R is a relation over set of real numbers and it is given by nm ge 0 . Then R is

English

Gujarati

Hindi

1.Relation and Function

easy

$R$ is a relation over the set of real numbers and it is given by $nm \ge 0$. Then $R$ is

Symmetric and transitive

Reflexive and symmetric

A partial order relation

An equivalence relation

Solution

Given, $m R n$ iff $m n \geq 0$

Reflexivity:

We know that $m^2 \geq 0$

$\Rightarrow mm \geq 0$

$\Rightarrow mRm$

Hence, $r$ is reflexive.

Symmetry:

Let $m R n$

$\Rightarrow m n \geq 0$

$\Rightarrow nm \geq 0$ (product of real numbers is commutative.)

$\Rightarrow nRm$

Hence, $R$ is symmetric.

Transitive

Let $m R n, n R p$

$\Rightarrow mn \geq 0 ; np \geq 0$

$\Rightarrow m , n , p$ are of same signs

$\Rightarrow mp \geq 0$

$\Rightarrow mRp$

Hence, $R$ is transitive.

Hence, $R$ is an equivalence relation.

Standard 12

Mathematics

Consider set $A = \{1,2,3\}$ . Number of symmetric relations that can be defined on $A$ containing the ordered pair $(1,2)$ & $(2,1)$ is

normal

Let $X$ be a family of sets and $R$ be a relation on $X$ defined by $‘A$ is disjoint from $B’$. Then $R$ is

medium

Let $\mathrm{T}$ be the set of all triangles in a plane with $\mathrm{R}$ a relation in $\mathrm{T}$ given by $\mathrm{R} =\left\{\left( \mathrm{T} _{1}, \mathrm{T} _{2}\right): \mathrm{T} _{1}\right.$ is congruent to $\left. \mathrm{T} _{2}\right\}$ . Show that $\mathrm{R}$ is an equivalence relation.

easy

Let a relation $R$ be defined by $R = \{(4, 5); (1, 4); (4, 6); (7, 6); (3, 7)\}$ then ${R^{ – 1}}oR$ is

medium

The relation $R =\{( x , y ): x , y \in z$ and $x + y$ is even $\}$ is :

medium

(JEE MAIN-2025)

$R$ is a relation over the set of real numbers and it is given by $nm \ge 0$. Then $R$ is

Solution

Similar Questions

Consider set $A = \{1,2,3\}$ . Number of symmetric relations that can be defined on $A$ containing the ordered pair $(1,2)$ & $(2,1)$ is

Let $X$ be a family of sets and $R$ be a relation on $X$ defined by $‘A$ is disjoint from $B’$. Then $R$ is

Let a relation $R$ be defined by $R = \{(4, 5); (1, 4); (4, 6); (7, 6); (3, 7)\}$ then ${R^{ – 1}}oR$ is

The relation $R =\{( x , y ): x , y \in z$ and $x + y$ is even $\}$ is :

Start a Free Trial Now