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
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)
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.