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Asked by jhaprabhat920 | 05 Feb, 2024, 10:33: AM

If A = {1, 2, 3, ..., 9} and R is the relation in A x A defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in A x A. Prove that R is an equivalence relation. Also, obtain an equivalence class [(2, 5)].

Solution:

Reflexive:

a + b = b + a

i.e. (a, b) R (a, b)

So, R is reflexive.

Symmetric:

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

i.e. a + d = b + c

i.e. c + b = d + a

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

So, R is symmetric.

Transitive:

Let (a, b) R (c, d)  & (c, d) R (e, f)

i.e. a + d = b + c     &     c + f = d + e

Since, a + d = b + c

Therefore, a + d + f = b + c + f

i.e. (a + f) + d = b + (c + f)

i.e. (a + f) + d = b + d + e

i.e. a + f = b + e

So, (a, b) R (e, f).

Thus, R is transitive.

Hence, R is an equivalence relation.

To find an equivalence class [(2, 5)], consider (2, 5) R (a, b)

i.e. 2 + b = 5 + a

i.e. b = 3 + a

As A = {1, 2, 3, ...} and (a, b) in A x A.

[(2, 5)] = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}

Answered by Renu Varma | 06 Feb, 2024, 11:20: AM

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