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Admissions tests / TMUA / Paper 2 / The logic of arguments

Stretch. 12 questions, 12 marks, about 55 minutes.

TMUA Paper 2: The logic of arguments, set 4

True and false, and, inclusive or, not, conditional statements in all their English forms, converse and contrapositive, necessary and sufficient, quantifiers, and negation.

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  • Answer all questions. No calculator.
  • Each question has exactly one correct answer.
  • This is a stretch set: several questions combine two specification points, and the fastest route is rarely the first one you see.
  1. 11 mark

    Consider the statement: 'For every positive integer n, there exists a prime number p such that p > n.' Which of the following is the correct negation of this statement?

    1. A For every positive integer n, there exists a prime number p such that p <= n.
    2. B There exists a positive integer n such that, for every prime number p, p <= n.
    3. C There exists a positive integer n such that, for every prime number p, p > n.
    4. D For every prime number p, there exists a positive integer n such that p <= n.
  2. 21 mark

    Consider the statement: 'For every integer n, if n is even and n is a multiple of 3, then n is a multiple of 6.' Which of the following is the correct negation of this statement?

    1. A For every integer n, if n is even and n is a multiple of 3, then n is not a multiple of 6.
    2. B There exists an integer n such that n is odd and n is not a multiple of 3, and n is not a multiple of 6.
    3. C There exists an integer n such that n is even and n is a multiple of 3, and n is a multiple of 6.
    4. D There exists an integer n such that n is even and n is a multiple of 3, and n is not a multiple of 6.
  3. 31 mark

    Take as given the true statement: 'For a positive integer n greater than 2, n is prime only if n is odd.' Which of the following correctly evaluates whether 'n is odd' is necessary and/or sufficient for 'n is prime', for n greater than 2?

    1. A 'n is odd' is necessary but not sufficient for 'n is prime': every prime greater than 2 is odd, but not every odd number greater than 2 is prime, for instance n = 9 is odd but not prime.
    2. B 'n is odd' is sufficient but not necessary for 'n is prime': it guarantees primality whenever it holds, but a prime greater than 2 need not be odd.
    3. C 'n is odd' is both necessary and sufficient for 'n is prime': the word 'only if' establishes a full two-way equivalence between the two.
    4. D 'n is odd' is neither necessary nor sufficient for 'n is prime': since 9 is odd but not prime, being odd does not connect reliably to being prime in either direction.
  4. 41 mark

    Consider the statement: 'If a positive integer n is a multiple of 9, then n is a multiple of 27.' This statement is false, since n = 9 is a multiple of 9 but not of 27. Given only that this original statement is false, what can we conclude about its contrapositive and its converse?

    1. A Both the contrapositive and the converse must also be false, since a statement, its converse and its contrapositive always share the same truth value.
    2. B The converse must also be false, since a false statement always makes its converse false; the contrapositive, however, could be either true or false and needs to be checked separately.
    3. C The contrapositive must also be false, because a statement and its contrapositive always share the same truth value. The converse is not automatically determined by this; checking directly, the converse ('if n is a multiple of 27, then n is a multiple of 9') is in fact true, since every multiple of 27 is automatically a multiple of 9.
    4. D Nothing can be concluded about either the contrapositive or the converse without checking both directly, since knowing the original statement is false does not by itself fix the truth of any related statement.
  5. 51 mark

    Take as given the true statement: 'For a real number x, x^2 - 2x - 3 = 0 if and only if x = 3 or x = -1.' Using this, which of the following is sufficient, but not necessary, for x^2 - 2x - 3 = 0 to hold?

    1. A x = 3 or x = -1
    2. B x is a real number
    3. C x = 3
    4. D x^2 - 2x - 3 != 0
  6. 61 mark

    The following statement is false: 'For all integers n, n^3 >= n.' Which option gives the correct negation of this statement, together with a genuine counterexample confirming the negation is true?

    1. A For all integers n, n^3 < n; for instance, n = -2 gives (-2)^3 = -8, which is less than -2.
    2. B There exists an integer n such that n^3 < n; for instance, n = -2 gives (-2)^3 = -8, which is less than -2.
    3. C There exists an integer n such that n^3 < n; for instance, n = 2 gives 2^3 = 8, which is less than 2.
    4. D There exists an integer n such that n^3 <= n; for instance, n = -2 gives (-2)^3 = -8, which is less than or equal to -2.
  7. 71 mark

    It is a necessary and sufficient condition for three positive lengths a, b and c to form a triangle that all three of a + b > c, b + c > a and a + c > b hold simultaneously. Given a = 5, b = 6 and c = 10, what can we conclude?

    1. A Since all three inequalities hold (5 + 6 = 11 > 10, 6 + 10 = 16 > 5, 5 + 10 = 15 > 6), and the condition is both necessary and sufficient, these lengths can definitely form a triangle.
    2. B The condition given is only necessary, so even though all three inequalities hold, this does not guarantee that a triangle can be formed.
    3. C Since a + b = 5 + 6 = 10, which is not greater than c = 10, the necessary condition fails and these lengths cannot form a triangle.
    4. D Since c = 10 is larger than both a = 5 and b = 6 individually, these lengths cannot form a triangle.
  8. 81 mark

    Take as given the true statement: 'If a positive integer n is a multiple of 12, then n is a multiple of 3.' What can we conclude about its converse ('if n is a multiple of 3, then n is a multiple of 12')?

    1. A The converse must also be true, since a statement and its converse always share the same truth value.
    2. B The converse must be false, since a converse is never guaranteed by the original statement.
    3. C The converse is true whenever n = 12 or any other multiple of 12, and only false for other multiples of 3, so it cannot be classified as simply true or false.
    4. D The converse does not have to be true just because the original is; and in fact it is false here, since n = 3 is a multiple of 3 but not a multiple of 12.
  9. 91 mark

    Exactly one of the following four statements is true. Which one is it?

    1. A The sum of any two prime numbers is always even.
    2. B The square of any odd number is always odd.
    3. C If a number is divisible by 2 and by 3, then it is divisible by 5.
    4. D The difference between any two consecutive square numbers is always even.
  10. 101 mark

    Consider the statement: 'A whole number's square ends in the digit 5 if the number itself ends in the digit 5.' Which of the following is the correct negation of this statement?

    1. A There exists a whole number that ends in the digit 5, whose square does not end in the digit 5.
    2. B There exists a whole number whose square ends in the digit 5, but the number itself does not end in the digit 5.
    3. C Every whole number that ends in the digit 5 has a square that does not end in the digit 5.
    4. D There exists a whole number that does not end in the digit 5, whose square ends in the digit 5.
  11. 111 mark

    Take as given the true statement: 'For a real number x, |x| < 5 if and only if -5 < x < 5.' Given a real number x satisfies |x| >= 5, which of the following must be true, using the biconditional above and its contrapositive?

    1. A x < -5 or x > 5
    2. B -5 <= x <= 5
    3. C x <= -5 or x >= 5
    4. D x <= -5 and x >= 5
  12. 121 mark

    The following statement is true: 'There exists a positive integer n such that n^2 is a multiple of 4 and n is not a multiple of 4' (n = 2 is one such number: 2^2 = 4 is a multiple of 4, but 2 is not a multiple of 4). Which option gives the correct negation of this statement, together with a correct judgement of whether the negation is true?

    1. A For all positive integers n, n^2 is not a multiple of 4 or n is a multiple of 4; the negation is true, since checking odd values of n (where n^2 is always odd, hence never a multiple of 4) always satisfies the first part.
    2. B For all positive integers n, n^2 is not a multiple of 4 and n is a multiple of 4; the negation is false, since n = 2 satisfies the original statement.
    3. C For all positive integers n, n^2 is a multiple of 4 or n is not a multiple of 4; the negation is false, since n = 2 satisfies the original statement.
    4. D For all positive integers n, n^2 is not a multiple of 4 or n is a multiple of 4; the negation is false, since n = 2 satisfies the original statement.

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: B

    1. To negate 'for all n, there exists p such that p > n', swap the quantifiers (for all becomes there exists, and there exists becomes for all) and negate the inner condition.
    2. Swapping the quantifiers gives 'there exists a positive integer n such that, for every prime number p, ...'.
    3. Negating p > n gives p <= n, since these are the only two mutually exclusive possibilities for real numbers.
    4. Putting the two steps together gives 'there exists a positive integer n such that, for every prime number p, p <= n'.
    5. So the answer is B.
    • Why not A: Keeps 'for every positive integer n, there exists a prime number p' completely unchanged and only flips the inequality; negating a 'for all ... there exists ...' statement requires swapping each quantifier's type, not just the comparison inside it.
    • Why not C: Correctly swaps 'for all' and 'there exists' to get 'there exists n such that for every p, ...', but leaves the inequality p > n unchanged instead of negating it to p <= n.
    • Why not D: Swaps which variable is attached to 'for all' and which to 'there exists' without actually reversing either quantifier's type, giving an unrelated (and in fact true) statement about primes rather than the genuine negation, which needs the outer quantifier to become existential.

  2. Question 2Answer: D

    1. Negating 'for all n, if A then B' requires two changes: switch 'for all' to 'there exists', and replace 'if A then B' with 'A and not B', where A stays exactly as it was and only B is negated.
    2. Here A is 'n is even and n is a multiple of 3' and B is 'n is a multiple of 6'; A must be carried across untouched.
    3. Negating B gives 'n is not a multiple of 6'.
    4. Combining both changes gives 'there exists an integer n such that n is even and n is a multiple of 3, and n is not a multiple of 6'.
    5. So the answer is D.
    • Why not A: Correctly negates the conclusion to 'not a multiple of 6' but leaves the quantifier as 'for every'; negating a 'for all' statement must switch it to 'there exists'.
    • Why not B: Wrongly changes the hypothesis 'n is even and n is a multiple of 3' as well as the conclusion. Negating 'if A then B' keeps A completely unchanged and negates only B, so the hypothesis should never be touched.
    • Why not C: Correctly switches the quantifier to 'there exists' and correctly keeps the hypothesis unchanged, but fails to negate the conclusion at all, simply restating 'n is a multiple of 6' rather than 'n is not a multiple of 6'.

  3. Question 3Answer: A

    1. 'n is prime only if n is odd' translates to 'if n is prime, then n is odd' (for n > 2); this direction alone is what 'only if' gives.
    2. This shows 'n is odd' is necessary for 'n is prime': whenever n is prime (and greater than 2), n must be odd.
    3. It is not sufficient: n = 9 is odd (and greater than 2), but 9 = 3 x 3 is not prime, so oddness alone does not force primality.
    4. Since it is necessary but not sufficient, the answer is A.
    • Why not B: Reverses the necessary and sufficient roles: it treats oddness as guaranteeing primality (false, since 9 is odd but not prime) while denying that primes greater than 2 must be odd (also false, since every one of them is).
    • Why not C: Mistakes 'only if', which gives a one-directional implication, for 'if and only if', which gives a full two-way equivalence; 'only if' alone never establishes the reverse direction.
    • Why not D: Correctly spots that oddness fails to guarantee primality (so it is not sufficient), but wrongly concludes this also breaks necessity, missing that every actual prime greater than 2 genuinely is odd.

  4. Question 4Answer: C

    1. A statement and its contrapositive are logically equivalent and always share the same truth value, so since the original statement is false, its contrapositive ('if n is not a multiple of 27, then n is not a multiple of 9') is false too.
    2. The converse, 'if n is a multiple of 27, then n is a multiple of 9', has no such guarantee; its truth has to be checked on its own.
    3. Checking directly: 27 = 3 x 9, so every multiple of 27 is automatically a multiple of 9 as well, making the converse true in this case.
    4. This shows the converse can end up true even when the original statement is false, confirming that only the contrapositive is fixed by the original's truth value.
    5. So the answer is C.
    • Why not A: Over-generalises the shared-truth-value rule: only a statement and its contrapositive are guaranteed to match; the converse has no such guarantee and can come out either way.
    • Why not B: Swaps which of the two carries the automatic guarantee, attaching it to the converse instead of the contrapositive.
    • Why not D: Fails to apply the guaranteed rule for the contrapositive at all, treating it the same as the converse, when the contrapositive's truth value is in fact logically forced by the original's.

  5. Question 5Answer: C

    1. 'Sufficient, not necessary' means: whenever the condition holds, the equation definitely holds too, but the equation can also hold in other cases where the condition is false.
    2. Substituting x = 3 into x^2 - 2x - 3 gives 9 - 6 - 3 = 0, so 'x = 3' does guarantee the equation holds: it is sufficient.
    3. But x = -1 also satisfies the equation, since (-1)^2 - 2(-1) - 3 = 1 + 2 - 3 = 0, and there 'x = 3' is false, so 'x = 3' is not necessary.
    4. By contrast, 'x = 3 or x = -1' is both necessary and sufficient, and 'x is a real number' is necessary but not sufficient, the reverse of what was asked.
    5. So the condition that is sufficient but not necessary is 'x = 3', option C.
    • Why not A: This condition is both necessary and sufficient by the given biconditional, since it is exactly equivalent to the equation; it does not fall short of necessity the way the question asks for.
    • Why not B: Given the stated domain this is necessary (trivially, since x must be real for the equation to make sense) but is not sufficient by itself, since most real numbers do not satisfy the equation; this reverses which property was asked for.
    • Why not D: Confuses the equation with its own negation: substituting the actual solutions x = 3 or x = -1 makes x^2 - 2x - 3 equal to 0, so this condition is false exactly where the equation holds, making it neither necessary nor sufficient for the equation.

  6. Question 6Answer: B

    1. The negation of 'for all integers n, n^3 >= n' is 'there exists an integer n such that n^3 < n': the quantifier flips to 'there exists', and '>=' negates to the strict '<'.
    2. Testing n = -2: (-2)^3 = -8.
    3. Is -8 < -2? Yes, since -8 lies further in the negative direction, making it a smaller number than -2.
    4. This confirms 'there exists n such that n^3 < n' is true, with n = -2 as a genuine counterexample, which is exactly why the original 'for all' statement was false.
    5. So the answer is B.
    • Why not A: Correctly negates the inequality and picks a genuine counterexample, but keeps the quantifier as 'for all' instead of switching it to 'there exists', which negating a 'for all' statement requires.
    • Why not C: Correctly forms the negation but checks the wrong counterexample: at n = 2, 2^3 = 8, which is not less than 2, so n = 2 does not actually confirm the negation.
    • Why not D: Negates '>=' incorrectly as '<=' instead of the strict '<'; the correct negation of 'a >= b' is 'a < b', since '>=' already includes the equality case.

  7. Question 7Answer: A

    1. The stated condition (all three of a + b > c, b + c > a, a + c > b) is both necessary and sufficient for three positive lengths to form a triangle, so satisfying all three is enough on its own to guarantee a triangle exists.
    2. Check each inequality with a = 5, b = 6, c = 10: a + b = 11 > 10 is true; b + c = 16 > 5 is true; a + c = 15 > 6 is true.
    3. All three inequalities hold.
    4. Because the condition is sufficient as well as necessary, this is enough to conclude these lengths can definitely form a triangle.
    5. So the answer is A.
    • Why not B: Misremembers the stated condition as only necessary, when it was explicitly given as necessary AND sufficient; satisfying it therefore does guarantee a triangle, not merely fail to rule one out.
    • Why not C: Makes an arithmetic slip: 5 + 6 = 11, not 10, so a + b actually is greater than c.
    • Why not D: Jumps to a conclusion based on c being the largest individual side without checking whether it exceeds the SUM of the other two, which is the actual test; 10 is less than 5 + 6 = 11, so this does not rule out a triangle.

  8. Question 8Answer: D

    1. A statement and its converse do not automatically share a truth value; that guarantee only holds between a statement and its contrapositive.
    2. So the converse here is not settled by the truth of the original statement; as a 'for all n' claim it is false as soon as one counterexample is found.
    3. Testing n = 3: 3 is a multiple of 3, but 3 is not a multiple of 12.
    4. This single counterexample is enough to make the whole converse false.
    5. So the answer is D.
    • Why not A: Applies the rule that only holds for the contrapositive (which always shares the original statement's truth value) to the converse instead, which has no such guarantee.
    • Why not B: Over-corrects the opposite way: having correctly noticed a converse is not automatically guaranteed, wrongly concludes it can therefore never be true, when it is simply undetermined by the logic alone.
    • Why not C: Misunderstands that the converse is itself a 'for all n' claim, not something to be evaluated case by case; since it fails for at least one n, the whole universal converse statement is simply false, with no need to qualify it by cases.

  9. Question 9Answer: B

    1. Check each statement against a genuine example rather than a general impression.
    2. 'The sum of any two prime numbers is always even' fails because 2 is prime and even, and 2 + 3 = 5 is odd.
    3. 'If a number is divisible by 2 and by 3, then it is divisible by 5' fails because 12 is divisible by both 2 and 3 but not by 5.
    4. 'The difference between any two consecutive square numbers is always even' fails because (n + 1)^2 - n^2 = 2n + 1 is always odd, for example 4 - 1 = 3.
    5. 'The square of any odd number is always odd' holds: writing an odd number as 2k + 1 gives (2k + 1)^2 = 4k^2 + 4k + 1, which is one more than an even number, so it is always odd.
    6. Since A, C and D are all false and B is true, the answer is B.
    • Why not A: Overlooks that 2 is prime and even; adding it to another prime such as 3 gives 2 + 3 = 5, an odd sum, which is enough to disprove the 'always even' claim.
    • Why not C: Finds this plausible from a vague impression but fails to test a genuine counterexample: 12 is divisible by both 2 and 3, yet 12 is not divisible by 5.
    • Why not D: Assumes consecutive squares must differ by an even amount; in fact (n + 1)^2 - n^2 = 2n + 1 is always odd, as shown directly by 4 - 1 = 3 or 9 - 4 = 5.

  10. Question 10Answer: A

    1. 'A if B' translates to 'if B then A': here B is 'the number ends in the digit 5' and A is 'its square ends in the digit 5', so the statement becomes 'if a whole number ends in 5, then its square ends in 5'.
    2. Negating 'if B then A' gives 'B and not A': keep the hypothesis (ends in 5) unchanged, and negate only the conclusion (square ends in 5).
    3. This gives 'a whole number ends in the digit 5, whose square does not end in the digit 5'.
    4. Also switch the quantifier: the negation of an implicit 'for all' is 'there exists'.
    5. Combining both, the correct negation is 'there exists a whole number that ends in the digit 5, whose square does not end in the digit 5', option A.
    • Why not B: Translates 'A if B' the wrong way round, as though it read 'if A then B' rather than 'if B then A', and then negates that reversed conditional instead of the correct one.
    • Why not C: Correctly identifies which part to negate (the conclusion) but leaves the quantifier as universal, when negating an implicit 'for all' claim must switch it to 'there exists'.
    • Why not D: Negates the hypothesis instead of the conclusion, keeping 'square ends in 5' unchanged and flipping 'ends in 5' instead, when the hypothesis should be carried across untouched.

  11. Question 11Answer: C

    1. The biconditional says |x| < 5 exactly when -5 < x < 5, i.e. 'x > -5 and x < 5'.
    2. Given |x| >= 5, this is precisely the negation of |x| < 5, so by the biconditional it must also be the negation of 'x > -5 and x < 5'.
    3. Negating 'x > -5 and x < 5' with De Morgan's law gives 'not(x > -5) or not(x < 5)', i.e. 'x <= -5 or x >= 5', since negating a strict inequality gives the opposite non-strict one, and 'and' flips to 'or'.
    4. So the correct conclusion is 'x <= -5 or x >= 5'.
    5. The answer is C.
    • Why not A: Drops the equality case: negating the strict inequality x > -5 gives x <= -5, not the strict x < -5, since '>=' rather than '>' is what should be excluded.
    • Why not B: Does not negate anything at all; this simply adds the endpoints to the original interval instead of describing what lies outside it.
    • Why not D: Gets the boundary values right but uses the wrong connective: negating 'x > -5 and x < 5' by De Morgan's law should switch 'and' to 'or', and 'x <= -5 and x >= 5' cannot be satisfied by any real number at all.

  12. Question 12Answer: D

    1. Negating 'there exists n such that A and B' gives 'for all n, not A or not B': the quantifier flips to 'for all', the connective flips from 'and' to 'or', and each inner part is individually negated.
    2. Here A is 'n^2 is a multiple of 4' and B is 'n is not a multiple of 4'; negating A gives 'n^2 is not a multiple of 4', and negating B gives 'n is a multiple of 4'.
    3. Combining these correctly gives 'for all positive integers n, n^2 is not a multiple of 4 or n is a multiple of 4'.
    4. To judge its truth, test n = 2: 2^2 = 4 is a multiple of 4, so the first disjunct is false there; and 2 is not a multiple of 4, so the second disjunct is also false there; both disjuncts fail at n = 2, so the 'for all' claim is false.
    5. This is expected, since the original statement (with n = 2 as a witness) was true, and a true statement always has a false negation. So the answer is D.
    • Why not A: Forms the correct negation but checks only odd n, where the first disjunct is indeed satisfied; this never tests an even value such as n = 2, which is exactly where the 'for all' claim breaks down.
    • Why not B: Correctly flips the quantifier but keeps the connective as 'and' instead of switching it to 'or', which negating an 'and' statement (via De Morgan's law) requires.
    • Why not C: Correctly flips the quantifier and the connective but forgets to negate the two inner parts themselves, leaving 'n^2 is a multiple of 4' and 'n is not a multiple of 4' exactly as they were in the original.

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