TMUA Paper 2: Identifying errors in proofs, set 3

• Answer all questions. No calculator.
• Each question has exactly one correct answer.
• This is a demanding set: the flawed step may be further into the argument, more lines will look plausible, and there is less signposting than in the real exam.

1. 11 mark

   A student claims to prove that (x^3 - 27)/(x - 3) = x^2 + 3x + 9 for every real number x.

   Step 1: Factorise the numerator using the difference-of-cubes identity a^3 - b^3 = (a - b)(a^2 + ab + b^2): x^3 - 27 = (x - 3)(x^2 + 3x + 9).

   Step 2: So (x^3 - 27)/(x - 3) = (x - 3)(x^2 + 3x + 9)/(x - 3).

   Step 3: Cancel the common factor (x - 3) from the numerator and denominator: (x^3 - 27)/(x - 3) = x^2 + 3x + 9.

   Step 4: Therefore (x^3 - 27)/(x - 3) = x^2 + 3x + 9 for every real number x.

   Which step contains the invalid deduction?

   1. A Step 1 is invalid: multiplying out (x - 3)(x^2 + 3x + 9) gives x^3 - 3x^2 + 27, not x^3 - 27, so the stated factorisation does not match the numerator.
   2. B Step 2 is invalid: substituting the factorised numerator changes the fraction's value unless the denominator is rewritten to match, the same way multiplying only a numerator by a constant would.
   3. C Step 3 is invalid, because cancelling the factor (x - 3) assumes x != 3, and the resulting identity fails to hold at x = 3, so it cannot be true for every real number x as claimed.
   4. D Step 4 is invalid: it relies on a limit as x approaches 3 to extend the identity to the excluded value, but no such limiting argument is ever stated.
2. 21 mark

   A student claims to prove: for every integer n, if n^2 is divisible by 3, then n is divisible by 3.

   Step 1: Suppose n^2 is divisible by 3, so n^2 = 3m for some integer m.

   Step 2: Write n = 3k for some integer k.

   Step 3: Then n^2 = (3k)^2 = 9k^2 = 3(3k^2), which is divisible by 3, consistent with step 1.

   Step 4: Hence n is divisible by 3.

   Which step contains the flaw in this argument?

   1. A Step 1 is invalid: writing n^2 = 3m already treats 'n^2 divisible by 3' as an established fact rather than the hypothesis under discussion, making the argument circular from its first line.
   2. B Step 2 is invalid: writing n = 3k assumes n is divisible by 3, which is exactly the conclusion the argument is trying to establish, making it circular.
   3. C Step 3 is invalid: squaring a product distributes only over addition, not multiplication, so (3k)^2 should expand to 3k^2, not 9k^2.
   4. D Step 4 is invalid: step 3 shows n^2 = 9k^2, divisible by 9, so the argument has actually proved a stronger fact than the conclusion drawn in step 4.
3. 31 mark

   A student claims to prove: there exists an integer M such that M > n for every integer n.

   Step 1: Let n be any integer. Then m = n + 1 is an integer with m > n.

   Step 2: Since n was arbitrary, this shows that for every integer n, there exists an integer m (which may depend on n) with m > n.

   Step 3: In particular, there exists a single integer M such that M > n for every integer n: simply take M to be the value of m found above.

   Step 4: Therefore an integer M exists that is greater than every integer n.

   Which step contains the invalid deduction?

   1. A Step 3 is invalid: it wrongly treats 'for every n there is an m depending on n' as equivalent to 'a single M works for every n', which are different statements and the second is false.
   2. B Step 1 is invalid: closure under addition can fail for negative integers, so for n < 0 the sum m = n + 1 need not stay within the integers.
   3. C Step 2 is invalid, because generalising a result shown for an arbitrary integer n to a statement about every integer n is not a legitimate step.
   4. D Step 4 is invalid: it treats the M built in step 3 as a fixed value, but step 3 only exhibited M for one particular n, so a different n would give a different M.
4. 41 mark

   A student solves the equation sin x cos x = (1/2) sin x for x in the range 0 <= x < 360 degrees.

   Step 1: sin x cos x = (1/2) sin x.

   Step 2: Divide both sides by sin x: cos x = 1/2.

   Step 3: In the range 0 <= x < 360 degrees, cos x = 1/2 gives x = 60 degrees or x = 300 degrees.

   Step 4: Therefore the solutions are x = 60 degrees and x = 300 degrees.

   Which step contains the invalid deduction?

   1. A Step 3 is invalid: cos x = 1/2 is satisfied only in the first quadrant within 0 <= x < 360 degrees, at x = 60 degrees, so listing x = 300 degrees as a second solution is wrong.
   2. B Step 1 is invalid: sin x cos x and (1/2) sin x need a domain check before being equated, since a product of two functions is not always defined.
   3. C Step 4 is invalid: it calls x = 60 and x = 300 'the' solutions, but step 3 never showed these were the only values of x satisfying the reduced equation.
   4. D Step 2 is invalid, because dividing both sides by sin x assumes sin x != 0, which silently discards the genuine solutions x = 0 and x = 180 degrees.
5. 51 mark

   A student claims to prove that for all real numbers a and b, if a^3 < b^3 then a < b.

   Step 1: Suppose a and b are real numbers with a^3 < b^3.

   Step 2: Then b^3 - a^3 > 0.

   Step 3: Factorise using the difference-of-cubes identity: b^3 - a^3 = (b - a)(b^2 + ab + a^2).

   Step 4: The factor b^2 + ab + a^2 can be rewritten as (a + b/2)^2 + (3/4)b^2, a sum of two squares, so b^2 + ab + a^2 >= 0 for all real a and b, with equality only when b = 0 and a + b/2 = 0, that is, only when a = 0 and b = 0 simultaneously.

   Step 5: If a^3 < b^3 then a and b are not both 0 (since 0^3 is not less than 0^3), so b^2 + ab + a^2 > 0 strictly. Dividing both sides of (b - a)(b^2 + ab + a^2) > 0 by this positive quantity gives b - a > 0, that is, a < b.

   Step 6: Since a and b were arbitrary real numbers satisfying a^3 < b^3, this shows a < b for all real numbers a and b with a^3 < b^3.

   Which of the following is the correct assessment of this proof?

   1. A Step 6 is invalid: step 5 only established a < b for the particular a and b of step 1, so extending this to 'all real a and b' needs a further generalisation step.
   2. B Step 4 is invalid: the identity b^2 + ab + a^2 = (a + b/2)^2 + (3/4)b^2 only holds when a and b share the same sign, so for opposite signs the expression can be negative.
   3. C Step 5 is invalid: dividing the inequality by b^2 + ab + a^2 assumes this quantity is positive, but its sign was never pinned down earlier in the argument.
   4. D No step is invalid: the proof is genuinely complete, since b^2 + ab + a^2 is provably positive whenever a and b are not both 0, which the hypothesis guarantees.
6. 61 mark

   A student claims to prove that if a and b are irrational numbers, then a + b is irrational.

   Step 1: Let a and b be irrational numbers.

   Step 2: It is a standard fact that the sum of two irrational numbers is always irrational.

   Step 3: By the fact stated in step 2, a + b is irrational.

   What is wrong with this argument?

   1. A Step 2 asserts a false general fact: two irrational numbers need not sum to an irrational number, since sqrt(2) + (-sqrt(2)) = 0 is a counterexample.
   2. B Step 1 is invalid, because a and b cannot both be assumed irrational without first proving that irrational numbers exist.
   3. C Step 3 is invalid: it applies step 2's 'fact' to the specific a and b of step 1 without checking any extra conditions that fact might secretly require.
   4. D The argument is valid, since irrational numbers, by definition, can never combine under addition to give a rational result.
7. 71 mark

   A student wants to disprove the statement: 'for every real number x, x^2 >= x'.

   Step 1: To disprove 'for every real number x, x^2 >= x', it suffices to show that for every real number x, x^2 < x.

   Step 2: Take x = 2: then x^2 = 4 and x = 2, so x^2 < x is false at x = 2.

   Step 3: Since 'for every real number x, x^2 < x' fails, already at x = 2, the student concludes that the original statement 'for every real number x, x^2 >= x' must be true after all.

   What is wrong with this argument?

   1. A Step 2 is invalid: squaring means multiplying a number by itself twice more, so 2^2 should be 2 x 2 x 2 = 8, not 4.
   2. B Step 3 is invalid: showing 'for every x, x^2 < x' fails at x = 2 only shows it fails for that one value; every real number would need checking to settle the universal claim.
   3. C No step is invalid; the student's conclusion that the original statement is true is correct.
   4. D Step 1 misstates the negation of the original universal statement: the true negation is an existence claim, not another 'for every x' statement.
8. 81 mark

   A student claims to prove that log(x^2) = 2 log(x) for every nonzero real number x, where log denotes logarithm to base 10.

   Step 1: For every nonzero real number x, x^2 > 0, so log(x^2) is defined.

   Step 2: By the power rule for logarithms, log(x^2) = 2 log(x).

   Step 3: Therefore log(x^2) = 2 log(x) for every nonzero real number x.

   What is wrong with this argument?

   1. A Step 1 is invalid: for a very small nonzero x, squaring shrinks it further, so x^2 is not guaranteed to stay strictly positive as x approaches zero.
   2. B Step 2 is invalid, because the power rule log(a^n) = n log(a) requires a > 0, but for negative x, log(x) is undefined even though log(x^2) is perfectly well defined.
   3. C Step 3 is invalid: step 2's rule was stated with no reference to a particular x, so asserting it holds 'for every nonzero real number x' needs extra justification.
   4. D The argument is valid: log(x^2) = 2 log(x) holds for every nonzero real x, because the power rule for logarithms applies uniformly to positive and negative bases alike.
9. 91 mark

   A student gives the following argument to show that n(n + 1)(n + 2) is divisible by 6 for every integer n.

   Step 1: Among any three consecutive integers n, n + 1 and n + 2, at least one is divisible by 2, since consecutive integers alternate between even and odd.

   Step 2: Among any three consecutive integers n, n + 1 and n + 2, exactly one is divisible by 3, since the remainders on division by 3 cycle through 0, 1 and 2 as the integers increase by 1.

   Step 3: Since 2 and 3 are distinct primes, and some factor of the product is divisible by 2 while some factor (possibly the same one) is divisible by 3, the whole product n(n + 1)(n + 2) is divisible by 2 x 3 = 6.

   Step 4: Therefore n(n + 1)(n + 2) is divisible by 6 for every integer n.

   Which of the following is the correct assessment of this proof?

   1. A Step 1 is invalid: three consecutive integers can all be odd, for example 1, 3 and 5, so it is not guaranteed that one of n, n+1, n+2 is even.
   2. B Step 2 is invalid: three consecutive integers can include two multiples of 3, for example 3, 4 and 5, so the count need not always be exactly one.
   3. C Step 3 is invalid, because having one factor divisible by 2 and another divisible by 3 does not guarantee the whole product is divisible by 6 unless it is the same factor that is divisible by both.
   4. D No step is invalid; the product is genuinely divisible by both 2 and 3, and since 2 and 3 are coprime, it follows that the product is divisible by 6.
10. 101 mark

    A student claims to prove: for all nonzero real numbers a and b, if a < b then 1/a > 1/b.

    Step 1: Suppose a and b are nonzero real numbers with a < b.

    Step 2: Taking the reciprocal of both sides of an inequality always reverses its direction, so 1/a > 1/b.

    Step 3: Therefore 1/a > 1/b whenever a < b.

    What is wrong with this argument?

    1. A Step 2 is false as a general rule: taking reciprocals reverses an inequality only when a and b share the same sign, and fails whenever their signs differ.
    2. B Step 1 is invalid, because a and b cannot be assumed nonzero without first proving that nonzero real numbers exist.
    3. C The claim becomes true provided a and b are restricted to integers, since the only exceptional pair is a = -1 and b = 1.
    4. D Step 3 is invalid: step 2's rule never explicitly mentions the particular a and b from step 1, so concluding something about THIS a and b in step 3 needs a separate justification.
11. 111 mark

    A student claims to prove that the solutions of |x - 1| = 2x - 5 are x = 4 and x = 2.

    Step 1: Case 1: x - 1 = 2x - 5, giving x = 4.

    Step 2: Case 2: x - 1 = -(2x - 5) = -2x + 5, giving x = 2.

    Step 3: Since both cases produce a value of x, both x = 4 and x = 2 solve the original equation.

    Step 4: Therefore the solutions of |x - 1| = 2x - 5 are x = 4 and x = 2.

    What is wrong with this argument?

    1. A Step 1 contains an algebraic error: rearranging gives 5 + 1 = 2x + x, that is 6 = 3x, so x should equal 2, not 4.
    2. B Step 3 wrongly accepts both case solutions as genuine without checking either one back in the original equation, which an absolute-value case split always requires.
    3. C Step 2 contains an algebraic error: distributing the minus sign should only flip the first term, giving -2x - 5, not -2x + 5.
    4. D Step 4 is invalid: step 3 accepts both case solutions without deriving this from anything beyond the case equations, so step 4 is not entitled to call them 'the' solutions.
12. 121 mark

    A student claims to prove that for all real numbers a and b, a + b >= 2 sqrt(ab).

    Step 1: Let a and b be real numbers.

    Step 2: By the AM-GM inequality, the arithmetic mean of a and b is at least their geometric mean: (a + b)/2 >= sqrt(ab).

    Step 3: Multiplying both sides by 2 gives a + b >= 2 sqrt(ab).

    Step 4: Therefore a + b >= 2 sqrt(ab) for all real numbers a and b.

    What is wrong with this argument?

    1. A Step 1 is invalid, because a and b cannot both be treated as general real numbers within the same argument.
    2. B No step is invalid; the AM-GM inequality holds for all real numbers a and b, not just nonnegative ones.
    3. C Step 2 applies the AM-GM inequality outside its valid range, since AM-GM in this square-root form requires a and b to both be nonnegative.
    4. D Step 3 contains an algebraic error: multiplying by 2 doubles the left side but should turn sqrt(ab) into 2ab on the right, not 2 sqrt(ab).

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

1. Question 1Answer: C

   1. The factorisation x^3 - 27 = (x - 3)(x^2 + 3x + 9) is correct: expanding the right-hand side gives x^3 + 3x^2 + 9x - 3x^2 - 9x - 27 = x^3 - 27, confirming steps 1 and 2.
   2. Cancelling the common factor (x - 3) in step 3 is only a valid algebraic step when x != 3, since dividing by (x - 3) requires (x - 3) to be nonzero.
   3. At x = 3, the left-hand side (x^3 - 27)/(x - 3) is 0/0, which is undefined, while the right-hand side x^2 + 3x + 9 equals 9 + 9 + 9 = 27; since the left side does not even exist at x = 3, the two sides cannot be equal there, so the claimed identity fails to hold 'for every real number x'.
   4. The correct statement is that (x^3 - 27)/(x - 3) = x^2 + 3x + 9 holds for every real x except x = 3, and the flaw in the argument is the unqualified cancellation in step 3 that ignores this exclusion. The answer is C.
   • Why not A: Expanding (x - 3)(x^2 + 3x + 9) term by term gives x^3 + 3x^2 + 9x - 3x^2 - 9x - 27; the 3x^2 and 9x terms cancel exactly, leaving x^3 - 27, which does match the numerator, so step 1's factorisation is correct.
   • Why not B: Step 2 does not multiply anything: it substitutes x^3 - 27 with the exactly equal expression (x - 3)(x^2 + 3x + 9) from step 1, leaving the denominator (x - 3) untouched. Substituting equals for equals leaves the fraction's value unchanged, so step 2 introduces no error.
   • Why not D: Step 4 involves no limit at all: it simply restates the equation from step 3 verbatim, adding the words 'for every real number x'. There is no limiting argument to be missing; the real fault is that step 3's unqualified cancellation is itself unjustified at x = 3.

2. Question 2Answer: B

   1. The statement being proved is: n^2 divisible by 3 implies n divisible by 3 (which happens to be true, since 3 is prime).
   2. A valid proof needs to start from 'n^2 divisible by 3' and deduce 'n divisible by 3', without assuming the conclusion anywhere along the way.
   3. Step 2 instead assumes n = 3k, that is, it assumes n is divisible by 3, which is exactly the conclusion the argument is meant to establish; step 3 then only shows that IF n is divisible by 3, n^2 is too, which is the converse direction and was not in question.
   4. So the argument never actually shows that n^2 being divisible by 3 forces n to be divisible by 3; it merely shows the reverse implication while assuming the very thing it needed to prove. The flaw is in step 2, and the answer is B.
   • Why not A: The statement being proved is a conditional: if n^2 is divisible by 3, then n is divisible by 3. Step 1 begins a direct proof by assuming the hypothesis and writing it as n^2 = 3m; this is exactly how such a proof should start, not circular reasoning, since the hypothesis is given, not being proved.
   • Why not C: Squaring a product does distribute over multiplication: (3k)^2 = 3^2 * k^2 = 9k^2 exactly, by the laws of multiplication. There is no arithmetic error in step 3; this option confuses the rule for squaring a product with the different rule for squaring a sum.
   • Why not D: Step 4's conclusion follows directly from the assumption n = 3k made in step 2, independent of how divisible n^2 happens to be; that n^2 = 9k^2 is divisible by 9 does not invalidate concluding n is divisible by 3. Step 4 is a valid restatement of step 2; the real problem is that step 2 assumed the conclusion in the first place.

3. Question 3Answer: A

   1. Steps 1 and 2 are valid: for any integer n, m = n + 1 is an integer greater than n, and since n was an arbitrary integer, this genuinely shows that for every integer n there is some integer (depending on n) exceeding it.
   2. Step 3 wrongly swaps the order of the quantifiers: 'for every n there exists an m (possibly different for each n) with m > n' does not mean 'there exists a single m that exceeds every n at once'. The value of m found in steps 1 and 2 depends on n (it is n + 1), so it cannot be pulled out as one fixed M that works for every n simultaneously.
   3. The resulting statement is in fact false: for any proposed integer M, taking n = M gives the requirement M > M, which is false. So no integer M can exceed every integer.
   4. The flaw is the quantifier swap in step 3, so the answer is A.
   • Why not B: The integers are closed under addition regardless of sign: adding 1 to any integer, positive, negative or zero, always gives another integer, for example (-5) + 1 = -4. There is no exception for negatives, so step 1's claim that m = n + 1 is an integer holds for every integer n.
   • Why not C: Proving a statement for an arbitrary, but fixed, integer n and then concluding it holds for every integer is exactly the standard way universal statements are proved; step 2 is a legitimate generalisation, not an error.
   • Why not D: Step 4 restates exactly the claim step 3 already reached, without relying on any particular numerical choice of M; whether M would look different from a different starting point does not change that step 3's own claim, once granted, implies step 4. The genuine error is the quantifier swap made in step 3 itself, not a further slip in step 4.

4. Question 4Answer: D

   1. Dividing both sides of sin x cos x = (1/2) sin x by sin x is only a valid step when sin x != 0; if sin x could be zero, dividing by it discards information.
   2. Within 0 <= x < 360 degrees, sin x = 0 exactly at x = 0 and x = 180 degrees. Substituting x = 0 into the original equation gives sin(0) cos(0) = 0 and (1/2) sin(0) = 0, so 0 = 0 holds; the same check works at x = 180 degrees, since sin(180) = 0 also.
   3. So x = 0 and x = 180 degrees are genuine solutions that step 2's division silently discards, alongside the two solutions x = 60 and x = 300 degrees correctly found from cos x = 1/2.
   4. The complete solution set is x = 0, 60, 180 and 300 degrees, not just x = 60 and x = 300 degrees, so the flaw is the unjustified division in step 2. The answer is D.
   • Why not A: Cosine is positive in both the first and fourth quadrants, so cos x = 1/2 has two solutions in 0 <= x < 360 degrees: x = 60 degrees and x = 360 - 60 = 300 degrees. Restricting to the first quadrant only misses the fourth-quadrant solution, so step 3's two solutions are both correct.
   • Why not B: sin x and cos x are defined for every real x with no domain restriction, so sin x cos x and (1/2) sin x are both defined throughout 0 <= x < 360 degrees. There is nothing to check before stating the equation, and step 1 simply records it, introducing no error.
   • Why not C: Step 3 does find every solution of cos x = 1/2 in the range, since cosine takes each value in (-1, 1) exactly twice per full turn; step 4's restatement is accurate for the reduced equation, and introduces no new deduction. The real fault is that step 2's division discarded other solutions of the ORIGINAL equation.

5. Question 5Answer: D

   1. Steps 1 to 3 are routine and correct: subtracting a^3 from both sides preserves the inequality, and b^3 - a^3 = (b - a)(b^2 + ab + a^2) is the standard difference-of-cubes factorisation.
   2. Step 4's algebra is exact: completing the square gives b^2 + ab + a^2 = (a + b/2)^2 + (3/4)b^2, a sum of two squares, so this quantity can never be negative, and it is 0 only when both squared terms vanish, which forces b = 0 and then a = 0.
   3. Step 5 correctly rules out the one case (a = b = 0) where b^2 + ab + a^2 could be 0, because a^3 < b^3 cannot hold when a = b = 0 (that would require 0 < 0, which is false). So under the given hypothesis, b^2 + ab + a^2 is strictly positive, and dividing the inequality (b - a)(b^2 + ab + a^2) > 0 by this strictly positive quantity to get b - a > 0 is entirely valid.
   4. Unlike the reciprocal and AM-GM traps elsewhere in this test, cubing (unlike squaring or taking reciprocals) is strictly increasing across all real numbers with no sign restriction, so the implication a^3 < b^3 => a < b genuinely holds for every pair of real numbers, and every step of this proof is sound. The answer is D.
   • Why not A: Step 1 introduces a and b as arbitrary real numbers, not particular fixed values, so any conclusion reached about them under the stated hypothesis genuinely holds for every such pair; step 6's generalisation is the standard closing step used throughout this test, and it introduces no new gap.
   • Why not B: Completing the square holds for every real a and b with no sign restriction: expanding (a + b/2)^2 + (3/4)b^2 gives a^2 + ab + b^2/4 + 3b^2/4 = a^2 + ab + b^2, matching b^2 + ab + a^2 exactly regardless of sign. Being a sum of two squares, it is nonnegative for every real a and b, so step 4 is correct as stated.
   • Why not C: This sounds like the sign-restriction traps that appear elsewhere in this test, but it does not apply here: step 4 has already shown b^2 + ab + a^2 >= 0 always, and step 5 rules out the one case where it could equal 0 (a = b = 0), using the fact that a^3 < b^3 cannot hold when a = b = 0. So the sign of b^2 + ab + a^2 is not actually unknown here; it is known, strictly positive, under the given hypothesis, and dividing by it is entirely legitimate.

6. Question 6Answer: A

   1. The claim itself is false in general, and the flaw is that step 2 invokes a 'standard fact' that is not actually true.
   2. A counterexample: sqrt(2) is irrational, and -sqrt(2) is also irrational, but sqrt(2) + (-sqrt(2)) = 0, which is rational. So the sum of two irrational numbers can be rational.
   3. A second counterexample confirms this is not a one-off: sqrt(2) and 5 - sqrt(2) are both irrational, but their sum is 5, which is rational.
   4. So step 2's claimed fact is unjustified and false, and the argument gives no genuine reason to believe a + b must be irrational whenever a and b are. The answer is A.
   • Why not B: Assuming a and b are irrational is simply stating the hypothesis of the claim being investigated; it introduces no error.
   • Why not C: Step 2's claimed fact is stated with no extra conditions attached, so applying it to any pair of irrational numbers, including the a and b from step 1, needs no further checking. The actual problem is that step 2's fact is simply false, not that step 3 misapplies it.
   • Why not D: This restates the same false 'fact' as though it were part of the definition of irrational numbers; no such guarantee exists, as the counterexamples sqrt(2) + (-sqrt(2)) = 0 and sqrt(2) + (5 - sqrt(2)) = 5 show.

7. Question 7Answer: D

   1. The negation of a 'for every x, P(x)' statement is 'there exists an x with not P(x)', not 'for every x, not P(x)'. Step 1 confuses these two, which have very different logical content.
   2. To disprove 'for every real x, x^2 >= x', a single counterexample suffices: x = 0.5 gives x^2 = 0.25, and 0.25 < 0.5, so the inequality x^2 >= x fails at x = 0.5.
   3. So the original statement is actually false, and the student's conclusion in step 3, that it is true, is wrong; the argument reached this wrong conclusion only because it tested the wrong (misnegated) statement in steps 1 and 2.
   4. The flaw is the mis-stated negation in step 1, so the answer is D.
   • Why not A: Squaring a number means multiplying it by itself exactly once, i.e. x^2 = x * x, not x * x * x (which is x cubed). So 2^2 = 2 * 2 = 4 exactly, and there is no arithmetic error in step 2; the invented rule in this option does not exist.
   • Why not B: To disprove a 'for every x' statement, exactly one counterexample suffices by definition: a single x for which the statement fails already shows it is not true of every x. Step 3 correctly uses x = 2 as such a counterexample, so no further checking is needed.
   • Why not C: The original statement 'for every real number x, x^2 >= x' is in fact false (x = 0.5 is a counterexample), so the student's conclusion that it is true is wrong, not correct.

8. Question 8Answer: B

   1. The power rule log(a^n) = n log(a) is only valid when a > 0, since log is only defined on positive real numbers.
   2. Step 1 correctly notes that x^2 > 0 for every nonzero x, so log(x^2) is always defined; but step 2 applies the power rule as if x itself, not x^2, were the base, which requires x > 0.
   3. When x is negative, log(x) simply does not exist as a real number, so the right-hand side 2 log(x) is undefined, while the left-hand side log(x^2) is perfectly well defined; the two sides cannot be compared, let alone equal, for negative x.
   4. The correct general identity, valid for every nonzero real x, is log(x^2) = 2 log|x|; step 2's version only holds for x > 0. The flaw is in step 2, so the answer is B.
   • Why not A: Squaring any nonzero real number, however small, always gives a strictly positive result: for example, (0.0001)^2 = 0.00000001, small but still greater than zero, and it can never actually reach zero for a nonzero input. So x^2 > 0 for every nonzero real x, and step 1 is correct.
   • Why not C: Step 2's rule is already stated generally, with x standing for an arbitrary nonzero real number throughout, so step 3's restatement adds no new content beyond what step 2 already claims. The actual flaw is that step 2's rule is not valid for every nonzero real x in the first place.
   • Why not D: The power rule for logarithms is not a general algebraic identity like multiplication or addition rules; it depends on logarithms only being defined for positive reals, and it fails to even make sense for negative x, since log(x) is then undefined. So the argument is not valid for every nonzero real x, only for positive x.

9. Question 9Answer: D

   1. Every integer is congruent to 0, 1 or 2 modulo 3, and these residues occur in order as consecutive integers increase, so among n, n+1, n+2 exactly one is a multiple of 3; step 2 is correct.
   2. Similarly, consecutive integers alternate in parity, so among n, n+1, n+2 at least one is even; step 1 is correct.
   3. Because 2 and 3 share no common factor, divisibility by each can be tracked independently: if some factor of the product supplies a factor of 2 and some factor supplies a factor of 3 (whether or not it is the same one), the overall product is divisible by 2 x 3 = 6.
   4. Every step is justified, so the proof is genuinely complete and valid; there is no error to identify. The answer is D.
   • Why not A: 1, 3 and 5 are not consecutive integers; they are consecutive odd numbers, spaced two apart. Consecutive integers strictly alternate in parity, so among any three in a row at least one is always even, and step 1 is correct.
   • Why not B: In 3, 4, 5 exactly one number, 3, is a multiple of 3; the example does not actually show two multiples of 3 among three consecutive integers. Since remainders on division by 3 cycle 0, 1, 2, exactly one of any three consecutive integers is always divisible by 3, and step 2 is correct.
   • Why not C: Since 2 and 3 are coprime, if one factor of a product is divisible by 2 and some factor (the same one or a different one) is divisible by 3, their contributions multiply: the product is divisible by 2 x 3 = 6 regardless of whether it is the same factor supplying both or two different factors, so step 3's combination is valid.

10. Question 10Answer: A

    1. Taking reciprocals does reverse an inequality between two numbers of the same sign: for positive a < b, 1/a > 1/b; for negative a < b, likewise 1/a > 1/b.
    2. But step 2 claims this always happens, with no restriction on sign. When a and b have opposite signs, this fails.
    3. Counterexample: a = -1 and b = 1 satisfy a < b (since -1 < 1). But 1/a = -1 and 1/b = 1, and -1 > 1 is false, so the claimed conclusion 1/a > 1/b does not hold here.
    4. So the rule in step 2 needs a and b to share the same sign, which the argument never establishes; the flaw is in step 2, and the answer is A.
    • Why not B: Assuming a and b are nonzero is simply stating part of the hypothesis of the claim under investigation; it introduces no error.
    • Why not C: Restricting to integers does not fix the problem: a = -2 and b = 3 are both integers with a < b, but 1/a = -0.5 and 1/b = 1/3, and -0.5 > 1/3 is false, so the rule still fails across a sign change even among integers.
    • Why not D: Step 2 already states its rule using the letters a and b from step 1, so it is not a generic rule waiting to be applied; step 3 simply restates what step 2 already concluded about the specific a and b under discussion. The genuine flaw is that step 2's rule is false as stated, not that step 3 fails to connect it to a and b.

11. Question 11Answer: B

    1. Splitting |x - 1| = 2x - 5 into the two cases x - 1 = 2x - 5 and x - 1 = -(2x - 5) is the standard way to remove an absolute value, and the algebra in steps 1 and 2 is correct, giving x = 4 and x = 2 respectively.
    2. However, solving each case equation only produces candidate solutions; each candidate must be checked in the original equation, since |x - 1| is always nonnegative, so it can never equal a negative right-hand side.
    3. Checking x = 4: |4 - 1| = 3, and 2(4) - 5 = 3, so 3 = 3 and x = 4 is genuine.
    4. Checking x = 2: |2 - 1| = 1, but 2(2) - 5 = -1, and 1 != -1, so x = 2 fails; it is an extraneous solution introduced by the case split. The only genuine solution is x = 4, and the flaw is step 3's unchecked assumption, so the answer is B.
    • Why not A: Rearranging x - 1 = 2x - 5 correctly, by adding 5 and subtracting x from both sides, gives 5 - 1 = 2x - x, that is 4 = x, so x = 4 exactly as step 1 states; moving terms to the wrong sides is what produces the wrong value 2 in this option.
    • Why not C: Distributing a minus sign across a bracket flips the sign of every term inside it: -(2x - 5) = -2x + 5, exactly as step 2 states. Flipping only the first term's sign is incorrect; a leading minus sign distributes over both terms.
    • Why not D: Step 4 does exactly what a valid closing step does: it restates step 3's claim as the final answer, adding nothing new. Step 4 follows from step 3 precisely as written; the actual fault lies in step 3, which wrongly accepts both case-solutions without checking either.

12. Question 12Answer: C

    1. The AM-GM inequality (a + b)/2 >= sqrt(ab) is a genuine theorem, but it is only stated, and only true, for nonnegative real numbers a and b.
    2. Step 2 applies it to arbitrary real numbers without this restriction. If a and b have opposite signs, ab < 0 and sqrt(ab) is not a real number at all, so the inequality does not even make sense.
    3. Even when a and b are both negative, so that ab > 0 and sqrt(ab) is real, the inequality can still fail: take a = b = -1. Then ab = 1, so sqrt(ab) = 1 and 2 sqrt(ab) = 2, but a + b = -2, and -2 >= 2 is false.
    4. So the claim as stated, for all real numbers, is false; the correct restriction is a, b >= 0. The flaw is applying AM-GM outside its valid domain in step 2, so the answer is C.
    • Why not A: Introducing two arbitrary real numbers a and b to state a general claim about them is standard practice, and is not itself an error.
    • Why not B: This is exactly the false claim the argument makes; AM-GM in this square-root form is only stated (and only true) for nonnegative a and b, as the counterexample a = b = -1 shows.
    • Why not D: Multiplying both sides of an inequality by the same positive constant multiplies every term by that constant, including root terms: sqrt(ab) multiplied by 2 is simply 2 sqrt(ab), not 2ab, which is a different operation (squaring), not multiplying by 2. Step 3's algebra is correct.