In a relational database, which one of the following is CORRECT:

Answer: A relation with only two attributes is always in BCNF.

Reasoning (proof):

• Let the relation have exactly two attributes, A and B. Any non-trivial functional dependency must be either A -> B, B -> A, or have the whole set {A, B} on the left.

• If A -> B holds, then A determines both attributes (A -> A is trivial), so A is a key (a superkey). The same argument applies if B -> A holds. If the left side is {A, B}, it is already a superkey.

• Therefore every non-trivial FD has a superkey on the left, so the relation satisfies BCNF.

Why the other statements are false (brief counterexamples):

• Statement: If all attributes are prime then the relation is in BCNF. Counterexample: R(A, B, C) with A -> B and B -> A (A and B determine each other) and no FD involving C. Candidate keys are {A, C} and {B, C}, so every attribute is prime, but A -> B violates BCNF because A is not a superkey.

• Statement: Every relation has at least one non-prime attribute. Counterexample: R(A, B) with A -> B and B -> A. Candidate keys are {A} and {B}, so both attributes are prime and there is no non-prime attribute.

• Statement: BCNF decomposition preserves functional dependencies. Counterexample: R(A, B, C) with FDs AB -> C and C -> B. Decomposing to (B, C) and (A, C) preserves C -> B but AB -> C is not preserved in any single decomposed relation; it can only be recovered by joining, so dependency preservation fails.

Summary: The two-attribute case forces any non-trivial FD to have a superkey on its left, so such relations are always BCNF. Other statements are false as shown by the counterexamples above.