Show that the relation R on N × N, defined by

(a, b) R (c, d) a + d = b + c


is an equivalent relation.



In order to show R is an equivalence relation we need to show R is Reflexive, Symmetric and Transitive.


Given that, R be the relation in N ×N defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in N ×N.


R is Reflexive if (a, b) R (a, b) for (a, b) in N ×N


Let (a,b) R (a,b)


a+b = b+a


which is true since addition is commutative on N.


R is reflexive.


R is Symmetric if (a,b) R (c,d) (c,d) R (a,b) for (a, b), (c, d) in N ×N


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


a+d = b+c


b+c = a+d


c+b = d+a [since addition is commutative on N]


(c,d) R (a,b)


R is symmetric.


R is Transitive if (a,b) R (c,d) and (c,d) R (e,f) (a,b) R (e,f) for (a, b), (c, d),(e,f) in N ×N


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


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


(a+d) – (d+e) = (b+c ) – (c+f)


a-e= b-f


a+f = b+e


(a,b) R (e,f)


R is transitive.


Hence, R is an equivalence relation.


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