Question

Show that the relation $R=\left\{ \left( a,b \right):a>b \right\}$ on N is transitive but neither reflexive nor symmetric.

Answer 1

Hint: To show that R is transitive prove that if $\left( a,b \right)\in R\ and\ \left( b,c \right)\in R\ then\ \left( a,c \right)\in R$. To show that R is not reflexive, find an example such that $a\in N\ and\ \left( a,a \right)\notin R$. 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\ $.

Complete step by step answer:
Let ‘A’ be a set, then:
Reflexivity, symmetry and transitive 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 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 $\begin{align}
  & \left( a,b \right)\in R\ and\ \left( b,c \right)\in R\ then\ \left( a,c \right)\in R \\
 & i.e.\ \ \left( a,b \right)\in R\ and\ \left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R \\
\end{align}$
We have to show that R is transitive. Let us assume that $\left( a,b \right)\in R\ and\ \left( b,c \right)\in R$. To prove that R is transitive, we have to prove that $\left( a,c \right)\in R$.
$\begin{align}
  & \text{since }\left( a,b \right)\in R,\ \ \ \ \ \ \ a>b........\left( 1 \right) \\
 & \text{since }\left( b,c \right)\in R,\ \ \ \ \ \ \ b>c........\left( 2 \right) \\
\end{align}$
From equation (1) and (2),
Since, ‘a’ is greater than ‘b’ and ‘b’ is greater than ‘c’, then ‘a’ must be greater than ‘c’.
i.e. a > c
$\Rightarrow \left( a,c \right)\in R$
$\therefore $R is transitive on N.
Next,
We have to show that R is not symmetric.
To show that R is not symmetric, we have to find an example such that $\left( a,b \right)\in R\ and\ \left( b,a \right)\notin R\ $.
Let us take a = 3 and b = 2
$\left( a,b \right)\in N$ [Both ‘2’ and ‘3’ are natural numbers]
$\begin{align}
  & And\ \left( a,b \right)\in R\ \left[ \text{since }a< b\ i.e.2<3 \right] \\
 & But\ \left( b,a \right)\notin R\ \left[ \text{since }b\nless a\ i.e.3\nless 2 \right] \\
\end{align}$
$\therefore $ R is not symmetric on set A.
Next, we have to show that R is not reflexive.
To show that R is not reflexive, we need to find an example such that $a\in N\ and\ \left( a,a \right)\notin R$.
Let us take a = 2.
\[\begin{align}
  & \Rightarrow a\in N\ \ \ \ \ \ \left[ '2'\text{ is a natural number} \right] \\
 & But\left( a,a \right)\notin R\ \ \ \ \left[ \text{since }2\nless 2 \right] \\
\end{align}\]
$\therefore $R is not reflexive on set ‘S’.

Hence, we have shown that R is transitive but neither reflexive nor symmetric.

Note: Be careful that to prove that the given relation is not symmetric or not reflexive, we need to just find an example which doesn’t satisfy the condition of symmetric or reflexive. But to prove that it is transitive, we can’t use an example, instead we need to prove the condition of transitivity for all possible cases.