Solve this

Solution:

Given that,

$A=\{1,2,3,4,5,6)$ and $R=\{(a, b): a, b \in A$ and $b=a+1\}$

$\therefore \mathrm{R}=\{(1,2),(2,3),(3,4),(4,5),(5,6)\}$

Now,

$\mathrm{R}$ is Reflexive if $(\mathrm{a}, \mathrm{a}) \in \mathrm{R} \forall \mathrm{a} \in \mathrm{A}$

Since, $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6) \notin \mathrm{R}$

Thus, $R$ is not reflexive.

$R$ is Symmetric if $(a, b) \in R \Rightarrow(b, a) \in R \forall a, b \in A$

We observe that $(1,2) \in \mathrm{R}$ but $(2,1) \notin \mathrm{R}$.

Thus, $R$ is not symmetric.

$R$ is Transitive if $(a, b) \in R$ and $(b, c) \in R \Rightarrow(a, c) \in R \forall a, b, c \in A$

We observe that $(1,2) \in R$ and $(2,3) \in R$ but $(1,3) \notin R$

Thus, $R$ is not transitive.