Question

On the set of all real numbers, define a relation R=\left\\{ \left( a,b \right):a < b \right\\}. Show that R is (i) not reflexive (ii) transitive (iii) not symmetric.

Answer 1

To show that R is reflexive, find an example such that $\left( a,a \right)\in R$ for $a\in S$ , To show that R is transitive , find an example such that $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$ then $\left( a,c \right)\in R$ for $a,b,c\in S$.To show that R is not symmetric, find an example such that $\left( a,b \right)\in R$ but $\left( b,a \right)\notin R$ for $a,b\in S$ .

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 relation R=\left\\{ \left( a,b \right):a < b \right\\}
\therefore R=\left\\{ \left( 1,2 \right),\left( 2,3 \right),\left( 2,4 \right),\left( 3,8 \right),\left( 1,3 \right),\left( 1,8 \right),\left( 1,4 \right),\left( 2,8 \right),.......... \right\\} .
Check reflexivity: If $\left( a,a \right)\in R$ , foe $a\in S$ , then R is reflexive. Since, R=\left\\{ \left( a,b \right):a < b \right\\}.
That means. $\left( a,a \right)\notin R$ , because a number is equal to itself but doesn’t belong to set R.
Therefore, R is not reflexive.
Check symmetricity: If $\left( a,b \right)\in R$ for $a.b\in S$ and $\left( b,a \right)\notin R$ , Then R is not symmetric.
Since R=\left\\{ \left( a,b \right):a < b \right\\}
If $\left( a.b \right)\in R$ , that means $\left( a < b \right)$ i.e. $\left( b > a \right)$ Therefore, $\left( b,a \right)\notin R$ .
It is very obvious that if $\left( a,b \right)\in R\Rightarrow a < b$ $\Rightarrow \text{ a is less than or equal to b }\Rightarrow \text{ b is not less than or equal to a}$ .
From set R, we can see that $\left( 1,2 \right)\in R$ But $\left( 2,1 \right)\notin R$
$\left( 2,3 \right)\in R,\text{ But }\left( 3,2 \right)\notin R$ .
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 $\left( a,c \right)\in R$ , then R is transitive.
From the set R, we can clearly see that –
$\begin{aligned} & \left( 1,2 \right)\in R\text{ and }\left( 2,3 \right)\in R\text{ then }\left( 1,3 \right)\in R \\\ & \left( 1,2 \right)\in R\text{ and }\left( 2,4 \right)\in R\text{ then }\left( 1,4 \right)\in R \\\ & \left( 1,3 \right)\in R\text{ and }\left( 3,8 \right)\in R\text{ then }\left( 1,8 \right)\in R \\\ \end{aligned}$
…………………………………….. and so on.
We can also notice that if $\left( a,b \right)\in R\Rightarrow a < b$…..(1)
And $\left( b,c \right)\in R\Rightarrow b < c$ ……………………. (2)
From (1) and (2) we can conclude that $a < c$ .
Therefore, R is transitive.

Hence we conclude that relation R is transitive but not reflexive and symmetric.

Note: To prove that R is transitive, be careful that if $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$ then $\left( a,c \right)\in R$ . If there is no pair such that $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$. Then we don’t need to check, R will always be transitive in this case.