**Domain and Range of Relations**

Domain and Range of Relations: Let R be a relation from A to B. The domain of R is the set of all those elements a ϵ A such that (a, b) ϵ R for some b ϵ R for some b ϵ B.

Domain of R = {a ϵ A: (a, b) ϵ R, ∀ b ϵ B}

And range of R is the set of all those elements b ϵ B such that (a, b) ϵ R for some b ϵ B such that (a, b) ϵ R for some a ϵ A.

Range of R = {b ϵ B: (a, b) ϵ R, ∀ a ϵ R},

Here, B is called the codomain of R

**Example: **Let A = {1, 2, 3} & B = {3, 5, 6}

aRb ⇒ a< b then, R = {(1, 5), (2, 5), (3, 5), (1, 6), (2, 6), (3, 6)}

Domain of R = {1, 2, 3}

Rage of R = {5, 6}

And codomain of R = {3, 5, 6}

**Note:
**Let A and B be two non empty finite sets having p and q
elements respectively

Total number of relations from A to B = 2^{pq}

**Example
1**: find the domain and range of the relation R defined by R = {x, x + 5}: x ϵ {0, 1,
2, 3, 4, 5}.

**Solution: **Given that R = {x, x+5}: x ϵ {0, 1,
2, 3, 4, 5} … (1)

Putting x = 0, 1, 2, 3, 4, 5, we get

Domain = {0, 1, 2, 3, 4, 5}

After putting the value of x from equation (1), we get

y = x + 5

= 5, 6, 7, 8, 9, 10

Range = {5, 6, 7, 8, 9, 10}

**Example 2:** The relation R defined on the best
of natural numbers as {(a, b): a differs from b by 3}, is given by

**Solution:** Given that

{(a, b): a differs from b by 3}

{(a, b): a – b = 3}

Put a = 4, 5, 6, . . .

a = 4

4 – b = 3

b = 1

(a, b) = (4, 1)

a = 5

5 – b = 3

b = 2

(a, b) = (5, 2)

a = 6

6 – b = 3

b = 3

(a, b) = (6, 3)

= {(4, 1), (5, 2), (6, 3), . . .}.