Question 1Identifying errors in proofs
TMUA Errors in Proofs — Practice Questions by Topic
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- ErrIdentifying errors in proofs
- 21 questions21 on Paper 2 only
- 8 free solutionsThe rest show the correct letter only
Covers: the first false step in an attempted solution, an unjustified move in an inequality, a proof that divides by zero or drops a case, which line of an attempt fails.
What this topic tests
Identifying errors in proofs asks which line of an attempt is the first false step: an unjustified move, a division by zero, a dropped case, or a step that uses the result being proved.
How it is assessed
This topic is examined on Paper 2 only. Each question has five options. Paper 2 is 20 questions in 75 minutes, with no calculator.
This note is written for SummitPapers. The official list of what can be examined is the specification, together with the Notes on Mathematics and, for Paper 2, the Notes on Logic and Proof.
Key methods
The first false line
Later lines may be false because they follow a false line. The question asks for the first line that does not follow. A true line that happens to look like the conclusion is not the error.
Division and lost cases
Dividing by an expression assumes it is not zero. Squaring both sides, or cancelling a common factor, can add or lose solutions. The line that does that is the error, even if a later line names the wrong root.
Common mistakes
Marking a line that is still valid
Multiplying an inequality by 2, or subtracting the same term from both sides, is valid. The error is the line that assumes the result, or that divides by zero, not the line that looks longest.
A counterexample to a different claim
A number that fails the hypothesis does not show that a later algebraic step is illegal. Check the step against the line before it.
Worked example
This question is also in the list below, with the solution folded. It is opened here so the method is on the page before the other questions.
Claim: the product of two odd integers is odd. Attempted proof. I: Let the integers be 2a + 1 and 2b + 1, where a and b are integers. II: Their product is (2a + 1)(2b + 1) = 4ab + 2a + 2b + 1 = 2(2ab + a + b) + 1. III: Since 2ab + a + b is an integer, the product is even. Which description is correct?
Answer: E. The proof is wrong, and the first error is on line III.
Worked solution. Lines I and II are correct: every odd integer has the form 2k + 1, and the product really is 2(2ab + a + b) + 1. A number of the form 2m + 1 is odd, not even. Line III draws the opposite conclusion from that form, so the first error is on line III. The claim itself is true, which is why a correct proof would have stopped at line II.
Why the other options look right. A accepts line III, which calls an odd number even. B says the claim is false, but the product of odd integers is odd. C rejects the standard form of an odd integer. D rejects the expansion, which multiplies correctly and factors out 2 correctly.
Paper 2 questions
Paper 2 is Mathematical Reasoning. This topic is examined on Paper 2 only.
Q1. Claim: the product of two odd integers is odd. Attempted proof. I: Let the integers be 2a + 1 and 2b + 1, where a and b are integers. II: Their product is (2a + 1)(2b + 1) = 4ab + 2a + 2b + 1 = 2(2ab + a + b) + 1. III: Since 2ab + a + b is an integer, the product is even. Which description is correct?
Free · worked solution included
Answer and worked solution
Answer: E. The proof is wrong, and the first error is on line III.
Worked solution. Lines I and II are correct: every odd integer has the form 2k + 1, and the product really is 2(2ab + a + b) + 1. A number of the form 2m + 1 is odd, not even. Line III draws the opposite conclusion from that form, so the first error is on line III. The claim itself is true, which is why a correct proof would have stopped at line II.
Why the other options look right. A accepts line III, which calls an odd number even. B says the claim is false, but the product of odd integers is odd. C rejects the standard form of an odd integer. D rejects the expansion, which multiplies correctly and factors out 2 correctly.
Q2. A student tries to prove that if a and b are real numbers with a > b, then a2 > b2. Where is the first error? Line 1. Suppose a > b. Line 2. Then a − b > 0. Line 3. Also a + b > 0, because a > b. Line 4. So (a − b)(a + b) > 0, as a product of two positive numbers. Line 5. Hence a2 − b2 > 0, so a2 > b2.
Free · worked solution included
Answer and worked solution
Answer: C. Line 3
Worked solution. Line 1 states the hypothesis, and Line 2 follows from it by subtracting b from both sides. Line 3 is false: a > b does not make a + b positive. For example, a = 1 and b = −2 satisfy a > b, but a + b = −1; indeed a2 = 1 < 4 = b2, so the claim itself is false. Line 4 would follow from Lines 2 and 3, because a product of two positive numbers is positive, and Line 5 uses the correct identity (a − b)(a + b) = a2 − b2. The first error is Line 3.
Why the other options look right. A rejects Line 1, but it is the hypothesis, which may simply be assumed. B rejects a − b > 0, which follows from a > b by subtracting b. D rejects the step that a product of two positive numbers is positive, which is valid once Lines 2 and 3 are accepted. E rejects the identity (a − b)(a + b) = a2 − b2, which is correct; the conclusion fails only because Line 3 is false.
Q3. A student was asked: given that cos x = 1/2, find the possible values of sin 2x. The student wrote: cos x = 1/2 so x = 60 degrees and 2x = 120 degrees, therefore sin 2x = √3/2. Which statement is correct?
Free · worked solution included
Answer and worked solution
Answer: C. √3/2 is not the only possible value, because other x with cos x = 1/2 give sin 2x = −√3/2
Worked solution. The solutions of cos x = 1/2 are x = ±60 degrees plus multiples of 360 degrees. Then 2x = ±120 degrees plus multiples of 720 degrees, so sin 2x is √3/2 or −√3/2. The student's answer checks only x = 60 degrees and misses the second value. The correct comment is option C.
Why the other options look right. A treats that single angle as a complete argument. B says the value is unique, but x = −60 degrees gives sin 2x = −√3/2. D looks for other x with sin 2x = √3/2, but those angles need not satisfy cos x = 1/2. E takes x = −60 degrees as the only solution of cos x = 1/2, but x = 60 degrees also works and gives sin 2x = √3/2, so −√3/2 is not the only value.
Q4. Consider this attempt to prove the true statement that there are no positive integers a and b with a2 + b2 = 3. Suppose a and b are positive integers with a2 + b2 = 3. (I) Then a2 = 3 − b2. (II) Hence a2 = (√3 − b)(√3 + b). (III) It follows that a = √3 − b and a = √3 + b, because the two factors of a2 must each equal a. (IV) Adding these equations gives 2a = 2√3, so a = √3. (V) Then a is not an integer. (VI) This contradicts the assumption that a is a positive integer, so no such integers exist. Where does the argument first fail?
Free · worked solution included
Answer and worked solution
Answer: D. The first mistake is on line III
Worked solution. Line I is a rearrangement, and line II is the difference of squares, since (√3 − b)(√3 + b) = 3 − b2. Line III says the factors must each be a. A factorisation of a2 can split in other ways, and the two factors here are not forced to equal a. That is the first false step. The answer is option D.
Why the other options look right. A accepts line III because the conclusion is true, but a2 can factorise as a product of two factors that are not each equal to a, so the argument is invalid even though the statement holds. B marks line I, but subtracting b2 from both sides of a2 + b2 = 3 is valid. C marks line II, but (√3 − b)(√3 + b) = 3 − b2 is a correct difference of squares. E is a later consequence of the false factorisation, so it is not where the argument first fails.
Q5. A student solves the inequality |x − 1| < |2x + 4| as follows. (I) Both sides are non-negative, so squaring gives (x − 1)² < (2x + 4)². (II) Expanding gives x² − 2x + 1 < 4x² + 16x + 16. (III) Rearranging gives 3x² + 18x + 15 > 0. (IV) Dividing by 3 and factorising gives (x + 1)(x + 5) > 0. (V) So x + 1 > 0 and x + 5 > 0, and the solution is x > −1. Which step, if any, is invalid?
Free · worked solution included
Answer and worked solution
Answer: E. Only step (V) is invalid
Worked solution. Step (I) is valid: both sides are non-negative, and for non-negative numbers u < v holds exactly when u² < v². Step (II) expands correctly. Step (III) moves every term to the right: 0 < 3x² + 18x + 15, which is the same inequality written the other way round. Step (IV) is valid because x² + 6x + 5 = (x + 1)(x + 5). Step (V) is invalid: a product is positive when both factors are positive or when both are negative, so the solution is x < −5 or x > −1. For example x = −6 gives |−7| = 7 < |−8| = 8, but the student's answer leaves it out. The answer is option E.
Why the other options look right. A misses the lost case: x = −6 satisfies the original inequality but not x > −1. B treats squaring an inequality as always unsafe; here both sides are non-negative, so squaring keeps the direction. C thinks moving the terms across reverses the inequality, but 3x² + 18x + 15 > 0 is just 0 < 3x² + 18x + 15 read from right to left. D finds the error in (V) but also rejects the valid squaring step (I).
Q6. Consider this attempt to solve the inequality (x + 4)/(x − 1) < 2. (I) Multiply both sides by x − 1: x + 4 < 2(x − 1). (II) Expand the bracket: x + 4 < 2x − 2. (III) Rearrange: 6 < x. (IV) So the solution set is x > 6. Which statement is true?
Free · worked solution included
Answer and worked solution
Answer: E. The solution set should also contain every x < 1, and the first error is in step (I).
Worked solution. Step (I) multiplies by x − 1 without knowing its sign. That keeps the inequality only when x − 1 > 0; when x − 1 < 0 the inequality sign must reverse, so step (I) is the first error. Correctly, (x + 4)/(x − 1) − 2 = (6 − x)/(x − 1), which is negative exactly when x < 1 or x > 6. For example, x = 0 gives 4/(−1) = −4 < 2. Every x > 6 does satisfy the inequality, so the attempt finds part of the answer but misses every x < 1.
Why the other options look right. A accepts x > 6 and misses the values x < 1, such as x = 0. B says some x > 6 fail, but for x > 1 step (I) is reversible, so every x > 6 is a solution. C reverses the inequality in step (I) as if x − 1 were negative, which gives x < 6, but then keeps only x > 1 so that the denominator is positive; these two assumptions contradict each other. The set 1 < x < 6 loses every x > 6 and includes values such as x = 2, where (2 + 4)/(2 − 1) = 6 is not less than 2. D sees that x < 1 is missing but blames the last line; steps (II) to (IV) are correct, and the values are lost when step (I) multiplies by a factor that may be negative.
Q7. An attempted proof of the conjecture “if 0 < θ < π and sin θ > cos θ, then tan θ > 1” runs as follows. Suppose 0 < θ < π and sin θ > cos θ. (I) Since 0 < θ < π, sin θ > 0. (II) The hypothesis gives sin θ − cos θ > 0. (III) Dividing both sides of (II) by cos θ gives tan θ − 1 > 0. (IV) Hence tan θ > 1. Which statement is the case?
Free · worked solution included
Answer and worked solution
Answer: D. The proof is incorrect, and the first error is in line (III).
Worked solution. Line (I) is true because the sine is positive on 0 < θ < π. Line (II) is the hypothesis rearranged. Line (III) divides an inequality by cos θ, which keeps its direction only if cos θ > 0. For π/2 < θ < π the cosine is negative, so the inequality reverses, and at θ = π/2 the division is impossible. For example θ = 2π/3 has sin θ = √3/2 > −1/2 = cos θ, but tan θ = −√3 < 1, so the conjecture is false. Line (IV) only adds 1 to both sides of (III). The proof is incorrect, and the first error is in line (III).
Why the other options look right. A accepts the proof, but θ = 2π/3 satisfies the hypothesis and has tan θ = −√3. B rejects (I), which holds because sin θ > 0 for 0 < θ < π. C rejects (II), which is the hypothesis with cos θ moved across. E rejects (IV), which adds 1 to both sides of (III) and is valid whenever (III) is.
Q8. A student attempts to solve 2 sin x tan x = 3, for 0 ≤ x ≤ 2π, excluding values where tan x is undefined. The attempt is: (I) 2 sin² x = 3 cos x (II) 2 − 2 cos² x = 3 cos x (III) 2 cos² x + 3 cos x − 2 = 0 (IV) (2 cos x − 1)(cos x + 2) = 0 (V) x = π/3 or x = 5π/3 Which description is best?
Free · worked solution included
Answer and worked solution
Answer: A. It is completely correct
Worked solution. Line (I) multiplies both sides by cos x, using tan x = sin x/cos x; this is reversible because cos x ≠ 0 wherever tan x is defined. Line (II) uses sin² x = 1 − cos² x. Line (III) moves every term to one side. Line (IV) expands back to line (III): (2 cos x − 1)(cos x + 2) = 2 cos² x + 4 cos x − cos x − 2 = 2 cos² x + 3 cos x − 2. Since −1 ≤ cos x ≤ 1, cos x + 2 ≠ 0, so cos x = 1/2, which gives x = π/3 or x = 5π/3 in [0, 2π]. Both check: at π/3, 2 × (√3/2) × √3 = 3, and at 5π/3, 2 × (−√3/2) × (−√3) = 3. Every line is correct.
Why the other options look right. B treats multiplying by cos x as unsafe, but cos x is non-zero wherever tan x is defined, so no solution is gained or lost. C reads 2 sin² x as 2 − cos² x, applying the 2 to only one term; in fact 2(1 − cos² x) = 2 − 2 cos² x. D expands (2 cos x − 1)(cos x + 2) with the sign of the middle term wrong, getting 2 cos² x − 3 cos x − 2; the correct expansion is 2 cos² x + 3 cos x − 2. E expects solutions from cos x = −2, but |cos x| ≤ 1, so that factor gives none.
Q9. A student claims that for every integer n, the expression 2((12n + 1)/2 − (2n − 1)/2) is divisible by 5. The argument is: (I) 2((12n + 1)/2 − (2n − 1)/2) = (12n + 1) − (2n − 1). (II) = 12n + 1 − 2n − 1. (III) = 10n. (IV) = 5(2n). (V) which is always a multiple of 5. Which one of the following is true?
Free · correct letter only
Answer
Answer: C. The argument is incorrect, and the first error is on line (II).
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Q10. A student differentiates the integral from x to 3x of t3 dt as follows. (I) The integral from x to 3x of t3 dt equals the integral from 0 to 3x of t3 dt minus the integral from 0 to x of t3 dt. (II) By the fundamental theorem, the derivative of the integral from 0 to x of t3 dt is x3. (III) By the fundamental theorem, the derivative of the integral from 0 to 3x of t3 dt is (3x)3 = 27x3. (IV) So the derivative of the original integral is 27x3 − x3. (V) This gives 26x3. Which description is correct?
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Answer
Answer: D. The calculation is incorrect, and the first error is on line (III).
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Q11. On which line is the first error in the following argument? A: For all real x, (x − 3)2 = x2 − 6x + 9. B: Therefore x2 − 6x + 9 = x − 3 for all real x. C: Hence x2 − 7x + 12 = 0 for all real x. D: Thus (x − 3)(x − 4) = 0 for all real x. E: Substituting x = 0 gives 12 = 0.
Free · correct letter only
Answer
Answer: B. B
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Q12. A student claims that 2x2 + 2x + 1 is prime for every positive integer x, and writes: I. 2x2 + 2x + 1 = x2 + (x + 1)2. II. The discriminant of 2t2 + 2t + 1 is −4. III. Therefore that quadratic has no factorisation into linear polynomials with integer coefficients. IV. Hence 2x2 + 2x + 1 is never a product of two integers greater than 1. Which description of this argument is correct?
Free · correct letter only
Answer
Answer: E. The first error is in line IV
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Q13. A student tries to prove that (a + b)2 ≥ a2 + b2 for all real a and b. Their argument is: (I) (a − b)2 ≥ 0; (II) a2 − 2ab + b2 ≥ 0; (III) a2 + b2 ≥ 2ab; (IV) therefore 2ab ≥ 0; (V) hence (a + b)2 = a2 + 2ab + b2 ≥ a2 + b2. Which line contains the first error?
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Answer
Answer: D. Line IV
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Q14. A student answers the question “Can three consecutive positive integers all be prime?” as follows. I: Any three consecutive positive integers can be written as n, n + 1 and n + 2. II: One of them is divisible by 2. III: One of them is divisible by 3. IV: A positive integer divisible by 2 or by 3 is never prime. V: Therefore no three consecutive positive integers are all prime. Which description is correct?
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Answer
Answer: B. The first error is on line IV.
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Q15. Here is an attempt to solve x² − 5x + 6 > 0. I: if and only if (x − 2)(x − 3) > 0. II: if and only if x − 2 > 0 and x − 3 > 0. III: if and only if x > 3. IV: if and only if x > 2. Which statement is true?
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Answer
Answer: D. The first error is in line II.
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Q16. Consider this attempt to solve an equation. The steps are numbered. (1) Start from √(2x + 3) = x and square both sides: 2x + 3 = x2. (2) Rearrange: x2 − 2x − 3 = 0. (3) Factorise: (x − 3)(x + 1) = 0, so x = 3 or x = −1. Which statement is true?
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Answer
Answer: C. One solution is correct, and the incorrect solution arises at step (1).
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Q17. A student solves the simultaneous equations sin x + cos x = √2 and sin x − cos x = 0 for 0 ≤ x ≤ π, as follows. (I) Adding the two equations gives 2 sin x = √2. (II) Therefore sin x = √2/2. (III) Therefore x = π/4 or x = 3π/4. (IV) Both values lie in the interval, so both solve the simultaneous equations. Which description of this attempt is correct?
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Answer
Answer: D. The first error is in line (IV).
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Q18. The following argument claims that every even integer greater than 2 is divisible by 4. On which line does the first mistake occur? Line 1: let n be an even integer greater than 2. Line 2: then n = 2k for some integer k greater than 1. Line 3: the integer k is even. Line 4: so k = 2m for some integer m, and n = 4m. Line 5: therefore n is divisible by 4.
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Answer
Answer: C. Line 3
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Q19. Consider this attempt to solve sin x + cos x = 1 for 0° ≤ x < 360°. I. (sin x + cos x)2 = 1 II. 1 + 2 sin x cos x = 1 III. sin(2x) = 0 IV. x = 0°, 90°, 180° or 270°, so the equation has four solutions in the interval. Which statement describes the attempt?
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Answer
Answer: E. It is incorrect, and the first mistake is on line IV
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Q20. A student solves log base 3 of (x2) = 2 as follows. (I) 2 × log base 3 of x = 2. (II) log base 3 of x = 1. (III) x = 3. (IV) The only solution is x = 3. Which statement best describes this attempt?
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Answer
Answer: B. It is incorrect, and the first error is on line (I).
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Q21. Here is an attempt to prove that 2 = 1. Line 1: let a = b, with a ≠ 0. Line 2: a² = ab. Line 3: a² − b² = ab − b². Line 4: (a − b)(a + b) = b(a − b). Line 5: a + b = b. Line 6: since a = b, 2b = b, so 2 = 1. Which statement is true?
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Answer
Answer: D. The proof is incorrect, and the first error is in line 5.
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