Solution

The insert operator is not from a relational algebra operations set.

Solution

¬(¬(P1)∨¬(P2))=P1∧P2

(By applying demorgan’s law)

Solution

All are the valid join expressions.

Solution

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

Solution

Complete set of operations:

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

Solution

Both Correct.

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

R⋈S=

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

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.

Solution

Here every tuple in Employee can join to at most one tuple in Department since Dno is primary key in Department table. So, the maximum number of tuples is 800.