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What normal form is the relation in?
A relation is in third normal form, if there is no transitive dependency for non-prime attributes as well as it is in second normal form.
What are the different normal forms explain with example?
The normal form is used to reduce redundancy from the database table….Types of Normal Forms.
| Normal Form | Description |
|---|---|
| 3NF | A relation will be in 3NF if it is in 2NF and no transition dependency exists. |
| 4NF | A relation will be in 4NF if it is in Boyce Codd normal form and has no multi-valued dependency. |
How do you determine if a relation is in 3NF?
A relation is in third normal form, if there is no transitive dependency for non-prime attributes as well as it is in second normal form. A relation is in 3NF if at least one of the following condition holds in every non-trivial function dependency X –> Y: X is a super key.
When is a relation in second normal form?
To be in second normal form, a relation must be in first normal form and relation must not contain any partial dependency. A relation is in 2NF if it has No Partial Dependency, i.e., no non-prime attribute (attributes which are not part of any candidate key) is dependent on any proper subset of any candidate key of the table.
When is a relation in the Boyce-Codd normal form?
Boyce-Codd Normal Form (BCNF) – A relation R is in BCNF if R is in Third Normal Form and for every FD, LHS is super key. A relation is in BCNF iff in every non-trivial functional dependency X –> Y, X is a super key. Step 1.
Which is a normal form in DBMS-geeksforgeeks?
X is a super key. Y is a prime attribute (each element of Y is part of some candidate key). Transitive dependency – If A->B and B->C are two FDs then A->C is called transitive dependency. For this relation in table 4, STUD_NO -> STUD_STATE and STUD_STATE -> STUD_COUNTRY are true.
Do you know all the keys of the relation C?
To answer to this question, you should know all the keys of the relation, and see if C belongs to one of them. And in this case this is very simple, since A is the unique key of this relation, so you can conclude that the relation is not in 3NF.