TMUA Paper 2: Identifying errors in proofs, set 1

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

1. 11 mark

   A student claims to prove that 2 = 1 for all real numbers a and b with a = b, as follows.

   Step 1: Let a and b be real numbers with a = b.

   Step 2: Multiply both sides by a: a^2 = ab.

   Step 3: Subtract b^2 from both sides: a^2 - b^2 = ab - b^2.

   Step 4: Factorise both sides: (a - b)(a + b) = b(a - b).

   Step 5: Divide both sides by (a - b): a + b = b.

   Step 6: Since a = b, this gives 2b = b, so 2 = 1.

   At which step does the argument first contain an invalid deduction?

   1. A Step 5, because a - b is zero here, and dividing an equation by zero is not valid.
   2. B Step 3, because subtracting the same quantity b^2 from both sides of an equation is not a legitimate algebraic step.
   3. C Step 4, because expanding (a - b)(a + b) leaves cross terms that do not cancel, so it cannot equal a^2 - b^2.
   4. D Step 6, because substituting a = b into 2b is not permitted once a and b have been treated as separate symbols.
2. 21 mark

   A student attempts to prove the general claim: for any angles A and B, if sin A = sin B then A = B.

   Step 1: Suppose sin A = sin B.

   Step 2: The sine function assigns a value to every angle, so if two angles give the same value of sine, those two angles cannot be different.

   Step 3: Since sin A = sin B, it follows from step 2 that A and B are not different angles.

   Step 4: Therefore A = B.

   Which step contains the error that makes this a purported, rather than a genuine, proof?

   1. A Step 1, because stating a hypothesis before it has been justified is not permitted in a valid proof.
   2. B Step 2, because the sine function is not one-to-one, so equal sine values need not come from equal angles.
   3. C Step 3, because the deduction from step 2 to step 3 does not follow the stated reasoning.
   4. D Step 4, because 'A and B are not different angles' is a weaker statement than 'A = B', so step 4 overreaches what step 3 established.
3. 31 mark

   A student claims to prove that the only real number x satisfying x^2 = 5x is x = 5.

   Step 1: Let x be a real number with x^2 = 5x.

   Step 2: Divide both sides by x: x = 5.

   Step 3: Therefore the only real number satisfying x^2 = 5x is x = 5.

   What is wrong with this argument?

   1. A Step 1 is invalid because x^2 = 5x is not a valid equation to state without first solving it.
   2. B Step 2 contains an arithmetic slip: dividing 5x by x should leave 5x remaining, not simplify to the constant 5.
   3. C Step 2 divides by x without checking x is nonzero, silently losing the valid solution x = 0.
   4. D Step 3 is invalid because it merely restates the conclusion of step 2 as a general claim, adding no justification of its own.
4. 41 mark

   A student claims to prove that the solutions of sqrt(x + 3) = x - 3 are x = 1 and x = 6.

   Step 1: Suppose x satisfies sqrt(x + 3) = x - 3.

   Step 2: Square both sides: x + 3 = (x - 3)^2 = x^2 - 6x + 9.

   Step 3: Rearranging gives x^2 - 7x + 6 = 0, so (x - 1)(x - 6) = 0, giving x = 1 or x = 6.

   Step 4: Therefore the solutions of sqrt(x + 3) = x - 3 are x = 1 and x = 6.

   Which step is the point at which the argument fails to be a valid proof?

   1. A Step 1, because writing down the equation before it has been justified is not a valid opening move in a proof.
   2. B Step 2, because squaring both sides of an equation is never a legitimate algebraic operation, whatever the signs involved.
   3. C Step 3, because multiplying out (x - 1)(x - 6) does not correctly reproduce the quadratic x^2 - 7x + 6.
   4. D Step 4, because squaring in step 2 is not reversible, and step 4 never checks x = 1 against the original equation.
5. 51 mark

   A student argues as follows.

   Step 1: It is known that if an integer n is divisible by 4, then n is divisible by 2.

   Step 2: Suppose n is divisible by 2.

   Step 3: By the fact stated in step 1, divisibility by 4 gives divisibility by 2. Since n is divisible by 2, this same fact tells us that n must be divisible by 4.

   Step 4: Therefore every integer divisible by 2 is divisible by 4.

   Which step contains the invalid deduction?

   1. A Step 1, because it is not true that every integer divisible by 4 is also divisible by 2.
   2. B Step 3, because it wrongly assumes the converse of the true implication stated in step 1.
   3. C Step 2, because supposing a fact about n before it has been proved is not a legitimate step in an argument.
   4. D Step 4, because it does not follow from step 3 as stated, treating an unjustified conclusion as newly established.
6. 61 mark

   A student claims to prove that n^2 - n + 11 is a prime number for every positive integer n.

   Step 1: When n = 1: 1 - 1 + 11 = 11, which is prime.

   Step 2: When n = 2: 4 - 2 + 11 = 13, which is prime.

   Step 3: When n = 3: 9 - 3 + 11 = 17, which is prime.

   Step 4: When n = 4: 16 - 4 + 11 = 23, which is prime.

   Step 5: Since the expression is prime for n = 1, 2, 3 and 4, it is prime for every positive integer n.

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

   1. A Step 5, because checking finitely many cases never establishes that a statement holds for every positive integer.
   2. B Step 2, because 4 - 2 + 11 works out to 12 rather than 13, an arithmetic slip in the subtraction.
   3. C Step 3, because 17 has factors other than 1 and itself, so it is not actually a prime number.
   4. D Step 4, because 16 - 4 + 11 comes out to 24 rather than 23, another arithmetic slip.
7. 71 mark

   A student claims to prove that for all real x != 0, if 1/x < 2 then x > 1/2.

   Step 1: Suppose 1/x < 2.

   Step 2: Multiply both sides by x: 1 < 2x.

   Step 3: Divide both sides by 2: 1/2 < x.

   Step 4: Therefore x > 1/2.

   Which step contains the invalid deduction?

   1. A Step 3, because dividing an inequality by the positive number 2 requires the inequality sign to be reversed.
   2. B Step 4, because '1/2 < x' and 'x > 1/2' are different statements that need separate justification to equate.
   3. C Step 1, because stating the hypothesis 1/x < 2 before it has been justified already assumes what is to be proved.
   4. D Step 2, because it multiplies both sides of the inequality by x without knowing whether x is positive or negative.
8. 81 mark

   A student writes the following proof that sqrt(2) is irrational.

   Step 1: Suppose, for contradiction, that sqrt(2) is rational, so sqrt(2) = p/q for integers p and q with no common factor and q != 0.

   Step 2: Then 2 = p^2/q^2, so p^2 = 2q^2.

   Step 3: So p^2 is even, and hence p is even (since the square of an odd number is odd). Write p = 2k for some integer k.

   Step 4: Substituting gives (2k)^2 = 2q^2, so 4k^2 = 2q^2, that is q^2 = 2k^2.

   Step 5: So q^2 is even, and hence q is even.

   Step 6: But then p and q are both even, contradicting the assumption that they have no common factor. So sqrt(2) is irrational.

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

   1. A Step 3 is invalid, because p^2 being even does not guarantee that p itself is even.
   2. B Step 4 is invalid, because substituting p = 2k into the equation assumes the very conclusion the proof is trying to reach.
   3. C No step is invalid; this is a valid proof by contradiction that sqrt(2) is irrational.
   4. D Step 6 is invalid, because the assumption that p and q share no common factor was never justified.
9. 91 mark

   A student claims to prove that n^2 - n is even for every integer n.

   Step 1: Case: n is even, so n = 2k for some integer k.

   Step 2: Then n^2 - n = (2k)^2 - 2k = 4k^2 - 2k = 2(2k^2 - k), which is even.

   Step 3: Therefore n^2 - n is even for every integer n.

   What is wrong with this argument?

   1. A The argument only considers n even, and never establishes the result for odd n, so the case analysis is incomplete.
   2. B Step 1 is invalid, because not every integer can be written in the form 2k for some integer k, so the case split is unjustified.
   3. C Step 2 contains an algebraic error: expanding 2(2k^2 - k) does not give back 4k^2 - 2k as claimed.
   4. D The conclusion in step 3 is false, because there exist odd integers n for which n^2 - n turns out to be odd rather than even.
10. 101 mark

    A student claims to prove that (x + 1)^2 = x^2 + 1 for every real number x.

    Step 1: Let x = 0.

    Step 2: Compute (x + 1)^2 = 1 and x^2 + 1 = 1, so the two expressions agree at x = 0.

    Step 3: Since they agree at this one value, (x + 1)^2 and x^2 + 1 must be equal for every real value of x.

    Step 4: Therefore (x + 1)^2 = x^2 + 1 for all real x.

    Which step is not validly justified?

    1. A Step 1, because choosing a specific value of x is not permitted when trying to prove a statement about all real x.
    2. B Step 2, because (x + 1)^2 does not equal 1 when x = 0, so the arithmetic in this step is already incorrect from the start.
    3. C Step 4, because the jump to the final conclusion does not follow from step 3 as stated, introducing an unjustified logical gap.
    4. D Step 3, because agreement at a single value of x is never enough to conclude that two expressions are equal for every real x.
11. 111 mark

    A student claims to disprove the statement: for every integer n > 3, n^2 - 4 is composite (that is, not prime).

    Step 1: Consider n = 3. Then n^2 - 4 = 9 - 4 = 5.

    Step 2: 5 is a prime number, not a composite number.

    Step 3: This contradicts the claim that n^2 - 4 is composite, so the statement 'for every integer n > 3, n^2 - 4 is composite' is false.

    What is wrong with this argument?

    1. A Step 1's arithmetic is wrong: 9 - 4 works out to a value other than 5, so the starting computation is flawed.
    2. B Step 3 draws an invalid conclusion, because n = 3 does not satisfy n > 3, so it cannot serve as a counterexample to the claim.
    3. C Step 2 wrongly claims that 5 is a prime number, when in fact 5 has a factor besides 1 and itself.
    4. D No step is invalid; the disproof is correct as it stands, and the original claim about n^2 - 4 is actually false.
12. 121 mark

    A student gives the following argument to show that n^2 + n + 1 is odd for every positive integer n, using induction.

    Step 1: Inductive step: assume k^2 + k + 1 is odd for some positive integer k. We show (k + 1)^2 + (k + 1) + 1 is odd.

    Step 2: (k + 1)^2 + (k + 1) + 1 = k^2 + 2k + 1 + k + 1 + 1 = (k^2 + k + 1) + (2k + 2).

    Step 3: Since k^2 + k + 1 is odd (by the inductive hypothesis) and 2k + 2 is even, their sum is odd + even = odd, so (k + 1)^2 + (k + 1) + 1 is odd.

    Step 4: Therefore, by induction, n^2 + n + 1 is odd for every positive integer n.

    What is the flaw in this argument?

    1. A Step 2 contains an algebraic error in expanding (k + 1)^2 + (k + 1) + 1 into the grouped form shown.
    2. B Step 3 wrongly claims that an odd number plus an even number is always odd, when the parity actually depends on which numbers are chosen.
    3. C No base case is ever verified, so the inductive step alone, however correct, cannot establish the statement for any starting value.
    4. D No step is invalid; the proof as written is a valid and complete induction with nothing missing from it.

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: A

   1. Steps 1 to 4 are all valid: multiplying both sides of an equation by the same quantity, subtracting the same quantity from both sides, and factorising a^2 - b^2 as (a - b)(a + b) are all legitimate operations.
   2. The error is in step 5: since a = b, the quantity (a - b) is exactly zero, and dividing both sides of an equation by zero is not a valid algebraic operation - it can turn a true equation into a false one.
   3. This is the same fallacy as claiming that ab = ac allows you to conclude b = c: it is only valid when a is known to be nonzero, since a x 0 = a x 0 is true for every value of the second factor.
   4. Once the invalid division is allowed, everything that follows, including 2 = 1, is meaningless, so the argument does not prove 2 = 1; it only shows that dividing by zero produces contradictions.
   • Why not B: Subtracting the same quantity from both sides of a true equation is always valid; step 3 introduces no error.
   • Why not C: The cross terms -ab and +ab in the expansion (a - b)(a + b) = a^2 + ab - ab - b^2 do cancel exactly, leaving a^2 - b^2; this is a standard, correct identity, and step 4's factorisation is legitimate.
   • Why not D: Substituting a known equality back into an expression is a standard and valid step; the error occurs earlier, at the point where a zero quantity is cancelled.

2. Question 2Answer: B

   1. The claim 'sin A = sin B implies A = B' is false in general: sin 30 degrees = 1/2 and sin 150 degrees = 1/2 also, but 30 and 150 are different angles.
   2. The purported proof's flaw is in step 2, which asserts that sine takes different values at different angles, that is, that sine is a one-to-one (injective) function. This is false: sine repeats values because sin(theta) = sin(180 - theta) for angles measured in degrees.
   3. Steps 3 and 4 are valid deductions from step 2, so if step 2 were true the rest of the argument would be too; the whole argument fails only because its second step asserts something false.
   4. This is exactly the kind of unjustified step the specification warns against: assuming that equal outputs of sin force equal inputs, when in fact they do not.
   • Why not A: Stating the hypothesis of an implication that is to be tested is standard practice in a proof and is not itself a deduction that could be invalid.
   • Why not C: Given step 2 as stated, step 3 follows from it exactly as claimed; the flaw lies in step 2 itself, not in the transition to step 3.
   • Why not D: For two specific angles, 'not different' and 'equal' mean exactly the same thing, so step 4 draws no stronger conclusion than step 3 already gave; there is no overreach here.

3. Question 3Answer: C

   1. x^2 = 5x can be rearranged as x^2 - 5x = 0, that is x(x - 5) = 0, so x = 0 or x = 5.
   2. Dividing both sides of x^2 = 5x by x is only valid when x is known to be nonzero; if x could be zero, dividing by it is not a legitimate step.
   3. Because x = 0 does satisfy the original equation (0^2 = 5 x 0), step 2 silently loses this valid solution by assuming x != 0 without ever establishing it.
   4. So the correct solution set is x = 0 or x = 5, not x = 5 alone, and the flaw in the argument is the unjustified division in step 2.
   • Why not A: Writing down the equation that x is assumed to satisfy is simply stating the hypothesis, and it introduces no invalid claim.
   • Why not B: Dividing 5x by x correctly gives 5 (when x is nonzero), so there is no arithmetic slip in the division itself; the problem is that the division was performed without checking whether x could be zero.
   • Why not D: Step 3 does follow from step 2 exactly as stated; the flaw is that step 2 itself is not fully justified, not that step 3 fails to follow from it.

4. Question 4Answer: D

   1. Squaring both sides of sqrt(x + 3) = x - 3 is valid in the sense that any x satisfying the original equation must also satisfy the squared equation, but the converse need not hold: squaring can introduce extra solutions that do not satisfy the original equation.
   2. Solving the squared equation correctly gives x = 1 or x = 6, so steps 2 and 3 contain no arithmetic error.
   3. Checking x = 6 in the original equation: sqrt(6 + 3) = sqrt(9) = 3, and 6 - 3 = 3, so x = 6 is genuinely a solution.
   4. Checking x = 1: sqrt(1 + 3) = sqrt(4) = 2, but 1 - 3 = -2, and 2 != -2, so x = 1 does not satisfy the original equation. It is an extraneous root introduced by squaring, and step 4 is wrong to include it without this check.
   • Why not A: Stating the equation a proof is about to investigate is simply setting up the problem, and cannot itself be an invalid deduction.
   • Why not B: Squaring both sides of a true equation always produces another true equation, so step 2 is a legitimate (though not reversible) step, whichever signs are involved.
   • Why not C: Multiplying out (x - 1)(x - 6) gives x^2 - 6x - x + 6 = x^2 - 7x + 6, which matches exactly, so this factorisation is correct.

5. Question 5Answer: B

   1. The true fact is: n divisible by 4 implies n divisible by 2. This does not mean the reverse also holds.
   2. Step 3 uses the true implication backwards, assuming that because n is divisible by 2 (the consequent of the true implication), n must also be divisible by 4 (the antecedent). 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 = 6 is divisible by 2 but not by 4.
   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: any integer of the form 4k is also of the form 2(2k), so divisibility by 4 does guarantee divisibility by 2.
   • Why not C: Introducing a hypothesis to be investigated is standard practice in a conditional argument and is not itself an invalid deduction.
   • Why not D: Step 4 follows exactly from step 3 as stated; the fault lies in step 3, which reaches an unjustified conclusion, not in the restatement of it in step 4.

6. Question 6Answer: A

   1. Each of the individual checks in steps 1 to 4 is arithmetically correct: 11, 13, 17 and 23 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 there are infinitely many values left unchecked.
   3. In fact the claim is false: at n = 11, n^2 - n + 11 = 121 - 11 + 11 = 121, and 121 = 11 x 11 is not prime.
   4. So the pattern that holds for n = 1 to 10 breaks down at n = 11, which is exactly the kind of case a hasty generalisation from a few examples misses.
   • Why not B: 4 - 2 + 11 = 13 exactly as stated; there is no arithmetic error here.
   • Why not C: 17's only positive divisors are 1 and 17; it has no other factors, so it is genuinely prime and step 3 contains no error.
   • Why not D: 16 - 4 + 11 = 23 exactly as stated; there is no arithmetic error here.

7. Question 7Answer: D

   1. The claimed implication is false in general: take x = -1. Then 1/x = -1, and -1 < 2 is true, so the hypothesis holds, but x = -1 is not greater than 1/2, so the conclusion fails.
   2. The error is in step 2: multiplying both sides of an inequality by x is only safe when the sign of x is known. If x is negative, multiplying reverses the direction of the inequality, so 1/x < 2 would give 1 > 2x instead of 1 < 2x.
   3. Since the argument never established that x is positive before multiplying by it, the deduction in step 2 is unjustified.
   4. Once that step is allowed unchecked, the rest of the argument follows validly from it, but from a false premise, so the conclusion cannot be relied on.
   • Why not A: Dividing an inequality by a positive number, such as 2, never requires reversing the sign, so step 3 is a valid step.
   • Why not B: '1/2 < x' and 'x > 1/2' are simply two ways of writing the same inequality, so there is nothing separate to justify: step 4 introduces no new claim and no error.
   • Why not C: Stating the hypothesis of an implication that is about to be examined is standard practice, not an assumption of the conclusion.

8. Question 8Answer: C

   1. Every step in this argument is a standard and valid part of the classical proof that sqrt(2) is irrational.
   2. Step 3's claim that an even square forces an even root is correct, because the square of any odd number is odd, so only an even number can square to an even number.
   3. Step 4 and step 5 repeat the same valid reasoning for q, showing q is also even.
   4. Since p and q were chosen to share no common factor, but both turn out to be even (sharing the factor 2), this is a genuine contradiction, so the original assumption that sqrt(2) is rational must be false: no step of this proof is invalid.
   • Why not A: This is false: if p were odd, p = 2m + 1, then p^2 = 4m^2 + 4m + 1 is odd, so an even p^2 does genuinely force p to be even. Step 3's deduction is correct.
   • Why not B: Substituting an already-established fact (p = 2k, from step 3) into an equation is a standard, valid algebraic step, not an assumption of the conclusion; nothing about the desired contradiction is assumed here.
   • Why not D: The assumption that p and q share no common factor was stated explicitly as part of the setup in step 1 (choosing p/q in lowest terms), so it was justified there, not left unjustified.

9. Question 9Answer: A

   1. The statement n^2 - n is even for every integer n is in fact true, since n^2 - n = n(n - 1) is a product of two consecutive integers, and one of any two consecutive integers is always even.
   2. However, the given proof only ever considers the case where n is even, correctly showing n^2 - n is even in that case.
   3. A complete proof by cases needs every case to be covered; here the case where n is odd is never examined, so the argument as written does not establish the result for odd n, even though the result happens to be true there too.
   4. So the flaw is not in the arithmetic of either case, but in the incompleteness of the case split: a genuine proof by cases must exhaust all possibilities.
   • Why not B: Step 1 does not claim every integer equals 2k; it correctly restricts attention to the case n even, which by definition means n = 2k for some integer k, so this step is a valid way to open one branch of a case split.
   • Why not C: Expanding 2(2k^2 - k) gives 4k^2 - 2k exactly, so 4k^2 - 2k = 2(2k^2 - k) is a correct factorisation and there is no algebraic error here.
   • Why not D: The conclusion is in fact true for every integer n, odd included: if n = 2k + 1, n^2 - n = (2k+1)^2 - (2k+1) = 4k^2 + 4k + 1 - 2k - 1 = 4k^2 + 2k = 2(2k^2 + k), which is even, so the statement itself is not false; the problem is that the given proof never shows this.

10. Question 10Answer: D

    1. Checking x = 0 correctly shows both (x + 1)^2 and x^2 + 1 equal 1 at that one point, but agreement at a single value never proves two expressions are identically equal.
    2. The claim (x + 1)^2 = x^2 + 1 is in fact false in general: expanding the left side gives (x + 1)^2 = x^2 + 2x + 1, which equals x^2 + 1 only when 2x = 0, that is, only at x = 0.
    3. For example, at x = 1: (x + 1)^2 = 4 but x^2 + 1 = 2, so the two sides disagree.
    4. So the flaw is 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 A: Testing a specific value is a legitimate way to explore or check a claim; the issue is not that a value was chosen, but what was concluded from it.
    • Why not B: (0 + 1)^2 = 1^2 = 1, so this arithmetic is entirely correct and step 2 contains no error.
    • Why not C: Step 4 restates exactly what step 3 already (wrongly) concluded, so it introduces no new deduction and no logical gap of its own; the fault lies in step 3.

11. Question 11Answer: B

    1. The claim under examination is only about integers n with n > 3, that is n = 4, 5, 6 and upwards; n = 3 does not satisfy this condition.
    2. Since n = 3 is outside the domain the claim is making a statement about, showing that n^2 - 4 is prime at n = 3 says nothing about whether the claim holds for n > 3.
    3. In fact, for n > 3, n^2 - 4 = (n - 2)(n + 2), and when n > 3, n - 2 > 1 and n + 2 > 1, so the product is a product of two factors each greater than 1, which makes it composite.
    4. So the claim (correctly restricted to its stated domain) is true, and the purported disproof fails because it tests a value the claim was never making an assertion about.
    • Why not A: 9 - 4 = 5 is correct arithmetic, so there is no error here.
    • Why not C: 5's only positive divisors are 1 and 5; it has no factor besides those, so it is genuinely prime and step 2 contains no mistake.
    • Why not D: The original claim, correctly restricted to n > 3, is in fact true (n^2 - 4 factorises as (n - 2)(n + 2), a product of two factors each greater than 1), so this option wrongly treats an invalid disproof as though it succeeded.

12. Question 12Answer: C

    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. This argument only ever gives the inductive step; the base case, such as checking n = 1 gives 1^2 + 1 + 1 = 3, which is odd, is never stated or verified.
    3. 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.
    4. So the proof as given is incomplete, even though every algebraic and parity claim inside the inductive step itself is correct, and a full proof needs the missing base case added before the induction is valid.
    • Why not A: Expanding gives (k + 1)^2 + (k + 1) + 1 = k^2 + 2k + 1 + k + 1 + 1 = k^2 + 3k + 3, and grouping as (k^2 + k + 1) + (2k + 2) = k^2 + 3k + 3 as well, so the algebra in step 2 is correct.
    • Why not B: Odd + even is always odd regardless of which specific numbers are chosen, since an odd number is 2m + 1 and an even number is 2j, and their sum 2(m + j) + 1 is always odd; step 3's parity claim is correct.
    • Why not D: The proof is missing a base case, so it is not in fact complete, even though the inductive step it does contain is valid.