AQA A-Level Mathematics Paper 3, June 2025: Question 12

3 marks · Medium difficulty · Proof

Use proof by contradiction to show that if the product of two real numbers is irrational, then at least one of the numbers must be irrational.

Practise this question

Question

Question 12 states: Given that the real numbers a, b, and c are such that a times b equals c. Use proof by contradiction to show that if c is irrational then at least one of a or b is irrational. The question is worth 3 marks.

Mark scheme

Show the mark scheme Mark scheme table for Question 12 with three marks: B1 for stating the assumption that both a and b are rational; M1 for setting a = m/n and b = p/q and forming the product c = (m/n)(p/q) = mp/nq; R1 for concluding that c is rational, which contradicts that c is irrational, therefore completing the proof by contradiction.

How to answer it

Proof by Contradiction: Products of Irrational Numbers

What this question tests

This question assesses your ability to construct a rigorous proof by contradiction (Specification AO2.1 & AO3.1a). You must understand the logical negation of a conditional statement ("if P then Q"), know the algebraic definition of a rational number as a quotient of two integers, manipulate fractions algebraically, and formulate an explicit concluding deduction.

Question 12 • 3 Marks Total

Full Question & Model Solution Walkthrough

Given that the real numbers a, b, and c are such that a × b = c. Use proof by contradiction to show that if c is irrational then at least one of a or b is irrational.

📐 Step-by-Step Proof Structure

  1. State the initial assumption (negation):
    Assume the statement is false. That is, assume that c is irrational , but neither a nor b is irrational (i.e. both a and b are rational). B1 [AO 2.1]
  2. Express rational numbers in algebraic fractional form:
    Since a and b are rational, define them as fractions of integers:
    Let a = m / n and b = p / q , where m, n, p, q are integers and n ≠ 0, q ≠ 0 .
  3. Form the product of the two numbers:
    Multiply a and b to express c :
    c = a × b = (m / n) × (p / q) = (mp) / (nq) M1 [AO 3.1a]
  4. Deduce the contradiction & complete the conclusion:
    Since m, p, n, q are integers, the product mp is an integer and nq is a non-zero integer. Therefore, c must be rational.
    This contradicts the given premise that c is irrational.
    Hence, the initial assumption is false, which proves that if c is irrational, at least one of a or b must be irrational. R1 [AO 2.1]

✅ Model Answer Box

"Assume that both a and b are rational.

Then let a = m / n and b = p / q, where m, n, p, q are integers (n, q ≠ 0).

Then c = ab = (m / n) × (p / q) = mp / nq.

Since mp and nq are integers, c is rational. This contradicts the given fact that c is irrational.

Therefore, by contradiction, if c is irrational then at least one of a or b must be irrational."

💡 Key Knowledge

  • Negating "At least one": The logical negation of "at least one of A or B is true" is "neither A nor B is true" (i.e. both are false). Thus, "at least one is irrational" becomes "both are rational".
  • Definition of Rational: Any number that can be expressed as a ratio of two integers x / y with y ≠ 0 .
  • Closure: The product of two integers is always an integer; the product of two rational numbers is always rational.

🧠 Exam Technique & Examiner Insight

  • The 3-part conclusion rule: To secure the final R1 mark, examiners look for three specific elements:
    1. Stating that c is rational.
    2. Explicitly stating that this contradicts that c is irrational.
    3. A concluding sentence restating the original conjecture.
  • Always show the single combined fraction mp / nq clearly. Leaving it as separate fractions does not fulfill the definition of a rational number.

❌ Common Student Traps

  • Incorrect Negation: Assuming that "both a and b are irrational" or "only one is rational". If you negate incorrectly at line 1, you lose all marks.
  • Using Examples: Trying to prove the statement with specific numbers like √2 or √3 . A proof must be general and use algebraic variables.
  • Missing Contradiction Statement: Showing that c = mp / nq but stopping without stating that this contradicts c being irrational.
Mark Scheme Breakdown:
  • B1 (AO 2.1): States that a and b are both rational (or "Let a and b be rational").
  • M1 (AO 3.1a): Forms the product of two general rational numbers, writing (m/n) × (p/q) or equivalent.
  • R1 (AO 2.1): Complete reasoned argument: must show mp/nq , state c is rational, note the contradiction with c being irrational, and state the final concluding sentence. Requires B1 and M1 to have been awarded.

Topics

Pure Mathematics · A: Proof

Question and mark scheme from the AQA A-Level Mathematics examination, Paper 3, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.