Question 14

Consider a relation with exactly two attributes, $R(A, B)$. The possible non-trivial functional dependencies are:

1. $A \to B$: Here $A$ uniquely determines all attributes of $R$, making $A$ a candidate key (and superkey). Thus $R$ satisfies BCNF.
2. $B \to A$: Similarly, $B$ is a candidate key and superkey, so $R$ satisfies BCNF.
3. Both $A \to B$ and $B \to A$: Both $A$ and $B$ are candidate keys, satisfying BCNF.
4. Neither holds: The only candidate key is the composite key $AB$, and there are no non-trivial functional dependencies. Hence $R$ is trivially in BCNF.

Therefore, every relation with only two attributes is always in BCNF, making Statement (A) TRUE.

Regarding the other options: (B) is false because a relation with all prime attributes is guaranteed to be in 3NF, but not necessarily BCNF (e.g., $R(A, B, C)$ with $AB \to C$ and $C \to B$). (C) is false as relations where all attributes are prime have no non-prime attributes. (D) is false because BCNF decomposition is not always dependency-preserving.