Answer-A relation R(A, B, C, D, E) has the functional dependencies AB→C, C→D, D→E, and E→A. Assuming AB is...

Computing the closure of AB ABC, CD, DE, EA, so AB A,B,C,D,E, confirming AB is a candidate key and all attributes are single-valued 1NF holds. Since no attribute depends on a proper subset of AB alone no partial dependency, 2NF holds. However, ABCD is a transitive dependency D depends on non-key attribute C, which is not a superkey, so 3NF is violated. The highest normal form satisfied is therefore 2NF.

Question

A relation R(A, B, C, D,

E has the functional dependencies AB→C, C→D, D→E, and E→A. Assuming AB is the only candidate key of R, which normal form does R satisfy at the highest level?

A R is already in Boyce-Codd Normal Form (BCNF), since every determinant shown is a candidate key

B R satisfies Third Normal Form (3NF) but not BCNF, because C acts as a partial determinant of the key

C R satisfies Second Normal Form (2NF) but violates 3NF, because non-prime attributes D and E are transitively dependent on key AB through the non-key attribute C

D R satisfies only First Normal Form (1NF), because AB does not functionally determine every other attribute in the relation

E R violates even First Normal Form due to the cyclic dependency chain formed by C→D→E→A

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A relation R(A, B, C, D,

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