Determine whether the following relation is reflexive,symmetric,and transitive:
Relation $R$ in the set $A$ of human beings in a town at a particular time is given by:
$R = \{(x, y) : x \text{ and } y \text{ live in the same locality}\}$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(A) $R = \{(x, y) : x \text{ and } y \text{ live in the same locality}\}$
$1.$ Reflexive: For any human being $x \in A$,$x$ lives in the same locality as $x$. Thus,$(x, x) \in R$. Therefore,$R$ is reflexive.
$2.$ Symmetric: Let $(x, y) \in R$. This means $x$ and $y$ live in the same locality. It follows that $y$ and $x$ also live in the same locality. Thus,$(y, x) \in R$. Therefore,$R$ is symmetric.
$3.$ Transitive: Let $(x, y) \in R$ and $(y, z) \in R$. This means $x$ and $y$ live in the same locality,and $y$ and $z$ live in the same locality. Consequently,$x$ and $z$ must live in the same locality. Thus,$(x, z) \in R$. Therefore,$R$ is transitive.
Conclusion: The relation $R$ is reflexive,symmetric,and transitive.

Explore More

Similar Questions

On the set $R$ of real numbers, we define $x P y$ if and only if $x y \geq 0$. Then, the relation $P$ is

If $R_{1}$ and $R_{2}$ are equivalence relations in a set $A$,show that $R_{1} \cap R_{2}$ is also an equivalence relation.

Let $r$ be a relation on the set $N \times N$ defined by $(a, b)r(c, d) \Rightarrow a + d = b + c$. Then $r$ is:

Let $R = \{(1,2), (2,3), (3,3)\}$ be a relation defined on the set $A = \{1, 2, 3, 4\}$. Then the minimum number of elements needed to be added to $R$ so that $R$ becomes an equivalence relation is:

Let $R$ be a relation on $\mathbb{Z} \times \mathbb{Z}$ defined by $(a, b) R (c, d)$ if and only if $ad - bc$ is divisible by $5$. Then $R$ is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo