6 6 votes Let R(ABCDE) be a relational schema and F ={AB->CD, ABC->E,C->E} BE A SET OF FUNCTIONAL dependencies. WHAT IS highest normal of R ? 1NF 2NF 3NF BCNF Databases database-normalization + – spriti1991 4.2k views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
Best answer 8 8 votes AB is candidate key.. AB-----> CD, AB is key so BCNF ABC ------> E , No partial dependency but E as well as C is not prime so 2NF.. C---->E, non key -----> non key so 2NF.. so overall relation is in 2NF.. Digvijay Pandey answered Apr 18, 2015 • selected Dec 18, 2015 by Pooja Palod Digvijay Pandey comment Share Follow See all 6 Comments 6 6 Comments reply spriti1991 commented Apr 18, 2015 reply Follow flag but here in a production ABC->E part of it is a prime attribute and part of it is a non prime which gives us non prime so is that allowed ? Because in 2nf whenever prime attribute gives non prime attribute that is not allowed which is nothing but partial dependency which we try to eliminate that in 2NF 0 0 replyShare spriti1991 commented Apr 18, 2015 reply Follow flag OH SORRY ABC will behave as super key so its fine yeah i got this ignore my previous comment !! 0 0 replyShare Digvijay Pandey commented Apr 18, 2015 reply Follow flag ABC ------>E is not Partial Dependency.. ABC is not proper subset of candidate key which is AB.. A relation is not in 2NF if Proper subset of candidate key determine non prime attribute but here ABC is not proper subset of AB.. ryt ?? 1 1 replyShare spriti1991 commented Apr 18, 2015 reply Follow flag yeah right ! 0 0 replyShare One commented Nov 18, 2016 reply Follow flag ABC---->E is BCNF because in BCNF if whenever a nontrivial functional dependency X---->A holds in relation ,then X is super key of the relation 1 1 replyShare skulliest commented Jun 12, 2024 reply Follow flag CAN ANYONE EXPLAIN WHY C IS NOT PRIME AB IS CK ABC IS THE SUPERSET OF AB THEN IT CAN BE SK SO C IS PRIME? 0 0 replyShare Please log in or register to add a comment.
