Let N be the set of all natural niumbers 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 .

Explanation

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 AA 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`
`implies (a " is a factor of b) and ( b is a factor of c)."`
`implies b-ad and c=be ` for some `d,ein N`
`implies c=(ad)e=a (de ) ` [ by associative law]
`implies a " is a factor of c"`
` implies (a,c) in R `
`therefore (A,b) in R and ( b,c) in R implies (a,c) in R.`
hence , R is transitive .
(iii) Nonsymmery
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) !in R.`
thus,`(2,6)in R and (6,2)!in R.`
hence ,R is not symmertric .