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:
Answer & explanation
Correct answer: option 1
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