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

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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 $R= \{(3, 3) (5, 5), (9, 9), (12, 12), (5, 12), (3, 9), (3, 12), (3, 5)\}$ be a relation on the set $A= \{3, 5, 9, 12\}.$ Then, $R$ is

hard

(JEE MAIN-2013)

Let $A=\{1,2,3, \ldots \ldots .100\}$. Let $R$ be a relation on A defined by $(x, y) \in R$ if and only if $2 x=3 y$. Let $R_1$ be a symmetric relation on $A$ such that $\mathrm{R} \subset \mathrm{R}_1$ and the number of elements in $\mathrm{R}_1$ is $\mathrm{n}$. Then, the minimum value of $n$ is……………………..

easy

(JEE MAIN-2024)

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medium

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