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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