GATE Computer Science Engineering(CSE) 2027 Test: Relational Algebra, Tuple

The expression

can be represented in relational algebra as

It says that select those rows from relation r where A = 10 union where B = 20.
Hence option (b) is correct.

According to entity integrity primary key of relation should not contain null values.

Five primitive operators of Codd’s algebra are the selection, the projection the Cartesian product, the set union and the set difference.

Option D is correct.

Union and Intersection are binary set operations and are commutative; that is, R ∪ S = S ∪ R and R ∩ S = S ∩ R.

Selection (σ) and Projection (π) are unary operators. Multiple selections commute with each other and multiple projections commute with each other, but projection does not in general commute with selection; changing their order can change the result.

For example, let R(A, B) be a relation. Compare π_A(σ_{B > 10}(R)) and σ_{B > 10}(π_A(R)). The first returns the A values of tuples where B > 10. The second is invalid or empty because B has been removed by the projection, so the selection condition B > 10 cannot be applied after projection. Thus the order matters.

Because the question asks for a single operation among the options that is not commutative in the usual sense used in relational-algebra discussions (i.e., it does not commute with selection in general), Projection (Option D) is chosen as the correct answer.

Loans (br_name, loan_no, amount)
For the query “find the loan number for each loan of an amount greater than 20000”.
We have to check for each loan to find a loan number which is greater than 20,000.

Hence (b) is the correct option.

Referential integrity constraint: In relational model, two relation are related to each other over the basis of attributes, Every value of referencing attribute must be null or be available in the referenced attribute.

Here d is the foreign key of S that refers to the primary key of R.
1. Insert into R will not cause any violation.
2. Insert into S may cause violation because for each entry in ‘S ’ it must be. in ‘R ’ .
3. Delete from R may cause violation because for the deleted entry in R there may be referenced entry in the reIation S.
4. Delete from S will not cause any violation. 
Hence (d) is the correct option.

The relational algebra expression:

Hence (c) is the required option.

  is not equivalent because θ join is not cumulative hence the entries in   may not be same. Hence option (d) is correct.