Question

Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set
A = {1, 2, 3, 4}. Then the relation R is-
A. Reflexive
B. Transitive
C. Not symmetric
D. A function

Answer 1

Hint: Any relation can be classified as reflexive, symmetric and transitive. If aRa exists in the relation, then it is said to be reflexive. If aRb and bRa both exist in the relation, then it is said to be symmetric. If aRb and bRc exist implies that aRc also exists, the relation is transitive.

Complete step-by-step answer:
It is given that R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} on A = {1, 2, 3, 4}

For R to be reflexive, (a, a) should exist in R, where a is an element of the set A. But,
$\left( {{\text{a}},{\text{a}}} \right) \notin {\text{R}}$. Hence, R is not a reflexive relation.

For R to be transitive, if (a, b) and (b, c) exist in R, then (a, c) also exists in R. That is,
$\left( {{\text{a}},{\text{b}}} \right),\;\left( {{\text{b}},{\text{c}}} \right) \in {\text{R}}\;then\;\left( {{\text{a}},{\text{c}}} \right) \in {\text{R}}$
We can see that (1, 3) and (3, 1) are elements of R. For R to be transitive, (1, 1) should also be an element of R, but this is not the case. Hence, R is not transitive.

For R to be symmetric, if (a, b) is an element in R, then (b, a) should also be present. We can clearly see that (2, 3) exists in R but (3, 2) does not. Hence, the relation R is not symmetric.

Function is a relation from a set of inputs to a set of possible outputs where each input is related to exactly one output.We can see that both (2, 3) and (2, 4) exist in R. That is, two values of the range of R exist for a single value of domain. For a function, every element has a unique pre-image or domain. Hence, R is not a function.

R is not symmetric, hence the correct option is C.

Note: It is important to check carefully for each condition. It is also recommended to check and verify each condition using a suitable example. Even if one case is false, the condition is not verified. Also if it is not possible to prove that relation is symmetric, reflexive or transitive, then use a suitable example to show that it is not.