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.
Clearly, $(x, y)$ $R (x, y)$, $\forall \,\,(x, y) \in A$, since $x y=y x .$ This shows that $R$ is reflexive. Further, $(x, y) R (u, v)$ $ \Rightarrow x v=y u$ $ \Rightarrow u y=v x$ and hence $(u, v) \,R (x, y) .$ This shows that $R$ is symmetric. Similarly, $(x, y) R (u, v)$ and $(u, v)$ $R$ $(a, b) \Rightarrow x v=y u$ and $u b=v a \Rightarrow $ $x v \frac{a}{u}=y u \frac{a}{u} $ $\Rightarrow x v \frac{b}{v}=$ $y u \frac{a}{u} \Rightarrow $ $x b=y a$ and hence $(x, y) \,R (a, b) .$ Thus, $R$ is transitive. Thus, $R$ is an equivalence relation.
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
${x^2} = xy$ is a relation which is
Let $A = \{1, 2, 3\}, B = \{1, 3, 5\}$. $A$ relation $R:A \to B$ is defined by $R = \{(1, 3), (1, 5), (2, 1)\}$. Then ${R^{ - 1}}$ is defined by
Let $R$ be the relation in the set $N$ given by $R =\{(a,\, b)\,:\, a=b-2,\, b>6\} .$ Choose the correct answer.
Let L be the set of all lines in a plane and $\mathrm{R}$ be the relation in $\mathrm{L}$ defined as $\mathrm{R}=\left\{\left(\mathrm{L}_{1}, \mathrm{L}_{2}\right): \mathrm{L}_{1}\right.$ is perpendicular to $\left. \mathrm{L} _{2}\right\}$. Show that $\mathrm{R}$ is symmetric but neither reflexive nor transitive.