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Question

Prove that the relation R on set Nx N (a,b) R(c,d) which implies a+d=b+c is an equivalence relation

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Solution

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We observe the following properties of relation R:Reflexivity: Let (a,b) be an arbitrary element of N×N.Then, (a,b)N×Na,bNa+b=b+a [By commutativity of add. on N](a,b)R(a,b)Thus, (a,b)R(a,b) for all (a,b)N×N.So, R is reflexive on N×N.Symmetry: Let (a,b),(c,d)N×N be such that (a,b)Rc,d.Then,(a,b)R(c,d)a+d=b+cc+d=d+a [By commutativity of add. on N](c,d)R(a,b) [By defn. of R]Thus, (a,b)R(c,d)(c,d)R(a,b) for all (a,b),(c,d)N×N.So, R is symmetric on N×N.Transitivity: Let (a,b),(c,d),(e,f)N×N such that (a,b)R(c,d) and (c,d)R(e,f).Then,a,bRc,da+d=b+c(c,d)R(e,f)c+f=d+ea+d+c+f=(b+c)+(d+e)a+f=b+e(a,b)R(e,f)Thus, (a,b)R(c,d) and (c,d)R(e,f)(a,b)R(e,f) for all (a,b),(c,d),(e,f)N×N.So, R is transitive on N×N.Hence, R being reflexive, symmetric and transitive,is an equivalence relation onN×N.
Regards

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