Final answer:
The key for the relation R is {A,B,D}. R is decomposed into 2NF by eliminating partial dependencies, resulting in five relations: R1, R2, R3, R4, and R5. These relations are already in 3NF since there are no transitive dependencies in the 2NF relations.
Step-by-step explanation:
Decomposing Relation R into 2NF and 3NF
The given relation R={A,B,C,D,E,F,G,H,I,J} has a set of functional dependencies given by G. To find the key for R, we need to look at the functional dependencies and determine the minimal set of attributes that functionally determine all other attributes in the relation.
The provided key is {A,B,D}. This means that these three attributes functionally determine all the other attributes in the relation R.
Decomposition into 2NF
For 2NF, we eliminate partial dependencies, which are dependencies on a part of a composite key. Looking at the functional dependencies:
- {A,B} → {C}
- {B,D} → {E, F}
- {A,D} → {G, H}
- {A} → {I}
- {H} → {J}
We can decompose R into the following 2NF relations:
- R1 = {A, B, C}
- R2 = {B, D, E, F}
- R3 = {A, D, G, H}
- R4 = {A, I}
- R5 = {H, J}
Decomposition into 3NF
For 3NF, we eliminate transitive dependencies. From the 2NF relations, there don't seem to be transitive dependencies as each non-prime attribute is fully functionally dependent on either the candidate key or a part of it. Therefore, the relations in 2NF are also in 3NF in this case.
The final set of 3NF relations will be the same as for 2NF:
- R1 = {A, B, C}
- R2 = {B, D, E, F}
- R3 = {A, D, G, H}
- R4 = {A, I}
- R5 = {H, J}