TMUA Paper 2: Identifying errors in proofs, set 2

• Answer all questions. No calculator.
• Each question has exactly one correct answer.

1. 11 mark

   A student claims to prove the general algebraic law: for all real numbers x, y and z, if xy = xz then y = z.

   Purported proof: Suppose xy = xz. Divide both sides by x: y = z.

   What is wrong with this argument?

   1. A The proof is valid: subtracting gives x(y - z) = 0, and a product equal to zero means y - z = 0, so y = z follows regardless of the value of x.
   2. B The proof is invalid, but only because dividing an equation by a negative number requires reversing the direction of the equation; no such reversal was carried out here.
   3. C The proof is invalid because dividing an equation by x is only valid when x is known to be nonzero; when x = 0 the claim itself is false in general.
   4. D The proof is invalid because it should have subtracted xz from both sides before dividing; dividing directly, without first subtracting, is not a legitimate step.
2. 21 mark

   A student is solving the equation cos(2x) = cos(40 degrees) for x in the range 0 <= x <= 180 degrees.

   Step 1: cos(2x) = cos(40 degrees).

   Step 2: Since cosine assigns exactly one output to each input, cos(2x) = cos(40 degrees) forces 2x = 40 degrees.

   Step 3: So x = 20 degrees.

   Step 4: Therefore x = 20 degrees is the only solution in the range 0 <= x <= 180 degrees.

   Which step contains the invalid deduction?

   1. A Step 2, because cosine is not one-to-one: cos(320 degrees) = cos(40 degrees) too, so 2x = 40 degrees is not the only possibility, and a genuine second solution in the stated range is silently ruled out.
   2. B Step 3, because dividing 2x = 40 degrees by 2 to get x = 20 degrees is an arithmetic error, and correcting that division would change the numerical value of x reported in step 3.
   3. C Step 1, because cos(2x) cannot be compared to cos(40 degrees) unless x is already known to equal 20 degrees, which makes writing the equation down presuppose the very answer being sought.
   4. D Step 4, because the domain 0 <= x <= 180 degrees stated in the question was never justified by the student, so no conclusion may legitimately be drawn within it.
3. 31 mark

   A student claims to prove that sqrt(x^2) = x for every real number x, arguing as follows: squaring a number and then taking its square root are inverse operations, so applying one after the other returns the original number, for any real x.

   What is wrong with this argument?

   1. A The argument is invalid because sqrt(x^2) is not defined for negative values of x, so the identity cannot even be stated once x is allowed to be negative.
   2. B The argument is valid: squaring then taking the square root always returns the original number, since the two operations undo one another for every real number.
   3. C The argument is invalid because sqrt(x^2) = x^2 / x, which is undefined when x = 0, so the identity sqrt(x^2) = x cannot hold at that one value.
   4. D The argument is invalid because squaring is not always reversible: sqrt(x^2) actually equals |x|, not x, so the claim fails whenever x is negative.
4. 41 mark

   A student claims to prove: for every real number x, if x^2 > 9 then x > 3.

   Step 1: Suppose x^2 > 9.

   Step 2: Take the (positive) square root of both sides: x > 3.

   Step 3: Therefore x > 3.

   Which step contains the invalid deduction?

   1. A Step 2, because taking square roots of an inequality always reverses its direction, the same rule that applies when multiplying or dividing an inequality by a negative number.
   2. B Step 2, because sqrt(x^2) equals |x|, not x, so x^2 > 9 gives |x| > 3, meaning x > 3 or x < -3; the argument only captures one of these two branches.
   3. C Step 1, because squaring an inequality does not preserve its truth value when x could be negative, so the hypothesis itself cannot be trusted before any root is taken.
   4. D The argument is valid: since x^2 > 9 means x is a number whose square exceeds 9, and only numbers greater than 3 have squares that large.
5. 51 mark

   A student argues as follows.

   Step 1: It is known that if an integer n is a perfect square, then n >= 0 (since the square of any integer is never negative).

   Step 2: Suppose n >= 0.

   Step 3: By the fact stated in step 1, being a perfect square gives n >= 0. Since n >= 0, this same fact tells us that n must be a perfect square.

   Step 4: Therefore every nonnegative integer is a perfect square.

   Which step contains the invalid deduction?

   1. A Step 1, because it is false that every perfect square is nonnegative: the square of a negative integer is itself negative, so step 1's claim already breaks down.
   2. B Step 2, because supposing n >= 0 without first proving it is not a legitimate step, and no conclusion built on an unproven assumption can ever be trusted.
   3. C Step 3, because it assumes the converse of the true statement in step 1: from 'perfect square implies nonnegative' it wrongly infers 'nonnegative implies perfect square'.
   4. D Step 4, because it draws a universal conclusion, 'every nonnegative integer is a perfect square', from step 3's reasoning about the single generic integer n.
6. 61 mark

   A student claims to prove that 6n + 1 is a prime number for every positive integer n.

   Step 1: When n = 1: 6(1) + 1 = 7, which is prime.

   Step 2: When n = 2: 6(2) + 1 = 13, which is prime.

   Step 3: When n = 3: 6(3) + 1 = 19, which is prime.

   Step 4: Since the expression is prime for n = 1, 2 and 3, it is prime for every positive integer n.

   Which step is the point where the argument stops being a valid proof?

   1. A Step 2, because 6(2) + 1 is 12, not 13: multiplying 6 by 2 gives 12, and if the final addition is mistakenly skipped, that wrong total is what gets carried forward.
   2. B Step 3, because 19 is not a prime number: a candidate rushing through the sequence might confuse it with 21, which does factor as 3 x 7, and wrongly carry that composite status onto 19 instead.
   3. C Step 1, because 6(1) + 1 is 6, not 7: this follows if the '+1' is mistakenly read as multiplying by 1 rather than adding it, collapsing the expression down to 6(1).
   4. D Step 4, because checking finitely many cases can never establish a statement for every positive integer; indeed at n = 4, 6n + 1 = 25 = 5^2, which is not prime, so the claim is false.
7. 71 mark

   A student claims to prove that |x - 3| = x - 3 for every real number x.

   Step 1: Case x >= 3: then x - 3 >= 0, so |x - 3| = x - 3.

   Step 2: Therefore |x - 3| = x - 3 for every real number x.

   What is wrong with this argument?

   1. A The argument only establishes the case x >= 3 and never examines x < 3, where |x - 3| equals 3 - x instead of x - 3, leaving the case split incomplete.
   2. B Step 1 contains an algebraic error: when x >= 3, x - 3 is not necessarily nonnegative, since subtracting a fixed number from a large enough x could still leave a result below zero.
   3. C Step 2 is invalid because it restates step 1 without adding new information, and repeating a case that has already been shown can never establish a claim about every real number.
   4. D The claim is true and the argument is valid, since |x - 3| is defined as x - 3 whenever the quantity inside is being measured for its size, which step 1 already shows.
8. 81 mark

   A student gives the following argument to show that n! > 2^n for every integer n >= 4, using induction.

   Step 1: Inductive step: assume k! > 2^k for some integer k >= 4. We show (k + 1)! > 2^(k + 1).

   Step 2: (k + 1)! = (k + 1) x k! > (k + 1) x 2^k, using the inductive hypothesis and the fact that k + 1 > 0.

   Step 3: Since k >= 4, k + 1 >= 5 > 2, so (k + 1) x 2^k > 2 x 2^k = 2^(k + 1).

   Step 4: Therefore, by induction, n! > 2^n for every integer n >= 4.

   What is the flaw in this argument?

   1. A Step 2 contains an algebraic error: (k + 1) x k! is not equal to (k + 1)!, because expanding the factorial that way drops a factor somewhere in the multiplication.
   2. B No base case is ever verified, so however correct the inductive step is, it never establishes a starting point, and the proof as given is incomplete.
   3. C Step 3 is invalid because k + 1 > 2 does not imply (k + 1) x 2^k > 2 x 2^k, since multiplying both sides of an inequality by a large power of 2 can sometimes reverse its direction.
   4. D No step is invalid; the proof is a valid and complete induction, since the inductive step alone is always sufficient to establish a statement for every integer above the starting point.
9. 91 mark

   A student claims to prove that (2^m)^n = 2^(m + n) for all positive integers m and n, as follows.

   Step 1: By the law of indices, 2^(m + n) = 2^m x 2^n.

   Step 2: Also by definition, (2^m)^n is the product of n copies of 2^m, that is 2^m x 2^m x ... x 2^m (n factors).

   Step 3: Both (2^m)^n and 2^(m + n) are products of powers of 2, so they must be the same product: (2^m)^n = 2^m x 2^n = 2^(m + n).

   Step 4: Therefore (2^m)^n = 2^(m + n) for all positive integers m and n.

   What is wrong with this argument?

   1. A Step 1 is invalid, because 2^(m + n) is not equal to 2^m x 2^n in general: adding exponents like this only works when the two powers involved have different bases, not the same one.
   2. B Step 2 is invalid, because (2^m)^n is not equal to a product of n copies of 2^m: raising to the power n should instead mean multiplying 2^m by n, giving n lots of 2^m added together rather than multiplied.
   3. C The argument is valid: since both (2^m)^n and 2^(m + n) are built as products of powers of 2, any two such products must agree, so the identity holds for every positive integer m and n.
   4. D Step 3 is invalid: (2^m)^n has n factors of 2^m, giving 2^(mn), not the two-factor product 2^m x 2^n = 2^(m + n); the number of factors on each side is different, so equating them is unjustified.
10. 101 mark

    A student claims to prove: for every nonzero real number x, if x^2 > x then x > 1.

    Step 1: Suppose x^2 > x (with x != 0).

    Step 2: Divide both sides by x: x > 1.

    Step 3: Therefore x > 1.

    Which step contains the invalid deduction?

    1. A Step 2 contains an arithmetic slip: dividing x^2 by x should give x^2, not x, as if the division left the power completely unchanged rather than reducing it by one.
    2. B Step 1 is invalid because stating the hypothesis x^2 > x already assumes what is to be proved, since writing down any inequality in x presupposes a value of x has been fixed.
    3. C Step 2 is invalid because dividing both sides of an inequality by x is only safe when the sign of x is known; a negative x would reverse the inequality's direction, which step 2 ignores entirely.
    4. D Step 3 is invalid because 'x > 1' was already concluded in step 2, so restating it cannot serve as a legitimate closing step of any proof.
11. 111 mark

    A student gives the following proof that the product of two consecutive integers is always even.

    Step 1: Let n be an integer, and consider the two consecutive integers n and n + 1.

    Step 2: Case n is even: n = 2k for some integer k. Then n(n + 1) = 2k(n + 1), which is even.

    Step 3: Case n is odd: then n + 1 is even, so n + 1 = 2k for some integer k. Then n(n + 1) = n(2k) = 2kn, which is even.

    Step 4: In both cases, n(n + 1) is even, so the product of two consecutive integers is always even.

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

    1. A Step 2 is invalid because n = 2k does not represent every even integer, since some even integers cannot be written in that form for any choice of k.
    2. B No step is invalid; the proof correctly covers both possible cases, n even and n odd, and is a complete, valid proof.
    3. C Step 3 is invalid because n odd does not guarantee that n + 1 is even, since for some odd values of n adding 1 can still leave an odd result.
    4. D Step 4 is invalid because the two cases in steps 2 and 3 do not together cover every integer, leaving some integers neither even nor odd unaccounted for.
12. 121 mark

    A student claims to prove that (a + b)^2 = a^2 + b^2 for all real numbers a and b.

    Step 1: Let a = 5 and b = 0.

    Step 2: Compute (a + b)^2 = 5^2 = 25, and a^2 + b^2 = 25 + 0 = 25, so the two expressions agree.

    Step 3: Since they agree, (a + b)^2 = a^2 + b^2 for all real numbers a and b.

    Step 4: Therefore (a + b)^2 = a^2 + b^2 for all real a and b.

    What is wrong with this argument?

    1. A The test case chosen (b = 0) makes both sides reduce to a^2 automatically, since (a + 0)^2 = a^2 and a^2 + 0^2 = a^2 for any value of a; this test can never distinguish a true identity from a false one.
    2. B Step 2 contains an arithmetic error: 5^2 is not equal to 25, because squaring 5 means doubling it rather than multiplying it by itself.
    3. C Step 1 is invalid because choosing specific numerical values for a and b is never permitted in an argument about all real numbers, since any single substitution automatically narrows the claim to that one case only.
    4. D Step 4 is invalid because it does not follow from step 3 as stated: step 3 only asserts a claim about the single test case a = 5, b = 0, and step 4 introduces a universal claim with no new justification.

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 claimed law is false in general: take x = 0, y = 1, z = 2. Then xy = 0 x 1 = 0 and xz = 0 x 2 = 0, so xy = xz holds, but y = 1 and z = 2 are not equal.
   2. The purported proof's flaw is dividing both sides of xy = xz by x without first establishing that x is nonzero. Dividing an equation by a quantity is only a valid step when that quantity is known not to be zero.
   3. Rearranging gives x(y - z) = 0, and a product of two real numbers is zero exactly when at least one of the factors is zero: either x = 0 (in which case y and z can be anything) or y - z = 0 (in which case y = z). Only the second case gives the claimed conclusion.
   4. So the correct statement is: if xy = xz and x != 0, then y = z. The unrestricted version, without the condition x != 0, is false, and the fallacy of quietly dividing by a quantity that might be zero is exactly the error the specification calls out for 'if ab = ac, then b = c'.
   • Why not A: A product equal to zero means at least one factor is zero, not necessarily the second one: here it could be x that is zero, which is exactly what makes the original claim false, so this option's reasoning does not establish y = z at all.
   • Why not B: Dividing an equation by a negative number never requires reversing anything; that rule applies only to inequalities, so this identifies a rule from the wrong context.
   • Why not D: Subtracting before dividing and dividing directly reach the same place; requiring a specific order of steps is not itself a mathematical error, so this option does not identify the actual flaw.

2. Question 2Answer: A

   1. Cosine is an even, periodic function, so cos(theta) = cos(-theta) and cos(theta) = cos(360 degrees - theta) for every angle theta; it is not one-to-one, and cos A = cos B does not force A = B.
   2. For cos(2x) = cos(40 degrees), the general solutions are 2x = 40 degrees + 360k degrees or 2x = -40 degrees + 360k degrees, for any integer k.
   3. Restricting to 0 <= x <= 180 degrees means 0 <= 2x <= 360 degrees. Within that range, 2x = 40 degrees gives x = 20 degrees, and 2x = 320 degrees (from -40 + 360) gives x = 160 degrees; checking, cos(320 degrees) = cos(360 - 40) = cos(40 degrees), confirming x = 160 degrees is a genuine second solution.
   4. So the argument's step 2 wrongly assumes cosine is injective, and misses x = 160 degrees; the correct solution set in this range is x = 20 degrees or x = 160 degrees, not x = 20 degrees alone.
   • Why not B: 40 degrees divided by 2 is exactly 20 degrees; there is no arithmetic error in this division.
   • Why not C: Setting two expressions equal in order to solve for an unknown is exactly how such equations are normally approached; nothing about writing down cos(2x) = cos(40 degrees) assumes the answer.
   • Why not D: The domain is given as part of the problem, not something the argument needs to justify; the actual flaw is that the argument misses a genuine solution that lies inside that stated domain.

3. Question 3Answer: D

   1. By convention, sqrt denotes the nonnegative square root, so sqrt(t) >= 0 for every t >= 0, and in particular sqrt(x^2) >= 0 for every real x.
   2. When x >= 0, sqrt(x^2) = x, since x itself is already the nonnegative number whose square is x^2.
   3. When x < 0, x is negative, so sqrt(x^2) cannot equal x (a negative number); instead sqrt(x^2) = -x, which is positive. For example, sqrt((-3)^2) = sqrt(9) = 3 = -(-3), not -3.
   4. So the correct general identity is sqrt(x^2) = |x| for every real x, which agrees with x only when x >= 0; the claim that squaring and square-rooting are always mutual inverses ignores that squaring first destroys the sign of a negative number, so the whole original value cannot be recovered.
   • Why not A: sqrt(x^2) is defined for every real x, since x^2 is never negative; the flaw is not a domain restriction but which value the square root actually returns.
   • Why not B: Squaring is not injective, since both 3 and -3 square to 9, so square-rooting a squared number does not always return the original value; the two operations only genuinely undo one another when x is already known to be nonnegative.
   • Why not C: sqrt(x^2) is not the same expression as x^2 / x; this manipulation is not how the square root function is defined or evaluated, and it does not identify the real flaw.

4. Question 4Answer: B

   1. x^2 > 9 rearranges to x^2 - 9 > 0, that is (x - 3)(x + 3) > 0, which holds exactly when x > 3 or x < -3.
   2. Taking square roots correctly gives sqrt(x^2) > sqrt(9), that is |x| > 3, since sqrt(x^2) = |x| for every real x, not x itself.
   3. |x| > 3 means x > 3 or x < -3, but step 2 only keeps the first branch and silently drops the second.
   4. A concrete counterexample confirms the claimed implication is false: x = -4 satisfies x^2 = 16 > 9, but x = -4 is not greater than 3, so the argument's conclusion does not hold for every x satisfying the hypothesis.
   • Why not A: Taking the square root of a larger positive quantity does not automatically reverse an inequality; the actual issue is which square root value applies, not the direction of the inequality.
   • Why not C: Step 1 states the hypothesis x^2 > 9 itself; no squaring operation is performed by the student in this step, so there is nothing here to have preserved or failed to preserve.
   • Why not D: This restates the false claim as if it were self-evidently true; x = -4 is a direct counterexample, since (-4)^2 = 16 > 9 but -4 is not greater than 3.

5. Question 5Answer: C

   1. The true fact is: n a perfect square implies n >= 0. This does not mean the reverse also holds.
   2. Step 3 uses the true implication backwards, assuming that because n satisfies the consequent (n >= 0), it must also satisfy the antecedent (n is a perfect square). This is assuming the converse of a true statement, which need not itself be true.
   3. A single counterexample shows the conclusion is false: n = 2 is a nonnegative integer, but there is no integer whose square equals 2 (1^2 = 1 and 2^2 = 4, and no integer lies between 1 and 2), so 2 is not a perfect square.
   4. So the argument's flaw is in step 3, where a valid implication is used in the wrong direction.
   • Why not A: The statement in step 1 is true: squaring any real number, positive or negative, always gives a nonnegative result (e.g. (-3)^2 = 9, not -9), since a negative sign does not survive being squared; so this step contains no error.
   • Why not B: Introducing a hypothesis to be investigated is standard practice in a conditional argument and is not itself an invalid deduction.
   • Why not D: n was introduced in step 2 as an arbitrary nonnegative integer, not one specific value, so restating the conclusion for every nonnegative integer is a faithful restatement of step 3, not an unjustified leap; step 4 follows exactly from step 3, and the actual flaw lies in step 3 itself.

6. Question 6Answer: D

   1. Each of the individual checks in steps 1 to 3 is arithmetically correct: 7, 13 and 19 are all genuinely prime.
   2. The flaw is entirely in the final step: verifying a formula for finitely many values of n, however many, never proves it holds for every positive integer, because infinitely many values remain unchecked.
   3. In fact the claim is false: at n = 4, 6n + 1 = 6(4) + 1 = 25, and 25 = 5 x 5 is not prime.
   4. So the pattern that holds for n = 1, 2 and 3 breaks down at n = 4, which is exactly the kind of case a hasty generalisation from a few examples misses.
   • Why not A: 6(2) + 1 = 12 + 1 = 13 exactly as stated, with an explicit addition step that is not skipped anywhere in the argument; there is no arithmetic error here.
   • Why not B: 19 is a prime number, since its only positive divisors are 1 and 19; confusing it with the unrelated number 21 = 3 x 7 does not change what 19 itself factors as, so this step contains no error.
   • Why not C: 6(1) + 1 = 6 + 1 = 7 exactly as stated: the '+1' is an addition performed after the multiplication 6(1) = 6, not folded into it, so there is no arithmetic error here.

7. Question 7Answer: A

   1. The true fact is that |x - 3| = x - 3 when x - 3 >= 0, that is when x >= 3, and |x - 3| = -(x - 3) = 3 - x when x - 3 < 0, that is when x < 3.
   2. Step 1 correctly establishes the case x >= 3, but the argument never considers x < 3.
   3. A counterexample in the missing case shows the claim as stated is false: at x = 0, |x - 3| = |-3| = 3, but x - 3 = 0 - 3 = -3, and 3 != -3.
   4. So the flaw is an incomplete case split: a genuine proof about |x - 3| for all real x needs both branches of the absolute value definition, and this argument only ever proves one of them.
   • Why not B: When x >= 3, subtracting 3 from both sides gives x - 3 >= 0 directly from the definition of >=, with no exception for a 'large enough' x; there is no algebraic error in this step.
   • Why not C: Step 2 does add something beyond step 1: it generalises a result shown only for x >= 3 into a claim about every real x, which is exactly the unjustified leap that breaks the argument, not merely an empty restatement.
   • Why not D: Absolute value is defined as |t| = t when t >= 0 and |t| = -t when t < 0; the claim ignores the second case entirely, so the argument is not valid and the claim is false for x < 3.

8. Question 8Answer: B

   1. A valid proof by induction needs two parts: a base case, verifying the statement for the smallest relevant value of n, and an inductive step, showing the statement for k + 1 assuming it for k.
   2. The algebra in steps 2 and 3 is correct: (k + 1)! does equal (k + 1) x k! by definition, and multiplying the inductive hypothesis by the positive quantity k + 1, then bounding k + 1 below by 2, does validly give (k + 1)! > 2^(k + 1).
   3. But the base case, checking that the statement holds at the smallest value n = 4 (namely 4! = 24 and 2^4 = 16, and 24 > 16), is never stated or verified anywhere in the argument.
   4. Without a verified base case, the inductive step has nothing to build from: assuming the statement is true for some k and deducing it for k + 1 tells you nothing unless you know it is true for at least one starting value, so the proof as given is incomplete.
   • Why not A: (k + 1)! = (k + 1) x k! is exactly the definition of factorial applied to k + 1, so this step contains no algebraic error.
   • Why not C: Multiplying both sides of the true inequality k + 1 > 2 by the positive quantity 2^k always preserves the inequality's direction, however large that power of 2 is, so (k + 1) x 2^k > 2 x 2^k does follow validly.
   • Why not D: The inductive step itself is valid, but a base case is never verified anywhere in the argument, so the proof is not in fact complete.

9. Question 9Answer: D

   1. By the law of indices, 2^(m + n) = 2^m x 2^n, which has exactly two factors: one factor of 2^m and one factor of 2^n. Step 1 states this correctly.
   2. By definition, (2^m)^n is the product of n factors, each equal to 2^m, that is 2^m x 2^m x ... x 2^m (n factors). Multiplying these n equal factors together, using the rule that multiplying powers of the same base adds the exponents, gives 2^(m + m + ... + m) = 2^(mn).
   3. Step 3 wrongly assumes that because both (2^m)^n and 2^(m + n) can be written as products of powers of 2, they must be the same product. But (2^m)^n has n factors of 2^m, while 2^(m + n) has exactly two factors (2^m and 2^n); these are generally different quantities, so the number of factors cannot simply be ignored.
   4. A concrete counterexample confirms the two expressions genuinely disagree: with m = 2 and n = 3, (2^2)^3 = 4^3 = 64 = 2^6, while 2^(m + n) = 2^(2 + 3) = 2^5 = 32, and 64 != 32. The correct general identity is (2^m)^n = 2^(mn), not 2^(m + n).
   • Why not A: 2^(m + n) = 2^m x 2^n is the standard law of indices for multiplying powers with the SAME base, 2, and it holds for every m and n; the claim that it needs different bases has the rule backwards, so step 1 states the correct law and introduces no error.
   • Why not B: Raising 2^m to the power n means, by definition, multiplying n copies of 2^m together, exactly as step 2 states; adding n copies of 2^m instead would be repeated addition, a different operation entirely from what a power denotes, so step 2 introduces no error.
   • Why not C: Two products can each be built from factors that are powers of 2 and still be unequal, since the number and identity of the factors differ; a concrete check, such as m = 2 and n = 3, shows the two sides genuinely disagree, so the argument is not valid.

10. Question 10Answer: C

    1. x^2 > x rearranges to x^2 - x > 0, that is x(x - 1) > 0, which holds exactly when x > 1 or x < 0 (the product of two factors is positive when both are positive, or both are negative).
    2. Dividing both sides of an inequality by x is only safe when the sign of x is already known: dividing by a positive x preserves the direction, but dividing by a negative x reverses it.
    3. The argument never establishes that x is positive before dividing by it in step 2, so the deduction is unjustified.
    4. A concrete counterexample confirms the claimed implication is false: x = -1 gives x^2 = 1 and x = -1, so x^2 = 1 > -1 = x holds, but x = -1 is not greater than 1.
    • Why not A: x^2 divided by x gives x (for x != 0), not x^2, so there is no arithmetic slip in the division itself; the actual problem is that the direction of the division depends on the unknown sign of x.
    • Why not B: Stating the hypothesis of an implication that is about to be tested is standard practice in a conditional argument, not an assumption of the conclusion.
    • Why not D: Restating a conclusion already reached does not itself introduce an error; the flaw lies earlier, in step 2's unjustified division.

11. Question 11Answer: B

    1. Every integer n is either even or odd, so a proof that splits into exactly these two cases has covered every possibility, provided each case is argued correctly.
    2. In the case n even, n = 2k, and n(n + 1) = 2k(n + 1) is manifestly a multiple of 2, hence even.
    3. In the case n odd, n + 1 is even (an odd number plus one is always even), so n + 1 = 2k for some integer k, and n(n + 1) = 2kn is again a multiple of 2, hence even.
    4. Since these two cases are exhaustive and each is argued validly, the proof genuinely establishes that n(n + 1) is even for every integer n; no step contains an error, unlike an incomplete case split that omits one of the possibilities.
    • Why not A: n = 2k for an integer k represents exactly the even integers, since every even integer can be written this way and every number of this form is even, so this step introduces no error.
    • Why not C: If n is odd, then n + 1 is always exactly one more than an odd number, and so is always even; this is a standard and correct fact about consecutive integers.
    • Why not D: Every integer is either even or odd, with no other possibility, so the two cases in steps 2 and 3 are exhaustive and do cover every integer.

12. Question 12Answer: A

    1. Setting b = 0 makes (a + b)^2 = a^2 and a^2 + b^2 = a^2 automatically, for any value of a whatsoever, regardless of whether the general claim is true; this choice provides no genuine evidence either way.
    2. Expanding correctly, (a + b)^2 = a^2 + 2ab + b^2, which equals a^2 + b^2 only when 2ab = 0, that is, only when a = 0 or b = 0.
    3. For a non-degenerate check, take a = 1 and b = 1: (a + b)^2 = 2^2 = 4, but a^2 + b^2 = 1 + 1 = 2, and 4 != 2, so the two expressions disagree.
    4. So the flaw is a test case that trivially satisfies the claim no matter what, combined with the leap in step 3 from 'equal at one point' to 'equal everywhere', which is never a valid deduction for two different expressions.
    • Why not B: 5^2 = 25 exactly as stated: squaring means multiplying a number by itself (5 x 5 = 25), not doubling it (which would give 10); there is no arithmetic error here.
    • Why not C: Testing specific values as exploration is a legitimate way to investigate a claim; the issue is not that values were chosen, but that this particular choice can never distinguish a true identity from a false one.
    • Why not D: Step 3 already states the universal claim, 'for all real numbers a and b', in full; step 4 introduces nothing new and simply restates it, so the fault lies entirely in step 3's unjustified generalisation, not in step 4.