105 views

I feel that answer is none of the above , because in question it is given AB is key , so any partial dependency will not be accepted , so none will be in 2NF.

what you say , anyways the answer in booklet is D

| 105 views

A. $A \to C$, $C$ is non-key attribute. $AB$ is candidate key and so is a partial FD. Hence, not in 2NF.

B. Here, $C\to B$ makes $AC$ also a candidate key. So, $B \to C$ and $C \to B$ are not violating 2NF as they are detrmining "key" attribute. But $A \to D$ violate 2NF.

C. $AB$ and $DB$ are candidate keys here. Due to $C \to D$, $CB$ also becomes a candidate key. So, no other attribute present other than prime(key) attributes. So, $R$ is in 2NF as well as 3NF.

D. Only $AB$ is the candidate key. $BC$ is not a super key, is also not a proper subset of candidate key. $D$ is not a prime attribute. So, $BC \to D$ violate 3NF but due to no partial dependency, $R$ is in 2NF.

by Veteran (430k points)
selected
+1

in d option what he( Who asked)  think in BC B is prime then  he think BC combine is prime. :)

0
but sir question says AB is primary key ,so how BC->D doesnot violate the 2NF?
0
Because BC is not a proper subset of AB as C is not there is AB. Only B and A are proper subset of AB