DBMS normalization concept problem

Question

R(ABCDEF)

FD set{

AB->CDE. CD->E, E->C

}

False statement?

A. 2 NF decomposition possible

B.3 NF decomposition possible

C.BCNF decomposition possible

D. ALL

Comments

2NF possible .

3NF not possible since non - key values determine non key terms

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pls verify my answer. and post a screenshot of the solution if possible.

Answer 1

ABF is the candidate key in the relation. And I think 3NF decomposition is possible which would be ABF, ABCD, CDE. in relation CDE the FD E-> C will not satisfy BCNF condition.

Comments

How 3NF possible ??
CD -> E NonPrime Attribute is determine Non Prime Attribute which is not allowed in 3NF

Moreover,
CD->  C is Trivial FD

CD->E

and E->C should not be present due to transitivity .

The given Relation is 2NF only

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CD and ED are candidate keys of the decomposition CDE. You must find the FD's of each decomposition by using closure, then find the candidate keys.And finally check for PDs and TDs within the decomposition. By practise you'll be able to do this intuitively and very quickly.

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Can u explain me the question ?
Doesn't it meant to find the highest normal form of the relation given ?
Im confused with the word "decomposed " in the question .

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When a relation doesn't satisfy a normal form, we can decompose it(split into multiple tables) in such a way that the resultant relations satisfy it. For eg: For  1NF if there's a multivalued attribute in a relation we decompose it into two tables. One with primary key and multivalued attribute and the other table will have the key and remaning attributes in the original relation. And referential relationship will exsist between the two new tables. Refer Navathe. best resource for DBMS. or refer techtud videos in youtube.

Answer 2

since they have only asked if a decomposition is possible and not about lossless or FD preserving hence all are true it is always possible to decompose to BCNF

Comments

Not a valid logical reason my friend. They need not explicitly specify every condition. your argument is similar to saying that a linear equation can be proven by multiplying by zero on both sides as the quesion does not mention that such techniques cannot be used.

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i think it is true because we don''t always need FD preserving decomposition we care about lossless decomposition and BCNF is always lossless