IB · Mathematics: Analysis & Approaches SL & HL · Number & algebra

Induction, contradiction & counterexample: IB Mathematics: Analysis & Approaches questions with answers

Free Induction, contradiction & counterexample questions for IB Mathematics: Analysis & Approaches (SL & HL), HL only, part of Number & algebra. Each question has the answer and the marking points an examiner would look for.

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Induction, contradiction & counterexample practice questions

Question 1Structured · 5 marks

Prove by contradiction that log₂3 is irrational.

[5]

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Part	Answer and marking points	Marks
	proof assume log₂3 = pq, p, q ∈ ℤ⁺ M1 2^p/q = 3 ⇒ 2^p = 3^q A1 2^p is even and 3^q is odd R1 so 2^p ≠ 3^q, a contradiction R1 hence log₂3 is irrational AG A1	5

Question 2Structured · 7 marks

Prove by mathematical induction that n³ + 2n is divisible by 3 for all n ∈ ℤ⁺.

[7]

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Part	Answer and marking points	Marks
	proof n = 1: 1 + 2 = 3 A1 assume k³ + 2k = 3M M1 (k + 1)³ + 2(k + 1) = k³ + 3k² + 3k + 1 + 2k + 2 M1 = (k³ + 2k) + 3k² + 3k + 3 (A1) = 3M + 3(k² + k + 1) A1 = 3(M + k² + k + 1), divisible by 3 A1 true for n = 1 and P(k) ⇒ P(k + 1), so true for all n ∈ ℤ⁺ R1	7

Question 3Structured · 3 marks

Show, using a counterexample, that the statement "n² + n + 41 is prime for all n ∈ ℕ" is false.

[3]

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Part	Answer and marking points	Marks
	proof choose n = 40 (or n = 41) M1 40² + 40 + 41 = 1681 A1 = 41², which is not prime R1	3

Question 4Structured · 7 marks

Prove by mathematical induction that 5ⁿ + 3 is divisible by 4 for all n ∈ ℤ⁺.

[7]

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Part	Answer and marking points	Marks
	proof n = 1: 5 + 3 = 8, which is divisible by 4 A1 assume true for n = k: 5^k + 3 = 4M, M ∈ ℤ M1 consider n = k + 1: 5^k+1 + 3 = 5·5^k + 3 M1 = 5(4M − 3) + 3 (A1) = 20M − 12 = 4(5M − 3) A1 which is divisible by 4 A1 since true for n = 1, and true for n = k implies true for n = k + 1, true for all n ∈ ℤ⁺ by induction R1	7

Question 5Structured · 4 marks

Prove by contradiction that there is no largest even integer.

[4]

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Part	Answer and marking points	Marks
	proof assume N is the largest even integer M1 N + 2 is also even A1 N + 2 > N R1 contradiction, so there is no largest even integer R1	4

Question 6Structured · 7 marks

Prove by mathematical induction that Σ_r=1ⁿr·r! = (n + 1)! − 1 for all n ∈ ℤ⁺.

[7]

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Part	Answer and marking points	Marks
	proof n = 1: LHS = 1·1! = 1, RHS = 2! − 1 = 1 A1 assume Σ_r=1^kr·r! = (k + 1)! − 1 M1 Σ_r=1^k+1r·r! = (k + 1)! − 1 + (k + 1)(k + 1)! M1 = (k + 1)!(1 + k + 1) − 1 (A1) = (k + 2)(k + 1)! − 1 A1 = (k + 2)! − 1 A1 true for n = 1 and P(k) ⇒ P(k + 1), so true for all n ∈ ℤ⁺ R1	7

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Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in