# Domain and Range of a Relation

In domain and range of a relation, if R be a relation from set A to set B, then

• The set of all first components of the ordered pairs belonging to R is called the domain of R.
Thus, Dom(R) = {a ∈ A: (a, b) ∈ R for some b ∈ B}.

• The set of all second components of the ordered pairs belonging to R is called the range of R.

Thus, range of R = {b ∈ B: (a, b) ∈R for some a ∈ A}.

Therefore, Domain (R) = {a : (a, b) ∈ R} and Range (R) = {b : (a, b) ∈ R}

Note:

The domain of a relation from A to B is a subset of A.

The range of a relation from A to B is a subset of B.

For Example:

If A = {2, 4, 6, 8)   B = {5, 7, 1, 9}.

Let R be the relation ‘is less than’ from A to B. Find Domain (R) and Range (R).

Solution:

Under this relation (R), we have

R = {(4, 5); (4, 7); (4, 9); (6, 7); (6, 9), (8, 9) (2, 5) (2, 7) (2, 9)}

Therefore, Domain (R) = {2, 4, 6, 8} and Range (R) = {1, 5, 7, 9}

### Solved examples on domain and range of a relation:

1. In the given ordered pair (4, 6); (8, 4); (4, 4); (9, 11); (6, 3); (3, 0); (2, 3) find the following relations. Also, find the domain and range.

(a) Is two less than

(b) Is less than

(c) Is greater than

(d) Is equal to

Solution:

(a) R₁ is the set of all ordered pairs whose 1ˢᵗ component is two less than the 2ⁿᵈ component.

Therefore, R₁ = {(4, 6); (9, 11)}

Also, Domain (R₁) = Set of all first components of R₁ = {4, 9} and Range (R₂) = Set of all second components of R₂ = {6, 11}

(b) R₂ is the set of all ordered pairs whose 1ˢᵗ component is less than the second component.

Therefore, R₂ = {(4, 6); (9, 11); (2, 3)}.

Also, Domain (R₂) = {4, 9, 2} and Range (R₂) = {6, 11, 3}

(c) R₃ is the set of all ordered pairs whose 1ˢᵗ component is greater than the second component.

Therefore, R₃ = {(8, 4); (6, 3); (3, 0)}

Also, Domain (R₃) = {8, 6, 3} and Range (R₃) = {4, 3, 0}

(d) R₄ is the set of all ordered pairs whose 1ˢᵗ component is equal to the second component.

Therefore, R₄ = {(3, 3)}

Also, Domain (R) = {3} and Range (R) = {3}

2. Let A = {2, 3, 4, 5} and B = {8, 9, 10, 11}.

Let R be the relation ‘is factor of’ from A to B.

(a) Write R in the roster form. Also, find Domain and Range of R.

(b) Draw an arrow diagram to represent the relation.

Solution:

(a) Clearly, R consists of elements (a, b) where a is a factor of b.

Therefore, Relation (R) in the roster form is R = {(2, 8); (2, 10); (3, 9); (4, 8), (5, 10)}

Therefore, Domain (R) = Set of all first components of R = {2, 3, 4, 5} and Range (R) = Set of all second components of R = {8, 10, 9}

(b) The arrow diagram representing R is as follows:

3. The arrow diagram shows the relation (R) from set A to set B. Write this relation in the roster form.

Solution:

Clearly, R consists of elements (a, b), such that ‘a’ is square of ‘b’
i.e., a = b².

So, in roster form R = {(9, 3); (9, -3); (4, 2); (4, -2); (16, 4); (16, -4)}

### Worked-out problems on domain and range of a relation:

4. Let A = {1, 2, 3, 4, 5} and B = {p, q, r, s}. Let R be a relation from A in B defined by
R = {1, p}, (1, r), (3, p), (4, q), (5, s), (3, p)}

Find domain and range of R.

Solution:

Given R = {(1, p), (1, r), (4, q), (5, s)}

Domain of R = set of first components of all elements of R = {1, 3, 4, 5}

Range of R = set of second components of all elements of R = {p, r, q, s}

5. Determine the domain and range of the relation R defined by

R = {x + 2, x + 3} : x ∈ {0, 1, 2, 3, 4, 5}

Solution:

Since, x = {0, 1, 2, 3, 4, 5}

Therefore,

x = 0 ⇒ x + 2 = 0 + 2 = 2 and x + 3 = 0 + 3 = 3

x = 1 ⇒ x + 2 = 1 + 2 = 3 and x + 3 = 1 + 3 = 4

x = 2 ⇒ x + 2 = 2 + 2 = 4 and x + 3 = 2 + 3 = 5

x = 3 ⇒ x + 2 = 3 + 2 = 5 and x + 3 = 3 + 3 = 6

x = 4 ⇒ x + 2 = 4 + 2 = 6 and x + 3 = 4 + 3 = 7

x = 5 ⇒ x + 2 = 5 + 2 = 7 and x + 3 = 5 + 3 = 8

Hence, R = {(2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 8)}

Therefore, Domain of R = {a : (a, b) ∈R} = Set of first components of all ordered pair belonging to R.

Therefore, Domain of R = {2, 3, 4, 5, 6, 7}

Range of R = {b : (a, b) ∈ R} = Set of second components of all ordered pairs belonging to R.

Therefore, Range of R = {3, 4, 5, 6, 7, 8}

6. Let A = {3, 4, 5, 6, 7, 8}. Define a relation R from A to A by

R = {(x, y) : y = x - 1}.

• Depict this relation using an arrow diagram.

• Write down the domain and range of R.

Solution:

By definition of relation

R = {(4, 3) (5, 4) (6, 5)}

The corresponding arrow diagram is shown.

We can see that domain = {4, 5, 6} and Range = {3, 4, 5}

7. The adjoining figure shows a relation between the sets A and B.

Write this relation in

• Set builder form

• Roster form

• Find the domain and range

Solution:

We observe that the relation R is 'a’ is the square of ‘b'.

In set builder form R = {(a, b) : a is the square of b, a ∈ A, b ∈ B}

In roster form R = {(4, 2) (4, -2)(9, 3) (9, -3)}

Therefore, Domain of R = {4, 9}

Range of R = {2, -2, 3, -3}

Note: The element 1 is not related to any element in set A.

Relations and Mapping

Ordered Pair

Cartesian Product of Two Sets

Relation

Domain and Range of a Relation

Functions or Mapping

Domain Co-domain and Range of Function

Relations and Mapping - Worksheets

Worksheet on Math Relation

Worksheet on Functions or Mapping

Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.

## Recent Articles

1. ### Estimating Sum and Difference | Reasonable Estimate | Procedure | Math

May 22, 24 06:21 PM

The procedure of estimating sum and difference are in the following examples. Example 1: Estimate the sum 5290 + 17986 by estimating the numbers to their nearest (i) hundreds (ii) thousands.

2. ### Round off to Nearest 1000 |Rounding Numbers to Nearest Thousand| Rules

May 22, 24 06:14 PM

While rounding off to the nearest thousand, if the digit in the hundreds place is between 0 – 4 i.e., < 5, then the hundreds place is replaced by ‘0’. If the digit in the hundreds place is = to or > 5…

3. ### Round off to Nearest 100 | Rounding Numbers To Nearest Hundred | Rules

May 22, 24 05:17 PM

While rounding off to the nearest hundred, if the digit in the tens place is between 0 – 4 i.e. < 5, then the tens place is replaced by ‘0’. If the digit in the units place is equal to or >5, then the…

4. ### Round off to Nearest 10 |How To Round off to Nearest 10?|Rounding Rule

May 22, 24 03:49 PM

Round off to nearest 10 is discussed here. Rounding can be done for every place-value of number. To round off a number to the nearest tens, we round off to the nearest multiple of ten. A large number…