The Relational Model and Normalization Questions and Answers

Q: In a schema with attributes A, B, C, D and E, with the following set of functional dependencies {A → B, A → C, CD → E, B → D, E → A}, which of the following functional dependencies are not implied by the above set ?

The dependency not implied is BD → CD.Test each by computing the closure of its left side under {A → B, A → C, CD → E, B → D, E → A}.\(\{B, D\}^{+}\): B → D adds nothing new, and no other dependency applies, because each needs A, E or C. So \(\{B, D\}^{+} = \{B, D\}\), which lacks C: BD → CD is not implied.\(\{C, D\}^{+}\): CD → E, E → A, A → B gives ABCDE, so CD → AC holds.\(\{B, E\}^{+}\): E → A, B → D, so BE → AD holds.\(\{A, B\}^{+}\): A → C, B → D, so AB → CD holds.

Q: Which of the following statements is not true for Relational Algebra ?

The statement that is not true is Natural join is a fundamental operation in relational algebra.The fundamental operations of relational algebra are select, project, union, set difference, Cartesian product and rename. Natural join is a derived operation: it can be written as a Cartesian product followed by a selection on the common attributes and a projection to remove duplicate columns.The other statements are true. Relational algebra is procedural, since a query states the sequence of operations to apply. Selection is written with σ, and its predicate appears as a subscript, as in \(\sigma_{marks \gt 60}(student)\).

Q: To get a database in 3NF, which of the following steps are essential ?(i) Converting composite attributes to individual attributes(ii) Removal of partial dependencies(iii) Removal of transitive dependencies

All three steps are needed, so the answer is (i), (ii) and (iii).Normalisation to 3NF goes through the earlier forms in order:1NF: every attribute must hold atomic values, so composite and multi-valued attributes are broken into individual attributes (i).2NF: partial dependencies of non-key attributes on part of a composite key are removed (ii).3NF: transitive dependencies of non-key attributes on other non-key attributes are removed (iii).Leaving out any step leaves the relation short of 3NF, so the options naming only some of them are wrong.

Q: Every table with two single valued attributes is in :

Such a table is in all of the above: 2NF, 3NF and BCNF.Take a relation R(A, B) with two single-valued (atomic) attributes, so it is in 1NF. The only non-trivial functional dependencies it can have are A → B or B → A. Whichever holds, its left side determines the whole relation and so is a key.BCNF requires the left side of every non-trivial dependency to be a superkey, which is satisfied. Since BCNF implies 3NF, and 3NF implies 2NF, the table is in all three, and picking any single form understates it.

Q: In the context of functional dependency, Armstrong inference rules refer to :

Armstrong's inference rules are reflexivity, augmentation and transitivity.Reflexivity: if Y is a subset of X, then X → Y.Augmentation: if X → Y, then XZ → YZ.Transitivity: if X → Y and Y → Z, then X → Z.These three axioms are sound and complete for deriving every functional dependency in the closure. Decomposition (and union, pseudotransitivity) are additional rules derived from them, so options naming decomposition list a derived rule instead of augmentation.

Q: Consider the relation scheme (M, N, O, P, Q, R) with the following set of functional dependencies :F = {MO → PQ, MPR → NO}Which of the following is the trivial functional dependencies in F+, where F+ is closure of F ?

The trivial dependency is MP → P.A functional dependency X → Y is trivial when Y is a subset of X, because it holds in every relation regardless of the data. P is part of MP, so MP → P is trivial; it is in F+ by reflexivity.MO → PQ and MPR → N are in F+ but are not trivial, since PQ and N are not contained in their left sides. MO → OP also follows from F, but P is not in MO, so it is not trivial either.

Q: Among the following, which is foreign key constraint ?

The foreign key constraint is referential integrity.Referential integrity requires that a foreign key value in one relation either matches an existing primary key value in the relation it refers to, or is null. It stops a row from referring to a record that does not exist.Domain integrity restricts attribute values to their allowed domain. Entity integrity requires primary key values to be unique and not null. Key integrity concerns uniqueness of keys within a relation. None of these is the constraint a foreign key enforces.

Q: A set of one or more attributes which, taken collectively, allow us to uniquely identify an entity in the entity set is called ................. .

Such a set of attributes is a super key.A super key is any set of one or more attributes that, taken together, uniquely identify an entity in the entity set. It may contain extra attributes that are not needed for uniqueness.A candidate key is a minimal super key, with no unnecessary attributes. A primary key is the candidate key chosen by the designer. A foreign key refers to the key of another relation and need not be unique at all. The definition in the question, without any minimality condition, is that of a super key.

Q: A relation in this form is free of all modification anomalies.

Correct answer: Domain/key normal formFagin proved that a relation in domain/key normal form has no modification anomalies of any kind, and conversely that any relation free of them is in DK/NF. It is defined by requiring that every constraint on the relation follow from its domain and key definitions alone. The difficulty is that there is no general procedure for putting a relation into it.

Q: A relation is in first normal form if every attribute value in it is atomic.

Correct answer: TrueFirst normal form forbids repeating groups and multi-valued entries, requiring each cell to hold a single indivisible value. A column storing a comma-separated list of phone numbers violates it.

11.

In a schema with attributes A, B, C, D and E, with the following set of functional dependencies {A → B, A → C, CD → E, B → D, E → A}, which of the following functional dependencies are not implied by the above set ?

A.CD → AC
B.BD → CD
C.BE → AD
D.AB → CD

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The dependency not implied is BD → CD.

Test each by computing the closure of its left side under {A → B, A → C, CD → E, B → D, E → A}.

\(\{B, D\}^{+}\): B → D adds nothing new, and no other dependency applies, because each needs A, E or C. So \(\{B, D\}^{+} = \{B, D\}\), which lacks C: BD → CD is not implied.
\(\{C, D\}^{+}\): CD → E, E → A, A → B gives ABCDE, so CD → AC holds.
\(\{B, E\}^{+}\): E → A, B → D, so BE → AD holds.
\(\{A, B\}^{+}\): A → C, B → D, so AB → CD holds.

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12.

Which of the following statements is not true for Relational Algebra ?

A.Relational Algebra is a procedural query language
B.Natural join is a fundamental operation in relational algebra
C.The symbol σ is used for select operation in relational algebra
D.In relational algebra, for select operation, predicates appear in the subscript

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The statement that is not true is Natural join is a fundamental operation in relational algebra.

The fundamental operations of relational algebra are select, project, union, set difference, Cartesian product and rename. Natural join is a derived operation: it can be written as a Cartesian product followed by a selection on the common attributes and a projection to remove duplicate columns.

The other statements are true. Relational algebra is procedural, since a query states the sequence of operations to apply. Selection is written with σ, and its predicate appears as a subscript, as in \(\sigma_{marks \gt 60}(student)\).

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13.

To get a database in 3NF, which of the following steps are essential ?

(i) Converting composite attributes to individual attributes
(ii) Removal of partial dependencies
(iii) Removal of transitive dependencies

A.(i) and (ii)
B.(iii) only
C.(ii) and (iii)
D.(i), (ii) and (iii)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

All three steps are needed, so the answer is (i), (ii) and (iii).

Normalisation to 3NF goes through the earlier forms in order:

1NF: every attribute must hold atomic values, so composite and multi-valued attributes are broken into individual attributes (i).
2NF: partial dependencies of non-key attributes on part of a composite key are removed (ii).
3NF: transitive dependencies of non-key attributes on other non-key attributes are removed (iii).

Leaving out any step leaves the relation short of 3NF, so the options naming only some of them are wrong.

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14.

Every table with two single valued attributes is in :

A.2NF
B.3NF
C.BCNF
D.All of the above

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

Such a table is in all of the above: 2NF, 3NF and BCNF.

Take a relation R(A, B) with two single-valued (atomic) attributes, so it is in 1NF. The only non-trivial functional dependencies it can have are A → B or B → A. Whichever holds, its left side determines the whole relation and so is a key.

BCNF requires the left side of every non-trivial dependency to be a superkey, which is satisfied. Since BCNF implies 3NF, and 3NF implies 2NF, the table is in all three, and picking any single form understates it.

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15.

In the context of functional dependency, Armstrong inference rules refer to :

A.Reflexivity, Decomposition and Transitivity
B.Decomposition and Transitivity
C.Transitivity, Augmentation and Reflexivity
D.Decomposition and Reflexivity

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

Armstrong's inference rules are reflexivity, augmentation and transitivity.

Reflexivity: if Y is a subset of X, then X → Y.
Augmentation: if X → Y, then XZ → YZ.
Transitivity: if X → Y and Y → Z, then X → Z.

These three axioms are sound and complete for deriving every functional dependency in the closure. Decomposition (and union, pseudotransitivity) are additional rules derived from them, so options naming decomposition list a derived rule instead of augmentation.

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16.

Consider the relation scheme (M, N, O, P, Q, R) with the following set of functional dependencies :

F = {MO → PQ, MPR → NO}

Which of the following is the trivial functional dependencies in F⁺, where F⁺ is closure of F ?

A.MO → PQ
B.MO → OP
C.MP → P
D.MPR → N

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The trivial dependency is MP → P.

A functional dependency X → Y is trivial when Y is a subset of X, because it holds in every relation regardless of the data. P is part of MP, so MP → P is trivial; it is in F⁺ by reflexivity.

MO → PQ and MPR → N are in F⁺ but are not trivial, since PQ and N are not contained in their left sides. MO → OP also follows from F, but P is not in MO, so it is not trivial either.

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17.

Among the following, which is foreign key constraint ?

A.Domain integrity
B.Entity integrity
C.Referential integrity
D.Key integrity

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The foreign key constraint is referential integrity.

Referential integrity requires that a foreign key value in one relation either matches an existing primary key value in the relation it refers to, or is null. It stops a row from referring to a record that does not exist.

Domain integrity restricts attribute values to their allowed domain. Entity integrity requires primary key values to be unique and not null. Key integrity concerns uniqueness of keys within a relation. None of these is the constraint a foreign key enforces.

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18.

A set of one or more attributes which, taken collectively, allow us to uniquely identify an entity in the entity set is called ................. .

A.Primary key
B.Foreign key
C.Super key
D.Candidate key

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

Such a set of attributes is a super key.

A super key is any set of one or more attributes that, taken together, uniquely identify an entity in the entity set. It may contain extra attributes that are not needed for uniqueness.

A candidate key is a minimal super key, with no unnecessary attributes. A primary key is the candidate key chosen by the designer. A foreign key refers to the key of another relation and need not be unique at all. The definition in the question, without any minimality condition, is that of a super key.

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19.

A relation in this form is free of all modification anomalies.

A.First normal form
B.Second normal form
C.Third normal form
D.Domain/key normal form

Answer: D

Correct answer: Domain/key normal form

Fagin proved that a relation in domain/key normal form has no modification anomalies of any kind, and conversely that any relation free of them is in DK/NF. It is defined by requiring that every constraint on the relation follow from its domain and key definitions alone. The difficulty is that there is no general procedure for putting a relation into it.

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20.

A relation is in first normal form if every attribute value in it is atomic.

A.True
B.False

Answer: A

Correct answer: True

First normal form forbids repeating groups and multi-valued entries, requiring each cell to hold a single indivisible value. A column storing a comma-separated list of phone numbers violates it.

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