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  • 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)].

Let A = {1, 2, 3, … 9} and R be the relation in A \times A defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in A \times A. Prove that R is an equivalence relation and also obtain the equivalent class [(2, 5)].

Answers (1)

\\Here, A = \{ 1, 2, 3, \ldots 9 \} \: \: and \: \: (a, b) R (c, d)\: \: if a + d = b + c for (a, b), (c, d) \in A \times A.\\ Say (a, b) R (a, b)\\ Therefore, a + b = b + a, \forall a, b \in A .

This must be true for any a, b \in A.\\

Hence, R is reflexive.

Say, (a, b) R (c, d)

Then,

\\a + d = b + c\\ c + b = d + a\\ (c, d) R (a, b)\\

Therefore, 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\\ Or, (a + d) - (d + e = (b + c) - (c + f)\\ Or, a - e = b - f\\ Or, a + f = b + e\\ Or, (a, b) R(e, f)\\

So, R is transitive.

Therefore, R is an equivalence relation.

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