# Mark the tick against the correct answer in the following:

Question:

Mark the tick against the correct answer in the following:

Let $S$ be the set of all real numbers and let $R$ be a relation on $S$, defined by $a R b \Leftrightarrow(1+a b)>0$. Then, $R$ is

A. reflexive and symmetric but not transitive

B. reflexive and transitive but not symmetric

C. symmetric and transitive but not reflexive

D. none of these

Solution:

According to the question,

Given set $S=\{\ldots \ldots,-2,-1,0,1,2 \ldots \ldots\}$

And $R=\{(a, b): a, b \in S$ and $(1+a b)>0\}$

Formula

For a relation $R$ in set $A$

Reflexive

The relation is reflexive if $(a, a) \in R$ for every $a \in A$

Symmetric

The relation is Symmetric if $(a, b) \in R$, then $(b, a) \in R$

Transitive

Relation is Transitive if $(a, b) \in R \&(b, c) \in R$, then $(a, c) \in R$

Equivalence

If the relation is reflexive, symmetric and transitive, it is an equivalence relation.

Check for reflexive

Consider, $(a, a)$

$\therefore(1+a \times a)>0$ which is always true because a $\times$ a will always be positive.

Ex_if $\mathrm{a}=2$

$\therefore(1+4)>0 \Rightarrow(5)>0$ which is true.

Therefore , R is reflexive ……. (1)

Check for symmetric

$a R b \Rightarrow(1+a b)>0$

b $\mathrm{R} \mathrm{a} \Rightarrow(1+\mathrm{ba})>0$

Both the equation are the same and therefore will always be true.

$E x_{-}$If $a=2$ and $b=1$

$\therefore(1+2 \times 1)>0$ is true and $(1+1 \times 2)>$ which is also true.

Therefore , R is symmetric ……. (2)

Check for transitive

$a R b \Rightarrow(1+a b)>0$

$b R c \Rightarrow(1+b c)>0$

$\therefore(1+\mathrm{ac})>0$ will not always be true

But $(1+-1 \times 2)>0$ is false.

Therefore , R is not transitive ……. (3)

Now , according to the equations (1) , (2) , (3)