ICSE Class 8 Maths Chapter 20 Relations and Mappings

Relations And Mappings

In Class VII, you learnt the basic ideas about relations and mappings. In this chapter, we shall strengthen these concepts and introduce an idea of equation form of functions.

Relations

Let A and B be two (non-empty) sets, then a relation R from A to B is a rule which associates elements of set A to elements of set B.

If an element a of A is associated by R to an element b of B, we say that a is related to b and we write it as a R b.

For Example

(i) Let A = {2, 3, 5, 7}, B = {3, 6} and R be the relation 'is less than' from A to B, then we get

2 R 3, 2 R 6, 3 R 6, 5 R 6.

(ii) Let A = {-1, 0, 3, -4, -5, 5}, B = {0, -3, -12, -15, 15, 8, 10} and R be the relation 'is one-third of' from A to B, then we get

-1 R -3, 0 R 0, -4 R -12, -5 R -15, 5 R 15.

Representation Of A Relation

We already know two methods of representing a relation.

Roster Form

In this form, a relation R from A to B is represented by the set of all ordered pairs (a, b) where a ∈ A, b ∈ B which satisfy the given relation R i.e.

R = {(a, b) : a ∈ A, b ∈ B, a R b}.

In the above example, relations R from A to B in the roster form can be written as

(i) R = {(2, 3), (2, 6), (3, 6), (5, 6)}

(ii) R = {(-1, -3), (0, 0), (-4, -12), (-5, -15), (5, 15)}.

Arrow Diagram

In this form, a relation R from A to B is represented by drawing arrows from first components to second components of all ordered pairs which satisfy the given relation R.

In the above example, relations R from A to B can be represented by arrow diagrams shown in the figures given below:

(i) A with elements {2, 3, 5, 7} connected by relation R to B with elements {3, 6}.

(ii) A with elements {-1, 0, 3, -4, -5, 5} connected by relation R to B with elements {0, -3, -12, -15, 15, 8, 10}.

Teacher's Note

Relations help us understand how different objects or numbers are connected. For example, a student's roll number is related to their test score - it's a real-world mapping of one set to another.

Domain And Range Of A Relation

Let A and B be two (non-empty) sets and R be a relation from A to B, then domain of R = the set of first components of all ordered pairs which belong to R, and range of R = the set of second components of all ordered pairs which belong to R.

In the above example,

(i) Domain of R = {2, 3, 5} and range of R = {3, 6}

(ii) Domain of R = {-1, 0, -4, -5, 5} and range of R = {-3, 0, -12, -15, 15}.

Let A = {2, 3, 4, 5}, B = {4, 8, 10, 11, 15, 21} and R be the relation 'is a prime factor of' from A to B, then

(i) write R in the roster form.

(ii) find domain and range of R.

(iii) represent R by an arrow diagram.

Solution. Given A = {2, 3, 4, 5}, B = {4, 8, 10, 11, 15, 21} and R is the relation 'is a prime factor of' from A to B.

(i) To write R in roster form, we form ordered pairs (a, b), a ∈ A, b ∈ B such that a is a prime factor of b. We get

R = {(2, 4), (2, 8), (2, 10), (3, 15), (3, 21), (5, 10), (5, 15)}.

(ii) Domain of R = {2, 3, 5} and range of R = {4, 8, 10, 15, 21}.

(iii) The given relation R from A to B can be represented by an arrow diagram shown in the adjoining figure.

Let A = {2, 3, 4, 5, 7, 9}, B = {0, 1, 2, 3, 4, 5, 6} and R = {(a, b) : a ∈ A, b ∈ B and a + b = 10}, then

(i) find R in roster form. (ii) find domain and range of R.

Solution. Given A = {2, 3, 4, 5, 7, 9}, B = {0, 1, 2, 3, 4, 5, 6} and R = {(a, b) : a ∈ A, b ∈ B and a + b = 10}

(i) To find R in roster form, we note that

4 + 6 = 10, 5 + 5 = 10, 7 + 3 = 10, 9 + 1 = 10

∴ R = {(4, 6), (5, 5), (7, 3), (9, 1)}.

(ii) Domain of R = {4, 5, 7, 9} and range of R = {6, 5, 3, 1}.

Mapping

Let A and B be two (non-empty) sets, then a mapping from A to B is a rule which associates to every element of set A, a unique element of set B.

The unique element b of B associated with an element a of A is called the image of a.

Thus, a mapping from A to B is a special relation from A to B such that each element of A has a unique image in B.

Teacher's Note

Mappings are like assigning a seat to each student in a classroom - every student gets exactly one seat, establishing a one-to-one correspondence between students and seating positions.

Conditions For A Relation To Be A Mapping

(A) When the relation is represented in roster form

Let A and B be two (non-empty) sets and R be a relation from A to B, then R is a mapping if

(i) domain of R = A and

(ii) first components of different ordered pairs of R are not repeated.

Note that second components of different ordered pairs of R may be repeated, and some elements of B may not appear as a second component in ordered pairs of R.

For Example

(1) Let A = {0, -1, 2, 5}, B = {0, -1/2, 1/2, 5, 2, 5} and R be the relation 'is reciprocal of' from A to B, then

R = {(-1, -1), (2, 1/2), (5, 1/5)}.

Here, domain of R = {-1, 2, 5} ≠ A

Therefore, R is not a mapping from A to B.

(2) Let A = {1, -1, 2, 3, 4, 5}, B = {1, 1, 4, 9, 16, 25, 36} and R be the relation 'has as its square' from A to B, then

R = {(1, 1), (-1, 1), (2, 4), (3, 9), (4, 16), (5, 25)}.

Here, domain of R = {1, -1, 2, 3, 4, 5} = A and the first components of different ordered pairs of R are not repeated.

Therefore, R is a mapping from A to B.

(3) Let A = {6, 8, 14, 20}, B = {3, 4, 5, 6, 7, 8, 9} and R be the relation 'is a multiple of' from A to B, then

R = {(6, 3), (6, 6), (8, 4), (8, 8), (14, 7), (20, 4), (20, 5)}.

Here, domain of R = {6, 8, 14, 20} = A, but the first components of different ordered pairs (6, 3) and (6, 6) of R are repeated, therefore, R is not a mapping from A to B.

(B) When the relation is represented by an arrow diagram

Let A and B be two (non-empty) sets and R be a relation from A to B, then R is a mapping if

(i) each element of A is connected with some element of B i.e. each element of A must have an arrow emanating from it, and

(ii) no element of A should be connected with more than one element of B i.e. no element of A should have more than one arrow emanating from it.

Note that arrows from different elements of A may be pointing to the same element of B, and some elements of B may not have arrows pointing at them.

For Example

(1) The relation represented by the adjoining arrow diagram is a mapping from A to B because each element of A is connected to a unique element of B.

(2) In the adjoining arrow diagram, an element b of A is not connected to any element of B i.e. element b of A is not associated to any element of B. Therefore, the relation represented by the adjoining arrow diagram is not a mapping from A to B.

(3) The relation represented by the adjoining arrow diagram is a mapping from A to B because each element of A is connected to a unique element of B. Note that the elements a and d of A are associated to the same element 2 of B.

(4) In the adjoining arrow diagram, an element b of A is connected to two elements 1 and 4 of B. So, each element of A is not associated to a unique element of B. Therefore, the relation represented by the adjoining arrow diagram is not a mapping from A to B.

Let A = {1, 2, 3, 4}, B = {1, 3, 4, 5} and R be the relation 'is greater than or equal to' from A to B, then

(i) write R in roster form.

(ii) find domain and range of R.

(iii) represent R by an arrow diagram.

(iv) Is R a mapping from A to B? Give reason for your answer.

Solution. Given A = {1, 2, 3, 4}, B = {1, 3, 4, 5} and R is the relation 'is greater than or equal to' from A to B.

(i) To write R in roster form, we form ordered pairs (a, b), a ∈ A, b ∈ B such that a ≥ b. We get

R = {(1, 1), (2, 1), (3, 1), (3, 3), (4, 1), (4, 3), (4, 4)}

(ii) Domain of R = {1, 2, 3, 4} and range of R = {1, 3, 4}

(iii) The given relation R from A to B can be represented by an arrow diagram shown in the adjoining figure.

(iv) The given relation R from A to B is not a mapping because element 3 of A is associated to two elements 1 and 3 of B.

Let A = {4, 5, 6, 7, 8}, B = {0, 1, 2, 3} and R = {(a, b) : a ∈ A, b ∈ B and a - b is divisible by 3}, then

(i) describe R in roster form.

(ii) find domain and range of R.

(iii) represent R by an arrow diagram.

(iv) Is R a mapping from A to B? Give reason for your answer.

Solution. Given A = {4, 5, 6, 7, 8}, B = {0, 1, 2, 3} and R = {(a, b) : a ∈ A, b ∈ B and a - b is divisible by 3}

(i) R = {(4, 1), (5, 2), (6, 0), (6, 3), (7, 1), (8, 2)}.

(ii) Domain of R = {4, 5, 6, 7, 8} and range of R = {0, 1, 2, 3}.

(iii) The given relation R from A to B can be represented by an arrow diagram shown in the adjoining figure.

(iv) The given relation R from A to B is not a mapping because the element 6 of A is associated to two elements 0 and 3 of B.

Teacher's Note

Understanding mappings helps in computer programming where functions map inputs to outputs - each input must produce exactly one output for the function to work correctly.

This is a preview of the first 3 pages. To get the complete book, click below.