Database-Management-System
August 29, 2024Functional-Dependency
August 29, 2024Database-Management-System
| Question 455 |
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.
→ Assume that given table R is decomposed in two tables
Which of the following option is true w.r.t. given decomposition?
| Dependency preservation property is followed | |
| R1 and R2 are both in 2 NF | |
| R2 is in 2 NF and R3 is in 3 NF | |
| R1 is in 3 NF and R2 is in 2 NF |
Question 455 Explanation:
Since In R1 and R2 BC can’t determine BC → D of relation “R”. Hence R1 and R2
are not following the Dependency preservation property.
Candidate key of R1 is “A”. And since KHS of R1 contains only “A” so R1 is in 3NF.
Candidate key of R2 is “A” , But Since D→E neither have Super key in its LHS nor have a prime key attribute in its RHS, So R2 is in 2NF but not in 3NF.
are not following the Dependency preservation property.
Candidate key of R1 is “A”. And since KHS of R1 contains only “A” so R1 is in 3NF.
Candidate key of R2 is “A” , But Since D→E neither have Super key in its LHS nor have a prime key attribute in its RHS, So R2 is in 2NF but not in 3NF.
Correct Answer: D
Question 455 Explanation:
Since In R1 and R2 BC can’t determine BC → D of relation “R”. Hence R1 and R2
are not following the Dependency preservation property.
Candidate key of R1 is “A”. And since KHS of R1 contains only “A” so R1 is in 3NF.
Candidate key of R2 is “A” , But Since D→E neither have Super key in its LHS nor have a prime key attribute in its RHS, So R2 is in 2NF but not in 3NF.
are not following the Dependency preservation property.
Candidate key of R1 is “A”. And since KHS of R1 contains only “A” so R1 is in 3NF.
Candidate key of R2 is “A” , But Since D→E neither have Super key in its LHS nor have a prime key attribute in its RHS, So R2 is in 2NF but not in 3NF.