Consider the schema R= ( S, T, U, V ) and the dependencies S→T, T→U, U→V and V→S. Let R (R1 and R2) be a decomposition such that R1∩R2 ≠ Ø. The decomposition is:
(A) Not in 2NF
(B) In 2NF but not in 3NF
(C) In 3NF but not in 2NF
(D) In both 2NF and 3NF
Explanation: R1∩R2 ≠ Ø means there is common attribute in R1 and R2.
Now if we choose a decomposition positively then we can choose something like R1(S, T, U) and R2(U, V) then we can say that decomposition is lossless because common attribute is U and LHS of every FDs are candidate key, therefore it is in 2NF as well as 3NF.
Option (D) is correct.
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