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

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

2. Lossless, dependency-preserving decomposition into $3NF$ is always possible

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

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

edited | 2.7k views

option $C$
by Boss (11k points)
edited by
+5

why dependency preserving decomposition into 3NF is always possible?

I found that indeed it is true that a dependency preserving, lossless join decomposition into 3NF is always possible. here is link. (2nd point)

+1
but, option (c) is about BCNF, which is not true always in case of BCNF
0
How option D is true?
+2
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
by Active (4.4k points)
+1 vote

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.

by (107 points)

If relation is in 3NF then both lossless and dependency preserving are Guaranteed

and If relation is in BCNF lossless is Guaranteed but dependency preserving is Not Guaranteed.

So, clearly option (C) is false.

by Active (1.7k points)

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