GATE | GATE-CS-2014-(Set-1) | Question 40

Given the following two statements:

  S1: Every table with two single-valued 
      attributes is in 1NF, 2NF, 3NF and BCNF.

  S2: AB->C, D->E, E->C is a minimal cover for 
      the set of functional dependencies 
      AB->C, D->E, AB->E, E->C. 

Which one of the following is CORRECT?

(A) S1 is TRUE and S2 is FALSE.
(B) Both S1 and S2 are TRUE.
(C) S1 is FALSE and S2 is TRUE.
(D) Both S1 and S2 are FALSE.


Answer: (A)

Explanation:

 
S1: Every table with two single-valued 
      attributes is in 1NF, 2NF, 3NF and BCNF.

A relational schema R is in BCNF iff in Every non-trivial Functional Dependency X->Y, X is Super Key. If we can prove the relation is in BCNF then by default it would be in 1NF, 2NF, 3NF also.

Let R(AB) be a two attribute relation, then



  1. If {A->B} exists then BCNF since {A}+ = AB = R
  2. If {B->A} exists then BCNF since {B}+ = AB = R
  3. If {A->B,B->A} exists then BCNF since A and B both are Super Key now.
  4. If {No non trivial Functional Dependency} then default BCNF.

Hence it’s proved that a Relation with two single – valued attributes is in BCNF hence its also in 1NF, 2NF, 3NF.

Hence S1 is true.

S2: AB->C, D->E, E->C is a minimal cover for 
      the set of functional dependencies 
      AB->C, D->E, AB->E, E->C.

As we know Minimal Cover is the process of eliminating redundant Functional Dependencies and Extraneous attributes in Functional Dependency Set.

So each dependency of F = {AB->C, D->E, AB->E, E->C} should be implied in minimal cover.

As we can see AB->E is not covered in minimal cover since {AB}+ = ABC in the given cover {AB->C, D->E, E->C}

Hence, S2 is false.

This explanation has been contributed by Manish Rai.

Learn more about Normal forms here:

Database Normalization | Introduction
Database Normalization | Normal Forms

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