An organization needs to maintain database having five attributes A, B, C, D,…
2019
An organization needs to maintain database having five attributes A, B, C, D, E. These attributes are functionally dependent on each other for which functionality 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
- A.
1 NF - B.
2 NF - C.
3 NF - D.
BCNF
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Correct answer: B
Answer: The relation is in Second Normal Form (2NF).
Find candidate keys: compute closure of A.
From A→BC and A→D we get B, C, D. From D→E we get E. So A+ = {A, B, C, D, E}.
Therefore A is a candidate key (and the only candidate key).
Check Second Normal Form (2NF):
Since the candidate key is a single attribute (A), there are no partial dependencies of non-prime attributes on part of a composite key. Thus the relation satisfies 2NF.
Check Third Normal Form (3NF):
Examine each functional dependency:
• A→BC and A→D: A is a superkey, so these meet 3NF requirement.
• BC→D: BC is not a superkey and D is not a prime attribute, so this violates 3NF.
• D→E: D is not a superkey and E is not a prime attribute, so this also violates 3NF.
Conclusion:
The relation satisfies 2NF but fails 3NF (and therefore fails BCNF) because of the dependencies BC→D and D→E where determinants are not superkeys and the dependents are non-prime attributes. Hence the highest normal form for R is 2NF.
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