In the context of relational database normalization, which of the following statements is/are true?

The correct statements are:
1. It is always possible to obtain a dependency-preserving 3NF decomposition of a relation
2. It is always possible to obtain a dependency-preserving 1NF decomposition of a relation
3. It is not always possible to obtain a dependency-preserving BCNF decomposition of a relation

Let us analyze each statement step-by-step:

1. Dependency Preservation in 3NF (Third Normal Form):
For any relational schema $R$ and a set of functional dependencies $F$, there is a standard algorithm (often called the 3NF Synthesis Algorithm) that guarantees a decomposition into schemas $R_1,R_2,...,R_k$ such that:
• Each $R_i$ is in 3NF.
• The decomposition is lossless-join.
• The decomposition preserves all functional dependencies in $F$ (i.e., the union of the projections of $F$ onto each $R_i$ is equivalent to $F$).
Thus, a dependency-preserving 3NF decomposition is always possible.

2. Dependency Preservation in 1NF (First Normal Form):
First Normal Form (1NF) only requires that the domain of each attribute contains only atomic values and that the value of each attribute in a tuple is a single value from its domain. It does not impose any restrictions on functional dependencies. Since a relation is already in 1NF or can trivially be decomposed/flattened into 1NF without losing any attributes or functional dependencies, obtaining a dependency-preserving 1NF decomposition is always possible.

3. Dependency Preservation in BCNF (Boyce-Codd Normal Form):
Boyce-Codd Normal Form is stricter than 3NF. It requires that for every non-trivial functional dependency $X\to Y$, the determinant $X$ must be a superkey. Due to this strict requirement, we cannot always guarantee dependency preservation.
For example, consider the relation $R(A,B,C)$ with functional dependencies:
$AB\to C$
$C\to A$
Here, the candidate keys are $AB$ and $BC$. The dependency $C\to A$ violates BCNF because $C$ is not a superkey. If we decompose the relation to satisfy BCNF, we get $R_1(A,C)$ and $R_2(B,C)$. However, in this decomposition, the functional dependency $AB\to C$ cannot be checked within any single decomposed relation without performing a join. Hence, the dependency is lost. This proves that it is not always possible to obtain a dependency-preserving BCNF decomposition.