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Solution :

Here R is satisfies the following properties : <br> (i) Reflexivity <br> Let A be an arbitrary element of N. <br> then ,` a = (axx1)` shows that a is a multiple of a . <br> ` therefore (a,a) in R R AA a in N.` <br> So ,R is reflexive <br>(ii) Transitivity <br> Let ` a,b,c in N` such that `(a,b) in R and (b,c) in R.` <br> NOw ,`(a,b) in R and ( b, c) in R.` <br> `implies (" a is multiple of b ) and ( b is a multiple of c)"` <br> `implies a=bd and b= ce ` for some ` d in N and e in N ` <br> `implies a = (ce) d` <br> `implies a= c (ed) ` <br> `implies " a is a multiple of c"` <br> `implies (A,c) in R ` <br> ` therefore (a,b) in R and ( b,c ) in R implies (a,c)in R.` <br> hence ,R is transitive . <br> (iii) Nonsymmetry <br> Clearly . 6 and 2 are natural numbers and 6 is a multiple of 2. <br>` therefore (6,2) in R.` <br> But , 2 is not a multiple of 6. <br> `therefore (2,6) !in R .` <br> thus ,`(6,2) in R and (2,6) !in R.` <br> Hence ,R is not symmetric.