TMUA Mathematical Proof — Questions and Methods

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.

A student is asked to show that if a and b are positive real numbers, then (a + b)/2 ≥ √(ab). The attempt is: (I) (a + b)/2 ≥ √(ab); (II) so a + b ≥ 2√(ab); (III) so a − 2√(ab) + b ≥ 0; (IV) so (√a − √b)² ≥ 0, which is true. Which description is best?

1. A. It is incorrect, but the student has correctly proved the converse.
2. B. It is completely correct.
3. C. It is incorrect, but it would be correct if written in the reverse order.
4. D. It is incorrect because there is an error in line (II).
5. E. It is incorrect because there is an error in line (III).

Answer: C. It is incorrect, but it would be correct if written in the reverse order.

Worked solution. The attempt starts from the inequality that has to be proved and deduces a statement that is always true. Deducing a true statement from the claim does not prove the claim, so the attempt as written is not a proof. Each step can be reversed, however: (√a − √b)² ≥ 0 expands to a − 2√(ab) + b ≥ 0 because (√a)² = a and (√b)² = b for positive a and b; adding 2√(ab) gives a + b ≥ 2√(ab); dividing by 2 gives (a + b)/2 ≥ √(ab). Written in the reverse order, starting from the true line (IV), the argument is a correct proof.

Why the other options look right. A is wrong because the lines do not prove the converse 'if (a + b)/2 ≥ √(ab) then a and b are positive'; they only deduce a true inequality from the claim. B accepts an argument that assumes what it is meant to prove. D is wrong because line (II) is line (I) multiplied by 2, which is valid. E is wrong because line (III) is line (II) with 2√(ab) subtracted from both sides, which is valid.

Paper 2 questions

Paper 2 is Mathematical Reasoning. This topic is examined on Paper 2 only.

1. Q1. Positive numbers a, b and c, with a ≠ 1 and b ≠ 1, satisfy log base b of c = (log base a of b)². Which list is a correct order of steps in a proof that log base a of c = (log base a of b)³? (1) Let x = log base a of b, so b = a^x. (2) (b^x)² = b^2x. (3) The given equation says c = b^x2. (4) Therefore log base a of c = x³. (5) c = (a^x)^x2 = a^x3. (6) c = a^2x. (7) a^x3 = (a^x2)^x.

   Free · worked solution included

   A(1), (2), (6), (4)B(2), (3), (5), (4)C(1), (3), (7), (4)D(1), (3), (5), (4)E(4), (5), (3), (1)

   Answer and worked solution

   Answer: D. (1), (3), (5), (4)

   Worked solution. Set x = log base a of b, which is line (1), so b = a^x. The given equation is log base b of c = x², hence c = b^x2, which is line (3). Substituting the expression for b gives c = (a^x)^x2 = a^x3, which is line (5). Therefore log base a of c = x³, which is line (4). The correct order is (1), (3), (5), (4).

   Why the other options look right. A uses (2) and (6). Squaring the exponent gives b^2x, not b^x2. B starts from (2), which is not a step in this proof. C goes from (3) straight to (7): replacing the base b by a^x produces (5), and (7) does not follow from (3) alone. E begins at (4), which is the equation being proved, so it is not yet available.

2. Q2. Consider the statement S: for every integer n, if n² is a multiple of 12, then n is a multiple of 12. A student writes: suppose that n is a multiple of 12, so n = 12k for some integer k. Then n² = 144k² = 12(12k²), which is a multiple of 12. Therefore S is true. Which of the following is correct?

   Free · worked solution included

   AS is false, and the student has proved the converse of S, not S.BS is true, but the student has proved the converse of S, not S.CS is true, and the student's proof is correct.DS is false, because n = 4 gives n² = 16, which is not a multiple of 12.ES is true, because a prime that divides n² must also divide n.

   Answer and worked solution

   Answer: A. S is false, and the student has proved the converse of S, not S.

   Worked solution. The student starts from 'n is a multiple of 12' and ends with 'n² is a multiple of 12'. That is a correct proof of the converse of S, not of S. S itself is false: n = 6 gives n² = 36 = 12 × 3, a multiple of 12, but 6 is not a multiple of 12. The reason is that 12 = 2² × 3 is not square-free: n² is a multiple of 12 exactly when n is a multiple of 6. So S is false, and the proof shows the converse.

   Why the other options look right. B describes the proof correctly but misses the counterexample n = 6, for which n² = 36 is a multiple of 12 and n is not. C accepts the argument as a proof of S, but it assumes the conclusion of S and derives the hypothesis. D uses n = 4, which does not satisfy the hypothesis because 16 is not a multiple of 12, so n = 4 cannot be a counterexample. E applies a fact about primes to 12, which is not prime: n = 6 has n² divisible by 12 although 4 does not divide 6.

3. Q3. A student is asked to show that if a and b are positive real numbers, then (a + b)/2 ≥ √(ab). The attempt is: (I) (a + b)/2 ≥ √(ab); (II) so a + b ≥ 2√(ab); (III) so a − 2√(ab) + b ≥ 0; (IV) so (√a − √b)² ≥ 0, which is true. Which description is best?

   Free · worked solution included

   AIt is incorrect, but the student has correctly proved the converse.BIt is completely correct.CIt is incorrect, but it would be correct if written in the reverse order.DIt is incorrect because there is an error in line (II).EIt is incorrect because there is an error in line (III).

   Answer and worked solution

   Answer: C. It is incorrect, but it would be correct if written in the reverse order.

   Worked solution. The attempt starts from the inequality that has to be proved and deduces a statement that is always true. Deducing a true statement from the claim does not prove the claim, so the attempt as written is not a proof. Each step can be reversed, however: (√a − √b)² ≥ 0 expands to a − 2√(ab) + b ≥ 0 because (√a)² = a and (√b)² = b for positive a and b; adding 2√(ab) gives a + b ≥ 2√(ab); dividing by 2 gives (a + b)/2 ≥ √(ab). Written in the reverse order, starting from the true line (IV), the argument is a correct proof.

   Why the other options look right. A is wrong because the lines do not prove the converse 'if (a + b)/2 ≥ √(ab) then a and b are positive'; they only deduce a true inequality from the claim. B accepts an argument that assumes what it is meant to prove. D is wrong because line (II) is line (I) multiplied by 2, which is valid. E is wrong because line (III) is line (II) with 2√(ab) subtracted from both sides, which is valid.

4. Q4. Which of these claims fails for some integer n?

   Free · worked solution included

   AIf 6 divides n, then 3 divides n².BIf n is even, then 4 divides n² + 2n.CIf 3 divides n², then 9 divides n².DIf 6 divides n(n + 1), then 3 divides n.EIf 4 divides n, then 16 divides n².

   Answer and worked solution

   Answer: D. If 6 divides n(n + 1), then 3 divides n.

   Worked solution. The claim in D fails for n = 2: n(n + 1) = 6 is divisible by 6, but 3 does not divide 2. (In general n(n + 1) is always even, so 6 divides it exactly when 3 divides n or n + 1.) The other four claims hold for every integer n. If n = 6k, then n² = 36k², a multiple of 3. If n = 2k, then n² + 2n = 4k² + 4k = 4k(k + 1). If 3 divides n², then 3 divides n because 3 is prime, so n = 3k and n² = 9k². If n = 4k, then n² = 16k².

   Why the other options look right. A, B, C and E hold for every integer n, so none of them is the claim that fails. A follows by writing n = 6k. B follows because n² + 2n = 4k(k + 1) when n = 2k. C follows because 3 is prime, so 3 divides n and 9 divides n². E follows by writing n = 4k.

5. Q5. Let p be a polynomial, and let a < b. Let (*) say that p(c) = 0 for some number c with a < c < b. Which description of the condition p(a)p(b) ≤ 0 is correct?

   Free · worked solution included

   AIt is necessary and sufficient for (*)BIt is neither necessary nor sufficient for (*)CIt is necessary but not sufficient for (*)DIt is sufficient but not necessary for (*)EIt is necessary and sufficient when p has odd degree, and neither otherwise

   Answer and worked solution

   Answer: B. It is neither necessary nor sufficient for (*)

   Worked solution. The condition is not sufficient for (*). For p(x) = x, with a = 0 and b = 1, the product p(a)p(b) is 0, but x = 0 is not strictly between 0 and 1, and there is no root in the open interval. It is not necessary either. For p(x) = (x − 1/2)², with a = 0 and b = 1, there is a root in (0, 1), but p(0)p(1) = 1/16 > 0. So the condition is neither necessary nor sufficient.

   Why the other options look right. A allows a root at an endpoint to count and treats a non-strict product as a sign change. C keeps the counterexample p(x) = x, but still treats a sign change as required for an interior root. D is the right classification of the strict inequality p(a)p(b) < 0, not of p(a)p(b) ≤ 0. E fails for the odd-degree polynomial (x − 1/2)²(x − 10): on (0, 1) it has a root, but the values at the endpoints have the same sign, and the endpoint example p(x) = x already shows that the non-strict product is not sufficient.

6. Q6. Which one of the following pairs of numbers a and b is a counterexample to the claim 'if a and b are both irrational, then a + b is irrational'?

   Free · worked solution included

   Aa = √2, b = √3Ba = √2, b = 3 − √2Ca = √2, b = √8Da = 1/2, b = 5/2Ea = √2 − 1, b = √2 + 1

   Answer and worked solution

   Answer: B. a = √2, b = 3 − √2

   Worked solution. A counterexample must make the hypothesis true and the conclusion false: a and b must both be irrational, and a + b must be rational. For a = √2 and b = 3 − √2, both numbers are irrational (if 3 − √2 were rational, then √2 = 3 − (3 − √2) would be rational too), and a + b = 3 is rational. So this pair refutes the claim. In each other pair, either a + b is irrational (√2 + √3, √2 + √8 = 3√2 and (√2 − 1) + (√2 + 1) = 2√2), or the hypothesis fails because 1/2 and 5/2 are rational.

   Why the other options look right. A satisfies the hypothesis, but √2 + √3 is irrational, so this pair agrees with the claim. C confuses the sum with the product: √2 × √8 = 4 is rational, but √2 + √8 = 3√2 is irrational. D has the rational sum 3, but a and b are themselves rational, so the hypothesis of the claim is false and the pair tests nothing. E confuses the sum with the difference: (√2 + 1) − (√2 − 1) = 2 is rational, but the sum 2√2 is irrational.

7. Q7. A proof by contradiction is to be given for the statement: for all integers a and b, if ab is even, then a is even or b is even. The proof begins: ‘Suppose that there are integers a and b such that …’. Which of the following correctly completes this sentence?

   Free · worked solution included

   Aab is odd, and a and b are both oddBab is even, and at least one of a and b is oddCab is odd, and at least one of a and b is evenDa and b are both oddEab is even, and a and b are both odd

   Answer and worked solution

   Answer: E. ab is even, and a and b are both odd

   Worked solution. A proof by contradiction assumes the negation of the statement. The statement says that for all a and b, if P then Q, where P is ‘ab is even’ and Q is ‘a is even or b is even’. Its negation is that for some a and b, P is true and Q is false. The negation of ‘a is even or b is even’ is ‘a is odd and b is odd’. So the proof assumes that ab is even and that a and b are both odd. (The contradiction then follows quickly: with a = 2m + 1 and b = 2k + 1, ab = 2(2mk + m + k) + 1 is odd.) The correct completion is ‘ab is even, and a and b are both odd’.

   Why the other options look right. A negates the hypothesis ‘ab is even’ as well as the conclusion, but a proof by contradiction keeps the hypothesis and negates only the conclusion. B negates ‘a is even or b is even’ as ‘a is odd or b is odd’ instead of ‘a is odd and b is odd’. C negates the hypothesis instead of the conclusion. D drops the hypothesis that ab is even, so it is not the negation of the statement, and on its own it leads to no contradiction (a = 3, b = 5).

8. Q8. Let n be an integer. A student proves the statement ‘if n² − 6n + 5 is even, then n is odd’ by proving its contrapositive instead. Which statement does the student prove?

   Free · worked solution included

   AIf n is odd, then n² − 6n + 5 is even.BIf n is even, then n² − 6n + 5 is odd.CIf n is even, then n² − 6n + 5 is even.DIf n is odd, then n² − 6n + 5 is odd.EThere is an even integer n for which n² − 6n + 5 is even.

   Answer and worked solution

   Answer: B. If n is even, then n² − 6n + 5 is odd.

   Worked solution. The contrapositive of ‘if P, then Q’ is ‘if not Q, then not P’. Here P is ‘n² − 6n + 5 is even’ and Q is ‘n is odd’, so not Q is ‘n is even’ and not P is ‘n² − 6n + 5 is odd’. The contrapositive is ‘if n is even, then n² − 6n + 5 is odd’, and it is logically equivalent to the original statement. It is also easy to prove: n² − 6n + 5 = (n − 1)(n − 5), and if n is even then n − 1 and n − 5 are both odd, so their product is odd. The student proves ‘If n is even, then n² − 6n + 5 is odd.’

   Why the other options look right. A swaps the two parts without negating them, which gives the converse. C swaps the parts and negates ‘n is odd’, but forgets to negate ‘n² − 6n + 5 is even’. D swaps the parts and negates ‘n² − 6n + 5 is even’, but forgets to negate ‘n is odd’. E writes the assumption of a proof by contradiction (the negation of the statement) instead of the contrapositive.

9. Q9. To prove that the equation x³ + x² + 1 = 0 has no rational solution, a student supposes that x = p/q is a solution, where p and q are integers, q ≠ 0, and p and q have no common factor greater than 1. Multiplying the equation by q³ gives p³ + p²q + q³ = 0. The student then shows, in each case of a list, that p³ + p²q + q³ is odd, which contradicts its being 0. Which list of cases is enough for a complete proof?

   Free · correct letter only

   Ap odd and q odd; p odd and q evenBp even and q odd; p odd and q evenCp odd and q odd; p even and q odd; p odd and q evenDp even and q even; p odd and q oddEp even and q odd; p odd and q odd

   Answer

   Answer: C. p odd and q odd; p even and q odd; p odd and q even

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10. Q10. Consider the claim: if the diagonals of a quadrilateral are perpendicular and one diagonal passes through the midpoint of the other, then the quadrilateral is a rhombus. Each option gives the vertices of a quadrilateral PQRS, taken in order around the quadrilateral. Which quadrilateral is a counterexample to the claim?

    Free · correct letter only

    AP(−1, 0), Q(0, 1), R(1, 0), S(0, −3)BP(−2, 0), Q(0, 1), R(2, 0), S(0, −1)CP(−1, 0), Q(0, 1), R(2, 0), S(0, −3)DP(0, 0), Q(4, 0), R(4, 2), S(0, 2)EP(0, 0), Q(2, 0), R(3, 2), S(1, 2)

    Answer

    Answer: A. P(−1, 0), Q(0, 1), R(1, 0), S(0, −3)

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11. Q11. Which one of the following is a correct proof that x⁴ − 2x² + 2 > 0 for every real number x?

    Free · correct letter only

    Ax⁴ − 2x² + 2 = (x² − 1)² + 1, and (x² − 1)² ≥ 0, so x⁴ − 2x² + 2 ≥ 1 > 0.BSubstituting x = −2, −1, 0, 1, 2 gives the values 10, 1, 2, 1, 10, which are all positive, so x⁴ − 2x² + 2 > 0 for every x.CSuppose that x⁴ − 2x² + 2 > 0. Then x⁴ − 2x² + 1 > −1, so (x² − 1)² > −1, which is true. Hence x⁴ − 2x² + 2 > 0.Dx⁴ ≥ 0 and 2 − 2x² > 0, so their sum x⁴ − 2x² + 2 is positive.Ex⁴ − 2x² + 2 = x²(x² − 2) + 2, and x²(x² − 2) ≥ 0, so x⁴ − 2x² + 2 ≥ 2 > 0.

    Answer

    Answer: A. x⁴ − 2x² + 2 = (x² − 1)² + 1, and (x² − 1)² ≥ 0, so x⁴ − 2x² + 2 ≥ 1 > 0.

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12. Q12. Here is a proof that √2 + √3 is irrational. ‘Suppose that √2 + √3 = r, where r is rational. Then √3 = r − √2. Squaring gives 3 = r² − 2r√2 + 2, so √2 = (r² − 1)/(2r). Since r is rational and r ≠ 0, the right-hand side is rational. This is a contradiction.’ Which one of the following facts does the proof use to reach the contradiction?

    Free · correct letter only

    A√3 is irrational.B√2 is irrational.C√6 is irrational.DThe sum of two irrational numbers is irrational.EEvery rational number can be written as p/q, where p and q are integers with no common factor greater than 1.

    Answer

    Answer: B. √2 is irrational.

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13. Q13. Which list gives the lines of a correct proof, in order, that x³ + 1 > x² + x for every real number x > 1? (1) x³ − x² − x + 1 = (x − 1)²(x + 1). (2) x³ − x² − x + 1 = (x − 1)(x + 1)². (3) Since x > 1, (x − 1)² > 0 and x + 1 > 0. (4) So x³ − x² − x + 1 > 0. (5) Hence x³ + 1 > x² + x. (6) Suppose that x³ + 1 > x² + x. (7) Since x > 1, x³ > x² and x² > x.

    Free · correct letter only

    A(6), (4), (3), (1)B(2), (3), (4), (5)C(1), (3), (4), (5)D(7), (5)E(3), (4), (5)

    Answer

    Answer: C. (1), (3), (4), (5)

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14. Q14. Consider the following argument. ‘Let n = 4k + 1, where k is an integer. Then n² + 3 = 16k² + 8k + 4 = 4(4k² + 2k + 1). Since 4k² + 2k + 1 = 2(2k² + k) + 1 is odd, n² + 3 is divisible by 4 but not by 8.’ Which one of the following statements does this argument prove?

    Free · correct letter only

    AFor every odd integer n, n² + 3 is divisible by 4 but not by 8.BFor every integer n, n² + 3 is divisible by 4 but not by 8.CIf n² + 3 is divisible by 4 but not by 8, then n leaves remainder 1 when divided by 4.DIf n leaves remainder 1 when divided by 4, then n² + 3 is divisible by 4 but not by 8.EIf n leaves remainder 1 when divided by 4, then n² + 3 is divisible by 8.

    Answer

    Answer: D. If n leaves remainder 1 when divided by 4, then n² + 3 is divisible by 4 but not by 8.

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15. Q15. Here is an incomplete proof that there are no integers x, y and n with x² + y² = 4n + 3. ‘Consider remainders on division by 4. [gap] Hence x² + y² leaves remainder 0, 1 or 2 when divided by 4. But 4n + 3 leaves remainder 3, so x² + y² ≠ 4n + 3.’ Which one of the following facts, inserted at [gap], makes the argument a complete proof?

    Free · correct letter only

    AThe square of every integer leaves remainder 0 or 1 when divided by 4.BThe square of every odd integer leaves remainder 1 when divided by 4.CThe square of every integer leaves remainder 0, 1 or 2 when divided by 4.DThe square of every integer leaves remainder 0 or 1 when divided by 3.EEvery integer leaves remainder 0, 1, 2 or 3 when divided by 4.

    Answer

    Answer: A. The square of every integer leaves remainder 0 or 1 when divided by 4.

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16. Q16. A proof by contradiction is to be given for the statement: among any five integers, there are two whose difference is divisible by 4. The proof begins: ‘Suppose that there are five integers such that …’. Which of the following correctly completes this sentence?

    Free · correct letter only

    Asome two of them have a difference that is not divisible by 4Bevery two of them have a difference that is divisible by 4Cnone of them is divisible by 4Dno two of them are equalEno two of them have a difference that is divisible by 4

    Answer

    Answer: E. no two of them have a difference that is divisible by 4

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17. Q17. A student wants to prove that no perfect square (the square of a whole number) has last digit 2, 3, 7 or 8. The only fact the student may use is that the last digit of n² depends only on the last digit of n. For which list of values of n must the student work out the last digit of n² to give a complete proof?

    Free · correct letter only

    An = 2, 3, 7, 8Bn = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9Cn = 1, 2, 3, 4, 5, 6, 7, 8, 9Dn = 0, 1, 2, 3, 4, 5En = 0, 1, 4, 5, 6, 9

    Answer

    Answer: B. n = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

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18. Q18. Each of the following claims is about all real numbers a and b, and each claim is false. For which claim is the pair a = −3, b = 2 a counterexample?

    Free · correct letter only

    AIf a > b, then a² > b².BIf a < b, then 1/a < 1/b.CIf a + b < 0, then a < b.DIf ab < 0, then a < b.EIf a < b, then a² < b².

    Answer

    Answer: E. If a < b, then a² < b².

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19. Q19. Three arguments are offered as proofs that if n is an integer that is not divisible by 3, then n² − 1 is divisible by 3. I: Such an n can be written as 3k + 1 or 3k − 1 for some integer k. Then n² − 1 = 9k² ± 6k = 3(3k² ± 2k), which is divisible by 3. II: n² − 1 = (n − 1)(n + 1). Exactly one of the three consecutive integers n − 1, n and n + 1 is divisible by 3, and it is not n. So 3 divides n − 1 or n + 1, and hence 3 divides n² − 1. III: Such an n can be written as 3k + 1 for some integer k. Then n² − 1 = 9k² + 6k = 3(3k² + 2k), which is divisible by 3. Which of these arguments are valid proofs?

    Free · correct letter only

    AI onlyBII onlyCI and II onlyDI and III onlyEI, II and III

    Answer

    Answer: C. I and II only

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20. Q20. In Euclid’s proof that there are infinitely many primes, a finite list of distinct primes p₁, p₂, …, p_k is taken and the number N = p₁ × p₂ × … × p_k + 1 is formed. Which one of the following statements is true for every finite list of distinct primes, and is the fact the proof relies on?

    Free · correct letter only

    AN is prime.BNo p_i divides N, so every prime factor of N is a prime that is not in the list.CN has a prime factor that is larger than every prime in the list.DN is odd.EN is not prime whenever the list has at least two primes.

    Answer

    Answer: B. No p_i divides N, so every prime factor of N is a prime that is not in the list.

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21. Q21. P and Q are the ends of a diameter of a circle with centre O, and R is another point on the circle. Here is a proof that angle PRQ = 90°. ‘OP = OR = OQ, since all three are radii. So angle ORP = angle OPR = x and angle ORQ = angle OQR = y, say. The angles of triangle PQR are x, y and x + y, so 2x + 2y = 180°. Hence angle PRQ = x + y = 90°.’ Which of the following facts does the proof use? I: The base angles of an isosceles triangle are equal. II: The angles of a triangle add up to 180°. III: The angle that an arc subtends at the centre of a circle is twice the angle it subtends at the circumference.

    Free · correct letter only

    AI onlyBII onlyCI and III onlyDI and II onlyEI, II and III

    Answer

    Answer: D. I and II only

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22. Q22. Consider the statement: there is a real number x such that x² + 1 < 2x. Which one of the following arguments shows that this statement is false?

    Free · correct letter only

    AWhen x = 0, x² + 1 = 1 and 2x = 0, so x² + 1 < 2x does not hold.BWhen x = 1, x² + 1 = 2 = 2x, so x² + 1 < 2x does not hold.CFor every integer x, x² + 1 − 2x = (x − 1)² ≥ 0, so x² + 1 < 2x holds for no integer x.DFor every real number x, x² + 1 > 0, so x² + 1 < 2x holds for no real number x.EFor every real number x, x² + 1 − 2x = (x − 1)² ≥ 0, so x² + 1 < 2x holds for no real number x.

    Answer

    Answer: E. For every real number x, x² + 1 − 2x = (x − 1)² ≥ 0, so x² + 1 < 2x holds for no real number x.

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23. Q23. Let n be an integer. A student argues: ‘Suppose that n is odd and n³ + 5 is odd. Since n is odd, n³ is odd, so n³ + 5 is even. This contradicts n³ + 5 being odd.’ Which one of the following has the student proved?

    Free · correct letter only

    AIf n³ + 5 is odd, then n is even.Bn is even.CIf n is even, then n³ + 5 is odd.Dn³ + 5 is even for every integer n.EIf n³ + 5 is odd, then n is odd.

    Answer

    Answer: A. If n³ + 5 is odd, then n is even.

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24. Q24. The following argument is meant to show that √k is irrational, where k is a positive integer. ‘Suppose √k = p/q, where p and q are positive integers with no common factor greater than 1. Then p² = kq², so k divides p². Hence k divides p, say p = km. Then k²m² = kq², so q² = km², so k divides q², and hence k divides q. So k is a common factor of p and q greater than 1, which is a contradiction.’ For which of the following values of k is every deduction in the argument justified? I: k = 12 II: k = 6 III: k = 5

    Free · correct letter only

    AI onlyBII onlyCIII onlyDII and III onlyEI, II and III

    Answer

    Answer: D. II and III only

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25. Q25. A student argues: ‘Let x and y be real numbers with x > y. Dividing both sides of x > y by xy gives 1/y > 1/x.’ Which one of the following conditions, if it is added to the condition x > y, makes the argument correct?

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    Axy > 0Bx > 0Cx ≠ 0 and y ≠ 0Dx + y > 0Ex² > y²

    Answer

    Answer: A. xy > 0

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26. Q26. Let n be an integer. To prove that ‘3 divides n if and only if 3 divides n²’, it is enough to prove both statements in which one of the following pairs?

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    AIf 3 divides n, then 3 divides n². If 3 does not divide n², then 3 does not divide n.BIf 3 divides n², then 3 divides n. If 3 does not divide n, then 3 does not divide n².CIf 3 divides n, then 3 divides n². If 3 does not divide n, then 3 does not divide n².DIf 3 divides n², then 3 divides n. If 3 divides n², then 9 divides n².EIf 3 divides n, then 9 divides n². If 9 divides n², then 3 divides n.

    Answer

    Answer: C. If 3 divides n, then 3 divides n². If 3 does not divide n, then 3 does not divide n².

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27. Q27. A proof that log₂(3) is irrational begins: ‘Suppose that log₂(3) = p/q, where p and q are positive integers (possible because log₂(3) > 0). Then 2^p/q = 3, so 2^p = 3^q.’ Which one of the following correctly completes the proof?

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    ASince log₂(3) lies between 1 and 2, p/q is not an integer, which is a contradiction.BWe may take p and q with no common factor greater than 1, and 2^p = 3^q forces p and q to be even, which is a contradiction.CSince 2 < 3, 2^p < 3^q, which is a contradiction.DSince p ≥ 1, 2^p is even, while 3^q is odd, so 2^p ≠ 3^q, which is a contradiction.ETaking logarithms gives p log 2 = q log 3, so p = q, which is a contradiction.

    Answer

    Answer: D. Since p ≥ 1, 2^p is even, while 3^q is odd, so 2^p ≠ 3^q, which is a contradiction.

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28. Q28. Which one of the following is a correct proof that there exist irrational numbers a and b such that a^b is rational? (You may use the fact that √2 is irrational, but no other fact about irrational numbers.)

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    ATake a = √2 and b = 2. Then a^b = 2, which is rational.BTake a = b = √2. Then a^b = (√2)^√2, which is rational.CTake a = 2 and b = 1/2. Then a^b = √2, which is irrational.DIf (√2)^√2 is irrational, take a = (√2)^√2 and b = √2. Then a^b = (√2)^{√2 × √2} = (√2)² = 2, which is rational.EEither (√2)^√2 is rational, in which case take a = b = √2; or (√2)^√2 is irrational, in which case take a = (√2)^√2 and b = √2, so that a^b = (√2)^{√2 × √2} = (√2)² = 2.

    Answer

    Answer: E. Either (√2)^√2 is rational, in which case take a = b = √2; or (√2)^√2 is irrational, in which case take a = (√2)^√2 and b = √2, so that a^b = (√2)^{√2 × √2} = (√2)² = 2.

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