A. The candidate key for the relation R is ssn. This is because ssn can uniquely determine all other attributes in the relation.

B. The relation R is not in 2NF because the non-prime attribute 'city' is not fully functionally dependent on the candidate key 'ssn'. We can decompose R into two relations to achieve 2NF:
R1(ssn, name) with a functional dependency ssn - name
R2(ssn, city) with a functional dependency ssn - city

C. After decomposition, the relations R1 and R2 are in 3NF because every non-prime attribute is non-transitively dependent on every candidate key in the relation.

D. The relations R1 and R2 are also in BCNF because for every non-trivial functional dependency X - Y, X is a superkey. In both R1 and R2, the only non-trivial functional dependency is ssn - name and ssn - city, and ssn is a superkey in both relations.

Question

A. The candidate key for the relation R is ssn. This is because ssn can uniquely determine all other attributes in the relation.

B. The relation R is not in 2NF because the non-prime attribute 'city' is not fully functionally dependent on the candidate key 'ssn'. We can decompose R into two relations to achieve 2NF:
R1(ssn, name) with a functional dependency ssn -> name
R2(ssn, city) with a functional dependency ssn -> city

C. After decomposition, the relations R1 and R2 are in 3NF because every non-prime attribute is non-transitively dependent on every candidate key in the relation.

D. The relations R1 and R2 are also in BCNF because for every non-trivial functional dependency X -> Y, X is a superkey. In both R1 and R2, the only non-trivial functional dependency is ssn -> name and ssn -> city, and ssn is a superkey in both relations.

Knowee AI · Accepted Answer

A. The candidate key for the relation R is ssn. This is because ssn can uniquely determine all other attributes in the relation.

B. The relation R is not in 2NF because the non-prime attribute 'city' is not fully functionally dependent on the candidate key 'ssn'. We can decompose R into two relations to achieve 2NF:
R1(ssn, name) with a functional dependency ssn -> name
R2(ssn, city) with a functional dependency ssn -> city

C. After decomposition, the relations R1 and R2 are in 3NF because every non-prime attribute is non-transitively dependent on every candidate key in the relation.

D. The relations R1 and R2 are also in BCNF because for every non-trivial functional dependency X -> Y, X is a superkey. In both R1 and R2, the only non-trivial functional dependency is ssn -> name and ssn -> city, and ssn is a superkey in both relations.

1.R(name，ssn，phone#,city)，F={ssn® (name,City)} A.Please calculate all the candidate KEY.B.Is R in 2NF？if R in not in 2NF，Decompose to 2NF.C.Is R in 3NF？if R in not in 3NF，Decompose to 3NF.D.Is R in BCNF？if R in not in BCNF，Decompose to BCNF

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