1)Assuming no attribute has multiple values, the relation is in 1NF.

2) here the candidate key is AB (as B is not in the right side in any FD, and B comes with A from FD-1. So, candidate key is AB. D is derived transitively from C). Since C is fully functional dependent on AB , and D is transitively dependent on C. So this is in 2NF.

3) since, we can derive D and A from C , which is not a candidate key, this is not in 3NF.

To make it in 3NF, we make 2 relations R¹(A,B,C) having FDs {AB->C} and R² (C,D) having FDs { C->D}. Now, it is in 3NF.

4) Now, BCNF says that , right side of all FD should be a super key.For R¹ AB is the candidate key and hence, it is also a super key. C is dependent on AB. So R¹ is in BCNF.

For R², C->D, i.e., D is dependent on C, so C is made a foreign key in R², so this relation is also in BCNF.

Hence, breaking R in R¹ (A,B,C) and R²(C,D) makes both the relations in BCNF.

Yes. Since AB is the candidate key, A belongs to prime attribute. So any non-prime attribute functionally detrimining a prime attribute makes no sense. So we can eliminate that..also if you look closely, the dependencies are preserved.

1)Assuming no attribute has multiple values, the relation is in 1NF.

2) here the candidate key is AB (as B is not in the right side in any FD, and B comes with A from FD-1. So, candidate key is AB. D is derived transitively from C). Since C is fully functional dependent on AB , and D is transitively dependent on C. So this is in 2NF.

3) since, we can derive D and A from C , which is not a candidate key, this is not in 3NF.

To make it in 3NF, we make 2 relations R¹(A,B,C) having FDs {AB->C} and R² (C,D) having FDs { C->D}. Now, it is in 3NF.

4) Now, BCNF says that , right side of all FD should be a super key.For R¹ AB is the candidate key and hence, it is also a super key. C is dependent on AB. So R¹ is in BCNF.

For R², C->D, i.e., D is dependent on C, so C is made a foreign key in R², so this relation is also in BCNF.

Hence, breaking R in R¹ (A,B,C) and R²(C,D) makes both the relations in BCNF.

we eliminated A from relation R2 because it is in first relation is this the only reason ?

Yes. Since AB is the candidate key, A belongs to prime attribute. So any non-prime attribute functionally detrimining a prime attribute makes no sense. So we can eliminate that..also if you look closely, the dependencies are preserved.

thankyou for the answer