Difficulty: Medium
Correct Answer: (A,B) is a composite determinant.
Explanation:
Introduction / Context:Functional dependencies (FDs) express how attributes relate. Understanding what can and cannot be inferred from a given FD is central to normalization (e.g., 2NF, 3NF, BCNF).
Given Data / Assumptions:
Concept / Approach:A “determinant” is any attribute set on the left side of a functional dependency. If two attributes together determine another, the pair (A,B) is a determinant, and it is composite because it contains more than one attribute. From (A,B) → C, we cannot infer A → C or B → C unless additional FDs hold.
Step-by-Step Solution:
Start with (A,B) → C.Identify the determinant: the LHS, (A,B), is the determinant.Because there are two attributes on the LHS, it is a composite determinant.No inference allows A → C or B → C individually without extra information.C is on the RHS, so C is not a determinant by this FD alone.Verification / Alternative check:Armstrong’s axioms do not permit decomposing an FD's LHS like (A,B) → C into A → C or B → C in general.
Why Other Options Are Wrong:A → C / B → C: Not guaranteed; they require additional FDs. C is a determinant: Not from this FD; C is the dependent attribute. A alone is always a key: Not implied; keys require uniqueness over all tuples, which is not established.
Common Pitfalls:Assuming multi-attribute determinants imply each attribute determines the RHS—this is incorrect without further constraints.
Final Answer:(A,B) is a composite determinant.
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