Solution

In ER model, the weak entities are represented using double outlined rectangle symbol.

Solution

Only {student_id}, {student_name, student_street} are candidate keys.

Solution

As PQR is part of the super key, hence PQR may be a candidate key.

Solution

(AB)+ = ABCDEFGHIJ

(BC)+ = BCFGH ✖

(CD)+ = CDIJ ✖

(EF)+ = EFGH ✖

Solution

• Two tables are required to represent the entities E1 and E2.
• The cardinality of the relationship R is 1:M. Hence, a separate table is not required for R

Solution

¬(¬(P1)∨¬(P2))=P1∧P2

(By applying demorgan’s law)

Solution

The different number of projections possible on a relation R of degree d is 2^d-1.

Solution

These are the fundamental operations of the relational algebra that are sufficient to express any relational algebra query.

{ σ, π, ∪, ρ, –, × } → Complete set of RA operations

• ∩ (Intersection) can be implemented by ‘–’ operator
• A ∩ B = A – (A – B)
• Join (⨝ ) can be implemented by × and σ operators
• A ⨝ B = σcondition (A × B)
• Division(÷) can be implemented by ×, – and π operators
• R ÷ S = (π(R-S))-(π(R-S)((S×π(R-S)(R))-R))

Solution

TRS

T =

Solution

In relational algebra,

1. UNION and INTERSECTION are commutative operations.
2. UNION and INTERSECTION are associative operations
3. The MINUS (Set difference) is not commutative: R-S ≠ S-R

Solution

{t.FNAME, t.LNAME | EMPLOYEE(t) AND t.SALARY>20000} will retrieve the firstname, lastname of employees whose salary is more than 20000.

Solution

Both the statements I and II hold.

Solution

• Selection operator is commutative in nature i.e.

σ _B (σ _A(R)) = σ _A (σ _B(R))

• Projection operator is not commutative in nature i.e.

π _<list2> (π _<list1> (R)) ≠ π _<list1> (π _<list2> (R))

Solution

Solution

If R has ‘x’ tuples and S has ‘y’ tuples then maximum and minimum size of the join respectively are xy and 0.