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A. No transitive dependencies remain
B. No join dependency remains
C. No multivalued dependency remains
D. All FDs have prime attributes
Explanation:
5NF (Projection-Join NF) addresses join dependencies — a relation is in 5NF when it cannot be non-trivially decomposed into smaller relations without loss.
A. Lossless join always
B. Loss of dependency preservation sometimes
C. Removal of multi-valued dependencies
D. Need for denormalization
Explanation:
BCNF ensures lossless decomposition, but it may break dependency preservation — some FDs might not be enforceable without joins.
A. 2NF
B. 3NF
C. BCNF
D. 4NF
Explanation:
If there are no partial dependencies (attributes depend on whole key) and no transitive dependencies (non-keys don't determine others), the relation is at least in 3NF.
A. Insertion anomaly
B. Deletion anomaly
C. Update anomaly
D. All of these
Explanation:
Transitive dependencies cause redundancy and can result in insertion, deletion and update anomalies — so all of these.
A. 1NF only
B. 2NF only
C. 3NF
D. BCNF
Explanation:
OrderID → CustomerID and CustomerID → CustomerPhone creates a transitive dependency OrderID → CustomerPhone via CustomerID. This violates 3NF.
A. Partial dependency
B. Multivalued dependency
C. Transitive dependency
D. Join dependency
Explanation:
Two independent sets of attributes cause multivalued dependency. A non-trivial MVD suggests need for 4NF decomposition.
A. Presence of MVDs
B. Presence of transitive dependencies
C. FD LHS not superkey
D. Repeating groups exist
Explanation:
BCNF handles functional dependencies. MVDs are a different redundancy source, handled only in 4NF.
A. Free from partial dependency
B. Free from transitive dependency
C. Free from MVDs
D. Free from join dependency
Explanation:
A table in 2NF must first be in 1NF and then remove partial dependencies on composite keys.
A. 1NF
B. 2NF
C. 3NF
D. 4NF
Explanation:
3NF eliminates transitive dependencies where non-key determines non-key.
A. A
B. D
C. AD
D. B
Explanation:
D → A and A → BC implies D → ABC. With D itself present, it determines all — so D is a key.
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