Consider a schema R(A,B,C,D) and functional dependencies A->B and C->D.

Then the decomposition of R into R1(AB) and R2(CD) is**(A)** dependency preserving and lossless join**(B)** lossless join but not dependency preserving**(C)** dependency preserving but not lossless join**(D)** not dependency preserving and not lossless join**Answer:** **(C)****Explanation:** **Dependency Preserving Decomposition:**

Decomposition of R into R1 and R2 is a dependency preserving decomposition if closure of functional dependencies after decomposition is same as closure of of FDs before decomposition.

A simple way is to just check whether we can derive all the original FDs from the FDs present after decomposition.

In the above question R(A, B, C, D) is decomposed into R1 (A, B) and R2(C, D) and there are only two FDs A -> B and C -> D. So, the decomposition is dependency preserving

**Lossless-Join Decomposition:**

Decomposition of R into R1 and R2 is a lossless-join decomposition if at least one of the following functional dependencies are in F+ (Closure of functional dependencies)

R1 ∩ R2 → R1 OR R1 ∩ R2 → R2

In the above question R(A, B, C, D) is decomposed into R1 (A, B) and R2(C, D), and R1 ∩ R2 is empty. So, the decomposition is not lossless.

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