Prove that the relation R on the set `NxN` defined by `(a , b)R(c , d) a+d=b+c` for all `(a , b),(c , d) in NxN` is an equivalence relation.

Explanation

Let (a, b) in N `xx` N We know that
`a + b = b + a`
`. ""_("(a,b)")R_("(a,b)")`
`rArr` R is reflexive.
(ii) Let `(a,b), (c, d)in NxxN and _("(a,b)")R_("(c,d)")`
`rArr a+d=b+c`
`rArrb+c=a+d`
`rArrc+b=d+a`
`rArr _("(c,d)")R_("(a,b)")`
`:.` R is symmetric.
(iii) Let `rArr a+d=b+c`
`rArrb+c=a+d`
`rArrc+b=d+a`
`rArr _("(c,d)")R_("(a,b)")`
`rArr a + d = b+c and c+f=d+e`
`rArr a+d+c+f = b+c+d+e`
`rArr a+ f = b +e `
`rArr "" _("(a,b)")R_("(e,f)")`
`rArr` R is transitive.
`:.` R is reflexive, symmetric and transitive.
`rArr` R is an equivalence relation . Hence Proved .