# The following relations are defined on the set of real numbers.

Question:

The following relations are defined on the set of real numbers.

(i) $a R b$ if $a-b>0$

(ii) $a R b$ if $1+a b>0$

(iii) $a R b$ if $|a| \leq b$

Find whether these relations are reflexive, symmetric or transitive.

Solution:

(i)
Reflexivity: Let be an arbitrary element of R. Then,

$a \in R$

But $a-a=0 \ngtr 0$

So, this relation is not reflexive.

Symmetry:

Let $(a, b) \in R$

$\Rightarrow a-b>0$

$\Rightarrow-(b-a)>0$

$\Rightarrow b-a<0$

So, the given relation is not symmetric.

Transitivity:

Let $(a, b) \in R$ and $(b, c) \in R$. Then,

$a-b>0$ and $b-c>0$

$a-b+b-c>0$

$\Rightarrow a-c>0$

$\Rightarrow(a, c) \in R$

So, the given relation is transitive.

(ii)
Reflexivity: Let a be an arbitrary element of R. Then,

$a \in R$

$\Rightarrow 1+a \times a>0$

i. e. $1+a^{2}>0$                                              [Since, square of any number is positive]

So, the given relation is reflexive.

Symmetry:

Let $(a, b) \in R$

$\Rightarrow 1+a b>0$

$\Rightarrow 1+b a>0$

$\Rightarrow(b, a) \in R$

So, the given relation is symmetric.

Transitivity:

Let $(a, b) \in R$ and $(b, c) \in R$

$\Rightarrow 1+a b>0$ and $1+b c>0$

But $1+a c \ngtr 0$

$\Rightarrow(a, c) \notin R$

So, the given relation is not transitive.

(iii)
Reflexivity: Let a be an arbitrary element of R. Then,

$a \in R$

$\Rightarrow|a| \nless a$            $[$ Since, $|a|=a]$

So, $R$ is not reflexive

Symmetry:

Let $(a, b) \in R$

$\Rightarrow|a| \leq b$

$\Rightarrow|b| \notin a$ for all $a, b \in R$

$\Rightarrow(b, a) \notin R$

So, $R$ is not symmetric.

Transitivity:

Let $(a, b) \in R$ and $(b, c) \in R$

$\Rightarrow|a| \leq b$ and $|b| \leq c$

Multiplying the corresponding sides, we get

$|a||b| \leq b c$

$\Rightarrow|a| \leq c$

$\Rightarrow(a, c) \in R$

Thus, $R$ is transitive.