LEARN MATH STEP BY STEP THROUGH VERY EASY PROCESS

RELAION & FUNCTION - INVERSE OF A RELATION

**The Inverse of a Relation –**

**For any binary relation R, a
second relation can be constructed by merely interchanging first and second
components in every ordered pair. The relation thus obtained is called the
inverse of the first one and is designed by R‾¹. Symbolically –**

** R‾¹ = {(y, x),
(x, y) ∈ R} **

**Thus, domain (R‾¹) = Range (R)
and Range (R‾¹) = Domain (R)**

__Illustration.-__

**(i) The inverse
of the husband-wife relation is the wife husband relation.**

**(ii) Let, R = {(2, 1), (3, 2),
(4, 3), (4, 5)}**

**Domain (R) = (2, 3, 4), Range
(R) = {1, 2, 3, 5}**

**=> Domain (R‾¹) = Range (R) =
{1, 2, 3, 5}, and **

**Range (R‾¹) = Domain (R) = {2,
3, 4}**

**Hence, R‾¹ = {(1, 2), (2, 3),
(3, 4), (5, 4)}**

**(iii) If, R = {(x, y) ǀ x = y + 2} and S = {(x, y) ǀ y
= x²} then**

**R‾¹ = {(x, y) ǀ y = x + 2} and
S‾¹ = {(x, y) ǀ x = y²}**

**It can be easily shown that
(R‾¹)‾¹ = R**

**Example.1) Let, R be the relation
from A = {1, 2, 3, 4, 5, 6}, to B = {1,
3, 5} which is defined by the open sentence “x is less than y”**

**(i) Find the solution set R, that
is, write R as a set of ordered pairs.**

**(ii) Plot R on a co-ordinate diagram
of A X B.**

**(iii) State the domain, range and
co-domain of R.**

**Ans.) **

**(i) R consists of those ordered pairs (a, b) ∈
A X B for which a < b, hence **

** R = {(1, 3), (1, 5), (2, 3), (2,
5), (3, 5), (4, 5)}**

**(ii) R is
sketched on the co-ordinate diagram of A
X B as shown in the figure**

**(iii) Domain of R = {1, 2, 3, 4}**

** Range of R = {3, 5}**

** Codomain of R = {1, 3, 5}**

**Example.2) Let A = {a, b, c, d}
and B = {x, y, x}. Which of the
following are relations from A to B ?**

**(i) {(a, y), (a, z), (c, x), (d, y)}**

**(ii) {(a, x), (b, y), (c, x), (a,
d)}**

**(iii) {(a, x), (y, d), (x, c)}**

**(iv) {(y, a), (z, a), (z, c), (y, d)}**

**(v) {(a, x), (x, a), (b, y), (y, b)}**

**(vi) {(a, x), (b, y), (c, z), z}**

**(vii) {a, b, x, y, z}**

**Ans.) (i) Yes,**

**(ii) No, because in the ordered
pair (a, d), a ∈ A and d ∉ B**

**(iii) No, because in (y, d), y ∈
B.**

**(iv) No, because here the first
entries in all the ordered pairs are in the set B.**

**(v) No.**

**(vi) No, because the element z
is not an ordered pair.**

**(vii) No, because the elements
of the set are not ordered pairs.**