Q44.Marks: +2.0UGC NET Paper 2: Computer Sc 6th Jan 2025 Shift 1
Which of the following/s is/are FALSE statement?
A. An all key relation is always in BCNF since it has no FDs.
B. A relation that is not in 4NF due to nontrivial MVD must be decomposed to convert it into a set of relations in 4NF.
C. The decomposition removes the redundancy by the MVD.
D. 3NF is stronger than BCNF.
Choose the correct answer from the options given below:
1.A only
2.A, B only
3.D only✓ Correct
4.C only
Solution
The correct answer is: option 3: D only
Concept & Statement Analysis:
Statement A: ✅ True
An all-key relation means every attribute is part of the candidate key. Hence, no non-trivial Functional Dependencies (FDs) exist. Since BCNF only cares about non-trivial FDs and their compliance with superkeys, the relation is automatically in BCNF.
Statement B: ✅ True
If a relation is not in 4NF due to the presence of a non-trivial Multivalued Dependency (MVD) that violates 4NF, it must be decomposed to eliminate redundancy and achieve 4NF compliance.
Statement C: ✅ True
Yes, the decomposition in 4NF removes redundancy that arises due to MVDs. That's the goal of 4NF: to eliminate non-trivial MVDs that are not supported by superkeys.
Statement D: ❌ False
This is incorrect. In fact, BCNF is stronger than 3NF. Every relation in BCNF is in 3NF, but not every 3NF relation is in BCNF. 3NF allows some redundancy in special cases where BCNF does not.