Choose the correct answerWhich one of the following statements about normal forms is FALSE?OptionsLossless, dependency–preserving decomposition into BCNF is always possibleLossless, dependency-preserving decomposition into 3 NF is always possibleBCNF is stricter than 3 NFAny relation with two attributes is BCNF

The false statement is: "Lossless, dependency–preserving decomposition into BCNF is always possible".

This is because, while it is always possible to decompose a relation into a collection of BCNF relations, it is not always possible to do this in a way that is both lossless and dependency-preserving. In some cases, a lossless, dependency-preserving decomposition into BCNF may not exist.

To determine which statement about normal forms is false, let's analyze each option:

1. Lossless, dependency-preserving decomposition into BCNF is always possible: This statement is true. It is possible to decompose a relation into BCNF while preserving dependencies and without losing any information.

2. Lossless, dependency-preserving decomposition into 3NF is always possible: This statement is also true. It is possible to decompose a relation into 3NF while preserving dependencies and without losing any information.

3. BCNF is stricter than 3NF: This statement is true. BCNF is a stricter normal form than 3NF because it eliminates all partial dependencies, whereas 3NF only eliminates transitive dependencies.

4. Any relation with two attributes is BCNF: This statement is false. A relation with two attributes may or may not be in BCNF. BCNF requires that every non-trivial functional dependency is a dependency on a candidate key, which may not be the case for a relation with only two attributes.

Therefore, the false statement is: "Any relation with two attributes is BCNF."

The false statement is: "Lossless, dependency–preserving decomposition into BCNF is always possible".

This is because, while it is always possible to decompose a relation into a set of BCNF relations, it is not always possible to do this in a way that is both lossless and dependency-preserving. There are some cases where a relation can be decomposed into BCNF in a lossless way, but not in a way that preserves all functional dependencies.