n + 1) = 2mn$ , which is divisible by 2.

In both cases $n(n + 1)$ is even, so $n^2 + n$ is even for all $n \in \mathbb{Z}$ .

For the deduction: $n(n + 1)(n + 2) = (n^2 + n)(n + 2)$ . By the result just proved, $n^2 + n = 2t$ for some $t \in \mathbb{Z}$ . Therefore:

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

Since $t(n + 2) \in \mathbb{Z}$ by closure, $2t(n + 2)$ is divisible by 2 for all $n \in \mathbb{Z}$ , as required.

Examiner commentary: Full marks (6/6). The case split is labelled and each case produces an explicit divisibility witness ( $2k$ , $2m$ ). The "by the result just proved" linking phrase makes the deductive structure visible to the examiner. Closure of $\mathbb{Z}$ is invoked twice and named once. The final "as required" closes the universal claim. This is examination craft — the maths is the same as the Grade B response, but every reasoning step is visibly justified.

9-mark question

Question: (a) Prove by deduction that for any integer $n$ , the expression $n^3 - n$ is divisible by 6. (6) (b) Hence, or otherwise, prove that the product of any three consecutive integers is divisible by 6. (3)

Grade A response (~360 words):*

(a) Factor: $n^3 - n = n(n^2 - 1) = n(n - 1)(n + 1) = (n - 1)n(n + 1)$ .

This is the product of three consecutive integers.

Divisibility by 2: Among any two consecutive integers, exactly one is even. So among $(n - 1), n, (n + 1)$ at least one is even, meaning the product has a factor of 2.

Divisibility by 3: Among any three consecutive integers, exactly one is divisible by 3 (since the residues modulo 3 cycle through $0, 1, 2$ ). So among $(n - 1), n, (n + 1)$ exactly one is a multiple of 3, meaning the product has a factor of 3.

Since the product is divisible by both 2 and 3, and $\gcd(2, 3) = 1$ , it is divisible by $2 \times 3 = 6$ .

Hence $n^3 - n$ is divisible by 6 for all $n \in \mathbb{Z}$ , as required.

(b) Let the three consecutive integers be $m, m + 1, m + 2$ for some $m \in \mathbb{Z}$ . Set $n = m + 1$ . Then:

$(n - 1)n(n + 1) = m(m + 1)(m + 2)$

By part (a), $(n - 1)n(n + 1) = n^3 - n$ is divisible by 6. Hence $m(m + 1)(m + 2)$ is divisible by 6 for all $m \in \mathbb{Z}$ , as required.

Examiner commentary: Full marks (9/9). Part (a) uses the cleanest deductive route — factor first, then apply two independent divisibility arguments (the 2-argument and the 3-argument), then combine using coprimality. The citation $\gcd(2, 3) = 1$ is the AO2 mark — without it, "divisible by 2 and by 3 implies divisible by 6" is asserted, not proved. (Counter-example: 4 is divisible by 2 and by 4, but not by 8.) Part (b) honours the "Hence" by reusing part (a) rather than re-proving from scratch — the substitution $n = m + 1$ is the load-bearing move. Total: 9/9.

A-Level-depth misconceptions

The errors that distinguish A from A* on proof-by-deduction questions:

Example-based "proof". Substituting $n = 1, 2, 3$ and checking the pattern is verification, not proof. A finite check cannot establish a universal claim. Examiners award zero marks for the deductive structure when this is the only argument offered.
Circular reasoning. Assuming what you're trying to prove and then deriving it is logically vacuous. The classic version: "prove $\sin^2\theta + \cos^2\theta = 1$ " answered by " $\sin^2\theta + \cos^2\theta = 1$ by Pythagoras, so $\sin^2\theta + \cos^2\theta = 1$ ".
Missing case analysis. Proofs about parity, divisibility, or modular structure often require splitting into cases ( $n$ even / $n$ odd, or $n \equiv 0, 1, 2 \pmod 3$ ). Failing to enumerate all cases leaves a gap in the argument.
Same-letter slip for arbitrary objects. Writing "let two odd integers be $2n + 1$ and $2n + 1$ " (same $n$ ) only proves the result for equal odd integers, not arbitrary pairs. Use distinct letters for distinct arbitrary objects.
Missing closing statement. A proof without a closing statement ("hence ... as required" or "QED") leaves the examiner uncertain whether the candidate has recognised the universal claim is established. The A1 reasoning mark is regularly lost here.
Confusing deduction with contradiction. Beginning a deduction proof with "assume the result is false" is the wrong technique — that's proof by contradiction. The opening of a deduction is "let $n$ be an arbitrary integer" or "consider the expression ...".
"And" vs "implies" slip. In divisibility, "divisible by 2 and divisible by 3" implies "divisible by 6" only because $\gcd(2, 3) = 1$ . Without this citation, the inference is asserted not proved. The same trap appears for divisibility by 4 and 6 (which does not imply divisibility by 24).

Common errors and mark-loss patterns on proof-by-deduction questions

Four patterns repeatedly cost candidates marks on Paper 1 proof questions. They are all about presentation and logical structure, not algebraic technique.

Verification masquerading as proof. Candidates substitute three numerical values, observe the pattern holds, and write "hence proved". This earns zero marks for the deduction. The cure: open with "let $n \in \mathbb{Z}$ " or equivalent and proceed algebraically from the start.
Hidden case omissions. When the proof requires a case split, candidates often handle the "obvious" case ( $n$ even) and forget the other ( $n$ odd), or vice versa. The cure: label cases explicitly ("Case 1: $n$ even", "Case 2: $n$ odd") and check both before concluding.
Closure-property silence. Stating " $3(n + 1)$ is divisible by 3" is fine if $n + 1$ is named as an integer. Without that citation, the divisibility claim is incomplete. The cure: write "since $n \in \mathbb{Z}$ , $n + 1 \in \mathbb{Z}$ " before invoking divisibility.
Closing-statement silence. Even a perfect chain of deductions loses the final A1 if there is no closing line returning to the original claim. The cure: end every proof with a line of the form "hence $P(n)$ holds for all $n \in \mathbb{Z}$ , as required".

This pattern is endemic to Paper 1 proof questions: candidates know the algebra, lose marks on logical scaffolding.

Going further — university and academic signposting

Proof by deduction points directly toward several undergraduate trajectories:

Formal logic (Year 1): the deductive system is formalised as natural deduction or sequent calculus, with rules of inference (modus ponens, universal generalisation) replacing intuitive reasoning. Every A-Level proof can in principle be rendered as a tree of formal inference steps.
Peano axioms and the foundations of arithmetic: the closure properties of $\mathbb{Z}$ that A-Level proofs cite implicitly ( $a + b \in \mathbb{Z}$ when $a, b \in \mathbb{Z}$ ) are theorems derived from the Peano axioms for the natural numbers. The construction $\mathbb{N} \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{R}$ is the spine of a first-year analysis course.
Euclid's Elements: the model for axiomatic deduction is over two thousand years old. Euclid's Book I begins with five postulates and derives 48 propositions deductively. Reading Proposition 1 (constructing an equilateral triangle on a given segment) shows the same deductive scaffolding A-Level rewards in compact form.
Galois theory (Year 3): classical impossibility theorems — that $\sqrt[3]{2}$ is not constructible, that the general quintic has no formula in radicals — are deductive proofs of impossibility, building on the same logical structure as A-Level deduction but with much richer algebraic machinery.

Oxbridge interview prompt: "Prove that there are infinitely many primes. Now extend the argument to primes of the form $4k + 3$ . What goes wrong if you try to extend to primes of the form $4k + 1$ ?"

Additional A* practice: divisibility under stress

A common A* trap on AQA 7357 Paper 1 is to give a divisibility prompt that looks like a single-step factorisation but requires combining two coprime divisibility arguments.

Worked example: Prove by deduction that $n^3 - n$ is divisible by 6 for all integers $n$ .

Factor: $n^3 - n = n(n^2 - 1) = (n - 1)n(n + 1)$ — the product of three consecutive integers.

Argument 1 (divisibility by 2): Among any two consecutive integers, exactly one is even. Since $(n - 1), n, (n + 1)$ contains two pairs of consecutive integers, at least one of them is even. Hence the product is divisible by 2.

Argument 2 (divisibility by 3): The integers $(n - 1), n, (n + 1)$ have residues modulo 3 that cover $\{0, 1, 2\}$ in some order (since they are three consecutive integers). Hence exactly one of them is $\equiv 0 \pmod 3$ , and the product is divisible by 3.

Combining: Since $\gcd(2, 3) = 1$ , divisibility by both 2 and 3 implies divisibility by $2 \times 3 = 6$ . Hence $n^3 - n$ is divisible by 6 for all $n \in \mathbb{Z}$ , as required.

Why A candidates spot this immediately:* the structure "product of $k$ consecutive integers" signals divisibility by $k!$ — a result derivable by combining $\lfloor k/p \rfloor$ -type counting arguments for each prime $p \leq k$ . The same trick proves the product of four consecutive integers is divisible by 24, and the product of five is divisible by 120. Recognising the pattern across these contexts is the synoptic skill AQA rewards.

A subtlety: the coprimality citation $\gcd(2, 3) = 1$ is essential. "Divisible by 2 and by 4" does not imply "divisible by 8" — the integer 12 is divisible by 2 and 4 but not by 8. Always check the divisors are coprime before multiplying.

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["Statement to prove<br/>(universal claim)"] --> B{"Choose method"}
    B -->|"Direct argument<br/>from definitions"| C["Proof by deduction"]
    B -->|"Show one failure"| D["Disproof by<br/>counter-example"]
    B -->|"Assume negation,<br/>derive contradiction"| E["Proof by<br/>contradiction"]
    C --> F["Step 1:<br/>introduce arbitrary object<br/>’let n in Z’"]
    F --> G["Step 2:<br/>algebraic manipulation<br/>using definitions"]
    G --> H{"Case split<br/>needed?"}
    H -->|"Yes"| I["Enumerate cases<br/>(parity, mod p, etc.)"]
    H -->|"No"| J["Continue chain<br/>of deductions"]
    I --> J
    J --> K["Step 3:<br/>cite closure properties<br/>of Z, Q, R"]
    K --> L["Step 4:<br/>closing statement<br/>’hence ... as required’"]

    style C fill:#27ae60,color:#fff
    style L fill:#3498db,color:#fff

Proof by Deduction

Proof by Deduction

The Structure of a Deductive Proof

Algebraic Representations for Proof

Proving Results About Even and Odd Numbers

Example 1: Prove that the sum of two even numbers is always even.

Example 2: Prove that the sum of two odd numbers is always even.

Example 3: Prove that the product of two odd numbers is always odd.

Proving Results About Divisibility

Example 4: Prove that for any integer n, n³ − n is divisible by 6.

Example 5: Prove that for any integer n, n² + n is always even.

Proving Algebraic Identities

Example 6: Prove that (a + b)² − (a − b)² ≡ 4ab.

Example 7: Prove that (n + 3)² − (n + 1)² = 4(n + 2).

Proving Results About Squares

Example 8: Prove that if n is even, then n² is even.

Example 9: Prove that if n is odd, then n² is odd.

Common Pitfalls in Deductive Proof

Summary

A-Level Deep Dive: Proof by Deduction

Spec mapping

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 on proof-by-deduction questions

Going further — university and academic signposting

Additional A* practice: divisibility under stress

AQA A-Level alignment footer

Visual summary

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