OCR A Level Mathematics A: Proof: deduction, exhaustion, counter-example, logical connectives and contradiction (H240) – 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.

These questions cover section 1.01 Proof (1.01a to 1.01d) of the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018. Questions 8 to 10 use proof by contradiction, which is A Level only content (1.01d). The rest is Stage 1 content, shared with AS.

Learn the methods first in the study guide and the revision notes. Course links: the OCR A Level Mathematics hub and the printable checklist. The pure mathematics practice set mixes proof with other pure topics.

Questions

1. In each part, write ⇒, ⇐ or ⇔ in the gap. The variable x is real and n is an integer.

(a) x² = 49 … x = −7 [1] (b) 3x − 5 < 10 … x < 5 [1] (c) n is a multiple of 12 … n is a multiple of 4 [1]

2. Disprove each statement by counter-example.

(a) “3n² + 3n + 1 is prime for every positive integer n.” [2] (b) “If a and b are irrational numbers, then ab is irrational.” [2]

3. Prove that the sum of any three consecutive even integers is a multiple of 6. [3]

4. Prove that 3x² − 12x + 16 > 0 for all real values of x. [3]

5. Prove that the square of an integer never leaves remainder 2 or remainder 3 when divided by 5. [5]

6. Let m and n be integers. Prove that m² + n² is even if and only if m + n is even. [5]

7. A student solves √(2x + 3) = x by squaring both sides.

Solve the equation, and explain why the student must check the answers in the original equation. [4]

8. Use proof by contradiction to show that if a and b are positive real numbers with ab = 100, then a + b ≥ 20. [4]

9. Prove by contradiction that √6 is irrational. You may use the fact that the square of an integer is even only if the integer is even. [7]

10. A proof that there are infinitely many primes assumes that p₁, p₂, …, pₖ is a complete list of primes. It then forms N = p₁ × p₂ × … × pₖ + 1, and a student writes “N is prime, which is a contradiction”.

(a) To test the student’s claim, use the list 2, 7, 11 (not a full list of primes). Show that N is not prime, and give a prime factor of N that is not on the list. [3] (b) Explain what the proof actually needs from N, and where the contradiction arises. [2]

11. (a) Disprove the statement: “If p is prime, then 2p + 1 is prime.” [2]

(b) Prove that, if p is a prime number greater than 3, then p² − 1 is a multiple of 24. [6]

(c) Write down the converse of the statement in part (b), and show that it is false. [2]

Answers

1. (a) If x = −7 then x² = 49, so x = −7 implies x² = 49. But x² = 49 also holds for x = 7. The correct symbol is ⇐ [1]. (b) Add 5 to get 3x < 15, then divide by 3 to get x < 5. Each step reverses, so the symbol is ⇔ [1]. (c) Any multiple of 12 is 4 × 3k, which is a multiple of 4. But 8 is a multiple of 4 and not of 12. The symbol is ⇒ [1]. [3] Examiner insight: Answers usually gain more credit when each choice has a one-line reason, such as a value that breaks the other direction. A bare symbol is hard to credit if it is wrong.

2. (a) For n = 5, 3n² + 3n + 1 = 75 + 15 + 1 = 91 [1]. 91 = 7 × 13 is not prime, so n = 5 is a counter-example [1]. [2] (b) Take a = √2 and b = √2, both irrational [1]. Then ab = 2, which is rational, so the statement is false [1]. [2] Examiner insight: A counter-example needs two things shown: that the starting condition holds and that the conclusion fails. Say that √2 is irrational when you use it, not just that ab = 2.

3. Let the integers be 2n, 2n + 2 and 2n + 4, where n is an integer [1]. Their sum is 2n + 2n + 2 + 2n + 4 = 6n + 6 [1]. This equals 6(n + 1) with n + 1 an integer, so 6 divides the sum, i.e. the sum is a multiple of 6 [1]. [3] Examiner insight: Define the letters and finish with a sentence. Writing the three integers as n, n + 1 and n + 2 would describe consecutive integers, not even ones.

4. Complete the square: 3x² − 12x + 16 = 3(x² − 4x) + 16 = 3[(x − 2)² − 4] + 16 [1] = 3(x − 2)² + 4 [1]. A real square is never negative, so 3(x − 2)² + 4 ≥ 4 > 0 for all real x, as required [1]. [3] Examiner insight: The reason matters as much as the algebra. Writing that (x − 2)² ≥ 0 because it is a square of a real number supports the conclusion. Checking x = 0 and x = 1 does not.

5. Every integer n has the form 5m, 5m + 1, 5m − 1, 5m + 2 or 5m − 2 for some integer m, so these five cases are exhaustive [1]. Case 5m: n² = 25m² = 5(5m²), remainder 0 [1]. Cases 5m ± 1: n² = 25m² ± 10m + 1 = 5(5m² ± 2m) + 1, remainder 1 [1]. Cases 5m ± 2: n² = 25m² ± 20m + 4 = 5(5m² ± 4m) + 4, remainder 4 [1]. Every square leaves remainder 0, 1 or 4. So no square leaves remainder 2 or 3 [1]. [5] Examiner insight: An exhaustion proof loses credit if the list of cases is incomplete or the cases are not shown to cover all integers. State that the cases cover everything.

6. (m + n)² = m² + 2mn + n², so m² + n² = (m + n)² − 2mn [1]. Since 2mn is even, m² + n² and (m + n)² have the same parity (both even or both odd) [1]. A square has the same parity as its base: an even number squared is even, and an odd number squared is odd [1]. So (m + n)² is even exactly when m + n is even [1]. Therefore m² + n² is even ⇔ (m + n)² is even ⇔ m + n is even. Each link works in both directions, so both directions are proved [1]. [5] Examiner insight: For “if and only if”, say that each step works in both directions, or prove the two directions separately. A one-way argument with ⇔ in the last line leaves half the proof missing.

7. Square both sides: 2x + 3 = x² [1]. Rearranged, this is x² − 2x − 3 = 0, which factorises as (x − 3)(x + 1) = 0, so x = 3 or x = −1 [1]. Check in the original: x = 3 gives √9 = 3, which works, and x = −1 gives √1 = 1, not −1, so reject it. x = 3 [1]. Squaring is a step with ⇒ only: x = −1 satisfies the squared equation but not the original, so the check is needed [1]. [4] Examiner insight: Giving both roots without a check is a common loss. Name the reason: squaring is not reversible.

8. Suppose that a and b are positive, ab = 100, and a + b < 20 [1]. Since a + b is positive, squaring gives (a + b)² < 400 [1]. But (a + b)² = (a − b)² + 4ab = (a − b)² + 400 ≥ 400, because (a − b)² ≥ 0 [1]. This contradicts (a + b)² < 400, so the assumption is false and a + b ≥ 20 [1]. [4] Examiner insight: The opposite of “a + b ≥ 20” is “a + b < 20”, not “a + b ≤ 20”. The wrong opposite makes the whole argument invalid.

9. Assume that √6 is rational, so √6 = p/q with p and q positive integers having no common factor [1]. Squaring gives p² = 6q² [1]. So p² is even, and therefore p is even [1]. Write p = 2r. Then 4r² = 6q², so 2r² = 3q² [1]. Now 3q² is even, and 3 is odd, so q² is even [1]. Therefore q is even [1]. Both p and q are even, so they share the factor 2, which contradicts the assumption of no common factor. Hence √6 is irrational [1]. [7] Examiner insight: The step from “p² is even” to “p is even” is the fact the question lets you use, so quote it each time. Answers lose credit when the second half, showing that q is even, is skipped as “similarly”.

10. (a) N = 2 × 7 × 11 + 1 = 155 [1]. 155 = 5 × 31, so N is not prime [1]. The prime factor 5 (or 31) is not on the list [1]. [3] (b) N leaves remainder 1 when divided by any prime on the list, and N > 1, so N has some prime factor outside the list [1]. That contradicts the assumption that the list contained every prime, so there are infinitely many primes [1]. [2] Examiner insight: The proof needs N to have a prime factor off the list. It does not need N to be prime, and a computed counter-example like 155 shows why the student’s claim is unsafe.

11. (a) For p = 7, 2p + 1 = 15 [1]. 15 = 3 × 5 is not prime, so p = 7 is a counter-example [1]. [2] (b) p² − 1 = (p − 1)(p + 1) [1]. Since p is prime and greater than 3, p is odd, so p − 1 = 2j and p + 1 = 2j + 2 for some integer j, and the product is 4j(j + 1) [1]. j(j + 1) is a product of consecutive integers, so it is even, and 4j(j + 1) is a multiple of 8 [1]. The integers p − 1, p and p + 1 are consecutive, so one is a multiple of 3 [1]. p is prime and greater than 3, so p is not a multiple of 3, which means p − 1 or p + 1 is [1]. So p² − 1 is a multiple of both 8 and 3. These have no common factor, so p² − 1 is a multiple of 24 [1]. [6] (c) The converse is “if p² − 1 is a multiple of 24, then p is a prime greater than 3”. For p = 25, p² − 1 = 624 = 24 × 26, which is a multiple of 24 [1]. 25 is not prime, so the converse is false [1]. [2] Examiner insight: Long proofs gain credit when they are split into clear claims, here “multiple of 8” and “multiple of 3”, before combining. State why 8 and 3 together give 24.

Where marks are usually lost

• Writing “hence proved” after substituting a few numbers.
• Not defining letters, or using the same letter for two integers that may differ.
• Leaving out the final sentence that states what has been proved.
• Claiming ⇔ after a step that is only ⇒, such as squaring both sides.
• Proving one direction of “if and only if” and stopping.
• Choosing a counter-example that does not satisfy the starting condition.
• Giving cases in an exhaustion proof without showing they cover every integer.
• Assuming the wrong opposite at the start of a contradiction proof.
• Asserting a fact such as “p² multiple of 3 means p multiple of 3” without a reason.

Next steps

• Review with the revision notes.
• Re-read the full explanations in the study guide.
• Visit the OCR A Level Mathematics hub and tick off topics on the checklist.
• Take any of the free 10-minute diagnostics.
• Want a teacher’s help? Book a free trial class.

Official syllabus

OCR, A Level in Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, section 1.01 Proof (references 1.01a to 1.01d), published by OCR.