Let X = R × R . Define a relation R on X as: left(a₁, b₁right) Rleft(a₂, b₂right) Leftright

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1.Relation and Function

medium

Let $X = R \times R$. Define a relation $R$ on $X$ as: $\left(a_1, b_1\right) R\left(a_2, b_2\right) \Leftrightarrow b_1=b_2 .$ Statement-$I: R$ is an equivalence relation. Statement$-II$: For some $( a , b ) \in X$, the set $S=\{(x, y) \in X:(x, y) R(a, b)\}$ represents a line parallel to $y = x$. In the light of the above statements, choose the correct answer from the options given below:

ABoth Statement$-I$ and Statement$-II$ are false.

BStatement$-I$ is true but Statement$-II$ is false.

CBoth Statement$-I$ and Statement$-II$ are true.

DStatement$-I$ is false but Statement$-II$ is true.

(JEE MAIN-2025)

Solution

Statement $- I :$
Reflexive : $\left(a_1, b\right) R\left(a_1, b_1\right) \Rightarrow b_1=b_1 \quad$ True
Symmetric: $\left.\begin{array}{rl} & \left(a_1, b_1\right) R\left(a_2, b_2\right) \Rightarrow b_1=b_2 \\ & \left(a_2, b_2\right) R\left(a_1, b_1\right) \Rightarrow b_2=b_1\end{array}\right\}$ True
Transitive :
$$\left.\begin{array}{l}
\left(a_1, b_1\right) R\left(a_2, b_2\right) \Rightarrow b_1=b_2 \\
\&\left(a_2, b_2\right) R\left(a_3, b_3\right) b_2=b_3 \\
\Rightarrow\left(a_1, b_1\right) R\left(a_3, b_3\right) \Rightarrow \text { True }
\end{array}\right\} b_1=b_3$$
Hence Relation $R$ is an equivence relation Statement$-I$ is true.
For statement $- II$ $\Rightarrow y = b$ so False

Standard 12

Mathematics

Show that the relation $\mathrm{R}$ in the set $\mathrm{Z}$ of integers given by $\mathrm{R} =\{(\mathrm{a}, \mathrm{b}): 2$ divides $\mathrm{a}-\mathrm{b}\}$ is an equivalence relation.

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 $R$ be a relation over the set $N × N$ and it is defined by $(a,\,b)R(c,\,d) \Rightarrow a + d = b + c.$ Then $R$ is

hard

Let $R\,= \{(x,y) : x,y \in N\, and\, x^2 -4xy +3y^2\, =0\}$, where $N$ is the set of all natural numbers. Then the relation $R$ is

hard

(JEE MAIN-2013)

Let $X =\{1,2,3,4,5,6,7,8,9\} .$ Let $R _{1}$ be a relation in $X$ given by $R _{1}=\{(x, y): x-y$ is divisible by $3\}$ and $R _{2}$ be another relation on $X$ given by ${R_2} = \{ (x,y):\{ x,y\} \subset \{ 1,4,7\} \} $ or $\{x, y\} \subset\{2,5,8\} $ or $\{x, y\} \subset\{3,6,9\}\} .$ Show that $R _{1}= R _{2}$.