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The logic of arguments — solutions

25 full solutions · spec Arg1-Arg4

Section 2: Mathematical reasoning (Paper 2 only)25 questions
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Every question on this topic with its answer, the working, and a full method: what the question is testing, the reason for each step, the mistakes that are easy to make and what they would give you, and the principle to take away.

1. LGC-01The logic of arguments · spec Arg1 · Paper 2 style · ★☆☆

Consider the statement: if \(n\) is a multiple of \(6\), then \(n\) is a multiple of \(3\). Which of the following is its contrapositive?

  1. AIf \(n\) is not a multiple of \(3\), then \(n\) is not a multiple of \(6\).
  2. BIf \(n\) is a multiple of \(3\), then \(n\) is a multiple of \(6\).
  3. CIf \(n\) is not a multiple of \(6\), then \(n\) is not a multiple of \(3\).
  4. D\(n\) is a multiple of \(6\) if and only if \(n\) is a multiple of \(3\).
  5. EIf \(n\) is a multiple of \(3\), then \(n\) is not a multiple of \(6\).
Answer: A. The contrapositive of 'if \(A\) then \(B\)' is 'if not \(B\) then not \(A\)'. It is logically equivalent to the original; B is the converse and C is the inverse.
Full method — how and why
The four relatives of an implication. Given \(A \Rightarrow B\), you must be able to write the converse, inverse and contrapositive without hesitating.
  1. Name the parts: \(A\) is '\(n\) is a multiple of \(6\)', \(B\) is '\(n\) is a multiple of \(3\)'.
    Always label before manipulating - most errors here are bookkeeping errors.
  2. The contrapositive is 'if not \(B\) then not \(A\)': if \(n\) is not a multiple of \(3\), then \(n\) is not a multiple of \(6\).
    Negate both parts and swap them. Doing only one of the two gives the wrong relative.
  3. Sanity-check the meaning: if \(3\) does not divide \(n\), then \(6\) certainly cannot, since every multiple of \(6\) is a multiple of \(3\).
    The contrapositive is logically equivalent to the original, so it must also be true - a free check.
Where it goes wrong
  • Choosing 'if \(n\) is a multiple of \(3\) then \(n\) is a multiple of \(6\)' - that is the converse (swap only), and it is false (\(n=9\)).
  • Choosing 'if \(n\) is not a multiple of \(6\) then \(n\) is not a multiple of \(3\)' - the inverse (negate only), also false (\(n=9\)).
Takeaway Converse: swap. Inverse: negate. Contrapositive: swap and negate - and only the contrapositive is equivalent to the original.
2. LGC-02The logic of arguments · spec Arg2 · Paper 2 style · ★☆☆

Let \(P\) be the statement '\(x > 2\)' and \(Q\) the statement '\(x^2 > 4\)', where \(x\) is a real number. Which one of the following is true?

  1. AP is necessary and sufficient for Q.
  2. BP is necessary but not sufficient for Q.
  3. CP is sufficient but not necessary for Q.
  4. DP is neither necessary nor sufficient for Q.
Answer: C. \(x > 2 \Rightarrow x^2 > 4\), so \(P\) is sufficient for \(Q\). But \(x = -3\) gives \(x^2 > 4\) without \(x > 2\), so \(P\) is not necessary.
Full method — how and why
Necessary/sufficient questions are two separate yes-no checks: does \(P \Rightarrow Q\)? does \(Q \Rightarrow P\)?
  1. Test sufficiency (\(P \Rightarrow Q\)): if \(x > 2\) then \(x^2 > 4\). True, so \(P\) is sufficient.
    Squaring preserves the inequality for positive numbers.
  2. Test necessity (\(Q \Rightarrow P\)): try to break it. \(x = -3\) gives \(x^2 = 9 > 4\) but \(x \not> 2\).
    One counterexample is enough to destroy an implication; negative numbers are the obvious place to look when squaring.
  3. So \(P\) is sufficient but not necessary.
    Sufficient \(=\) 'enough to guarantee'; necessary \(=\) 'can't do without'.
Where it goes wrong
  • Getting the direction backwards: '\(P\) is necessary for \(Q\)' means \(Q \Rightarrow P\), not \(P \Rightarrow Q\).
  • Only checking positive \(x\) and concluding the two are equivalent.
Takeaway \(P\) sufficient for \(Q\) means \(P \Rightarrow Q\); \(P\) necessary for \(Q\) means \(Q \Rightarrow P\). Check each direction separately, hunting for counterexamples.
3. LGC-03The logic of arguments · spec Arg4 · Paper 2 style · ★☆☆

Which of the following is the negation of the statement 'every student in the class scored more than \(50\)'?

  1. AAt least one student in the class scored more than \(50\).
  2. BEvery student in the class scored \(50\) or less.
  3. CEvery student in the class scored less than \(50\).
  4. DAt least one student in the class scored \(50\) or less.
  5. ENo student in the class scored more than \(50\).
Answer: D. The negation of 'for all \(x\), \(P(x)\)' is 'there exists \(x\) such that not \(P(x)\)'. 'Not more than \(50\)' means '\(50\) or less'.
Full method — how and why
Negating a universal statement. 'All' flips to 'at least one', and the property inside flips too.
  1. Write it formally: 'for all students \(s\), mark\((s) > 50\)'.
    Formalising stops the English from misleading you.
  2. Negate: 'there exists a student \(s\) with mark\((s) \not> 50\)'.
    \(\neg \forall x\, P(x) \equiv \exists x\, \neg P(x)\): the quantifier flips and the inside is negated.
  3. Translate \(\not> 50\) into '\(\leq 50\)', i.e. \(50\) or less.
    The negation of 'greater than' is 'less than or equal to' - the boundary belongs to the negation.
  4. So: at least one student scored \(50\) or less.
    One failure is all it takes to make 'every' false.
Where it goes wrong
  • Over-negating to 'every student scored \(50\) or less' - that is far stronger than the negation and is not implied.
  • Writing 'less than \(50\)' and losing the case of exactly \(50\).
Takeaway Negating 'all ... are' gives 'at least one ... is not' - and remember \(\neg(x > a)\) is \(x \leq a\).
4. LGC-04The logic of arguments · spec Arg1 · Paper 2 style · ★☆☆

Consider the statement about a real number \(x\): if \(x^2 = x\), then \(x = 1\). Which one of the following is true?

  1. AThe statement is true and its converse is true.
  2. BThe statement is true and its converse is false.
  3. CThe statement is false and its converse is true.
  4. DThe statement is false and its converse is false.
Answer: C. \(x = 0\) satisfies \(x^2 = x\) but not \(x = 1\), so the statement is false. The converse, 'if \(x = 1\) then \(x^2 = x\)', is true.
Full method — how and why
Judge a statement and its converse separately. Each needs either a proof or a counterexample.
  1. The statement: if \(x^2 = x\) then \(x = 1\). Solve \(x^2 = x\): \(x(x-1) = 0\), so \(x = 0\) or \(x = 1\).
    Never divide by \(x\) - that is exactly how the root \(x = 0\) gets lost.
  2. \(x = 0\) satisfies the hypothesis but not the conclusion, so the statement is false.
    A single counterexample settles it.
  3. The converse: if \(x = 1\) then \(x^2 = x\). Substituting gives \(1 = 1\). True.
    Verifying one value is enough because the hypothesis pins \(x\) down completely.
  4. So: false statement, true converse.
    Match to the option list.
Where it goes wrong
  • Dividing \(x^2 = x\) by \(x\), getting only \(x = 1\), and wrongly declaring the statement true.
  • Assuming a false statement must have a false converse - the two are independent.
Takeaway A statement and its converse have no logical connection; test each on its own, and never divide an equation by a variable.
5. LGC-07The logic of arguments · spec Arg2 · Paper 2 style · ★☆☆

Let \(P\): 'the quadrilateral \(ABCD\) is a square' and \(Q\): 'the quadrilateral \(ABCD\) has four equal sides'. Which one of the following is true?

  1. AP is necessary and sufficient for Q.
  2. BP is necessary but not sufficient for Q.
  3. CP is sufficient but not necessary for Q.
  4. DP is neither necessary nor sufficient for Q.
Answer: C. A square has four equal sides (\(P \Rightarrow Q\), sufficient), but a rhombus that is not a square shows \(Q\) does not imply \(P\) (not necessary).
Full method — how and why
The same two-direction test, now in geometry. Look for a shape satisfying one condition but not the other.
  1. Sufficiency: every square has four equal sides, so \(P \Rightarrow Q\). True.
    Straight from the definition of a square.
  2. Necessity: does four equal sides force a square? A rhombus with angles \(60^\circ\) and \(120^\circ\) has four equal sides and is not a square.
    A square also needs right angles; equal sides alone do not fix the angles.
  3. So \(P\) is sufficient but not necessary for \(Q\).
    \(P\) is the stronger condition, so it implies the weaker one but not conversely.
Where it goes wrong
  • Confusing which is the stronger condition: the more demanding property (square) is sufficient for the less demanding one (four equal sides).
  • Failing to produce a concrete counterexample and guessing.
Takeaway A stronger condition is sufficient for a weaker one; the weaker is necessary for the stronger. Name a counterexample to confirm.
6. LGC-08The logic of arguments · spec Arg4 · Paper 2 style · ★☆☆

Which of the following is the negation of 'there exists a positive integer \(n\) such that \(n^2 + n + 41\) is not prime'?

  1. AFor every positive integer \(n\), \(n^2 + n + 41\) is not prime.
  2. BThere exists a positive integer \(n\) such that \(n^2 + n + 41\) is prime.
  3. CFor every positive integer \(n\), \(n^2 + n + 41\) is prime.
  4. DFor some positive integer \(n\), \(n^2 + n + 41\) is prime.
  5. EThere is no positive integer \(n\) for which \(n^2 + n + 41\) is prime.
Answer: C. Negating 'there exists \(n\) with not \(P(n)\)' gives 'for all \(n\), \(P(n)\)'.
Full method — how and why
Negating an existential statement whose inside is already a negation - so two flips happen.
  1. Formalise: '\(\exists n\) such that \(n^2+n+41\) is not prime'.
    Note the inner 'not'.
  2. Negate the quantifier: \(\neg\exists n\, P(n) \equiv \forall n\, \neg P(n)\).
    'There is one' becomes 'there is none', i.e. 'every one fails'.
  3. Negate the inside: 'not (not prime)' \(=\) 'prime'.
    Double negation cancels.
  4. Result: for every positive integer \(n\), \(n^2+n+41\) is prime.
    Exactly one of the original and this is true; in fact this one is false, since \(n = 41\) gives \(41^2+41+41 = 41 \times 43\).
Where it goes wrong
  • Leaving the inner 'not' in place: 'for every \(n\), \(n^2+n+41\) is not prime' - that negates the quantifier but not the property.
  • Keeping 'there exists' and only flipping the property.
Takeaway \(\neg\exists x\,\neg P(x) \equiv \forall x\, P(x)\): flip the quantifier and the statement inside.
7. LGC-10The logic of arguments · spec Arg2 · Paper 2 style · ★☆☆

For real numbers \(a\) and \(b\), let \(P\): '\(a = b\)' and \(Q\): '\(a^2 = b^2\)'. Which one of the following is true?

  1. AP is necessary and sufficient for Q.
  2. BP is necessary but not sufficient for Q.
  3. CP is sufficient but not necessary for Q.
  4. DP is neither necessary nor sufficient for Q.
Answer: C. \(a = b \Rightarrow a^2 = b^2\) (sufficient), but \(a = 1\), \(b = -1\) gives \(a^2 = b^2\) without \(a = b\) (not necessary).
Full method — how and why
The squaring trap again, in its purest form.
  1. Sufficiency: if \(a = b\) then squaring both sides gives \(a^2 = b^2\). True.
    Doing the same thing to equal quantities keeps them equal.
  2. Necessity: does \(a^2 = b^2\) force \(a = b\)? Take \(a = 1\), \(b = -1\): \(1 = 1\) but \(a \neq b\).
    \(a^2 = b^2\) only gives \(a = \pm b\).
  3. So \(P\) is sufficient but not necessary.
    Squaring loses sign information, so it cannot be reversed.
Where it goes wrong
  • Treating squaring as reversible - it is exactly this that makes squaring an equation introduce extra solutions.
  • Only trying positive values.
Takeaway \(a = b \Rightarrow a^2 = b^2\), but \(a^2 = b^2 \Rightarrow a = \pm b\). Squaring is a one-way street.
8. LGC-11The logic of arguments · spec Arg1 · Paper 2 style · ★☆☆

The statement 'if a triangle is equilateral then it is isosceles' is true. Which of the following must therefore also be true?

  1. IIf a triangle is isosceles then it is equilateral.
  2. IIIf a triangle is not isosceles then it is not equilateral.
  3. IIIIf a triangle is not equilateral then it is not isosceles.
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Answer: C. Only the contrapositive (II) is logically equivalent to the original. I is the converse and III is the inverse; both are in fact false here.
Full method — how and why
Given a true implication, only one of the three relatives is guaranteed. Identify which of I, II, III is the contrapositive.
  1. I is 'isosceles \(\Rightarrow\) equilateral' - the converse. Not guaranteed, and here false (a \(3,3,5\) triangle).
    Swapped but not negated.
  2. II is 'not isosceles \(\Rightarrow\) not equilateral' - the contrapositive. Always guaranteed. True.
    Swapped and negated; logically equivalent to the original.
  3. III is 'not equilateral \(\Rightarrow\) not isosceles' - the inverse. Not guaranteed, and here false (the same \(3,3,5\) triangle).
    Negated but not swapped - and the inverse is just the contrapositive of the converse, so it carries the converse's truth value, not the original's.
  4. So II only.
    Match to the list.
Where it goes wrong
  • Accepting III because both its parts are negated - negation alone is not enough; the parts must also swap.
  • Accepting I because equilateral triangles are also isosceles - that supports the original, not the converse.
Takeaway From \(A \Rightarrow B\) you get exactly one free statement: \(\neg B \Rightarrow \neg A\).
9. LGC-13The logic of arguments · spec Arg4 · Paper 2 style · ★☆☆

Which of the following is the negation of '\(x > 3\) and \(y \leq 5\)'?

  1. A\(x < 3\) and \(y > 5\)
  2. B\(x < 3\) or \(y \geq 5\)
  3. C\(x \leq 3\) or \(y \geq 5\)
  4. D\(x \leq 3\) and \(y > 5\)
  5. E\(x \leq 3\) or \(y > 5\)
Answer: E. Not (\(A\) and \(B\)) is (not \(A\)) or (not \(B\)). Not \(x > 3\) is \(x \leq 3\); not \(y \leq 5\) is \(y > 5\).
Full method — how and why
De Morgan's law: negating an 'and' turns it into an 'or', with both parts negated.
  1. \(\neg(A \text{ and } B) \equiv (\neg A) \text{ or } (\neg B)\).
    For the conjunction to fail, it is enough that one part fails.
  2. Negate the first part: \(\neg(x > 3)\) is \(x \leq 3\).
    The boundary \(x = 3\) satisfies the negation, since \(3 > 3\) is false.
  3. Negate the second: \(\neg(y \leq 5)\) is \(y > 5\).
    Strict inequality, because \(y = 5\) satisfied the original.
  4. Combine with 'or': \(x \leq 3\) or \(y > 5\).
    The joining word flips too - this is the half most often forgotten.
Where it goes wrong
  • Keeping 'and': '\(x \leq 3\) and \(y > 5\)' is far too strong - it demands both failures.
  • Getting the boundaries wrong: writing \(x < 3\) or \(y \geq 5\), which mishandles \(x = 3\) and \(y = 5\).
Takeaway De Morgan: not(A and B) \(=\) notA or notB; not(A or B) \(=\) notA and notB. Negate the connective as well as the parts.
10. LGC-14The logic of arguments · spec Arg1 · Paper 2 style · ★☆☆

For a positive integer \(n\), consider the statement: if \(n\) is prime, then \(n\) is odd. Which one of the following is true?

  1. AThe statement is true and its converse is true.
  2. BThe statement is true and its converse is false.
  3. CThe statement is false and its converse is true.
  4. DThe statement is false and its converse is false.
Answer: D. \(n = 2\) is an even prime, so the statement is false. The converse, 'if \(n\) is odd then \(n\) is prime', fails for \(n = 9\).
Full method — how and why
Statement and converse again, with the classic \(n = 2\) counterexample.
  1. Statement: primes are odd? \(n = 2\) is prime and even. False.
    \(2\) is the only even prime - it is the standard counterexample and worth memorising.
  2. Converse: odd numbers are prime? \(n = 9 = 3 \times 3\) is odd and composite. False.
    Also \(n = 1\), which is odd and not prime by definition.
  3. So both are false.
    Match to the option list.
Where it goes wrong
  • Forgetting that \(2\) is prime and declaring the statement true.
  • Forgetting that \(1\) is not prime when hunting counterexamples.
Takeaway \(2\) is the even prime and \(1\) is not prime - the two facts behind most prime-related counterexamples.
11. LGC-17The logic of arguments · spec Arg1 · Paper 2 style · ★☆☆

Let \(x\) be a real number. Which of the following statements is equivalent to '\(x \geq 2\) if and only if \(x^3 \geq 8\)'?

  1. A\(x^3 \geq 8 \Rightarrow x \geq 2\)
  2. B(\(x \geq 2 \Rightarrow x^3 \geq 8\)) and (\(x^3 \geq 8 \Rightarrow x \geq 2\))
  3. C(\(x < 2 \Rightarrow x^3 \geq 8\)) or (\(x^3 < 8 \Rightarrow x \geq 2\))
  4. D\(x \geq 2 \Rightarrow x^3 \geq 8\)
  5. E\(x \geq 2\) or \(x^3 \geq 8\)
Answer: B. 'If and only if' is the conjunction of the implication and its converse.
Full method — how and why
Unpacking 'if and only if' into its two implications.
  1. '\(A\) if and only if \(B\)' means both '\(A\) if \(B\)' (\(B \Rightarrow A\)) and '\(A\) only if \(B\)' (\(A \Rightarrow B\)).
    The phrase is literally built from 'if' and 'only if' - each supplies one direction.
  2. With \(A\): \(x \geq 2\) and \(B\): \(x^3 \geq 8\), that is \((x \geq 2 \Rightarrow x^3 \geq 8)\) and \((x^3 \geq 8 \Rightarrow x \geq 2)\).
    The conjunction of the implication and its converse.
  3. (Both happen to be true here, since cubing is increasing - but the question only asks for the equivalent form.)
    Read what is asked: the structure, not the truth value.
Where it goes wrong
  • Giving only one implication - that is half the statement.
  • Joining the two with 'or' instead of 'and'.
Takeaway \(A \Leftrightarrow B\) is exactly \((A \Rightarrow B)\) and \((B \Rightarrow A)\).
12. LGC-19The logic of arguments · spec Arg2 · Paper 2 style · ★☆☆

Let \(f\) be a function defined for all real numbers. \(P\): '\(f'(a) = 0\)'; \(Q\): '\(f\) has a stationary point at \(x = a\)'. Which one of the following is true?

  1. AP is necessary and sufficient for Q.
  2. BP is necessary but not sufficient for Q.
  3. CP is sufficient but not necessary for Q.
  4. DP is neither necessary nor sufficient for Q.
Answer: A. A stationary point is by definition a point where the derivative is zero, so \(P\) and \(Q\) are equivalent.
Full method — how and why
Sometimes the two statements are the same by definition - recognise that and both directions are immediate.
  1. Recall the definition: \(f\) has a stationary point at \(x = a\) precisely when \(f'(a) = 0\).
    That is what 'stationary' means - the rate of change is zero.
  2. So \(P \Rightarrow Q\) and \(Q \Rightarrow P\) both hold trivially.
    Definitions are biconditional.
  3. Hence \(P\) is necessary and sufficient for \(Q\).
    Match to the option list.
Where it goes wrong
  • Confusing this with 'turning point': \(f'(a) = 0\) does not guarantee a maximum or minimum (\(y = x^3\) at \(x=0\)). But the question says stationary point, not turning point.
  • Worrying about functions that are not differentiable - the question states \(f\) is defined and \(f'(a)\) is referred to, so differentiability is assumed.
Takeaway A definition gives both implications at once; but stationary \(\neq\) turning - the extra condition is a sign change in \(f'\).
13. LGC-21The logic of arguments · spec Arg3 · Paper 2 style · ★☆☆

Consider the statement \(S\): for all real \(x\), \(x^2 - 6x + 10 > 0\). Which of the following is true?

  1. A\(S\) is false, because \(x = 3\) gives \(x^2 - 6x + 10 = 0\).
  2. B\(S\) cannot be decided without knowing \(x\).
  3. C\(S\) is false, because a quadratic can be negative.
  4. D\(S\) is true, because \(x^2 - 6x + 10 = (x - 3)^2 + 1\).
  5. E\(S\) is true, because \(x^2 - 6x + 10\) has two real roots.
Answer: D. Completing the square shows the expression is at least \(1\) for every \(x\).
Full method — how and why
A universal claim about a quadratic. Completing the square settles it, and the question also tests whether the reason given is sound.
  1. Complete the square: \(x^2 - 6x + 10 = (x-3)^2 - 9 + 10 = (x-3)^2 + 1\).
    Halve the coefficient of \(x\) to get \(-3\), then correct by \(-(-3)^2 = -9\).
  2. \((x-3)^2 \geq 0\), so the expression is \(\geq 1 > 0\) for every real \(x\). True.
    A square is never negative; the minimum value is \(1\) at \(x = 3\).
  3. Check the false reasons: \(x = 3\) gives \(9 - 18 + 10 = 1\), not \(0\); and the discriminant \(36 - 40 = -4 < 0\), so there are no real roots.
    The claim that it has two real roots is simply wrong, and would in any case not prove positivity.
Where it goes wrong
  • Assuming any quadratic takes negative values - one with a positive leading coefficient and negative discriminant never does.
  • Mis-completing the square as \((x-3)^2 + 10\) or \((x-6)^2+1\).
Takeaway Completing the square proves positivity outright: \(a(x-h)^2 + k\) with \(a>0, k>0\) is always positive.
14. LGC-22The logic of arguments · spec Arg2 · Paper 2 style · ★☆☆

For real numbers \(a, b\): \(P\): '\(ab = 0\)'; \(Q\): '\(a = 0\)'. Which one of the following is true?

  1. AP is necessary and sufficient for Q.
  2. BP is necessary but not sufficient for Q.
  3. CP is sufficient but not necessary for Q.
  4. DP is neither necessary nor sufficient for Q.
Answer: B. \(a = 0 \Rightarrow ab = 0\), so \(P\) is necessary for \(Q\). But \(ab = 0\) could be because \(b = 0\) with \(a \neq 0\), so \(P\) is not sufficient.
Full method — how and why
This one reverses the usual direction - the check must be done carefully rather than by pattern-matching.
  1. Test '\(P\) sufficient for \(Q\)', i.e. \(ab = 0 \Rightarrow a = 0\): take \(a = 5\), \(b = 0\). Then \(ab = 0\) but \(a \neq 0\). Not sufficient.
    The zero-product rule gives \(a = 0\) or \(b = 0\), not \(a = 0\).
  2. Test '\(P\) necessary for \(Q\)', i.e. \(Q \Rightarrow P\): if \(a = 0\) then \(ab = 0 \cdot b = 0\). True, so necessary.
    Anything times zero is zero.
  3. So \(P\) is necessary but not sufficient for \(Q\).
    Note the direction: the weaker condition (\(ab=0\)) is necessary for the stronger one (\(a=0\)).
Where it goes wrong
  • Answering 'sufficient but not necessary' by reflex - in this question the roles are swapped from the common pattern.
  • Thinking \(ab = 0\) requires both to be zero.
Takeaway \(ab = 0 \Rightarrow a = 0\) or \(b = 0\); and always check both directions explicitly rather than assuming the familiar answer.
15. LGC-23The logic of arguments · spec Arg1 · Paper 2 style · ★☆☆

Consider the statement: if \(a\) and \(b\) are irrational, then \(a + b\) is irrational. Which of the following pairs is a counterexample?

  1. A\(a = 1\), \(b = -1\)
  2. B\(a = \sqrt2\), \(b = -\sqrt2\)
  3. C\(a = \pi\), \(b = \sqrt2\)
  4. D\(a = \sqrt2\), \(b = 2\sqrt2\)
  5. E\(a = \sqrt2\), \(b = \sqrt3\)
Answer: B. A counterexample needs the hypothesis true (both irrational) and the conclusion false (\(a + b\) rational): \(\sqrt2 + (-\sqrt2) = 0\).
Full method — how and why
A counterexample must make the hypothesis true and the conclusion false. Check both halves for each candidate.
  1. Hypothesis: both \(a\) and \(b\) irrational. Conclusion to break: \(a + b\) irrational - so we need \(a+b\) rational.
    Only a case satisfying the hypothesis can refute the statement.
  2. \(a = \sqrt2\), \(b = -\sqrt2\): both irrational ✓, and \(a + b = 0\), which is rational ✓. Counterexample.
    \(-\sqrt2\) is irrational: if it were rational, so would its negative be.
  3. Reject \(a = 1, b = -1\): these are rational, so the hypothesis fails and nothing is refuted.
    A case that does not satisfy the hypothesis says nothing about the implication.
  4. Reject \(\sqrt2 + \sqrt3\), \(\sqrt2 + 2\sqrt2 = 3\sqrt2\) and \(\pi + \sqrt2\): each sum is irrational, so the conclusion holds and the statement is not broken.
    These are examples of the statement, not against it.
Where it goes wrong
  • Picking rational numbers whose sum is rational - the hypothesis is not satisfied.
  • Picking irrationals whose sum is irrational - the conclusion is not broken.
Takeaway A counterexample satisfies the 'if' and violates the 'then'. Check both halves every time.
16. LGC-24The logic of arguments · spec Arg4 · Paper 2 style · ★☆☆

Let \(S\) be the statement: every even number greater than \(2\) can be written as the sum of two primes. A single counterexample to \(S\) would be

  1. Aan odd number that is not the sum of two primes
  2. Ban even number greater than \(2\) that can be written as the sum of two primes in two different ways
  3. Ctwo primes whose sum is odd
  4. Dan even number that is the sum of two primes
  5. Ean even number greater than \(2\) that is not the sum of any two primes
Answer: E. To refute 'for all \(n\), \(P(n)\)' you exhibit one \(n\) with \(P(n)\) false.
Full method — how and why
What it takes to refute a universal statement - and what it takes to refute *this* one in particular.
  1. The statement has the form 'for all even \(n > 2\), \(P(n)\)'.
    Identify the quantifier and the domain: even numbers greater than \(2\).
  2. Its negation is 'there exists an even \(n > 2\) with \(P(n)\) false'.
    One failure suffices; you do not have to show the property fails often.
  3. So a counterexample is a single even number greater than \(2\) that cannot be written as a sum of two primes.
    It must lie in the stated domain and fail the property.
  4. (This is Goldbach's conjecture - no counterexample is known, which is why it remains a conjecture.)
    Verifying billions of cases is not a proof; one counterexample would be a disproof.
Where it goes wrong
  • Offering an odd number that is not a sum of two primes - outside the stated domain, so irrelevant.
  • Offering an even number that is a sum of two primes - that confirms the statement, it does not refute it.
  • Thinking many confirming examples add up to a proof.
Takeaway To disprove 'for all', produce one object in the domain for which the property fails; confirming instances never prove it.
17. X2-24The logic of arguments · spec Arg2 · Paper 2 style · ★☆☆

For a real number \(x\), \(P\): '\(x > 0\)'. \(Q\): '\(x^3 > 0\)'. Which one of the following is true?

  1. AP is necessary and sufficient for Q.
  2. BP is necessary but not sufficient for Q.
  3. CP is sufficient but not necessary for Q.
  4. DP is neither necessary nor sufficient for Q.
Answer: A. Cubing preserves sign, so \(x > 0 \Leftrightarrow x^3 > 0\).
Full method — how and why
Cubing preserves sign in both directions, so this is an equivalence - unlike squaring.
  1. If \(x>0\) then \(x^3 = x \cdot x \cdot x > 0\): sufficient.
    A product of three positives is positive.
  2. Conversely, if \(x \leq 0\) then \(x^3 \leq 0\) (negative cubed is negative, and \(0^3=0\)). So \(x^3>0\) forces \(x>0\): necessary.
    Proved by contrapositive - check the negative and zero cases rather than hunting for a counterexample.
  3. So \(P\) is necessary and sufficient.
    Cubing is a strictly increasing, sign-preserving operation.
Where it goes wrong
  • Importing the squaring counterexample (\(x=-2\)) - for cubes \((-2)^3 = -8 < 0\), so it does not apply.
  • Answering 'sufficient but not necessary' by pattern-matching to \(x^2\) questions.
Takeaway Odd powers preserve sign and order; even powers do not. \(x>0 \iff x^3>0\).
18. LGC-05The logic of arguments · spec Arg3 · Paper 2 style · ★★☆

Which of the following statements is/are true?

  1. IFor every real number \(x\), there exists a real number \(y\) such that \(y > x\).
  2. IIThere exists a real number \(y\) such that for every real number \(x\), \(y > x\).
  3. IIIFor every real number \(x\), there exists a real number \(y\) such that \(xy = 1\).
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Answer: B. I is true (take \(y = x + 1\)). II claims a single largest real number exists: false. III fails for \(x = 0\).
Full method — how and why
Quantifier order matters. '\(\forall x \exists y\)' and '\(\exists y \forall x\)' say completely different things.
  1. I: for each \(x\), can I find a bigger \(y\)? Take \(y = x + 1\). True.
    \(y\) is allowed to depend on \(x\), because \(x\) is chosen first.
  2. II: is there a single \(y\) bigger than every real \(x\)? That would be a largest real number. Taking \(x = y\) contradicts \(y > x\). False.
    Here \(y\) is fixed before \(x\), so it must work for all \(x\) at once - including \(x = y\).
  3. III: for each \(x\) is there \(y\) with \(xy = 1\)? For \(x = 0\), \(0 \cdot y = 0 \neq 1\). False.
    'For every \(x\)' includes the awkward cases; zero has no reciprocal.
  4. So I only.
    Match to the list.
Where it goes wrong
  • Reading I and II as the same statement - the difference is the order of the quantifiers, nothing else.
  • Forgetting \(x = 0\) in III and accepting \(y = \frac1x\) uncritically.
Takeaway In \(\forall x \exists y\) the \(y\) may depend on \(x\); in \(\exists y \forall x\) one \(y\) must serve every \(x\). And 'for every' always includes the edge cases.
19. LGC-06The logic of arguments · spec Arg1 · Paper 2 style · ★★☆

A teacher says: 'You will pass only if you revise.' Which of the following statements means the same thing?

  1. AIf you pass, then you revised.
  2. BIf you revise, then you will pass.
  3. CYou will pass if you revise.
  4. DIf you do not pass, then you did not revise.
  5. EYou will pass if and only if you revise.
Answer: A. '\(A\) only if \(B\)' means '\(A \Rightarrow B\)': passing implies revising. It does not promise that revising guarantees a pass.
Full method — how and why
The phrase 'only if' is the one piece of logical English almost everyone mistranslates. '\(A\) only if \(B\)' means \(A \Rightarrow B\).
  1. Let \(A\) = 'you pass', \(B\) = 'you revise'. The sentence is '\(A\) only if \(B\)'.
    Label first.
  2. '\(A\) only if \(B\)' says that \(A\) can happen only in the case that \(B\) holds - so \(A\) forces \(B\): \(A \Rightarrow B\).
    Compare with '\(A\) if \(B\)', which is the other direction \(B \Rightarrow A\). 'Only if' and 'if' point opposite ways.
  3. Translate back: if you pass, then you revised.
    Revising is necessary for passing, but the teacher promises no guarantee that it is enough.
Where it goes wrong
  • Reading it as 'if you revise, you will pass' - the converse, and a much stronger promise than the teacher made.
  • Reading it as 'if and only if'; 'only if' gives one direction only.
  • Choosing 'if you do not pass then you did not revise' - the inverse, not equivalent (the correct contrapositive is 'if you did not revise, you did not pass').
Takeaway '\(A\) if \(B\)' means \(B \Rightarrow A\); '\(A\) only if \(B\)' means \(A \Rightarrow B\). 'Only if' marks the necessary condition.
20. LGC-12The logic of arguments · spec Arg3 · Paper 2 style · ★★☆

Consider the statement: there exists a real number \(k\) such that for all real \(x\), \(x^2 + kx + 1 > 0\). Which of the following is true?

  1. AThe statement is true; for example \(k = 0\) works.
  2. BThe statement is false because \(x^2 + kx + 1\) can be negative for large \(x\).
  3. CThe statement is true; every value of \(k\) works.
  4. DThe statement is false because for every \(k\) there is an \(x\) with \(x^2 + kx + 1 \leq 0\).
  5. EThe statement is true; for example \(k = 2\) works.
Answer: A. Need discriminant \(k^2 - 4 < 0\), i.e. \(-2 < k < 2\). So some (not all) values of \(k\) work; \(k = 0\) gives \(x^2 + 1 > 0\). (\(k = 2\) gives \((x+1)^2\), which is \(0\) at \(x = -1\).)
Full method — how and why
An \(\exists \forall\) statement: you only need to produce one \(k\) that works for all \(x\). Then check why the offered reasons are right or wrong.
  1. Try \(k = 0\): \(x^2 + 1 > 0\) for every real \(x\), since \(x^2 \geq 0\).
    One successful witness proves an existential statement.
  2. So the statement is true. Now check the reason offered with each claim.
    In a 'true because...' question both the verdict and the justification must be right.
  3. Is it true for every \(k\)? No: the condition is discriminant \(< 0\), i.e. \(k^2 - 4 < 0\), so only \(-2 < k < 2\).
    A positive quadratic with no real roots never crosses the axis.
  4. And \(k = 2\) gives \((x+1)^2\), which equals \(0\) at \(x = -1\) - so that witness fails the strict inequality.
    The boundary case of the discriminant gives \(\geq 0\), not \(> 0\).
  5. So: true, with \(k = 0\) as the witness.
    Match to the list.
Where it goes wrong
  • Claiming it holds for all \(k\) - \(k = 5\) gives \(x^2+5x+1\), which is negative at \(x = -1\).
  • Offering \(k = 2\) as the witness, which fails only at the single point \(x = -1\) - easy to miss.
  • Thinking a quadratic with a positive \(x^2\) coefficient must eventually be negative; it is large positive for large \(|x|\).
Takeaway To prove \(\exists\), exhibit one witness - but check the witness satisfies the condition strictly where required.
21. LGC-16The logic of arguments · spec Arg3 · Paper 2 style · ★★☆

Which of the following statements is/are true about real numbers?

  1. IFor all \(x\) and all \(y\), if \(x < y\) then \(x^2 < y^2\).
  2. IIFor all \(x\) there exists \(y\) such that \(x < y\) and \(x^2 < y^2\).
  3. IIIThere exist \(x\) and \(y\) with \(x < y\) and \(x^2 > y^2\).
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Answer: G. I fails for \(x = -2\), \(y = 1\). II: take \(y = |x| + 1\). III: \(x = -2\), \(y = 1\) works.
Full method — how and why
Three quantified statements about squaring and ordering. Negative numbers decide all three.
  1. I: \(x < y \Rightarrow x^2 < y^2\) for all reals? Take \(x = -2\), \(y = 1\): \(-2 < 1\) but \(4 > 1\). False.
    Squaring is only order-preserving for non-negative numbers.
  2. II: for each \(x\), is there \(y\) with \(x < y\) and \(x^2 < y^2\)? Take \(y = |x| + 1\): then \(y > x\) and \(y > |x|\) so \(y^2 > x^2\). True.
    Choosing \(y\) larger than \(|x|\) guarantees both conditions at once - and \(y\) may depend on \(x\).
  3. III: do some \(x < y\) have \(x^2 > y^2\)? The counterexample from I works: \(x = -2\), \(y = 1\). True.
    Reuse the work - the same pair settles both.
  4. So II and III.
    Match to the list.
Where it goes wrong
  • Accepting I by testing only positive numbers.
  • Rejecting II by picking a bad \(y\); you get to choose \(y\), so choose a helpful one.
Takeaway \(x
22. LGC-18The logic of arguments · spec Arg4 · Paper 2 style · ★★☆

A statement reads: for every positive integer \(n\), there is a prime \(p\) with \(n < p < 2n\). Which of the following is its negation?

  1. AFor every positive integer \(n\), there is a prime \(p\) with \(p \leq n\) or \(p \geq 2n\).
  2. BThere is a positive integer \(n\) such that no prime \(p\) satisfies \(n < p < 2n\).
  3. CThere is a positive integer \(n\) and a prime \(p\) with \(p \leq n\) or \(p \geq 2n\).
  4. DThere is a prime \(p\) such that for every positive integer \(n\), \(n < p < 2n\) fails.
  5. EFor every positive integer \(n\), no prime \(p\) satisfies \(n < p < 2n\).
Answer: B. Negate the quantifiers in turn: 'for every \(n\), there exists \(p\) with ...' becomes 'there exists \(n\) such that for every prime \(p\), not ...', i.e. no prime lies strictly between \(n\) and \(2n\).
Full method — how and why
Negating a nested \(\forall \exists\) statement: work from the outside in, flipping one quantifier at a time.
  1. Formalise: \(\forall n\ \exists p\) (prime) with \(n < p < 2n\).
    Two quantifiers, so two flips.
  2. Flip the outer: \(\exists n\) such that \(\neg[\exists p\) with \(n
    'For all' fails exactly when one case fails.
  3. Flip the inner: \(\exists n\) such that \(\forall\) primes \(p\), \(\neg(nno prime lies strictly between \(n\) and \(2n\).
    'There exists' fails exactly when every candidate fails.
  4. So: there is a positive integer \(n\) for which no prime \(p\) satisfies \(n < p < 2n\).
    (The original is Bertrand's postulate and is in fact true, so this negation is false - but the question is about form.)
Where it goes wrong
  • Keeping 'for every \(n\)' and only negating the inside - that is a much stronger claim, not the negation.
  • Turning it into '\(\exists n\) and \(\exists p\) with \(p \leq n\) or \(p \geq 2n\)', which is trivially true and says nothing: the inner quantifier must become for all primes.
Takeaway Negate nested quantifiers outside-in: each \(\forall\) becomes \(\exists\), each \(\exists\) becomes \(\forall\), and the innermost statement is negated once.
23. LGC-20The logic of arguments · spec Arg1 · Paper 2 style · ★★☆

Which of the following statements is/are logically equivalent to 'if it is raining, then the ground is wet'?

  1. IIt is not raining, or the ground is wet.
  2. IIIf the ground is not wet, then it is not raining.
  3. IIIIf the ground is wet, then it is raining.
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Answer: E. '\(A \Rightarrow B\)' is equivalent to 'not \(A\), or \(B\)' (I) and to its contrapositive (II). III is the converse, which is not equivalent.
Full method — how and why
Two standard equivalents of \(A \Rightarrow B\): the contrapositive, and the disjunctive form 'not \(A\), or \(B\)'.
  1. I: 'not \(A\) or \(B\)'. An implication only fails when \(A\) is true and \(B\) false - exactly when '\(\neg A\) or \(B\)' fails. Equivalent.
    Check the one bad case: raining and ground dry makes both statements false; every other case makes both true.
  2. II: 'if the ground is not wet, it is not raining' - the contrapositive. Equivalent.
    Swap and negate.
  3. III: 'if the ground is wet, it is raining' - the converse. Not equivalent; the ground could be wet from a burst pipe.
    A concrete alternative cause is the cleanest refutation.
  4. So I and II.
    Match to the list.
Where it goes wrong
  • Rejecting I because it looks nothing like an implication - test it against the single case that breaks an implication.
  • Accepting III because wet ground is evidence of rain; evidence is not implication.
Takeaway \(A \Rightarrow B\) is equivalent both to 'not \(A\), or \(B\)' and to \(\neg B \Rightarrow \neg A\). The converse is never among the equivalents.
24. LGC-09The logic of arguments · spec Arg1 · Paper 2 style · ★★★

Consider the statement \(S\): for all real \(x\), if \(x > 0\) then \(x + \dfrac{1}{x} \geq 2\). Which of the following statements is/are true?

  1. I\(S\) is true.
  2. IIThe converse of \(S\) (for all real \(x\), if \(x + \frac1x \geq 2\) then \(x > 0\)) is true.
  3. IIIThe contrapositive of \(S\) is true.
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Answer: H. For \(x > 0\), \(x + \frac1x - 2 = \frac{(x-1)^2}{x} \geq 0\), so \(S\) is true, and so is its contrapositive (III). For the converse: if \(x < 0\) then \(x + \frac1x < 0 < 2\), so \(x + \frac1x \geq 2\) forces \(x > 0\) (II true).
Full method — how and why
One inequality, examined three ways. The contrapositive is free once the statement is settled; the converse must be tested separately.
  1. I: for \(x > 0\), write \(x + \dfrac1x - 2 = \dfrac{x^2 - 2x + 1}{x} = \dfrac{(x-1)^2}{x}\).
    Put over a common denominator - the numerator turns out to be a perfect square.
  2. Since \((x-1)^2 \geq 0\) and \(x > 0\), the fraction is \(\geq 0\), so \(x + \frac1x \geq 2\). True.
    A non-negative numerator over a positive denominator is non-negative. Equality at \(x = 1\).
  3. III: the contrapositive of a true statement is automatically true.
    No work needed - contrapositive and original are logically equivalent.
  4. II (the converse): suppose \(x + \frac1x \geq 2\); must \(x > 0\)? If \(x < 0\) then both \(x\) and \(\frac1x\) are negative, so \(x + \frac1x < 0 < 2\) - the hypothesis fails. And \(x = 0\) is not allowed. So \(x > 0\). True.
    Proved by eliminating the alternatives, which is itself a contrapositive argument.
  5. So all three.
    Match to the list.
Where it goes wrong
  • Assuming the converse is false because converses 'usually' are - here it genuinely holds.
  • Multiplying the inequality \(x + \frac1x \geq 2\) by \(x\) without knowing the sign of \(x\); that is precisely what is in question.
Takeaway \(x + \frac1x \geq 2\) for \(x>0\) is worth knowing; and a statement always shares its truth value with its contrapositive, never necessarily with its converse.
25. LGC-15The logic of arguments · spec Arg2 · Paper 2 style · ★★★

Let \(P\): 'the integer \(n\) is divisible by \(4\)' and \(Q\): 'the integer \(n^2\) is divisible by \(8\)'. Which one of the following is true?

  1. AP is necessary and sufficient for Q.
  2. BP is necessary but not sufficient for Q.
  3. CP is sufficient but not necessary for Q.
  4. DP is neither necessary nor sufficient for Q.
Answer: A. If \(4 \mid n\) then \(16 \mid n^2\), so \(8 \mid n^2\): \(P\) is sufficient. Conversely, if \(8 \mid n^2\) then \(n\) is even, say \(n = 2m\), and \(8 \mid 4m^2\) forces \(m\) to be even, so \(4 \mid n\): \(P\) is also necessary. (Tempting wrong answer: 'sufficient but not necessary' - but no counterexample exists.)
Full method — how and why
A genuine 'necessary and sufficient' case, dressed to look like the usual 'sufficient only'. You must prove both directions rather than assume a counterexample exists.
  1. Sufficiency: if \(4 \mid n\), write \(n = 4k\). Then \(n^2 = 16k^2\), which is divisible by \(8\). True.
    \(16\) is a multiple of \(8\), so divisibility by \(8\) follows immediately.
  2. Necessity: suppose \(8 \mid n^2\). Then \(n^2\) is even, so \(n\) is even - write \(n = 2m\).
    An odd number squared is odd, so \(n\) cannot be odd.
  3. Then \(n^2 = 4m^2\), and \(8 \mid 4m^2\) forces \(2 \mid m^2\), so \(m\) is even, say \(m = 2j\).
    Divide the divisibility through by the common factor \(4\).
  4. So \(n = 4j\), i.e. \(4 \mid n\). Both directions hold: necessary and sufficient.
    No counterexample exists, which is what distinguishes this from the usual pattern.
Where it goes wrong
  • Answering 'sufficient but not necessary' out of habit, without actually hunting for a counterexample. Test \(n = 2\): \(n^2 = 4\), not divisible by \(8\) - so it is not a counterexample; \(n = 6\): \(36\), not divisible by \(8\). The hunt keeps failing, which is the clue.
  • Assuming \(8 \mid n^2 \Rightarrow 8 \mid n\) - false (\(n = 4\) gives \(n^2 = 16\)).
Takeaway Before declaring a condition 'not necessary', actually look for a counterexample; repeated failure to find one is a signal to prove the converse instead.