2,604 views
5 votes
5 votes

Q)Which of the following is false??

1)Relation R in BCNF with atleast one simple candidate key is also in 4NF

2)Relation R in BCNF with all candidate key simple is also in 4NF

3)Relation R in 3NF with only one compound candidate key is also in BCNF

4)Relation R in 3NF with atleast one simple candidate key is also in 4NF

Explain with example....

 

 

1 Answer

6 votes
6 votes

1)Relation R in BCNF with atleast one simple candidate key is also in 4NF
     if multivalue attribute exist in the table then every candidate key will be compound key 
     so here there is one simple candidate key no multivalue attribute present so surely its in 4NF also.

2)Relation R in BCNF with all candidate key simple is also in 4NF
    same reason as above.

3)Relation R in 3NF with only one compound candidate key is also in BCNF
   
  if relation R is in 3NF and atmost one compound CK then it is also in BCNF
 if more than one compound Ckey will present then prime to prime will be possible which will     violate the BCNF property but with one compound CK there will be no such possiblity  

e.g. R(ABCDE) {AB->C C->D D->E E>A} 
       CK{AB EB DB CB} Here  C->D D->E E>A violates the BCNF

4)Relation R in 3NF with atleast one simple candidate key is also in 4NF

   There is no MVD in the relation 
  But there can be possiblity of two compound candidate key then it will not be in BCNF

 R(ABCDE) {AB->C C->BD D->E E>A} 
 CK{AB C EB DB } Here D->E E>A violates the BCNF so 4NF will also not satisfied.

option 4 is false.

 

Related questions

2 votes
2 votes
1 answer
1
0 votes
0 votes
0 answers
2
ayush201 asked Dec 7, 2018
1,039 views
Q.47. How many minimum relation tables are required which satisfy 1NF? A 2, 2, and 1 respectivelyB 2, 2, and 2 respectivelyC 1, 2, and 1 respectivelyD 1, 1, and 1 respect...
1 votes
1 votes
0 answers
3
Balaji Jegan asked Oct 23, 2018
301 views