# Let ( N ) be the set of all natural numbers and let ( R ) be a relation in ( N ), defined by ( R={(a, b)}: a ) is a factor of ( b} ) Then, show that ( R ) is reflexive and transitive but not symmetric.

Here, ( R ) satisfies the following properties:

(i) Reflexivity Let ( a ) be an arbitrary element of ( N ).

Then, clearly, ( a ) is a factor of ( a ).

( therefore(a, a) in R forall a in N )

So, ( R ) is reflexive.

(ii) Transitivity Let ( a, b, c in N ) such that ( (a, b) in R ) and ( (b, c) in R )

Now, ( (a, b) in R ) and ( (b, c) in R )

( Rightarrow quad(a ) is a factor of ( b) ) and ( (b ) is a factor of ( c) )

( Rightarrow quad b=a d ) and ( c=b e ) for some ( d, e in N )

( Rightarrow quad c=(a d) e=a(d e) )

[by associative law]

( Rightarrow a ) is a factor of ( c )

( Rightarrow(a, c) in R )

( therefore(a, b) in R ) and ( (b, c) in R Rightarrow(a, c) in R )

Hence, ( R ) is transitive.

(iii) Nonsymmetry Clearly, 2 and 6 are natural numbers and 2 is a factor of ( 6 . )

( therefore(2,6) in R )

But, 6 is not a factor of 2 .

( therefore(6,2) otin R )

Thus, (2,6)( in R ) and (6,2)( otin R )

Hence, ( R ) is not symmetric.