AQA A-Level Mathematics: A: Proof (7357) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

This set covers section 3.2 A: Proof (content reference A1) of the AQA A-level Mathematics (7357) specification, version 1.3 (31 January 2018), for A-level exams from June 2018 onwards: proof by deduction, proof by exhaustion, disproof by counter-example and proof by contradiction, including unfamiliar proofs. 7357 is a linear A-level with no AS/A split. Proof is listed in the Paper 1 content, and Papers 2 and 3 can assess any Paper 1 content. A calculator is required in all 7357 papers, but every mark here is for written reasoning.

Before you start, read the Proof study guide and the Proof revision notes. Course links: AQA A-level Mathematics hub and printable checklist. For notation and argument questions, try the Overarching themes practice questions.

Questions

1. A student claims: “If p is prime, then 2ᵖ − 1 is prime.” Find a counter-example to show that the claim is false. [2]

2. Prove that the sum of any three consecutive odd numbers is a multiple of 3. [3]

3. Prove that 4x² − 12x + 10 > 0 for all real values of x. [3]

4. By considering n even and n odd, prove that n² + n + 1 is odd for every integer n. [3]

5. Prove by exhaustion that every cube of an integer is of the form 9k, 9k + 1 or 9k − 1, where k ∈ ℤ. [5]

6. Prove by contradiction that if n² is odd, then n is odd (n ∈ ℤ). [3]

7. Prove by contradiction that the sum of a rational number and an irrational number is irrational. [4]

8. Prove by contradiction that log₂ 5 is irrational. [4]

9. Prove that there are no integers m and n such that m² − n² = 10. [5]

10.

(a) Prove by exhaustion that if n is an integer that is not a multiple of 3, then n² is not a multiple of 3. [3] (b) Hence prove by contradiction that √3 is irrational. [5]

11. For each statement, prove it if it is true, or give a counter-example if it is false.

(a) x² ≥ x for all real x. [2] (b) n³ − n is a multiple of 6 for every integer n. [3] (c) If a and b are both irrational, then ab is irrational. [2] (d) (a + b)/2 ≥ √(ab) for all positive real numbers a and b. [3]

12. A student writes this argument.

Claim: the sum of any two odd numbers is even.
Let the odd numbers be 2n + 1 and 2n + 1.
(2n + 1) + (2n + 1) = 4n + 2 = 2(2n + 1), which is even.
So the sum of any two odd numbers is even.

(a) Explain the error in the argument. [1] (b) Write a correct proof of the claim. [3]

Answers

1. Take p = 11, which is prime [1]. 2¹¹ − 1 = 2047 = 23 × 89, which is not prime, so the claim is false [1]. Examiner insight: the counter-example mark needs the factorisation (or a factor) shown; writing “2047 is not prime” with no factor is usually not enough to show the claim fails.

2. Let the numbers be 2n + 1, 2n + 3 and 2n + 5, where n ∈ ℤ [1]. Sum = 6n + 9 = 3(2n + 3) [1]. 2n + 3 is an integer, so the sum is a multiple of 3 [1]. Examiner insight: the final mark needs both the reason (2n + 3 is an integer) and the conclusion restated; stopping at 3(2n + 3) loses it.

3. 4x² − 12x + 10 = (2x − 3)² + 1 [1]. (2x − 3)² ≥ 0 for all real x [1], so (2x − 3)² + 1 ≥ 1 > 0, therefore 4x² − 12x + 10 > 0 for all real x [1]. Examiner insight: the method mark is for a correct completed-square form; the reasoning mark needs the explicit statement that a square is never negative.

4. If n is even, n = 2k: n² + n + 1 = 4k² + 2k + 1 = 2(2k² + k) + 1, which is odd [1]. If n is odd, n = 2k + 1: n² + n + 1 = 4k² + 6k + 3 = 2(2k² + 3k + 1) + 1, which is odd [1]. Every integer is even or odd, so n² + n + 1 is odd for every integer n [1]. Examiner insight: each case must be written in the form 2(integer) + 1; “4k² + 2k + 1 is odd” without the factorised form is an assertion, not a step.

5. Every integer n can be written as 3m, 3m + 1 or 3m − 1 (m ∈ ℤ) [1]. n = 3m: n³ = 27m³ = 9(3m³), of the form 9k [1]. n = 3m + 1: n³ = 27m³ + 27m² + 9m + 1 = 9(3m³ + 3m² + m) + 1, of the form 9k + 1 [1]. n = 3m − 1: n³ = 27m³ − 27m² + 9m − 1 = 9(3m³ − 3m² + m) − 1, of the form 9k − 1 [1]. These cases cover every integer, so every cube is of the form 9k, 9k + 1 or 9k − 1 [1]. Examiner insight: the first mark is for cases that clearly cover all integers; using 3m, 3m + 1, 3m + 2 is also fine, but then the last case must be rewritten as 9k − 1.

6. Assume n² is odd and n is even, so n = 2k [1]. Then n² = 4k² = 2(2k²), which is even [1]. This contradicts n² being odd, so if n² is odd, n is odd [1]. Examiner insight: the opening assumption must be the correct negation (n² odd and n even); assuming “n² is even” earns nothing.

7. Let a/b be rational (a ∈ ℤ, b ∈ ℤ⁺) and let x be irrational. Assume a/b + x is rational, equal to c/d (c ∈ ℤ, d ∈ ℤ⁺) [1]. Then x = c/d − a/b = (bc − ad)/(bd) [1]. bc − ad ∈ ℤ and bd ∈ ℤ⁺, so x is rational [1]. This contradicts x being irrational, so the sum is irrational [1]. Examiner insight: writing the difference as a single fraction with integer numerator and positive integer denominator is what earns the third mark; “rational minus rational is rational” may be accepted only if clearly stated.

8. Assume log₂ 5 is rational. log₂ 5 > 0, so log₂ 5 = a/b with a, b ∈ ℤ⁺ [1]. Then 2^(a/b) = 5, so 2ᵃ = 5ᵇ [1]. a ≥ 1, so 2ᵃ is even; 5ᵇ is odd [1]. An even number cannot equal an odd number, a contradiction, so log₂ 5 is irrational [1]. Examiner insight: the step from 2^(a/b) = 5 to 2ᵃ = 5ᵇ (raising both sides to the power b) must be shown; the parity argument then needs a ≥ 1 to be valid.

9. Assume there are integers m and n with m² − n² = 10. Then (m − n)(m + n) = 10 [1]. (m + n) + (m − n) = 2m is even, so m − n and m + n are both even or both odd [1]. If both are odd, their product is odd, but 10 is even [1]. If both are even, their product is a multiple of 4, but 10 is not [1]. Either way there is a contradiction, so there are no integers m and n with m² − n² = 10 [1]. Examiner insight: listing factor pairs of 10 only covers positive factors; the parity argument covers negative and zero values too, and the both-even case is the one most often missed.

10. (a) If n is not a multiple of 3, then n = 3k + 1 or n = 3k + 2 (k ∈ ℤ). n = 3k + 1: n² = 9k² + 6k + 1 = 3(3k² + 2k) + 1 [1]. n = 3k + 2: n² = 9k² + 12k + 4 = 3(3k² + 4k + 1) + 1 [1]. In both cases n² leaves remainder 1, so n² is not a multiple of 3 [1]. (b) Assume √3 is rational: √3 = a/b, a ∈ ℤ, b ∈ ℤ⁺, in lowest terms [1]. Then a² = 3b², so a² is a multiple of 3. By (a), a is a multiple of 3, so a = 3c [1]. Then 9c² = 3b², so b² = 3c² [1], and by (a) b is a multiple of 3 [1]. So 3 is a common factor of a and b, contradicting lowest terms, so √3 is irrational [1]. Examiner insight: “hence” means part (a) must be quoted to go from “a² is a multiple of 3” to “a is a multiple of 3”; a proof that does not use (a) may lose that mark.

11. (a) False [1]. Counter-example: x = 0.5 gives x² = 0.25 < 0.5 [1]. (b) True. n³ − n = (n − 1)n(n + 1), a product of three consecutive integers [1]. Among any three consecutive integers one is a multiple of 3 and at least one is even [1], so the product is a multiple of 2 and of 3, and so a multiple of 6 [1]. (c) False [1]. Counter-example: a = b = √2 are irrational, but ab = 2 is rational [1]. (d) True. For positive a, b: (√a − √b)² ≥ 0 [1]. Expanding: a − 2√(ab) + b ≥ 0 [1], so a + b ≥ 2√(ab), and (a + b)/2 ≥ √(ab) [1]. Examiner insight: each part must start with a clear decision (true or false); a correct counter-example with no statement that the claim is false, or a proof attached to the wrong decision, loses marks.

12. (a) Using 2n + 1 for both numbers means the two odd numbers are equal, so it only proves the result for an odd number added to itself [1]. (b) Let the odd numbers be 2m + 1 and 2n + 1, where m, n ∈ ℤ [1]. (2m + 1) + (2n + 1) = 2m + 2n + 2 = 2(m + n + 1) [1]. m + n + 1 ∈ ℤ, so the sum of any two odd numbers is even [1]. Examiner insight: in “critique” questions the mark is for naming the specific flaw (same letter, so the numbers are equal), not a general comment such as “it is not a proper proof”.

Where marks are usually lost

• Picking a counter-example but not showing the working that makes it fail.
• Not defining letters (n ∈ ℤ) at the start of a number proof.
• Using one letter for two independent numbers, as in Question 12.
• Factorising to 3(…) or 2(…) + 1 without saying the bracket is an integer.
• Missing a case in exhaustion, or not saying why the cases cover every integer.
• Writing the wrong negation as the opening assumption of a contradiction proof.
• Leaving out “in lowest terms” in irrationality proofs, so no contradiction is possible.
• Ignoring “hence” and proving part (b) without using part (a).
• Not stating “true” or “false” before proving or disproving.
• Leaving out the final sentence that restates the result.

Next steps

• Recap the methods: Proof revision notes.
• Full explanations and worked proofs: Proof study guide.
• AQA A-level Mathematics hub and printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

AQA A-level Mathematics (7357) specification, version 1.3, 31 January 2018 (A-level exams June 2018 onwards), published by AQA. Section 3.2 A: Proof, content reference A1.