Which one of the following statements about normal forms is FALSE? (A) BCNF is stricter than 3NF (B) Lossless, dependency-preserving decomposition into 3NF is always possible (C) Lossless, dependency-preserving decomposition into BCNF is always possible (D) Any relation with two attributes is in BCNF
The Correct Answer and Explanation is :
Correct Answer: (C) Lossless, dependency-preserving decomposition into BCNF is always possible
Explanation:
Normal Forms Overview:
- First Normal Form (1NF): A relation is in 1NF if all attributes contain only atomic (indivisible) values.
- Second Normal Form (2NF): A relation is in 2NF if it is in 1NF and has no partial dependencies (no attribute depends only on part of a candidate key).
- Third Normal Form (3NF): A relation is in 3NF if it is in 2NF and has no transitive dependencies (non-prime attributes depend only on candidate keys).
- Boyce-Codd Normal Form (BCNF): A stricter version of 3NF, where every determinant is a candidate key.
Why the Answer is False:
- Option (A): BCNF is stricter than 3NF
This is true. BCNF is a stronger normal form than 3NF because all relations in BCNF are also in 3NF, but not all 3NF relations are in BCNF. BCNF eliminates situations where a non-candidate key attribute acts as a determinant. - Option (B): Lossless, dependency-preserving decomposition into 3NF is always possible
This is true. A relation can always be decomposed into 3NF in a way that preserves all functional dependencies and ensures the decomposition is lossless. - Option (C): Lossless, dependency-preserving decomposition into BCNF is always possible
This is false. While a lossless decomposition into BCNF is always possible, it may not preserve all functional dependencies. For example, if a relation has a dependency that violates BCNF but is crucial for query optimization, decomposing into BCNF may lose that dependency. - Option (D): Any relation with two attributes is in BCNF
This is true. A two-attribute relation can always satisfy BCNF because any functional dependency in such a relation would involve both attributes, making both candidate keys.
Key Takeaway:
Lossless, dependency-preserving decomposition into BCNF is not always achievable, making option (C) false. This is an important limitation of BCNF compared to 3NF, which guarantees both properties.