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+18 votes
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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 $\text{BCNF}$

asked in Databases by Veteran (59.7k points)
edited by | 2.2k views

3 Answers

+25 votes
Best answer
option $C$
answered by Boss (11.6k 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
+2 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
answered by Active (3.5k points)
0 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.

answered by (123 points)
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