Let $A=\left\{ 1,2,3,4,5,6 \right\}$ and let $R=\left\{ \left( a,b \right):a.b\in A\text{ and b=a+1} \right\}$ .Show that R is (i) not reflexive, (ii) not symmetric and (iii) not transitive.

Hint:Find all sets formed by set A, which follows the relation R. To show that relation ‘R’ is not reflexive, find an example such that $\left( a,a \right)\notin R$ from $a\in S$ . To show that R is not symmetric, find an example such that $\left( a,b \right)\in R$ and $\left( b,a \right)\notin R$ for $a,b\in S$ . To show that R is not transitive, find an example such that $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$ but $\left( a,c \right)\notin R$ for $a,b,c\in S$ .

Complete step-by-step answer:
Let ‘A’ be a set then Reflexivity, Symmetry and transitivity of a relation on set ‘A’ is defined as follows:
Reflexive relation: - A relation R on a set ‘A’ is said to be reflexive if every element of A is related to itself i.e. for every element say (a) in set A, $\left( a,a \right)\in R$ .
Thus, R on a set ‘A’ is not reflexive if there exists an element $a\in A$ such that $\left( a,a \right)\notin R$.
Symmetric Relation: - A relation R on a set ‘A’ is said to be symmetric relation if $\left( a,b \right)\in R$ then $\left( b,a \right)$must be belong to R. i.e. $\left( a,b \right)\in R\Rightarrow \left( b,a \right)\in R$ For all $a,b\in A$.
Transitive relation:- A relation R on ‘A’ is said to be a transitive relation If $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$ then $\left( a,c \right)\in R$.
I.e. $\left( a,b \right)\in R$and $\left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R$.
Given set $A=\left\{ 1,2,3,4,5,6 \right\}$ relation $R=\left\{ \left( a,b \right):a.b\in A\text{ and b=a+1} \right\}$
Set formed by set A, Which follows the relation R is given by $S=R=\left\{ \left( 1,2 \right),\left( 2,3 \right),\left( 3,4 \right),\left( 4,5 \right),\left( 5,6 \right) \right\}$ ……………………… (1)
From set S, we can clearly check for reflexivity, symmetric and transitivity.
Check Reflexivity: If $\left( a,a \right)\notin R$ from $a\in S$, then R is not reflexive
From the set A, we can clearly see that $\left( 1,1 \right),\left( 2,2 \right),\left( 3,3 \right),\left( 4,4 \right),\left( 5,5 \right),\left( 6,6 \right)\notin R$ .
$\therefore R=\left\{ a,b:b=a+1 \right\},a,b\in R$ .
It is obvious that $\left( a,a \right)\notin R$, because a number cannot be equal to a number greater than itself by 1. Therefore, R is not reflexive.
Check symmetric: If $\left( a,b \right)\in R$ , for $a,b\in S$ and $\left( b,a \right)\notin R$ for $a,b\in S$ then R is not symmetric.
From the set R, we can clearly see that.
\[\left( 1,2 \right)\in R\] But $\left( 2,1 \right)\notin R$ .
\[\left( 2,3 \right)\in R\] But $\left( 3,2 \right)\notin R$ .
$\left( 3,4 \right)\in R$ But $\left( 4,3 \right)\notin R$ .
\[\left( 4,5 \right)\in R\] But $\left( 5,4 \right)\notin R$ .
$\left( 5,6 \right)\in R$ But $\left( 6,5 \right)\notin R$ .
Also, here $R=\left\{ \left( a,b \right):a.b\in A\text{ and b=a+1} \right\}$ It can be clearly notice that if $\left( a,b \right)\in R$ then $b=a+1$ , therefore $\left( b,a \right)\notin R$ , because here $b=a+1\Rightarrow a=b-1\ne b+1$ .
Therefore, R is not symmetric.
Check transitivity: if $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$ for $a,b,c\in S$ and $\left( a,c \right)\notin R$ , then R is not transitive. From the set R, we can clearly see that
\[\left( 1,2 \right)\in R\] and \[\left( 2,3 \right)\in R\] But $\left( 1,3 \right)\notin R$ .
\[\left( 2,3 \right)\in R\] and $\left( 3,4 \right)\in R$ But $\left( 2,4 \right)\notin R$ .
$\left( 3,4 \right)\in R$ and \[\left( 4,5 \right)\in R\] But $\left( 3,5 \right)\notin R$ .
\[\left( 4,5 \right)\in R\] and $\left( 5,6 \right)\in R$ But $\left( 4,6 \right)\notin R$ .
Also, here $R=\left\{ \left( a,b \right):a.b\in A\text{ and b=a+1} \right\}$ , it can be clearly notice that if $\left( a,b \right)\in R$ then $b=a+1$ and $\left( b,c \right)\in R$ then $c=b+1$ . That means $c=\left( a+1 \right)+1$ i.e. $c=a+2\ne a+1$ , Hence $\left( a,c \right)\notin R$ for $a,b,c\in R$ .
Therefore, R is not transitive.
We can conclude that R is not reflexive, not symmetric and not transitive.

Note: Students can also use just one example to show that relation R is not reflexive. One example to show that R is not transitive and one example to show that relation R is not symmetric. But these need to keep in mind that those examples must be taken from set A, which is defined as a set formed by A, which follows the relation R.
$R=\left\{ \left( 1,2 \right),\left( 2,3 \right),\left( 3,4 \right),\left( 4,5 \right),\left( 5,6 \right) \right\}$.