A mathematician conjectures: 'For any positive integers a and b, if a divides b² then a divides b.' Which of the following best disproves this conjecture?
AShow that the statement fails for infinitely many pairs of integers
BExhibit the single case a = 4, b = 6, since 4 divides 36 but 4 does not divide 6
CConstruct a formal proof by contradiction that the statement cannot hold universally
DFind a general pattern of values where the divisibility property fails
One counterexample is logically sufficient. For a = 4 and b = 6: 4 divides 36 (= 6²), but 4 does not divide 6. Exhibiting this single concrete case where P(4, 6) is false immediately establishes ∃a, b such that the statement fails — the disproof is complete. No additional cases, patterns, or general arguments are needed. This is the asymmetry: disproving a universal requires only one witness.
Question 2 Multiple Choice
What positive work does a good counterexample do beyond simply showing a conjecture is false?
AIt demonstrates the logical form of proof by contradiction
BIt calibrates the conjecture by revealing where the boundary between truth and falsity lies
CIt shows that no valid proof of the original conjecture can possibly be constructed
DIt establishes that the domain of the conjecture was improperly specified
Counterexamples are diagnostic tools, not just destroyers. If the counterexample exploits a boundary or degenerate case, it suggests the conjecture can be saved by restricting the domain. If it is generic, the conjecture is fundamentally wrong. In either case, the counterexample reveals exactly where truth ends and falsity begins — pointing toward the correct, restricted statement worth proving.
Question 3 True / False
To disprove 'Most prime numbers are odd,' you need to show that infinitely many even primes exist.
TTrue
FFalse
Answer: False
One counterexample is logically complete. The prime 2 is even, and that single case immediately refutes the universal claim. Gathering more even primes would be redundant — the disproof is finished the moment you exhibit one case where the predicate fails. The logical negation of 'For all primes p, p is odd' is 'There exists a prime p that is even,' and 2 proves the existential.
Question 4 True / False
A counterexample found in a degenerate or boundary case (such as x = 0 or the empty set) is weaker evidence against a conjecture than a counterexample using typical values.
TTrue
FFalse
Answer: False
Any counterexample has identical logical force — one case is one case. Logically, a boundary counterexample and a central counterexample are equally decisive. In practice, boundary cases (0, 1, the empty set, constant functions, disconnected graphs) are PREFERRED starting points for finding counterexamples precisely because many universal claims fail there. The force of the disproof is independent of how typical the witness is.
Question 5 Short Answer
Why does the logical structure of universal statements create such a strong asymmetry between proof and disproof?
Think about your answer, then reveal below.
Model answer: A universal statement 'For all x, P(x)' makes a claim that must hold for every element in the domain — no finite collection of confirming instances constitutes a proof, only a general argument covering all cases. But its logical negation is the existential claim 'There exists x such that ¬P(x),' which is proved by exhibiting a single witness. Proving the universal requires universality; disproving it requires only one concrete counterexample that satisfies the domain condition and fails the predicate.
This asymmetry is a direct consequence of the semantics of quantifiers. Universal and existential quantifiers are logical duals: the negation of ∀x P(x) is ∃x ¬P(x). A proof of an existential claim is always a witness — one concrete element. So a counterexample IS a proof of the negation, complete and rigorous. Understanding this makes disproofs feel less like failures and more like what they are: successful proofs of a different kind of statement.