Using Relational AlgebraYou learned in Module 9 about Relational algebra; the theoretical language for operating on one or more relations. It forms the theoretical basis for relational databases and f

Optional Lab 9-1

2. The following is the relational algebraic expression:

student_id = 3115(Student_Grade)

The following is the SQL equivalent:

SELECT *

FROM Student_Grade

WHERE student_id = 3115;

3. The following is the relational algebraic expression:

class_id = 'Hist201' and grade != 'A'(Student_Grade)

The following is the SQL equivalent:

SELECT *

FROM Student_Grade

WHERE class_id = 'Hist201'

AND NOT grade = 'A';

5. The following is the relational algebraic expression:

name, email(Student_Info)

The following is the SQL equivalent:

SELECT name, email

FROM Student_Info;

7. The following is the relational algebraic expression:

NL_Hitters X AL_Hitters

The following is the SQL equivalent:

SELECT NL_Hitters.*, AL_Hitters.*

FROM NL_Hitters, AL_Hitters;

8. The following is the relational algebraic expression:

NL_Hitters.Home_Runs > 50 and AL_Hitters.Home_Runs > 50(NL_Hitters X AL_Hitters)

The following is the SQL equivalent:

SELECT NL_Hitters.*, AL_Hitters.*

FROM NL_Hitters, AL_Hitters

WHERE NL_Hitters.Home_Runs > 50

AND AL_Hitters.Home_Runs > 50;

10. The following is the relational algebraic expression:

state(Location)  state(Property)

The following is the SQL equivalent:

SELECT state

FROM Location

UNION

SELECT state

FROM Property;

12. The following is the relational algebraic expression:

(cust_id,name (Customer))  (prop_id,cust_comment (Showings))

The following is the SQL equivalent:

SELECT Customer.cust_id, name, Showings.prop_id, Showings.cust_comment

FROM Customer, Showings

WHERE Customer.cust_id = Showings.cust_id;