Let R be a relation on the set N of naturalnumbers defined by `nRm n` is a factor of m(ie. nim) Then R isA. R is reflexive, symmetric but not transitiveB. R is transitive, symmetric but not reflexiveC. R is reflexive, transitive but not symmetricD. R is an equivalence relation

Let R be a relation on the set N of naturalnumbers defined by `nRm n`

1.	Let R be a relation on the set N of naturalnumbers defined by `nRm n` is a factor of m(ie. nim) Then R isA. R is reflexive, symmetric but not transitiveB. R is transitive, symmetric but not reflexiveC. R is reflexive, transitive but not symmetricD. R is an equivalence relation
Answer» Correct Answer - C nRm `iff` n is a factor of m. implies m is divisible by n. Reflexivity We know that n is divisible by `nAAn in N` `(n, n) inRAAn inN` R is reflexive. Symmetric `n, m in N` Let n=2, m=6 m is divisible by n but n is not divisible by m. Hence R is not symmetric. Transtivity Let `(n,m)in R and (m,p) in R " then "(n, m) in R and (m, p)inRimplies(n, p)in R` or If m is divisible by n and p is divisible by m. Hence p is divisible by n. `(n, p)inR AAn, p in N` R is transitive relation on N. Hence R is reflexive, transitive but not symmetric. `therefore underset(0)overset(2a)intf(x)dx=2underset(0)overset(a)intf(x)dx if f(2a-x)=fx`