## Sweta Kumari is teaching live on Unacademy Plus

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 R1nR2 . 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 A table has fields F1, F2, F3, F4, F5 with the following functional dependencies In terms of Normalization, this table is in (A)1 NF (B) 2 NF (C) 3 NF (D) none GATE 1999 GATE 2005 Relation R with an associated set of functional dependencies, F is decomposed into BCNF. The redundancy (arising out of functional dependencies) in the resulting set relations is. (A) Zero (B) More than zero but less than that of an equivalent 3NF decomposition (C) Proportional to the size of F+ (D) Indeterminate GATE 2002 NORMALIZATION GATE PROBLEMS

A table has fields F1, F2, F3, F4, F5 with the following functional dependencies In terms of Normalization, this table is in (A) 1 NF (B) 2 NF (C) 3 NF (D) none 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 R1nR2 . 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 Dependencies are GATE 1999 GATE 2005 1st Normal Form every attribute is singled valued attribute. Second Normal Form No Partial Dependency, i.e., no relation b/w 2 non-prime attributes. This table has Partial Dependencyf1->f3, f2->f4 given (F1,F2) is Key. S+ STUV UUVST V+ VSTU There is no Partial Dependencies here So it is in 2 NF RHS of every Dependencies is a Key as well as all are Prime Attributes.So it is in 3 NF. Relation R with an associated set of functional dependencies, F is decomposed into BCNF. The redundancy (arising out of functional dependencies) in the resulting set relations is. (A) Zero (B) More than zero but less than that of an equivalent 3NF decomposition (C) Proportional to the size of F+ (D) Indeterminate GATE 2002 NORMALIZATION GATE PROBLEMS