1

I have a relation like

R(A,B,C)

and dependencies like

A->B , B->C , A->C

I think the relation is in 2NF as candidate key is A. AFAIK when there is only one candidate key then the relation is in 2NF. So to decompose it into 3NF I broke the relation into

R1(A,B) R2(B,C)

and dependencies are

{A->B} and {B->C}

This is also in BCNF but A->C isn't preserved. So do I have to make another relation as R3(A,C) with A->C dependency?

Your Answer

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

Browse other questions tagged or ask your own question.