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3 votes

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Answer: $B$

$Pk(r)$ denotes the primary key attribute of relation $r$.

• The functional dependency $Pk(r_1) \rightarrow Pk(r_2)$ indicates a many-to-one relationship between $r_1,r_2,$ since any $r_1$ value which is repeated will have the same $r_2$ value, but many $r_1$ values may have the same $r_2$ value.

NOTE:

**Consider the statement:**

- A many-to-one relationship set exists between entity sets student and instructor.
- A many-to-one relationship set exists between entity sets instructor and student.

Read the above statements & think about what do they mean.

Statement 1 says that “one instructor may have many students, but one student learns from at most one instructor”.

Statement 2 says that “one student may have many instructors, but one instructor teaches at most one student”.

**Source of this UGC NET Question: Korth DBMS Book.**

https://www.db-book.com/db6/practice-exer-dir/8s.pdf (Page 10)

http://web.cs.ucla.edu/classes/fall04/cs143/solutions/ch7.pdf (Page 86)

**Same Question here:**

https://gateoverflow.in/225819/ugc-net-cse-july-2018-part-2-question-67

4 votes

For example consider the relation R_{1 }be employee (primary key eid} R_{2 }be department (primary key did)

The many to one relationship is many employees works in one department

given an employee we can determine his department

given a department there are n number of possilbe employees. so we cannot uniquely determine

now based on definition of functional dependency we can conclude

**pk(R _{1}) → pk(R_{2})**

3 votes

**Ans will be 2) pk(R _{1})→pk(R_{2})**

Consider the following case

A many-to-one relationship set exists between entity sets students and course (a relation r with primary key roll_no )

let it is decomposed in 2 relations r1(student with Pk roll_no ) and r2 (course with Pk course name )

Roll_no can uniquely identify course name but course name can not uniquely identify roll_no

**so the functional dependency Pk(student) → Pk(course) indicates a many-to-one relationship between r1 and r2 **

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