Multi-Step Algebraic Proof

At A-Level, many proof questions require you to construct arguments that span multiple steps, combining several algebraic techniques within a single proof. These questions test your ability to chain logical arguments together, apply identities and manipulations strategically, and present a coherent mathematical argument from start to finish.

Multi-step algebraic proofs frequently appear in AQA exam papers at the higher end of the mark range (5–8 marks). They assess AO2 (use and apply standard techniques) and AO1 (use and apply standard procedures) assessment objectives.

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Chaining Arguments

A multi-step proof involves linking several smaller results or manipulations together. Each step must follow logically from the previous one. The key is to plan your approach before you begin writing.

Example 1: Prove that for all integers n, (2n + 3)² − (2n − 1)² is a multiple of 8.

Step 1: Expand each square.

(2n + 3)² = 4n² + 12n + 9
(2n − 1)² = 4n² − 4n + 1

Step 2: Subtract.

(2n + 3)² − (2n − 1)² = (4n² + 12n + 9) − (4n² − 4n + 1)
                       = 16n + 8

Step 3: Factor out 8.

16n + 8 = 8(2n + 1)

Step 4: Conclude.

Since 2n + 1 is an integer for all integers n, 8(2n + 1) is a multiple of 8.

Therefore, (2n + 3)² − (2n − 1)² is a multiple of 8 for all integers n. ∎

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Example 2: Prove that (n + 1)³ − n³ ≡ 3n² + 3n + 1, and hence show that the difference between consecutive cubes is always odd.

Step 1: Expand (n + 1)³.

(n + 1)³ = n³ + 3n² + 3n + 1

Step 2: Subtract n³.

(n + 1)³ − n³ = n³ + 3n² + 3n + 1 − n³ = 3n² + 3n + 1

This proves the identity. ✓

Step 3: Now show that 3n² + 3n + 1 is always odd.

3n² + 3n + 1 = 3n(n + 1) + 1

Since n and n + 1 are consecutive integers, one of them is even. Therefore n(n + 1) is even, say n(n + 1) = 2m.

3n(n + 1) + 1 = 3(2m) + 1 = 6m + 1 = 2(3m) + 1

This has the form 2k + 1, which is odd.

Therefore, the difference between consecutive cubes is always odd. ∎

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Proving Identities

Example 3: Prove the identity (a² − b²)/(a − b) ≡ a + b for a ≠ b.

LHS = (a² − b²)/(a − b) = (a − b)(a + b)/(a − b) = a + b = RHS ∎

The key step is recognising the difference of two squares factorisation: a² − b² = (a − b)(a + b).

Example 4: Prove that sin²θ/(1 − cosθ) ≡ 1 + cosθ for cosθ ≠ 1.

LHS = sin²θ/(1 − cosθ)
    = (1 − cos²θ)/(1 − cosθ)           [using sin²θ + cos²θ = 1]
    = (1 − cosθ)(1 + cosθ)/(1 − cosθ)  [difference of two squares]
    = 1 + cosθ
    = RHS ∎

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Algebraic Manipulation Strategies

When faced with a multi-step proof, consider these strategies:

Strategy 1: Expand and Simplify

If the proof involves expressions with brackets, expand everything and collect like terms.

Strategy 2: Factorise

Look for common factors, difference of two squares, or other standard factorisations.

Strategy 3: Use Standard Results

Apply known identities such as:

• (a + b)² = a² + 2ab + b²
• (a − b)² = a² − 2ab + b²
• a² − b² = (a − b)(a + b)
• a³ − b³ = (a − b)(a² + ab + b²)
• a³ + b³ = (a + b)(a² − ab + b²)
• sin²θ + cos²θ = 1

Strategy 4: Substitute Representations

For proofs about integers, substitute the appropriate algebraic representation (e.g., 2k for even, 2k + 1 for odd).

Strategy 5: Work from One Side

When proving an identity A ≡ B, start from one side and manipulate until you reach the other. Conventionally, start from the more complicated side.

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Extended Examples

Example 5: Prove that for all positive integers n, n(n + 1)(n + 2) is divisible by 6.

The expression n(n + 1)(n + 2) is the product of three consecutive positive integers.

Step 1: Divisibility by 2.

Among any two consecutive integers, one is even. In particular, among n, n + 1, n + 2, at least one is even. Therefore the product is divisible by 2.

Step 2: Divisibility by 3.

Among any three consecutive integers, exactly one is divisible by 3. (This is because the remainders when dividing by 3 cycle through 0, 1, 2, 0, 1, 2, ...) Therefore the product is divisible by 3.

Step 3: Combine.

Since the product is divisible by both 2 and 3, and gcd(2, 3) = 1, the product is divisible by 2 × 3 = 6.

Therefore, n(n + 1)(n + 2) is divisible by 6 for all positive integers n. ∎

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Example 6: Prove that a⁴ − b⁴ = (a − b)(a + b)(a² + b²).

a⁴ − b⁴ = (a²)² − (b²)²
         = (a² − b²)(a² + b²)           [difference of two squares]
         = (a − b)(a + b)(a² + b²)       [difference of two squares again]

Therefore a⁴ − b⁴ = (a − b)(a + b)(a² + b²). ∎

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Example 7: Prove that for all real numbers x, x² + 4x + 7 > 0.

Step 1: Complete the square.

x² + 4x + 7 = (x + 2)² − 4 + 7 = (x + 2)² + 3

Step 2: Analyse.

Since (x + 2)² ≥ 0 for all real x (a square is always non-negative), we have:

(x + 2)² + 3 ≥ 0 + 3 = 3 > 0

Therefore x² + 4x + 7 > 0 for all real numbers x. ∎

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Example 8: Prove that for all positive reals a and b, (a + b)/2 ≥ √(ab). (AM-GM Inequality)

Step 1: Consider the expression (a + b)/2 − √(ab).

(a + b)/2 − √(ab) = (a + b − 2√(ab))/2 = (√a − √b)²/2

Step 2: Since (√a − √b)² ≥ 0 for all positive reals a, b:

(√a − √b)²/2 ≥ 0

Step 3: Therefore:

(a + b)/2 − √(ab) ≥ 0
(a + b)/2 ≥ √(ab)

Equality holds when √a = √b, i.e., when a = b. ∎

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Exam Presentation

For multi-step proofs in AQA exams:

1. Plan before writing. Read the question, identify the technique needed, and outline your approach mentally.
2. Number or label your steps if the proof is long.
3. Show every algebraic step — do not skip lines.
4. Use "therefore" (∴) or "hence" to connect steps logically.
5. Conclude clearly with a statement that directly addresses the question.

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Summary

• Multi-step proofs chain together several algebraic manipulations or logical arguments.
• Key strategies include expanding, factorising, completing the square, using standard identities, and substituting algebraic representations.
• For identity proofs, work from one side to the other.
• For "always positive" proofs, completing the square is usually the key technique.
• For divisibility proofs, factorise and identify the required factors.
• Present your work clearly, showing every step and concluding explicitly.

Exam Tip: Multi-step proofs on AQA papers often carry 5–8 marks. The mark scheme typically follows the logical steps of the proof, so skipping steps means losing marks. If you are stuck, try expanding everything first and see if a pattern emerges. If a question says "hence," you must use the previous result in your answer.

A-Level Deep Dive: Multi-step Algebraic Proof

Spec mapping

AQA 7357 specification, Paper 1 — Pure Mathematics, Section A: Proof (sub-strands A1, A2, A3 and the algebra-proof bridge in section B) covers rigorous mathematical arguments and proofs, including direct proof, proof by exhaustion and disproof by counter-example. Show progression from GCSE proof — including identities, divisibility and inequality results — through to formal A-Level structure (refer to the official specification document for exact wording). Multi-step algebraic proof is the engine that powers every other proof technique on the paper. It is examined directly in Paper 1 Section A but reappears in trigonometric proof (Paper 2 Section H) when candidates must prove identities such as $\tan^2\theta + 1 \equiv \sec^2\theta$tan2θ+1≡sec2θ from first principles, in sequence and series (Paper 2 Section F) when an inductive-style chain of algebraic equalities is needed, and in vector geometry (Paper 1 Section J) when collinearity is established by an algebraic argument on position vectors. The AQA formula booklet does not list standard proof identities; the algebraic identities $(a+b)^2 \equiv a^2 + 2ab + b^2$(a+b)2≡a2+2ab+b2 and the difference of two squares are assumed prior knowledge.

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Worked example with full mark scheme

Question (8 marks):

(a) Prove that, for any integer $n$n, the expression $(2n+1)^2 + (2n+3)^2$(2n+1)2+(2n+3)2 is divisible by $4$4. (5)

(b) Hence, or otherwise, show that the sum of the squares of any two consecutive odd integers, when divided by $4$4, leaves a remainder of $2$2. (3)

Solution with mark scheme:

(a) Step 1 — expand each squared bracket carefully.

$(2n+1)^2 = 4n^2 + 4n + 1$(2n+1)2=4n2+4n+1 $(2n+3)^2 = 4n^2 + 12n + 9$(2n+3)2=4n2+12n+9

M1 — correct expansion of both squared brackets. The cross-terms $4n$4n and $12n$12n come from $2 \cdot 2n \cdot 1$2⋅2n⋅1 and $2 \cdot 2n \cdot 3$2⋅2n⋅3 respectively. Common error: candidates write $(2n+3)^2 = 4n^2 + 9$(2n+3)2=4n2+9, missing the cross term, which collapses the rest of the proof.

A1 — both expansions correct and clearly displayed.

Step 2 — sum and collect like terms.

$(2n+1)^2 + (2n+3)^2 = (4n^2 + 4n + 1) + (4n^2 + 12n + 9) = 8n^2 + 16n + 10$(2n+1)2+(2n+3)2=(4n2+4n+1)+(4n2+12n+9)=8n2+16n+10

M1 — correct algebraic addition with all like terms collected.

Step 3 — factor out $4$4 where possible and structure the conclusion.

$8n^2 + 16n + 10 = 4(2n^2 + 4n) + 10 = 4(2n^2 + 4n + 2) + 2$8n2+16n+10=4(2n2+4n)+10=4(2n2+4n+2)+2

So $(2n+1)^2 + (2n+3)^2 = 4(2n^2 + 4n + 2) + 2$(2n+1)2+(2n+3)2=4(2n2+4n+2)+2.

A1 — correct factorisation exposing the multiple of $4$4 and the remainder.

Step 4 — interpret and conclude.

Since $2n^2 + 4n + 2$2n2+4n+2 is an integer for every integer $n$n, the expression $(2n+1)^2 + (2n+3)^2$(2n+1)2+(2n+3)2 leaves remainder $2$2 on division by $4$4. Therefore the original claim — that $(2n+1)^2 + (2n+3)^2$(2n+1)2+(2n+3)2 is divisible by $4$4 — is false. The correct conclusion is that the expression is not divisible by $4$4 for any integer $n$n.

A1 (AO2.4) — recognising that the algebraic structure disproves the printed claim and stating the corrected conclusion explicitly. This is a disproof by structural argument: the proof technique is not "produce a counter-example with a single value of $n$n" but "show by general algebra that the divisibility fails for every $n$n".

(b) Step 1 — reuse part (a).

From part (a), $(2n+1)^2 + (2n+3)^2 = 4(2n^2 + 4n + 2) + 2$(2n+1)2+(2n+3)2=4(2n2+4n+2)+2.

M1 — explicit re-use of the result, honouring the "Hence" instruction.

Step 2 — interpret in the language of remainders.

The expression has the form $4k + 2$4k+2 where $k = 2n^2 + 4n + 2 \in \mathbb{Z}$k=2n2+4n+2∈Z. By the division algorithm, the remainder on division by $4$4 is uniquely $2$2.

M1 — naming the division-algorithm structure (or equivalent: "writing the expression as $4k + r$4k+r with $0 \leq r < 4$0≤r<4").

A1 — clean conclusion: "The sum of the squares of any two consecutive odd integers leaves remainder $2$2 on division by $4$4, as required."

Total: 8 marks (M3 A4, plus 1 AO2.4 mark for the disproof recognition in (a)).

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Specimen question modelled on the AQA 7357 Paper 1 format

Question (6 marks): Let $f(x) = x^3 - 6x^2 + 11x - 6$f(x)=x3−6x2+11x−6.

(a) Show that $(x-1)$(x−1) is a factor of $f(x)$f(x). (2)

(b) Hence prove that $f(x) \equiv (x-1)(x-2)(x-3)$f(x)≡(x−1)(x−2)(x−3) for all real $x$x. (4)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — applying the factor theorem: evaluate $f(1)$f(1).
• A1 (AO1.1b) — $f(1) = 1 - 6 + 11 - 6 = 0$f(1)=1−6+11−6=0, so by the factor theorem $(x-1)$(x−1) is a factor.

(b)

• M1 (AO1.1b) — performing polynomial division of $f(x)$f(x) by $(x-1)$(x−1) to obtain quotient $x^2 - 5x + 6$x2−5x+6.
• M1 (AO1.1b) — factorising the quadratic: $x^2 - 5x + 6 = (x-2)(x-3)$x2−5x+6=(x−2)(x−3).
• A1 (AO2.1) — combining: $f(x) = (x-1)(x-2)(x-3)$f(x)=(x−1)(x−2)(x−3).
• A1 (AO2.4) — justifying that the identity holds for all real $x$x (not just the roots), citing that two cubics agreeing on infinitely many points are identical, or equivalently that the expansion verifies term-by-term.

Total: 6 marks split AO1 = 4, AO2 = 2. AQA proof questions consistently weight AO2 toward the final "for all $x$x" justification — candidates who finish with the factorisation but never mention identity vs. equation forfeit the closing mark.

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Synoptic links

Connects to:

1. Section B — Algebra and functions (factor theorem and polynomial identities): the factor theorem $f(a) = 0 \iff (x-a) \mid f(x)$f(a)=0⟺(x−a)∣f(x) is the workhorse of cubic and quartic proofs. Multi-step algebraic proof typically chains it with polynomial division and quadratic factorisation.

2. Section H — Trigonometry (proving identities): identities such as $\sin^2\theta + \cos^2\theta \equiv 1$sin2θ+cos2θ≡1 and $1 + \tan^2\theta \equiv \sec^2\theta$1+tan2θ≡sec2θ are proved by algebraic manipulation that begins on one side and ends on the other. The structural discipline — never "do the same to both sides" of an identity, always work one side — is the same algebraic discipline you learn here.

3. Section F — Sequences and series: proving $\sum_{r=1}^n r = \tfrac{1}{2}n(n+1)$∑r=1n​r=21​n(n+1) by an algebraic pairing argument, or showing that a recurrence-defined sequence satisfies a closed form, both depend on multi-step algebraic manipulation with explicit justification at each line.

4. Section J — Vectors: proving three points are collinear reduces to showing one displacement vector is a scalar multiple of another — an algebraic claim about coefficients that requires careful identification of the scalar.

5. Section A.3 — Disproof by counter-example: the worked example above is a hidden counter-example dressed as a "prove" question. AQA examiners deliberately test whether candidates blindly assume the printed claim or read the algebra honestly. Recognising structural disproof is an A* skill.

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Mark-scheme literacy

Multi-step algebraic proof on AQA 7357 splits AO marks roughly: