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 and also obtain the equivalent class [(2, 5)].
Given, A = {1, 2, 3, … 9} and (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) ∈ A ×A.
Let (a, b) R(a, b)
So, a + b = b + a, ∀ a, b ∈ A which is true for any a, b ∈ A.
Thus, R is reflexive.
Let (a, b) R(c, d)
Then,
a + d = b + c
c + b = d + a
(c, d) R(a, b)
Thus, R is symmetric.
Let (a, b) R(c, d) and (c, d) R(e, f)
a + d = b + c and c + f = d + e
a + d = b + c and d + e = c + f
(a + d) – (d + e = (b + c) – (c + f)
a – e = b – f
a + f = b + e
(a, b) R(e, f)
So, R is transitive.
Therefore, R is an equivalence relation.
And, [(2,5)=(1,4)(2,5),(3,6),(4,7),(5,8),(6,9)] is the equivalent class under relation R.
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