A = {1,2,3,....9,10}

(a, b) R(c, d) ≡ a + d  = b + c

Reflexivity: Let (a, b) \(\in\) A x A be any arbitrary element

\(\because\) Relation R is reflexive.

Symmetricity: Let (a, b), (c, d) \(\in\) A x A such that

(a, b) R (c, d)

Then a + d = b + c

⇒ a + d = c + d

⇒ c + b = a + d

⇒ c + b = d + a

⇒ (c, d) R (a, b)

\(\therefore\) Relation R is a symmetric relation.

Transitivity: Let (a, b), (c, d), (e, f) \(\in\) A x A such that

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

We have (a, b) R (c, d)

\(\therefore\) a + b = b + c---(1)

Also, (c, d) R (e, f)

\(\therefore\) c + f = d + e---(2)

By adding (1) & (2), we get

a + d + c + f = b + c + d + e

⇒ a + f = b + c

⇒ (a, b) R (e, f)

\(\therefore\) Relation R is a transitive relation.

Since, relation R is reflexive, symmetric & transitive relation.

\(\therefore\) Relation R is an equivalence relation.

Equivalence class of (3, 4)

\(\because\) (3, 4) R (3, 4)

(3, 4) R (1, 2) (\(\because\) 3 + 2 = 4 + 1)

(3, 4) R (2, 3) (\(\because\) 3 + 3 = 4 + 2)

(3, 4) R (4, 5) (\(\because\) 3 + 5 = 4 + 4)

(3, 4) R (5, 6) (\(\because\) 3 + 6 = 4 + 5)

(3, 4) R (6, 7) (\(\because\) 3 + 7 = 4 + 6)

(3, 4) R (7, 8) (\(\because\) 3 + 8 = 4 + 7)

(3, 4) R (8, 9) (\(\because\) 3 + 9 = 4 + 8)

(3, 4) R (9, 10) (\(\because\) 3 + 10 = 4 + 9)

\(\therefore\) Equivalence  class of (3, 4) is

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