Let $R=\{(a, b): a, b, \in N$ and $a Show that $\mathrm{R}$ is a binary relation on $\mathrm{N}$, which is neither reflexive nor symmetric. Show that $R$ is transitive.
N is the set of all the natural numbers.
$N=\{1,2,3,4,5,6,7 \ldots \ldots\}$
$R=\{(a, b): a, b, \in N$ and $a $R=\{(1,2),(1,3),(1,4) \ldots(2,3),(2,4),(2,5) \ldots \ldots\}$ For reflexivity, A relation R on N is said to be reflexive if (a, a) є R for all a є N. But, here we see that $\mathrm{a}<\mathrm{b}$, so the two co-ordinates are never equal. Thus, the relation is not reflexive. For symmetry, A relation $R$ on $N$ is said to be symmetrical if $(a, b) \in R \dot{e}(b, a) \in R$ Here, $(a, b) \in R$ does not imply $(b, a) \in R$. Thus, it is not symmetric. For transitivity A relation $R$ on $A$ is said to be transitive if $(a, b) \in R$ and $(b, c) \in R$ è $(a, c) \in R$ for all $(a$, b, c) $€ \mathrm{~N}$. Let's take three values $a, b$ and $c$ such that $a