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Show that the relation R defined by (a, b)R(c,d) ⇒ a + d = b + c on the A x A , where A = {1, 2,3,...,10} is an equivalence relation. Hence write the equivalence class [(3, 4)]; a, b, c,d ∈ A.

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#### Solution

Here (a, b)R(c,d) ⇒ a + d = b + c on A x A, where A = {1, 2,3,...,10} .

Reflexivity: Let (a, b) be an arbitrary element of A x A. Then, (a,b) ∈ A x A `forall` a, b ∈ A.

So, a + b = b + a

⇒ (a,b) R (a,b).

Thus, (a,b) R (a,b) `forall` (a,b) ∈ A x A.

Hence R is reflexive.

Symmetry: Let (a,b), (c,d) ∈ A x A be such that (a,b) R (c,d).

Then, a + d = b + c

⇒ c + b = d + a

⇒ (c,d ) R (a,b).

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

⇒ (c,d) R (a,b) `forall` (a,b), (c,d) ∈ A x A.

Hence R is symmetric.

Transitivity: Let (a,b),(c,d),(e,f) ∈ A x A be such that (a,b) R (c,d) R (e,f).

Then, a + d = b + c and c + f = d + e

⇒ (a+d) + (c+f)

= (b + c) + (d+e)

⇒ a + f = b + e

⇒ (a, b) R (e,f).

That is (a,b) R (c,d) and (c,d) R (e,f)

⇒ (a,b) R (e,f) `forall` (a,b), (c,d), (e,f) ∈ A x A.

Hence R is transitive.

Since R is reflexive, symmetric and transitive so, R is an equivalence relation as well.

For the equivalence class of [(3, 4)], we need to find (a,b) s.t. (a,b) R (3,4)

⇒ a + 4 = b + 3

⇒ b - a = 1.

So, [(3,4)] = {(1,2),(2,3),(3,4),(4,5),(5,6),(6,7),(7,8),(8,9),(9,10)}.

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