Answer :

Given: A = {1,2,3......10}

and R is defined as

(a,b) R (c,d) ⇒ a + b = b + c

__Reflexive:__

Consider (a,b) R (a,b) ⩝(a,b) ∈ A × A

⇒ a + b = b + a

Hence, R is reflexive

__Symmetric:__

Consider (a,b)R(c,d) given by (a,b)(c,d) ∈ A × A

(a,b) R (c,d) ⇒ a + d = b + c

⇒ b + c = a + d

⇒ c + b = d + a

⇒ (c,d) R (a,b)

Hence R is symmetric

__Transitive:__

Let (a,b) R (c,d) and (c,d) R (e,f) given by (a,b),(c,d),(e,f)∈ A × A

(a,b) R (c,d) ⇒ a + b = b + c

⇒ a – c = b – d …(1)

and (c,d) R (e,f) ⇒ c + f = d + e

& c + f = d + e …(2)

adding (1) and (2), we get

a – c + c + f = b – d + d + e

⇒ a + f = b + e

⇒ (a,b) R (e,f)

Hence, R is transitive

Hence, R is an equivalence relation

We need to find [(3,4)]

So, (3, 4) will go in and (c, d) will come out

This will be possible if

a + d = b + c

⇒ 3 + d = 4 + c

⇒ d – c = 4 – 3

⇒ d – c = 1

So, in our relation [(3,4)]

We need to find values of c and d which satisfy d – c = 1

Since (c, d) ∊ A × A

Both c and d are in set A = {1, 2, 3, …10}

[(3,4)]={(1,2)(2,3),(3,4),(4,5),(5,6),(6,7),(7,8),(8,9),(9,10)}is the equivalent class under relation R

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