GATE 2007
62. Which one of the following statements if FALSE?
(A) Any relation with two attributes is in BCNF
(B) A relation in which every key has only one attribute is in 2NF
(C) A prime attribute can be transitively dependent on a key in a 3 NF relation.
(D) A prime attribute can be transitively dependent on a key in a BCNF relation.
 

  • Explanation

    Ans: D. Let us see each option carefully.

    A) Any relation with 2 attributes will be in BCNF. How? Let's assume R(a,b).There can be types of FDs in this relation, 1)a->b ad 2) b->a . If there is no FD, we can assume trivial FDs as ab->a and ab->b.

    In all cases, FDs like X->Y, X is key.right. So they all will be BCNF, irrespective of the FD set.So Relation with 2 attributes will be in BCNF, for sure.It is TRUE.

    B) If in the relation each key is single attribute, then there is not chance of forming partial FDs for non prime attributes too. Then it will be in 2NF. TRUE statement.

    C) Yes it is TRUE. Prime attribute are allowed for transitive dependency.

    D) It is FALSE statement.

    Please let me know if I am unclear or wrong.

Responses

monty12's picture

(A) true, any relation with two attributes is in BCNF;
     See R(A,B)  { A--->B, B--->A} candidate key {A,B};
     BCNF thoery Every left part  of fd's are Super Key;

(B) true , A relation in which every  key has only one attribute is in 2NF;
      {A--->B,B--->DC} candidate key {A};
       there is no partial dependeny between candidate key and left part of fd's  attribute;

(C)  true, A prime attribute can be transitively dependent on a key in a 3 NF relation;
        R(ABCDE) { AB-->C, C-->D, D-->E, E---> A};
        key={AB, BE, DB, CB} ;
        C,E ,B,A,D  are prime attribute because it is a part of key ;
         in which C, D , E are  can be transitively dependent on a key ;
         and 3nf rule are  X--->Y ( X should  be super key or y  is prime attribute ) ;

(D) false, A prime attribute can be transitively dependent on a key in a BCNF relation.  
       BCNF rule is X--->Y ( X should  be super key ONLY);

And (D)        

            
         

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