Question

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Answer 1

$\large\underline{\sf{Solution-}}$

Give relation is

$\rm :\longmapsto\:R = \{(a,b) : a \leqslant b \} \: \forall \: a, \: b \: \in \: R$

Reflexive

We observe that

$\rm :\longmapsto\:0.1 \cancel \leqslant \: 0.01$

$\rm :\implies\:0.1 \: \cancel \leqslant \: {(0.1)}^{2}$

$\rm :\implies\:(0.1, \: 0.1) \: \cancel \in \: R$

$\bf\implies \:R \: is \: not \: reflexive$

Symmetric

We observe that

$\rm :\longmapsto\: - 1 < 4$

$\rm :\longmapsto\: - 1 < {2}^{2}$

$\rm :\implies\:( - 1,2) \: \in \: R$

But

$\rm :\longmapsto\:2 \: \cancel < {( - 1)}^{2}$

$\rm :\implies\:(2, - 1) \: \cancel \in \: R$

$\bf\implies \:R \: is \: not \: symmetric$

Transitive

We observe that

$\rm :\longmapsto\:3 < 16$

$\rm :\longmapsto\:3 < {( - 4)}^{2}$

$\rm :\implies\:(3, - 4) \: \in \: R$

Also,

$\rm :\longmapsto\: - 4 < 1$

$\rm :\longmapsto\: - 4 < {(1)}^{2}$

$\rm :\implies\:( - 4, 1) \: \in \: R$

Also,

$\rm :\longmapsto\:3 \: \cancel \leqslant \: 1$

$\rm :\longmapsto\:3 \: \cancel \leqslant \: {(1)}^{2}$

$\rm :\implies\:( 3, 1) \: \cancel\in \: R$

Now, we have

$\rm :\implies\:(3, - 4) \: \in \: R$

$\rm :\implies\:( - 4, 1) \: \in \: R$

but

$\rm :\implies\:( 3, 1) \: \cancel\in \: R$

$\bf\implies \:R \: is \: not \: transitive$

Basic Concept Used :-

Let R be a relation defined on set A then

1. Relation R is Reflexive if (a, a) ∈ R for all a ∈ A.

2. Relation R is symmetric if (a, b) ∈ R then (b, a) ∈ R for all a, b ∈ A.

3. Relation R is transitive if (a, b) ∈ R, (b, c) ∈ R then (a, c) ∈ R for all a, b, c ∈ A