UGC NET CS 2015 June Paper-3
August 29, 2024Functional-Dependency
August 29, 2024Database-Management-System
| Question 450 |
Comprehension:
Answer question (96-100) based on the problem statement given below:
An organization needs to maintain database having five attributes A, B, C, D, E. These attributes are functionally dependent on each other for which functionally dependency set F is given as : F: {A→ BC, D → E, BC → D, A →D}. Consider a universal relation R(A, B, C, D, E) with functional dependency set F. Also all attributes are simple and take atomic values only.
→ Identify the normal form in which relation R belong to
| 1 NF | |
| 2 NF | |
| 3 NF | |
| BCNF |
Question 450 Explanation:
Since “A” is the primary key or “R” and there is no partial dependency So “R” is in 2NF.
Since, D → E, BC → D neither have a super key in their LHS nor a prime key attribute in their RHS so “R” is not in 3NF.
Since “R” is not in 3NF it can’t be in BCNF.
Hence option(B) is correct
Since, D → E, BC → D neither have a super key in their LHS nor a prime key attribute in their RHS so “R” is not in 3NF.
Since “R” is not in 3NF it can’t be in BCNF.
Hence option(B) is correct
Correct Answer: B
Question 450 Explanation:
Since “A” is the primary key or “R” and there is no partial dependency So “R” is in 2NF.
Since, D → E, BC → D neither have a super key in their LHS nor a prime key attribute in their RHS so “R” is not in 3NF.
Since “R” is not in 3NF it can’t be in BCNF.
Hence option(B) is correct
Since, D → E, BC → D neither have a super key in their LHS nor a prime key attribute in their RHS so “R” is not in 3NF.
Since “R” is not in 3NF it can’t be in BCNF.
Hence option(B) is correct