Let N denote set of all natural numbers. Define two binary relations on N as R₁ = {(x,y) ∈

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1.Relation and Function

hard

Let $N$ denote the set of all natural numbers. Define two binary relations on $N$ as $R_1 = \{(x,y) \in N \times N : 2x + y= 10\}$ and $R_2 = \{(x,y) \in N\times N : x+ 2y= 10\} $. Then

Both $R_1$ and $R_2$ are transitive relations

Both $R_1$ and $R_2$ are symmetric relations

Range of $R_2$ is $\{1, 2, 3, 4\}$

Range of $R_1$ is $\{ 2, 4, 8\}$

(JEE MAIN-2018)

Solution

Here,

${R_1} = \left\{ {\left( {x,y} \right) \in N \times N:2x + y = 10} \right\}$

${R_2} = \left\{ {\left( {x,y} \right) \in N \times N:x + 2y = 10} \right\}$

For ${R_1};2x + y = 10$ and $x,y \in N$

So, possible values for $x$ and $y$ are:

$x = 1,y = 8$ i.e. $(1,8);$

$x = 2,y = 6\,$ i.e. $(2,6);$

$x = 3,y = 4$ i.e. $\,(3,4)$ and

$x = 4,y = 2\,$ i.e. $(4,2)$.

${R_1} = \left\{ {(1,8),(2,6),(3,4),(4,2)} \right\}$

Therefore, Range of ${R_1}$ is $\left\{ {2,4,6,8} \right\}$

${R_1}$ is not symmetric

Also,${R_1}$ is not transitive because

$(3,4),(4,2) \in {R_1}$ but $\,(3,2) \notin {R_1}$

Thus, options $A,B$ and $D$ are incorrect.

For ${R_2};x + 2y = 10$ and $x,y \in N$

So, possible values for $x$ and $y$ are:

$x = 8,y = 1\,$i.e.$\,(8,1);$

$x = 6,y = 2$i.e.$(6,2);$

$x = 4,y = 3$i.e.$(4,3)$ and

$x = 2,y = 4$i.e.$(2,4)$

${R_2} = \left\{ {(8,1),(6,2),(4,3),(2,4)} \right\}$

Therefore, Range of ${R_2}$ is $\left\{ {1,2,3,4} \right\}$

${R_2}$ is not symmetric and transitive.

Standard 12

Mathematics

Show that each of the relation $R$ in the set $A=\{x \in Z: 0 \leq x \leq 12\},$ given by $R =\{( a , b ): a = b \}$ is an equivalence relation. Find the set of all elements related to $1$ in each case.

medium

Given a non empty set $X$, consider $P ( X )$ which is the set of all subsets of $X$.

Define the relation $R$ in $P(X)$ as follows :

For subsets $A,\, B$ in $P(X),$ $ARB$ if and only if $A \subset B .$ Is $R$ an equivalence relation on $P ( X ) ?$ Justify your answer.

hard

Give an example of a relation. Which is Symmetric and transitive but not reflexive.

easy

Consider the relations $R_1$ and $R_2$ defined as $a R_1 b$ $\Leftrightarrow a^2+b^2=1$ for all $a, b, \in R$ and $(a, b) R_2(c, d)$ $\Leftrightarrow a+d=b+c$ for all $(a, b),(c, d) \in N \times N$. Then

medium

(JEE MAIN-2024)

Let $P ( S )$ denote the power set of $S =\{1,2,3, \ldots, 10\}$. Define the relations $R_1$ and $R_2$ on $P(S)$ as $A R_1 B$ if $\left( A \cap B ^{ c }\right) \cup\left( B \cap A ^{ c }\right)=\varnothing$ and $AR _2 B$ if $A \cup B ^{ c }=$ $B \cup A ^{ c }, \forall A , B \in P ( S )$. Then :

hard

(JEE MAIN-2023)

Let $N$ denote the set of all natural numbers. Define two binary relations on $N$ as $R_1 = \{(x,y) \in N \times N : 2x + y= 10\}$ and $R_2 = \{(x,y) \in N\times N : x+ 2y= 10\} $. Then

Solution

Similar Questions

Show that each of the relation $R$ in the set $A=\{x \in Z: 0 \leq x \leq 12\},$ given by $R =\{( a , b ): a = b \}$ is an equivalence relation. Find the set of all elements related to $1$ in each case.

Given a non empty set $X$, consider $P ( X )$ which is the set of all subsets of $X$. Define the relation $R$ in $P(X)$ as follows : For subsets $A,\, B$ in $P(X),$ $ARB$ if and only if $A \subset B .$ Is $R$ an equivalence relation on $P ( X ) ?$ Justify your answer.

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Given a non empty set $X$, consider $P ( X )$ which is the set of all subsets of $X$.

Define the relation $R$ in $P(X)$ as follows :

For subsets $A,\, B$ in $P(X),$ $ARB$ if and only if $A \subset B .$ Is $R$ an equivalence relation on $P ( X ) ?$ Justify your answer.