Consider the following relational schema along with all the functional…
2025
Consider the following relational schema along with all the functional dependencies that hold on them.
\(R1(A, B, C, D, E): { 𝐷 → 𝐸, 𝐸𝐴 → 𝐵, 𝐸𝐵 → 𝐶} \\\)
\(R2(A, B, C, D): { 𝐴 → 𝐷, 𝐴 → 𝐵, 𝐶 → 𝐴}\)
Which of the following statement(s) is/are TRUE?
- A.
\(R1\)is in 3NF - B.
\(R2\)is in 3NF - C.
\(R1\)is NOT in 3NF - D.
\(R2\)is NOT in 3NF
Attempted by 215 students.
Show answer & explanation
Correct answer: C, D
Answer: Both relations are NOT in Third Normal Form (3NF).
R1(A,B,C,D,E) with FDs: D → E, EA → B, EB → C
Candidate key reasoning: {A,D} is a key because D → E gives E, then EA → B gives B, and EB → C gives C, so A,D → A,B,C,D,E.
Prime attributes: A and D (members of the candidate key).
Check FDs against 3NF: D → E fails 3NF because D is not a superkey and E is not a prime attribute. EA → B and EB → C also fail because their left sides are not superkeys and their right-hand attributes are not prime. Therefore R1 is not in 3NF.
R2(A,B,C,D) with FDs: A → D, A → B, C → A
Candidate key reasoning: C is a key because C → A and A → B,A → D yield all attributes, so C → A,B,D and hence C → A,B,D,C.
Prime attribute: C only.
Check FDs against 3NF: A → B and A → D violate 3NF because A is not a superkey and B and D are not prime attributes. C → A is fine because C is a superkey. Overall R2 is not in 3NF.
Conclusion: The true statements are the ones claiming that R1 is NOT in 3NF and R2 is NOT in 3NF.
A video solution is available for this question — log in and enroll to watch it.