Database Concepts, 7e, Global Edition (Kroenke/Auer)

Chapter 5 Database Design

1) The first step in representing entities using the relational model is to determine which
identifier will be used as the key.
Answer: FALSE
Diff: 2 Page Ref: 305

2) Relations should always be normalized to the highest degree possible.

Answer: FALSE
Diff: 1 Page Ref: 309

3) If a weak entity is ID-dependent but not existence-dependent, it can be represented using the
same techniques as a strong entity.
Answer: FALSE
Diff: 3 Page Ref: 311-312

4) The key of the parent entity becomes part of the key of an ID-dependent entity.
Answer: TRUE
Diff: 1 Page Ref: 311-312

5) From a pragmatic standpoint, the only important rule of normalization is that the determinant
of every functional dependency must be a candidate key.
Answer: TRUE
Diff: 2 Page Ref: 306-307

6) An entity needs to be examined according to normalization criteria before creating a table

from it in the relational database design.
Answer: FALSE
Diff: 2 Page Ref: 305-307 Fig 5-1

7) When creating a table in the relational database design from an entity in the extended E-R
model, the attributes of the entity become the rows of the table.
Answer: FALSE
Diff: 1 Page Ref: 305-307 Fig 5-1

8) By default, the identifier of the entity becomes the foreign key of the corresponding table.
Answer: FALSE
Diff: 1 Page Ref: 305-307

9) The ideal primary key is short, numeric, and fixed.

Answer: TRUE
Diff: 2 Page Ref: 306

10) A surrogate key is appropriate when the primary key of a table contains a lengthy text field.
Answer: TRUE
Diff: 1 Page Ref: 306
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11) One of the important properties of an attribute is whether or not it is required.
Answer: TRUE
Diff: 2 Page Ref: 306

12) The technique for representing E-R relationships in the relational model is dependent on the
minimum cardinality.
Answer: FALSE
Diff: 1 Page Ref: 313

13) For a 1:1 relationship, the key of each table should be placed in the other table as the foreign
key.
Answer: FALSE
Diff: 2 Page Ref: 313-315

14) Relationships that are 1:1 do not require referential integrity constraints.
Answer: FALSE
Diff: 2 Page Ref: 313-315

15) In certain circumstances, there may be a preference as to which table in a 1:1 relationship
contains the foreign key.
Answer: TRUE
Diff: 1 Page Ref: 313-315

16) When applied to 1:N relationships, the term "parent" refers to the many side of the
relationship since a child may have many parents.
Answer: FALSE
Diff: 1 Page Ref: 315-317

17) To represent a 1:N relationship in the relational model, the key of the entity on the one side
of the relationship is placed as a foreign key in the entity on the many side of the relationship.
Answer: TRUE
Diff: 2 Page Ref: 315-317

18) To represent a 1:N relationship in the relational model, the key of either entity may be placed
as a foreign key in the other entity.
Answer: FALSE
Diff: 2 Page Ref: 315-317

19) In the relational model, many-to-many relationships cannot be directly represented by

relations the way 1:1 and 1:N relationships can.
Answer: TRUE
Diff: 3 Page Ref: 317-320

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20) To represent a M:N relationship in the relational model, an intersection relation is created to
represent the relationship itself.
Answer: TRUE
Diff: 1 Page Ref: 317-320

21) The key for an intersection relation is always the combination of the keys of the parent
entities.
Answer: TRUE
Diff: 2 Page Ref: 317-318

22) All recursive relationships are 1:1.

Answer: FALSE
Diff: 2 Page Ref: 322

23) Recursive relationships can be represented in the relational model using the same techniques
that are used for binary relationships.
Answer: TRUE
Diff: 2 Page Ref: 322-326

24) Microsoft Access uses the same pure N:M relationships that occur in data modeling.
Answer: FALSE
Diff: 2 Page Ref: 334

25) As far as Microsoft Access is concerned, there are no N:M relationships.

Answer: TRUE
Diff: 1 Page Ref: 334

26) As far as Microsoft Access is concerned, there are no 1:N relationships.

Answer: FALSE
Diff: 1 Page Ref: 334

27) By default, Microsoft Access creates 1:1 relationships between tables.

Answer: FALSE
Diff: 2 Page Ref: 337

28) To create a 1:1 relationship in Microsoft Access, the Indexed property of the foreign key
column must be set to Yes (No Duplicates).
Answer: TRUE
Diff: 3 Page Ref: 338

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29) The first step in transforming an extended E-R model into a relational database design is to
________.
A) create a table for each relationship
B) evaluate the entities against the normalization criteria
C) create a table for each entity
D) remove any recursive relationships
E) document referential integrity constraints
Answer: C
Diff: 1 Page Ref: 304-307 Fig 5-1

30) Each attribute of an entity becomes a(n) ________ of a table.

A) column
B) primary key
C) foreign key
D) alternate key
E) B or D
Answer: A
Diff: 1 Page Ref: 304-307 Fig 5-1

31) The identifier of the entity becomes the ________ of the corresponding table.
A) primary key
B) foreign key
C) supertype
D) subtype
E) either A or B
Answer: A
Diff: 1 Page Ref: 304-307

32) The ideal primary key is ________.

A) short
B) numeric
C) fixed
D) A and B
E) A, B, and C
Answer: E
Diff: 2 Page Ref: 306

33) A surrogate key should be considered when ________.

A) a relationship is M:N
B) a composite key is required
C) the key contains a lengthy text field
D) the key contains a number
E) an index needs to be created
Answer: C
Diff: 3 Page Ref: 306

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34) Which of the following is not true about surrogate keys?
A) They are identifiers that are supplied by the system, not the users.
B) They have no meaning to the users.
C) They are nonunique within a table.
D) They can be problematic when combining databases.
E) The DBMS will not allow their values to be changed.
Answer: C
Diff: 2 Page Ref: 306

35) One of the important properties of a column is whether or not it is ________.

A) found in more than one entity
B) required
C) character or numeric
D) subject to normalization
E) subject to denormalization
Answer: B
Diff: 2 Page Ref: 306

36) In a relational database design, all relationships are expressed by ________.

A) creating a primary key
B) creating a foreign key
C) creating a supertype
D) creating a subtype
E) creating a line between entities
Answer: B
Diff: 1 Page Ref: 304-307 Fig 5-1

37) Which of the following would be a reason to denormalize a relation?

A) Relax security
B) Lack of design time
C) End user preference
D) Improve performance
E) None of the above
Answer: D
Diff: 2 Page Ref: 309

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38) Which of the following is true about representing a weak entity with the relational model?
A) If the weak entity is existence-dependent, the key of the parent must be part of the key of the
weak entity.
B) If the strong entity has a minimum cardinality of 1, the key of the weak entity must be part of
the strong entity.
C) If the weak entity is ID-dependent, the key of the weak entity must be part of the key of the
parent entity.
D) If the weak entity is ID-dependent, the key of the parent entity must be part of the key of the
weak entity.
E) If the parent entity is existence-dependent, then the minimum cardinality of the weak entity is
zero.
Answer: D
Diff: 2 Page Ref: 311-312

39) Which of the following is true when representing a 1:1 binary relationship using the
relational model?
A) The key of the entity with the highest minimum cardinality must be placed in the other entity
as a foreign key.
B) The key of each entity must be placed in the other as a foreign key.
C) The key of either entity is placed in the other as a foreign key.
D) The key of the entity with the most attributes must be placed in the other entity as a foreign
key.
E) Both entities must have the same primary key.
Answer: C
Diff: 1 Page Ref: 315-317

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40) Given the tables

TABLE_A (Attribute1, Attribute2, Attribute3)

TABLE_B (Attribute4, Attribute5, Attribute6)

as shown in the figure below, which of the following would display the correct placement of
foreign keys in the relational model?

A) TABLE_A (Attribute1, Attribute2, Attribute3)

TABLE _B (Attribute4, Attribute5, Attribute6, Attribute1)
B) TABLE _A (Attribute1, Attribute2, Attribute3, Attribute4, Attribute5)
TABLE _B (Attribute4, Attribute5, Attribute6)
C) TABLE _A (Attribute1, Attribute2, Attribute3, Attribute4)
TABLE _B (Attribute4, Attribute5, Attribute6, Attribute1)
D) TABLE _A (Attribute1, Attribute2, Attribute3)
TABLE _B (Attribute4, Attribute5, Attribute6)
E) TABLE _A (Attribute1, Attribute2, Attribute3, Attribute6)
TABLE _B (Attribute4, Attribute5, Attribute6)
Answer: A
Diff: 2 Page Ref: 315-317

41) Which of the following is the correct technique for representing a 1:N relationship in the
relational model?
A) The key of the entity on the one side is placed into the relation for the entity on the many side.
B) The key of the child is placed into the relation of the parent.
C) The key of either relation can be placed into the other relation.
D) The key of the entity on the many side is placed into the relation for the entity on the one side.
E) An intersection relation is created, and the keys from both parent entities are placed as keys in
the intersection relation.
Answer: A
Diff: 2 Page Ref: 315-317

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42) Given the tables

PRODUCT (ProductID, Description, Cost)

SUPPLIER (SupplierID, ContactName, PhoneNumber)

as shown in the figure below, which of the following would represent the correct placement of
foreign keys?

A) PRODUCT (ProductID, Description, Cost)

SUPPLIER (SupplierID, ContactName, PhoneNumber)
B) PRODUCT (ProductID, Description, Cost)
SUPPLIER (SupplierID, ContactName, PhoneNumber, ProductID)
C) PRODUCT (ProductID, Description, Cost, SupplierID)
SUPPLIER (SupplierID, ContactName, PhoneNumber, ProductID)
D) PRODUCT (ProductID, Description, Cost, ContactName)
SUPPLIER (SupplierID, ContactName, PhoneNumber)
E) PRODUCT (ProductID, Description, Cost, SupplierID)
SUPPLIER (SupplierID, ContactName, PhoneNumber)
Answer: E
Diff: 2 Page Ref: 315-317

43) Which of the following is the correct technique for representing a M:N relationship using the
relational model?
A) An intersection relation is created, and the key of either entity is placed as a key in both the
intersection relation and in the other relation.
B) An intersection relation is created with a surrogate key, which is placed in each of the parent
entities.
C) An intersection relation is created, and the keys of both parent entities are placed as a
composite key in the intersection relation.
D) The key from either relation is placed as a foreign key in the other relation.
E) None of the above
Answer: C
Diff: 2 Page Ref: 317-319

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44) Given the tables

PRODUCT (ProductID, Description, Cost)

SUPPLIER (SupplierID, ContactName, PhoneNumber)

as shown in the figure below, which of the following would represent the correct placement of
foreign keys?

A) PRODUCT (ProductID, Description, Cost)

SUPPLIER (SupplierID, ContactName, PhoneNumber)
B) PRODUCT (ProductID, Description, Cost, SupplierID)
SUPPLIER (SupplierID, ContactName, PhoneNumber, ProductID)
C) PRODUCT (ProductID, Description, Cost)
SUPPLIER (SupplierID, ContactName, PhoneNumber)
PRODUCT_SUPPLIER (ProductID, SupplierID)
D) PRODUCT (ProductID, Description, Cost, SupplierID)
SUPPLIER (SupplierID, ContactName, PhoneNumber)
E) PRODUCT (ProductID, Description, Cost)
SUPPLIER (SupplierID, ContactName, PhoneNumber)
PRODUCT_SUPPLIER (ProductID, SupplierID)
Answer: E
Diff: 3 Page Ref: 317-319

45) In many-to-many relationships in a relational database design, ________.

A) the key of the child is placed as a foreign key into the parent
B) the key of the parent is placed as a foreign key into the child
C) the keys of both tables are placed in a third table
D) the keys of both tables are joined into a composite key
E) C and D
Answer: E
Diff: 2 Page Ref: 317-320

46) In many-to-many relationships in a relational database design, ________.

A) the intersection table is ID-dependent on one of the parents
B) the intersection table is ID-dependent on both of the parents
C) the minimum cardinality from the intersection table to the parents is always M
D) A and B
E) B and C
Answer: E
Diff: 3 Page Ref: 317-320

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47) In relational database design, ID-dependent entities are used to ________.
A) represent 1:1 relationships
B) represent 1:N relationships
C) represent N:M relationships
D) handle recursive relationships
E) eliminate the need for weak entities being converted to tables.
Answer: C
Diff: 3 Page Ref: 317-320

48) When transforming an E-R data model into a relational database design, the key of the parent
entity should be placed as part of the primary key into the child entity ________.
A) when the child entity is ID-dependent
B) when the child entity is non-ID-dependent
C) when the child entity has a 1:1 relationship with the parent entity
D) when the child entity has a 1:N relationship with the parent entity
E) when the child entity has a recursive relationship with the parent entity
Answer: A
Diff: 3 Page Ref: 313-326

49) When transforming an ID-dependent E-R data model relationship into a relational database
design and the child entity is designed to use a surrogate key, then ________.
A) the parent entity must also use a surrogate key
B) the relationship remains an ID-dependent relationship
C) the relationship changes to a non-ID-dependent relationship
D) A and B
E) A and C
Answer: C
Diff: 2 Page Ref: 313-326

50) What relationship pattern is illustrated in the following schema?

PRODUCT (ProductID, Description)

SUPPLIER (SupplierID, ContactName, PhoneNumber)
PRODUCT_SUPPLIER (ProductID, SupplierID, Cost)

ProductID in PRODUCT_SUPPLIER must exist in ProductID in PRODUCT

SupplierID in PRODUCT_SUPPLIER must exist in SupplierID in PRODUCT
A) Association relationship
B) Intersection relationship
C) Recursive relationship
D) Strong entity relationship
E) Supertype/subtype relationship
Answer: A
Diff: 2 Page Ref: 320-322

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51) What relationship pattern is illustrated in the following schema?

VEHICLE (VehicleID, Cost)

CAR (VehicleID, NumberOfSeats)
TRUCK (VehicleID, CargoCapacity)

VehicleID in CAR must exist in VehicleID in VEHICLE

VehicleID in TRUCK must exist in VehicleID in VEHICLE
A) Association relationship
B) Intersection relationship
C) Recursive relationship
D) Strong entity relationship
E) Supertype/subtype relationship
Answer: E
Diff: 2 Page Ref: 322

52) Which of the following is not true about representing subtypes in a relational database
design?
A) One table is created for the supertype and one for each subtype.
B) All of the attributes of the supertype are added to the subtype relations.
C) The key of the supertype is made the key of the subtypes.
D) A subtype and its supertype are representations of the same underlying table.
E) An instance of the supertype may be related to one instance each of several subtypes.
Answer: B
Diff: 2 Page Ref: 322-323

53) What relationship pattern is illustrated in the following schema?

EMPLOYEE (EmployeeID, OfficePhone, Manager)

Manager in EMPLOYEE must exist in EmployeeID in EMPLOYEE

A) Association relationship
B) Intersection relationship
C) Recursive relationship
D) Strong entity relationship
E) Supertype/subtype relationship
Answer: C
Diff: 2 Page Ref: 322-326

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54) Which of the following is not true of recursive relationships?
A) When the recursive relationship is M:N, an intersection table is created.
B) The rows of a single table can play two different roles.
C) The techniques for representing the tables are the same as for non-recursive relationships
except the rows are in the same table.
D) Recursive relationships can be 1:1, 1:N, or M:N relationships.
E) Even when the relationship is 1:N, a new table must be defined to represent the relationship.
Answer: E
Diff: 2 Page Ref: 322-326

55) Microsoft Access does not create N:M relationships because:

A) Microsoft Access creates databases based on database designs instead of data models.
B) Microsoft Access creates databases based on data models instead of database designs.
C) Microsoft Access cannot implement association relationships.
D) Microsoft Access cannot implement supertype/subtype relationships.
E) Microsoft Access cannot implement recursive relationships.
Answer: A
Diff: 3 Page Ref: 334

56) As far as Microsoft Access is concerned, there are no:

A) 1:1 relationships.
B) 1:N relationships.
C) N:1 relationships.
D) N:M relationships.
E) recursive relationships.
Answer: D
Diff: 2 Page Ref: 334

57) By default, when Microsoft Access creates a relationship between two tables, it creates a(n):
A) 1:1 relationship.
B) 1:N relationship.
C) N:M relationship.
D) association relationship.
E) recursive relationship.
Answer: B
Diff: 2 Page Ref: 337-338

58) To create a 1:1 relationship between two tables in Microsoft Access:

A) the Indexed property of the foreign key column must be set to No.
B) the Indexed property of the foreign key column must be set to Yes (Duplicates OK).
C) the Indexed property of the foreign key column must be set to Yes (No Duplicates).
D) the Data Type of the foreign key column must be set to AutoNumber.
E) the Smart Tag property of the foreign key column must be set to Foreign Key.
Answer: C
Diff: 2 Page Ref: 337-338

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59) After a 1:1 relationship has been created between two tables in Microsoft Access, the
Relationship Type of One-To-One appears:
A) in the Relationship Type property of the primary key column in table Design View.
B) in the Relationship Type property of the foreign key column in table Design View.
C) in the table object in the Relationships window.
D) in the Show Table dialog box.
E) in the Edit Relationships dialog box.
Answer: E
Diff: 2 Page Ref: 337-342

60) The first step of database design is to define a table for each ________.
Answer: entity
Diff: 1 Page Ref: 305

61) Once a table has been defined, it should be examined according to ________ criteria.
Answer: normalization
Diff: 2 Page Ref: 306-307

62) There are cases where it is possible to normalize a table too far, in which case there may be a
need for ________.
Answer: denormalization
Diff: 2 Page Ref: 309

63) To normalize a relation, the determinant of every functional dependency should be a(n)
________.
Answer: candidate key
Diff: 2 Page Ref: 307

64) To represent a many-to-many relationship in the relational model, a(n) ________ table is
used.
Answer: intersection
Diff: 1 Page Ref: 317-319

65) For a(n) ________ weak entity, it is necessary to add the key of the parent entity to the weak
entity's relation so that this added attribute becomes part of the weak entity's key.
Answer: ID-dependent
Diff: 2 Page Ref: 319-322

66) A(n) ________ is a relationship among entities of the same class.

Answer: recursive relationship
Diff: 2 Page Ref: 322

67) Microsoft Access does not create N:M relationships because Microsoft Access creates
databases based on ________.
Answer: database designs
Diff: 2 Page Ref: 334

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68) As far as Microsoft Access is concerned, there are no ________.
Answer: N:M relationships
Diff: 2 Page Ref: 334

69) By default, when Microsoft Access creates a relationship between two tables it creates a(n)
________ relationship.
Answer: 1:N
Diff: 2 Page Ref: 334

70) To create a 1:1 relationship between two tables in Microsoft Access, the Indexed property of
the foreign key column must be set to ________.
Answer: Yes (No Duplicates)
Diff: 3 Page Ref: 337-338

71) After a 1:1 relationship has been created between two tables in Microsoft Access, the
Relationship Type of One-To-One appears in the ________.
Answer: Edit Relationships dialog box
Diff: 3 Page Ref: 337-338

72) Explain the process of representing an entity using the relational model.
Answer: To represent an entity using the relational model, first define a relation for the entity.
The name of the entity is the name of the relation. Each attribute in the entity becomes a column
in the relation. If the entity had a unique identifier that would make an appropriate primary key,
then that attribute is made the key. If there is no such attribute, then the development team
discusses identifiers with the users to determine if an acceptable key attribute exists. If not, then
a surrogate key may be used. Finally, the relation is evaluated against the normalization criteria.
Changes to the design may be necessary to satisfy the normalization requirements.
Diff: 2 Page Ref: 305-307

73) Explain the pragmatic reason for using surrogate keys.

Answer: Primary keys are commonly included in indexes, and are used to identify records to be
retrieved by users. The ideal primary key is short, numeric and fixed. When the primary key
contains a lengthy text field, this creates a large amount of duplicated data that must be
frequently manipulated. For these reasons, it is often practical to use a surrogate key that is
generated by the system and is relatively small and easy to manipulate.
Diff: 1 Page Ref: 306

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74) What is denormalization and why can it be desirable?
Answer: Denormalization is the process of consolidating relations that are in a higher normal
form into a single relation that is in a lower normal form, thereby making it susceptible to more
anomalies. Denormalization can be desirable for two reasons. First, sometimes the act of
normalizing a table can cause it to become too cumbersome or contrived, which makes it
difficult to work with. Therefore denormalization may be preferable when the dangers of the
additional anomalies are considered to be of less importance than the ease of working with the
relations. Second, processing multiple tables requires more processing overhead than processing
a single table. Therefore, performance can often be improved by using a single, denormalized
table than multiple, normalized tables. When the performance gain outweighs the dangers of the
additional anomalies, the denormalized table may be preferred.
Diff: 2 Page Ref: 309-310

75) Explain the representation of ID-dependent weak entities using the relational model.
Answer: A weak entity is represented in the relational model similarly to the representation of a
strong entity. First, a relation is defined for the entity. Each attribute in the weak entity becomes
a column in the relation. The primary key of the strong entity on which the weak entity is ID-
dependent is added to the relation and is made a part of the primary key of the weak entity's
relation along with the weak entity's identifier. Finally, the relation is evaluated according to the
normalization criteria, and any necessary design changes are made to normalize the relation.
Diff: 2 Page Ref: 311-312

76) Explain the representation of a one-to-many strong entity relationship in a relational database
design.
Answer: One-to-many relationships are represented by placing the primary key of the table on
the one side of the relationship into the table on the many side of the relationship as a foreign
key. The term "parent" refers to the table on the one side of a 1:N relationship, and the term
"child" refers to the table on the many side of the 1:N relationship. Therefore, the rule for
representing a one-to-many relationship can be summarized as "Place the key of the parent table
in the child table as a foreign key."
Diff: 2 Page Ref: 315-317

77) Explain the representation of a many-to-many strong entity relationship in a relational

database design.
Answer: Many-to-many relationships cannot be directly represented in a relational database
design. Therefore, many-to-many relationships are essentially broken into two one-to-many
relationships by creating an intersection table that represents the relationship itself. The
intersection table takes its key as a combination of the keys of the two original, or parent,
entities. Each of the parent entities has a one-to-many relationship with the intersection table that
is represented by placing the keys of the parents into the intersection table.
Diff: 2 Page Ref: 317-320

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78) What is an association relationship, and how does it differ from an N:M relationship?
Answer: An association relationship is very similar to an N:M relationship except that the
intersection table has attributes of its own. This means that in addition to the foreign key fields
linking to the two strong entities, there is at least one additional field in what would otherwise be
called the intersection table but is now an association table. For example, the intersection table
ENROLLMENT for STUDENT and CLASS showing student enrollment in each class would
normally have two columns: StudentID and ClassID. However, we can turn this intersection
table into an association table by adding the column Grade, which records each student's grade in
each class.
Diff: 2 Page Ref: 320-322

79) Write the schema to represent the entities below, including tables, the proper placement of
the foreign key, and referential integrity constraint.

Answer:
STUDENT (StudentID, StuName, StuMajor, StuPhone, AdvisorID)
ADVISOR (AdvisorID, AdvName, AdvOffice, AdvPhone)

AdvisorID in STUDENT must exist in AdvisorID in ADVISOR.

Diff: 2 Page Ref: 310-317

80) How are one-to-one recursive relationships addressed using the relational model?
Answer: One-to-one recursive relationships are addressed just the same as one-to-one
nonrecursive relationships. The only difference is that both of the related entity instances are in
the same entity class. The key of either instance is placed in the other instance as a foreign key.
In the case of a recursive relationship, this means that a new attribute is added to the entity class
with the recursive relationship. For each instance, this new attribute will contain the value of the
key attribute of the instance that is related.
Diff: 2 Page Ref: 322-326

81) How are 1:1, 1:N and N:M relationships handled in Microsoft Access?
Answer: By default, when Microsoft Access creates a relationship between two tables, it creates
a 1:N relationship. N:M relationships are created in Microsoft Access, as in all other DBMS
products, as two 1:N relationships linking the two tables (based on the original two entities)
through an intersection table. As far as Microsoft Access is concerned, there are no N:M
relationships! To create a 1:1 relationship between two tables in Microsoft Access, the Indexed
property of the foreign key column in the table containing the foreign key must be set to Yes (No
Duplicates) before the relationship is created. With the property set, the relation is automatically
created as a 1:1 relationship.
Diff: 2 Page Ref: 334-340
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