Question

6 Show that the relation “greater thar" on the set of natural numbers N, is transitive but is neither
reflexive nor symmetric.​

Answer 1

Basic Concept :-

Reflexive :-

• Relation is reflexive, If (a, a) ∈ R for every a ∈ A.

Symmetric :-

• Relation is symmetric, If (a, b) ∈ R, then (b, a) ∈ R.

Transitive :-

• Relation is transitive, If (a, b) ∈ R & (b, c) ∈ R, then (a, c) ∈ R.

Let's solve the problem now!!

The given Relation R can be mathematically represent as

$\rm :\longmapsto\:R \: = \{(a, b) : a > b \: where \: a, b \in \: N \}$

Reflexive :-

$\rm :\longmapsto\:Let \: a \in \: N$

$\rm :\longmapsto\:We \: know, \: a \: = \: a$

$\rm :\implies\:a \: \cancel > \: a$

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

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

Symmetric :-

$\rm :\longmapsto\:Let \: a, b \: \in \: N \: such \: that \: (a, b) \in \: R$

$\rm :\implies\:a > b$

$\rm \: \cancel{\implies\:} \: b > a$

$\rm :\implies\:(a, b) \: \cancel \in \: R$

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

Transitive :-

$\rm :\longmapsto\:Let \: a,b,c \in \: N \: such \: that \: (a, b) \in \: R \: and \: (b, c) \in \: R$

$\rm :\longmapsto\:as \: (a, b) \in \: R \rm \: \implies\:a > b - - (1)$

$\rm :\longmapsto\:as \: (b, c) \in \: R \rm \: \implies\:b > c - - (2)$

From equation (1) and equation (2), we concluded

$\rm :\longmapsto\:a > c$

$\rm :\implies\:(a, c) \in \: R$

$\bf\implies \:R \: is \: transitive.$