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Let N denote the set of all natural numbers and R be the relation on N X N defined by

( a,b ) R ( c,d ) both sided arrow ad ( b + c ) = bc ( a + d )

Prove that R is an equivalence relation on N x N.

$We observe the following properties on Relation R Reflexicity: Let (a,b) be an arbitary element of N \times N.Then (a, b) \in N \times N \Rightarrow a, b \in N \Rightarrow ab(b+a)=ba(a+b) \Rightarrow (a, b) R (a, b) Thus , (a,b)R(a,b) for all (a,b) \in N \times N.So ,R is reflexive on N \times N Symmetry : Let (a,b),(c,d) \in N \times N be such that (a,b)R(c,d).Then (a, b) R (c, d) \Rightarrow ad(b+c)=bc(a+d) \Rightarrow cb(d+a)=da(c+b) \Rightarrow (c,d)R(a,b) Thus (a, b) R(c,d) \Rightarrow (c,d)R(a,b) for all (a,b),(c,d) \in N \times N So,R is symmetric on N \times N Transitiv ity: Let (a,b),(c,d),(e,f) \in N \times N such that(a,b)R(c,d) and (c,d)R(e,f) then (a,b)R(c,d) \Rightarrow ad(b+c)=bc(a+d) = \frac{b + c}{b c} = \frac{a + d}{a d} \Rightarrow \frac{1}{b} + \frac{1}{c} = \frac{1}{a} + \frac{1}{d} - - - - (1) and (c,d)R(e,f) \Rightarrow ef(d+e)=de(c+f) \Rightarrow \frac{d + c}{d e} = \frac{c + f}{c f} \Rightarrow \frac{1}{d} + \frac{1}{e} = \frac{1}{f} + \frac{1}{c} - - - - (2) adding (1) and (2) we get (\frac{1}{b} + \frac{1}{c}) + (\frac{1}{d} + \frac{1}{e}) = (\frac{1}{a} + \frac{1}{d}) + (\frac{1}{f} + \frac{1}{c}) \Rightarrow \frac{1}{b} + \frac{1}{e} = \frac{1}{a} + \frac{1}{f} \Rightarrow \frac{b + e}{b e} = \frac{a + f}{a f} \Rightarrow af(b+e)=be(a+f) \Rightarrow (a,b)R(e,f) Thus (a,b)R(c,d) and (c,d)R(e,f) \Rightarrow (a,b)R(e,f) for all (a,b),(c,d),(e,f) \in N \times N So R is transitive on N \times N. Hence R being Reflexive,Symmetric,Transitive is an Equivalence relation on N \times N$

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