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Question

Let N denote the set of all natural numbers and R be the relation on N×N defined by (a,b)R(c,d), if ad(b+c)=bc(a+d), then show that R is an equivalence relation.

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Solution

(a,b)R(c,d)ad(b+c)=bc(a+d)
ab(b+a)=ba(a+b)(a,b)R(a,b)
Thus, the given relation is reflexive
(a,b)R(c,d)ad(b+c)=bc(a+d)ad(b+c)=bc(a+d)cb(d+a)=da(c+b)(c,d)R(a,b)(a,b)R(c,d)(c,d)R(a,b)
Thus, the given relation is symmetric
(a,b)R(c,d)ad(b+c)=bc(a+d)1b+1c=1a+1d.....(1)(c,d)R(e,f)cf(d+e)=de(c+f)1f+1c=1e+1d....(2)(2)(1)1f1b=1e1a1f+1a=1e+1baf(b+e)=be(a+f)(a,b)R(e,f)(a,b)R(c,d)&(c,d)R(e,f)(a,b)R(e,f)
Thus, the given relation is transitive.
Thus, the given relation is an equivalence relation.

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