FINDING INVERSE OF A RELATION FROM THE GIVEN RELATION

Question 1 :

Let A be the set of first ten natural numbers and let R be a relation defined by (x, y) ∈ R <=> x + 2y  =  10 where R = {(x,y), x ∈ A, y ∈ A and x + 2y  =  10. Express R and R^-1 as the set of ordered pairs. Also determine 

(i)  Domain of R and R^-1

(ii)  Range of R and R^-1

Solution :

To find the set of ordered pairs in the relation, let us apply the values of x one by one in the given function.

x + 2y  =  10

y  =  (10 - x)/2

A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

R = {(2, 4) (4, 3) (6, 2) (8, 1) (10, 0)}

Domain of R  =  {2, 4, 6, 8, 10}

Range of R  =  {4, 3, 2, 1, 0}

R^-1  =  {(4, 2) (3, 4) (2, 6) (1, 8) (0, 10)}

Domain of R^-1  =  {4, 3, 2, 1, 0}

Range of R^-1  =  {2, 4, 6, 8, 10}

Question 2 :

A relation R is defined from a set A  =  {2, 3, 4, 5} to a set B  = {3, 6, 7, 10} as follows (x, y) ∈ R => x divides y. Express R as the set of ordered pairs and determine the domain and range of R also find R^-1.

Solution :

R  =  {(2, 6) (2, 10) (3, 3) (3, 6) (5, 10)}

Domain of R  =  {2, 3, 5}

Range of R  =  {3, 6, 10}

R^-1  =  {(6, 2) (10, 2) (3, 3) (6, 3) (10, 5)}

Question 3 :

Let R be relation in N defined by (x, y) ∈ R <=> x + 2y  =  8. Express R and R^-1 as a set of ordered pairs.

Solution :

x = 0, 1, 2, ...........

y  =  (8 - x)/2

R = {(2, 3) (4, 2) (6, 1)}

R^-1  =  {(3, 2) (2, 4) (1, 6)}

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