DBMS, 2NF

Question

Given a relation R( P, Q, R, S, T) and Functional Dependency set FD = { PQ → R, S → T }, determine whether the given R is in 2NF? If not convert it into 2 NF.

Comments

No it is not in 2NF because candidate key is $PQS$ and  $PQ → R$ and $S → T$ are both partial keys.

To convert into 2NF find $PQ^+$ and $S^+$

So R(P, Q, R, S, T) will decompose into $R_1(P,Q,S), R_2(P,Q,R), R_3(S,T)$

 

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Answer 1

Caption

Not in 2NF. (pqs) will be candidate key