Questions: Negation of Quantified Statements

Question 1

Which of the following is the correct negation of 'All students passed the exam'?

• A. No student passed the exam
• B. Some student did not pass the exam
• C. All students failed the exam
• D. Most students did not pass the exam

Answer: (B)

The statement 'All students passed' has the form ∀x P(x). Its negation is ∃x ¬P(x) — 'some student did not pass.' The negation is false only if every student passed; it becomes true the moment even one student fails. Option A ('no student passed') is ∀x ¬P(x), which overcorrects — it makes a much stronger claim than merely denying universality. The common error is swapping the quantifier AND over-negating to make all x fail, rather than asserting that only one x fails.

Question 2

What is the negation of 'For all ε > 0, there exists δ > 0 such that the condition holds'?

• A. For all ε > 0, for all δ > 0, the condition fails
• B. There exists ε > 0 such that for all δ > 0, the condition fails
• C. There exists ε > 0 such that there exists δ > 0 such that the condition fails
• D. For all ε > 0, there exists δ > 0 such that the condition fails

Answer: (B)

Working from outside in: ∀ε flips to ∃ε, then ∃δ flips to ∀δ, then the predicate negates. Result: 'There exists ε > 0 such that for all δ > 0, the condition fails.' Each quantifier flips exactly once; the predicate negates at the innermost level. Option C keeps ∃δ (fails to flip it). Option A keeps ∀ε (fails to flip the outer quantifier). The rule is purely mechanical: work outward to inward, flip each quantifier.

Question 3

The negation of 'There exists a prime number greater than 100' is 'There exists a prime number less than or equal to 100.'

• T. True
• F. False

Answer: False

The statement has the form ∃x [Prime(x) ∧ x > 100]. Its negation is ∀x ¬[Prime(x) ∧ x > 100], which by De Morgan's law is ∀x [¬Prime(x) ∨ x ≤ 100] — 'every number is either not prime or at most 100.' The negation quantifier flips from ∃ to ∀; it does not introduce a new existential. Introducing a new existential (as in the false option) would give a completely different statement that doesn't contradict the original.

Question 4

The negation of 'For all x, P(x)' is 'There exists an x such that not P(x).'

• T. True
• F. False

Answer: True

This is the fundamental quantifier negation rule: ¬(∀x P(x)) ≡ ∃x ¬P(x). It holds because ∀x P(x) is false exactly when at least one x makes P(x) false — and the existence of such an x is precisely what ∃x ¬P(x) asserts. The quantifier swaps from ∀ to ∃, and the predicate gains a negation.

Question 5

Why is the negation of 'For all x, P(x)' not 'For all x, not P(x)'?

Model answer: 'For all x, not P(x)' is a much stronger statement than the mere denial of 'For all x, P(x).' The original claim fails as soon as one x makes P(x) false — you only need one counterexample. 'For all x, not P(x)' asserts that P(x) fails for every single x, which goes far beyond disproving the universal.

Consider 'All students passed' (∀x Passed(x)). Its negation is satisfied the moment one student failed — just one. The overcorrected version, 'All students failed' (∀x ¬Passed(x)), makes a strong positive claim about every student. If 99 of 100 students passed and one failed, the original is false (correctly negated by ∃x ¬Passed(x)), but 'all students failed' is also false. The correct negation must be the logical opposite — exactly one must be true and the other false, which is satisfied by ∃x ¬P(x) but not by ∀x ¬P(x).