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Question
Let A = {1, 2, 3,......, 9} and R be the relation in A × A defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in A × A. Prove that R is an equivalence relation. Also, obtain the equivalence class [(2, 5)].
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Solution
A = {1, 2, 3, ..., 9} ⊂ ℕ, the set of natural numbers
Let R be the relation in A × A defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in A × A.
We have to show that R is an equivalence relation.
Reflexivity:
Let (a, b) be an arbitrary element of A × A. Then, we have:
(a, b) ∈ A × A
⇒ a, b ∈ A
⇒ a + b = b + a (by commutativity of addition on A ⊂ ℕ)
⇒ (a, b) R (a, b)
Thus, (a, b) R (a, b) for all (a, b) ∈ A × A.
So, R is reflexive.
Symmetry:
Let (a, b), (c, d) ∈ A × A such that (a, b) R (c, d).
a + d = b + c
⇒ b + c = a + d
⇒ c + b = d + a (by commutativity of addition on A ⊂ ℕ)
⇒ (c, d) R (a, b)
Thus, (a, b) R (c, d) ⇒ (c, d) R (a, b) for all (a, b), (c, d) ∈ A × A.
So, R is symmetric
Transitivity:
Let (a, b), (c, d), (e, f) ∈ A × A such that (a, b) R (c, d) and (c, d) R (e, f). Then, we have:
(a, b) R (c, d)
⇒ a + d = b + c ... (1)
(c, d) R (e, f)
⇒ c + f = d + e ... (2)
Adding equations (1) and (2), we get:
(a + d) + (c + f) = (b + c) + (d + e)
⇒ a + f = b + e
⇒ (a, b) R (e, f)
Thus, (a, b) R (c, d) and (c, d) R (e, f) ⇒ (a, b) R (e, f) for all (a, b), (c, d), (e, f) ∈ A × A.
So, R is transitive on A × A.
Thus, R is reflexive, symmetric and transitive.
∴ R is an equivalence relation.
To write the equivalence class of [(2, 5)], we need to search all the elements of the type (a, b) such that 2 + b = 5 + a.
∴ Equivalence class of [(2, 5)] = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}
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