Question

In the set of real numbers R, two relations R₁ and R₂ can be defined as

(i) aR₁b ⇔ a – b > 0

(ii) aR₂b ⇔ |a| ≤ b

Also test the reflexivity, symmetricity and transitivity of R₁ and R₂.

Answer 1

(i) Given set R = Set of real numbers

Reflexivity R aR₁b ⇔ a – b > 0 ∀ a, b ∈ R

Reflexivity : Let a ∈ R

a ∈ R

⇒ a – a = 0 > 0

⇒ a – a  0

⇒ (a, a) ∉ R₁ a ∈ R

R is not reflexive relation.

Symmetricity : Let a, b ∈ R are such that (a, b) ∈ R₁

(a, b) ∈ R₁

⇒ a – b > 0

⇒ b – a < 0

⇒ (b, a) ∉ R₁

So, (a, b) ∈ R₁

⇒ (b, a) ∉ R₁

R₁ is not symmetric.

Transitivity : Let a, b, c ∈ R are such that

(a, b) ∈ R₁, (b, c) ∈ R₁

(a, b) ∈ R₁

⇒ a – b > 0 …(1)

(b, c) ∈ R₁ ⇒ b – c > 0 …(2)

and a – c = (a – b) + (b – c) > 0 [∴ From (1) and (2)]

⇒ a – c > 0

⇒ (a, c) ∈ R₁

So, (a, b) ∈ R₁, (b, c) ∈ R₁

⇒ (a, c) ∈ R₁ ∀ a, b, c ∈ R

R₁ is transitive.

From above it is clear that R₁ is not reflexive and symmetric relation.

It is only a transitive relation.

(ii) Given: Relation R = set of real numbers

A relation R₂ in R is defined as

aR₂b ⇔ |a| ≤ b ∀ a, b ∈ R

Reflexivity: Let a ∈ R

a ∈ R ⇒ |a| ≤ a is not necessary

For example a = -2 ∈ R

and |-2| ≤ -2 ⇒ (-2, -2) ∈ R₂

So, R₂ is not a reflexive relation.

Symmetricity: 

Let (a, b) ∈ R are such that (a, b) ∈ R₂

(a, b) ∈ R₂ ⇒ |a| ≤ b

Then |b| ≤ a is not necessary.

For example : a = -2, b = 3

and (-2, 3) ∉ R₂ as |-2| ≤ 3

But (3, -2) ∉ R₂ because |3| ≤ -2

R₂ is not symmetric.

Transitivity : 

Let a, b, c ∈ R are such that

(a, b) ∈ R₂ and (b, c) ∈ R₂

(a, b) ∈ R₂ ⇒ |a| ≤ b …(1)

(b, c) ∈ R₂ ⇒ |b| ≤ c …(2)

From equation (1) and (2),

|a| ≤ b ≤ |b| ≤ c

⇒ |a| ≤ c

⇒ (a, c) ∈ R₂

So, (a, b) ∈ R₂, (b, c) ∈ R₂

⇒ (a, c) ∈ R₂ ∀ a, b, c ∈ R

R₂ is transitive.

Hence, from above it is clear that R₂ is not reflexive and symmetric relation.

It is only a transitive relation.