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32 votes
32 votes

Which one of the following statements about normal forms is $\text{FALSE}?$

  1. $\text{BCNF}$ is stricter than $\text{3NF}$

  2. Lossless, dependency-preserving decomposition into $\text{3NF}$ is always possible

  3. Lossless, dependency-preserving decomposition into $\text{BCNF}$ is always possible

  4. Any relation with two attributes is in $\text{BCNF}$

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Best answer
11 votes
11 votes

Option C is the only FALSE statement.

We can always have a lossless decomposition into $\textsf{BCNF}$ but not always we can have a lossless and dependency preserving decomposition. But this is always possible in the case of $\textsf{3NF}.$

Option A is true as the requirement of $\textsf{BCNF}$ required a relation schema to be in $\textsf{3NF}.$ Actually $\textsf{3NF}$ allows transitive dependency for prime attributes whereas $\textsf{BCNF}$ does not.

Option D is true as shown below.

Assume the two attributes to be $A$ and $B.$

Now, we can have three cases:

  1. Either $A$ or $B$ is the candidate key but not both. i.e., $A \to B$ or $B \to A.$ No other $\text{FD}$ is possible and $\text{LHS}$ of all $\text{FDs}$ are superkeys and so $\textsf{BCNF}$ requirement is satisfied.
  2. Both $A$ and $B$ are candidate keys. i.e., $A \to B$ and $B \to A.$ Like in above case $\textsf{BCNF}$ requirement is satisfied.
  3. Neither $A \to B$ nor $B \to A$ and so $AB$ is the key. So, no other $\text{FD}$ is possible and this case also satisfies $\textsf{BCNF}$ requirement.

Thus any relation with $2$ attributes is guaranteed to be in $\textsf{BCNF}.$

Ref: https://gatecse.in/demystifying-database-normalization/

33 votes
33 votes
option $C$
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7 votes
7 votes
option (d)

It is true . If there are only 2 attributes then relation is always in bcnf.

let R(A B) possible cases are:

(1) {A->B } here A is cand key so BCNF

(2) {B->A}  here B is cand key so BCNF

(3) { A->B  B->A} here A and B both are cand key so BCNF

​​​​​​​(4) no non-trivial FD's here AB is cand key so BCNF
3 votes
3 votes

As the FALSE statement is asked, therefore, Option C

It is not always possible  to have a dependency preserving BCNF decomposition. 

For example. this 3NF relation cant be transformed into BCNF

AB –> C, C –> B

Regarding other options, it is always possible to have a 3NF or lesser normalization to be lossless and dependency preserving.

Answer:

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