Let $f: X \rightarrow Y$ be a function. Define a relation $R$ in $X$ given by $R =\{(a, b): f(a)=f(b)\} .$ Examine if $R$ is an equivalence relation.
For every $a \in X ,(a, a) \in R ,$ since $f(a)=f(a),$ showing that $R$ is reflexive. Similarly, $(a, b) \in R \Rightarrow f(a)=f(b)$ $ \Rightarrow f(b)=f(a)$ $ \Rightarrow(b, a) \in$ $R$. Therefore, $R$ is symmetric. Further, $(a, b) \in R$ and $(b, c) \in R \Rightarrow$ $f(a)=f(b)$ and $f(b)=f(c) \Rightarrow f(a)$ $=f(c) \Rightarrow(a, c) \in R ,$ which implies that $R$ is transitive. Hence, $R$ is an equivalence relation.
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
Let a relation $R$ on $\mathbb{N} \times \mathbb{N}$ be defined as : $\left(\mathrm{x}_1, \mathrm{y}_1\right) \mathrm{R}\left(\mathrm{x}_2, \mathrm{y}_2\right)$ if and only if $\mathrm{x}_1 \leq \mathrm{x}_2$ or $\mathrm{y}_1 \leq \mathrm{y}_2$
Consider the two statements :
($I$) $\mathrm{R}$ is reflexive but not symmetric.
($II$) $\mathrm{R}$ is transitive
Then which one of the following is true?
Let $R$ be a relation on the set $A$ of ordered pairs of positive integers defined by $(x, y) R (u, v)$ if and only if $x v=y u .$ Show that $R$ is an equivalence relation.
Let $R = \{(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)\}$ be a relation on the set $A = \{1, 2, 3, 4\}$. The relation $R$ is
Give an example of a relation. Which is Symmetric but neither reflexive nor transitive.