Let S be set of all real numbers ; then on se | vedclass

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Question

$R=\{(a, b): 1+a b>0\}$

Checking for reflexive

If the relation is reflexive,

then $(a, a) \in R$

i.e. $1+a^{2}>0$

since square nu

Vedclass · Accepted Answer

Reflexive and symmetric but not transitive

Vedclass · Answer

$R=\{(a, b): 1+a b>0\}$

Checking for reflexive

If the relation is reflexive,

then $(a, a) \in R$

i.e. $1+a^{2}>0$

since square numbers are always positive

Hence, $1+a^{2}>0$ is true for all values of a.

So, the given relation it is reflexive.

Checking for symmetric,

To check whether symmetric or not,

If $(a, b) \in R,$ then $(b, a) \in R$

i.e., if $1+a b<0$

then $1+ ba >0$

since if $1+a b>0,$ then $1+b a>0$

is always true for all value of a $\&$ b

Hence, the given relation is symmetric

Checking transitive

To check whether transitive or not,

If $(a, b) \in R \&(b, c) \in R,$ then $(a, c) \in R$

i.e., if $1+a b>0, \& 1+b c>0,$ then $1+a c>0$

Let's take an example

$a=-8, b=-2, c=\frac{1}{4}$

$1+a b=1+(-8) 	imes(-2)=1+16=17>0$

$1+b c=1+(-2) 	imes \frac{1}{4} \quad=1-\frac{1}{2} \quad=\frac{1}{2}>0$

$1+a c=1+(-8) 	imes \frac{1}{4} \quad=1-2 \quad=-1>0$

since $1+\operatorname{ac} \gg 0$

when $1+a b>0$ and $1+b c>0$

$	herefore$ The condition is not true for all values of $a, b, c$

Hence, the given relation it is not transitive

Therefore, the given relation is reflexive and symmetric, but not transitive

Let $S$ be set of all real numbers ; then on set $S$ relation $R$ defined as $R = \{\ (a, b) : 1 + ab > 0\ \}$ is

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