Let S be the set of all points in a plane and let R be a relation in S

1.	Let S be the set of all points in a plane and let R be a relation in S defined by R = {(A, B) : d(A, B) < 2 units}, where d(A, B) is the distance between the points A and B. Show that R is reflexive and symmetric but not transitive.
Answer» Given that, ∀ A, B ∈ S, R = {(A, B) : d(A, B) < 2 units}. Now, R is Reflexive if (A,A) ∈ R ∀ A ∈ S For any A ∈ S, we have d(A,A) = 0, which is less than 2 units ⇒ (A,A) ∈ R Thus, R is reflexive. R is Symmetric if (A, B) ∈ R ⇒ (B,A) ∈ R ∀ A,B ∈ S (A, B) ∈ R ⇒ d(A, B) < 2 units ⇒ d(B, A) < 2 units ⇒ (B,A) ∈ R Thus, R is symmetric . R is Transitive if (A, B) ∈ R and (B,C) ∈ R ⇒ (A,C) ∈ R ∀ A,B,C ∈ S Consider points A(0,0),B(1.5,0) and C(3.2,0). d(A,B)=1.5 units < 2 units and d(B,C)=1.7 units < 2 units d(A,C)= 3.2 ≮ 2 ⇒ (A, B) ∈ R and (B,C) ∈ R ⇒ (A,C) ∉ R Thus, R is not transitive. Thus, R is reflexive, symmetric but not transitive.

Let S be the set of all points in a plane and let R be a relation in S defined by R = {(A, B) : d(A, B) < 2 units}, where d(A, B) is the distance between the points A and B. Show that R is reflexive and symmetric but not transitive.

Answer»

Given that, ∀ A, B ∈ S, R = {(A, B) : d(A, B) < 2 units}.

Now,

R is Reflexive if (A,A) ∈ R ∀ A ∈ S

For any A ∈ S, we have

d(A,A) = 0, which is less than 2 units

⇒ (A,A) ∈ R

Thus, R is reflexive.

R is Symmetric if (A, B) ∈ R ⇒ (B,A) ∈ R ∀ A,B ∈ S

(A, B) ∈ R

⇒ d(A, B) < 2 units

⇒ d(B, A) < 2 units

⇒ (B,A) ∈ R

Thus, R is symmetric .

R is Transitive if (A, B) ∈ R and (B,C) ∈ R ⇒ (A,C) ∈ R ∀ A,B,C ∈ S

Consider points A(0,0),B(1.5,0) and C(3.2,0).

d(A,B)=1.5 units < 2 units and d(B,C)=1.7 units < 2 units

d(A,C)= 3.2 ≮ 2

⇒ (A, B) ∈ R and (B,C) ∈ R ⇒ (A,C) ∉ R

Thus, R is not transitive.

Thus, R is reflexive, symmetric but not transitive.