Let r be a relation from R (set of real numbers) to R defined by r = {(a,b) ,

English

Gujarati

Hindi

1.Relation and Function

normal

Let $r$ be a relation from $R$ (set of real numbers) to $R$ defined by $r = \{(a,b) \, | a,b \in R$ and $a - b + \sqrt 3$ is an irrational number$\}$ The relation $r$ is

an equivalence relation

reflexive only

symmetric only

transitive only

Solution

$r=\{(a, b) | a, b \in R \text { and } a-b+\sqrt{3}$ is an irrational number$\}$, then $\mathrm{r}$ is reflexive as, $\mathrm{aRa}=\mathrm{a}-\mathrm{a}+\sqrt{3}=\sqrt{3}$ which is an irrational number $\sqrt{3} \mathrm{R} 1=\sqrt{3}-1+\sqrt{3}=2 \sqrt{3}-1 \mathrm{which}$ is an irrational number but $1R$ $\sqrt{3}=1-\sqrt{3}+\sqrt{3}=1$ which is not irrational number

$\therefore \sqrt{3} \mathrm{R} 1 \Rightarrow 2 \sqrt{3}-1$

$\therefore \mathrm{r}$ is not symmetric Also, $r$ is not transitive

$\because \sqrt{3} \mathrm{Rl}$ and $1 \mathrm{R} 2 \sqrt{3}$ but $\sqrt{3} \mathrm{R} 2 \sqrt{2}$

$\Rightarrow \sqrt{3}-2 \sqrt{3}+\sqrt{3}=0$

Hence, $r$ is not an equivalenece relation

Standard 12

Mathematics

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

medium

(JEE MAIN-2025)

Let $R$ be a relation from $A = \{2,3,4,5\}$ to $B = \{3,6,7,10\}$ defined by $R = \{(a,b) |$ $a$ divides $b, a \in A, b \in B\}$, then number of elements in $R^{-1}$ will be-

normal

Let $L$ be the set of all lines in $XY$ plane and $R$ be the relation in $L$ defined as $R =\{\left( L _{1}, L _{2}\right): L _{1} $ is parallel to $L _{2}\} .$ Show that $R$ is an equivalence relation. Find the set of all lines related to the line $y=2 x+4$

medium

Let $R_1$ be a relation defined by $R_1 =\{(a, b) | a \geq b, a, b \in R\}$ . Then $R_1$ is

normal

Let $I$ be the set of positve integers. $R$ is a relation on the set $I$ given by $R =\left\{ {\left( {a,b} \right) \in I \times I\,|\,\,{{\log }_2}\left( {\frac{a}{b}} \right)} \right.$ is a non-negative integer$\}$, then $R$ is

normal

Let $r$ be a relation from $R$ (set of real numbers) to $R$ defined by $r = \{(a,b) \, | a,b \in R$ and $a - b + \sqrt 3$ is an irrational number$\}$ The relation $r$ is

Solution

Similar Questions

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

Let $R$ be a relation from $A = \{2,3,4,5\}$ to $B = \{3,6,7,10\}$ defined by $R = \{(a,b) |$ $a$ divides $b, a \in A, b \in B\}$, then number of elements in $R^{-1}$ will be-

Let $L$ be the set of all lines in $XY$ plane and $R$ be the relation in $L$ defined as $R =\{\left( L _{1}, L _{2}\right): L _{1} $ is parallel to $L _{2}\} .$ Show that $R$ is an equivalence relation. Find the set of all lines related to the line $y=2 x+4$

Let $R_1$ be a relation defined by $R_1 =\{(a, b) | a \geq b, a, b \in R\}$ . Then $R_1$ is

Let $I$ be the set of positve integers. $R$ is a relation on the set $I$ given by $R =\left\{ {\left( {a,b} \right) \in I \times I\,|\,\,{{\log }_2}\left( {\frac{a}{b}} \right)} \right.$ is a non-negative integer$\}$, then $R$ is

Start a Free Trial Now