Let $A= \{1, 2, 3\}$, then the possible equivalence relations on A are:

(A) $\{(1, 1), (2, 2), (3, 3)\}$
(B) $\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$
(C) $\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3)\}$
(D) $\{(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)\}$

Choose the correct answer from the options given below:

(A), (B) and (D) only

The correct answer is Option (1) → (A), (B) and (D) only

(A) $\{(1, 1), (2, 2), (3, 3)\}$
(B) $\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$
(D) $\{(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)\}$

Set A = {1, 2, 3}

An equivalence relation must be:

• Reflexive: $(a, a)$ ∈ R ∀ a ∈ A
• Symmetric: If $(a, b)$ ∈ R, then $(b, a)$ ∈ R
• Transitive: If $(a, b)$ ∈ R and $(b, c)$ ∈ R, then $(a, c)$ ∈ R

(A) R = { (1,1), (2,2), (3,3) } 

Reflexive:  All (a,a) included

Symmetric:  Only (a,a) → symmetric

Transitive:  Holds trivially

⇒ Valid equivalence relation

(B) R = { (1,1), (2,2), (3,3), (1,2), (2,1) } 

Reflexive: 

Symmetric: 

Transitive:  Since (1,2), (2,1) ∈ R ⇒ (1,1) is already present, but (1,2), (2,1) ⇒ (2,2) ⇒ holds ⇒ transitive ✔️

⇒ Valid equivalence relation

(C) R = { (1,1), (2,2), (3,3), (1,2), (2,1), (1,3) } 

Symmetric: (1,3) ∈ R but (3,1) 

⇒ Not symmetric ⇒ Not an equivalence relation

(D) R = { (1,1), (2,2), (3,3), (1,3), (3,1) } 

Symmetric: 

Transitive: (1,3), (3,1) ⇒ (1,1) ∈ R 

Reflexive: 

Transitive: (1,3), (3,1) ⇒ (1,1) . No other violations ⇒ holds

⇒ Valid equivalence relation