Suppose we have R=(A,B,C,D) and FDs are {B->C, C->D}. The only candidate key is AB, then A and B are prime; C and D are not prime. B->C breaks the 2NF requirements because is a partial dependency. So we break the relation into R1(A,B,D) and R2(B,C). But now C->D is lost! Am I wrong? So this breaking is not "good"?


Yes, the breaking is not good.

The reason is that you FDs (B->C and C->D) also imply B->D. So B->CD (both B->C and B->D) break 2NF and should be used for the table breaking.

So, the relations, in order to be 2NF should be: (A,B) and (B,C,D).

(and then the C->D can be used to break the second further for 3NF.)

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.