Which of the following is the correct negation of '∀x ∃y R(x, y)'?
A∀x ∀y ¬R(x, y)
B∃x ∀y ¬R(x, y)
C∀x ∃y ¬R(x, y)
D∃x ∃y ¬R(x, y)
Apply the rule outside-in: the outer ∀ flips to ∃, the inner ∃ flips to ∀, and the predicate R becomes ¬R. So ¬(∀x ∃y R(x,y)) = ∃x ∀y ¬R(x,y). Option C is the most common mistake — it negates the predicate but leaves both quantifiers unchanged. Option A flips both quantifiers but wrongly — the ∀ should flip to ∃, not stay ∀.
Question 2 Multiple Choice
A student wants to disprove the claim 'Every continuous function on [0,1] achieves its maximum value at an interior point.' What form does their disproof take?
AThey must prove that no continuous function on [0,1] achieves its maximum anywhere
BThey must find one specific continuous function on [0,1] that does not achieve its maximum at an interior point
CThey must prove that for all continuous functions on [0,1], the maximum is at a boundary point
DThey must show the statement is true but only under certain conditions
The claim has the form ∀f P(f). Its negation is ∃f ¬P(f): there exists one function where P fails. To disprove a universal statement, you only need a single counterexample. For instance, f(x) = x achieves its maximum at x = 1, a boundary point, not an interior point — one example suffices to refute the universal claim. Option C would be proving a different universal statement, not disproving the original.
Question 3 True / False
The negation of '∃x P(x)' is '∃x ¬P(x)'.
TTrue
FFalse
Answer: False
This is the most common error: negating only the predicate while leaving the quantifier unchanged. The correct negation is '∀x ¬P(x)'. To show '∃x P(x)' is false, you must show that no x satisfies P — not just that some x fails P. For example, the negation of 'there exists a prime number less than 2' is 'all numbers less than 2 are non-prime' (∀x ¬P(x)), not 'there exists a number less than 2 that is non-prime' (∃x ¬P(x)).
Question 4 True / False
A single counterexample is sufficient to disprove the statement '∀x P(x)'.
TTrue
FFalse
Answer: True
Since ¬(∀x P(x)) = ∃x ¬P(x), showing that one specific x fails P is exactly what the negation asserts. One counterexample witnesses the truth of the negation and therefore establishes the falsity of the universal statement. This is why finding a single counterexample is such a powerful move in mathematics — it completely destroys a universal claim, no matter how many cases the claim was previously verified for.
Question 5 Short Answer
Explain in your own words why the negation of a universal statement is existential, and the negation of an existential statement is universal.
Think about your answer, then reveal below.
Model answer: A universal statement '∀x P(x)' claims P holds for every element. To show this is false, you only need to find one element where P fails — one counterexample. So the negation asserts the existence of such an element: ∃x ¬P(x). Conversely, an existential statement '∃x P(x)' claims at least one element satisfies P. To show this is false, you must show every element fails P — no exceptions. So the negation is a universal claim: ∀x ¬P(x). In each case, the quantifier flips because the standard of failure is the opposite of the standard of truth.
This reasoning grounds the mechanical rule in the semantics of quantifiers: what it takes to make each kind of statement false determines the form of its negation. Understanding why the rule works — not just memorizing it — is crucial for applying it correctly in complex nested cases and for using it in proof strategies like proof by contradiction.