Question 4432 – Mathematical-Reasoning
March 11, 2024Computer-Organization
March 11, 2024Database-Management-System
| Question 7 |
Consider a relation R ( A , B , C , D , E ) with the following three functional dependencies.
AB → C; BC → D; C → E;
The number of superkeys in the relation R is _____________.
AB → C; BC → D; C → E;
The number of superkeys in the relation R is _____________.
| 8 |
Question 7 Explanation:
R(A, B, C, D, E)
AB → C
BC → D
C → E
¯ The attributes A, B are not there in the right hand side of any of the given FDs.
¯ So, AB can be a candidate key.
(AB)^+ = ABCDE
¯ (AB)^+ is deriving all the attributes of R Hence, AB is a candidate key.
¯ The number of super keys possible for R with “AB” candidate kay is:
2^5 – 2 = 2^3
= 8
AB → C
BC → D
C → E
¯ The attributes A, B are not there in the right hand side of any of the given FDs.
¯ So, AB can be a candidate key.
(AB)^+ = ABCDE
¯ (AB)^+ is deriving all the attributes of R Hence, AB is a candidate key.
¯ The number of super keys possible for R with “AB” candidate kay is:
2^5 – 2 = 2^3
= 8
Correct Answer: A
Question 7 Explanation:
R(A, B, C, D, E)
AB → C
BC → D
C → E
¯ The attributes A, B are not there in the right hand side of any of the given FDs.
¯ So, AB can be a candidate key.
(AB)^+ = ABCDE
¯ (AB)^+ is deriving all the attributes of R Hence, AB is a candidate key.
¯ The number of super keys possible for R with “AB” candidate kay is:
2^5 – 2 = 2^3
= 8
AB → C
BC → D
C → E
¯ The attributes A, B are not there in the right hand side of any of the given FDs.
¯ So, AB can be a candidate key.
(AB)^+ = ABCDE
¯ (AB)^+ is deriving all the attributes of R Hence, AB is a candidate key.
¯ The number of super keys possible for R with “AB” candidate kay is:
2^5 – 2 = 2^3
= 8
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