Consider the universal relation R = {A, B, C, D, E, F, G, H, I, J} and the set of functional dependencies F = { {A, B}→{C}, {A}→{D, E}, {B}→{F}, {F}→{G,H}, {D}→{I, J} }. What is the key for R? Decompose R into 2NF and then 3NF relation

To find the key for R, we need to find the attribute(s) that can uniquely determine all other attributes in the relation.

Step 1: Find the closure of each attribute.

• Closure of A: {A, D, E, I, J}
• Closure of B: {B, F, G, H}
• Closure of A, B: {A, B, C, D, E, F, G, H, I, J}

From the above, we can see that {A, B} is the key for R because it can determine all other attributes in the relation.

Step 2: Decompose R into 2NF relations.

A relation is in 2NF if it is in 1NF and no non-prime attribute is dependent on any proper subset of any candidate key of the relation.

From the given functional dependencies, we can decompose R into the following 2NF relations:

• R1(A, B, C)
• R2(A, D, E)
• R3(B, F)
• R4(F, G, H)
• R5(D, I, J)

Step 3: Decompose the 2NF relations into 3NF relations.

A relation is in 3NF if it is in 2NF and every non-prime attribute of R is non-transitively dependent on every key of R.

All the 2NF relations are already in 3NF because there is no transitive dependency in any of the relations. So, the 3NF relations are the same as the 2NF relations:

• R1(A, B, C)
• R2(A, D, E)
• R3(B, F)
• R4(F, G, H)
• R5(D, I, J)