Show that the relation $R$ in the set $\{1, 2, 3\}$ given by $R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)\}$ is reflexive but neither symmetric nor transitive.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) relation $R$ on a set $A$ is reflexive if $(a, a) \in R$ for all $a \in A$. Here,$A = \{1, 2, 3\}$. Since $(1, 1), (2, 2), (3, 3) \in R$,the relation $R$ is reflexive.
$A$ relation $R$ is symmetric if $(a, b) \in R$ implies $(b, a) \in R$. Here,$(1, 2) \in R$,but $(2, 1) \notin R$. Therefore,$R$ is not symmetric.
$A$ relation $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R$ implies $(a, c) \in R$. Here,$(1, 2) \in R$ and $(2, 3) \in R$,but $(1, 3) \notin R$. Therefore,$R$ is not transitive.

Explore More

Similar Questions

Let $R$ be the relation in the set $N$ given by $R = \{(a, b) : a = b - 2, b > 6\}$. Choose the correct answer.

Let $N$ be the set of natural numbers and the relation $R$ on $N \times N$ is defined by $(a, b) R (c, d)$ if $ad(b + c) = bc(a + d)$. Then $R$ is:

Difficult
View Solution

Let $R$ and $S$ be two relations on a set $A$. Then which of the following is true?

The relation $R$ defined on a set $A$ is antisymmetric if $(a, b) \in R$ and $(b, a) \in R$ implies $a = b$ for all $a, b \in A$. Based on this definition,the relation $R$ is antisymmetric if $(a, b) \in R$ and $(b, a) \in R$ implies $a = b$,which is equivalent to saying that if $a \neq b$,then it is not possible for both $(a, b) \in R$ and $(b, a) \in R$ to be true. Therefore,the condition is that for $a \neq b$,we cannot have both $(a, b) \in R$ and $(b, a) \in R$.

On the set $N$ of natural numbers,the relation $R$ is defined by $nRm$ if $n$ is a factor of $m$ (i.e.,$n|m$). 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