In a schema R(A, B, C, D, E, F, G, H), each field of R contains only atomic…
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In a schema R(A, B, C, D, E, F, G, H), each field of R contains only atomic values. F = {CH→G, A→BC, B→CFH, E→A, F→EG} is a set of functional dependencies so that F+ is exactly the set of FDs that holds R. The relation R is :
Answer: A. In 1NF, but not in 2NF — To determine the Normal Form, first identify the Candidate Keys. Since attribute D never appears on the right-hand side of any dependency, it must be part of…
- A.
In 1NF, but not in 2NF
- B.
In 2NF, but not in 3NF
- C.
In 3NF, but not in BCNF
- D.
In BCNF
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Correct answer: A
To determine the Normal Form, first identify the Candidate Keys. Since attribute D never appears on the right-hand side of any dependency, it must be part of every key. Combining D with other attributes reveals that AD, BD, ED, and FD are the candidate keys.
A relation is in 2NF only if there are no partial dependencies (where a proper subset of a candidate key determines a non-prime attribute). Here, the dependency A→BC violates 2NF because A is a proper subset of the key AD, and it determines C (which is a non-prime attribute). Since it fails the check for 2NF, the relation is in 1NF but not in 2NF.
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