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3 Answers

2 votes
2 votes
It is possible for a relation to be in 3NF and have only one candidate key and still not be in BCNF. In order for a relation to be in BCNF, every determinant in the relation must be a candidate key. In other words, if there are any non-trivial functional dependencies in the relation, at least one of the determinants must be a candidate key.

For example, consider the relation R(A, B, C, D) with the functional dependencies A -> B, B -> C, and C -> D. This relation is in 3NF because no non-prime attribute is transitively dependent on any candidate key. However, it is not in BCNF because the determinant C -> D is not a candidate key.

In general, it is not true that a relation in 3NF with only one candidate key is automatically in BCNF.
1 votes
1 votes
True.

If a relation R is in 3NF and has only one candidate key, then it must be in BCNF (Boyce-Codd Normal Form). This is because a relation is in BCNF if and only if every determinant of every non-trivial functional dependency in the relation is a candidate key.

Since R has only one candidate key, every determinant of every non-trivial functional dependency in R must be the candidate key. Therefore, every determinant is a superkey, and R is in BCNF.
0 votes
0 votes
True.

Why?

Because we cannot create any relation with 1 candidate key and its BCNF. other than R(A,B) { A -→ B }.

If we try to create FD set with more than one FD in it , we will never be able to create one satisfying given conditions.

But since, R(A,B) { A --> B }.  exist the answer is true.

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