Consider the following relations:
`R={(x,y)|(x,y " are real numbers and " x=wy " for some rational number" w}, `
` S={((m)/(n),(p)/(q))m, n,p, and q` are integers such that `n, q ne 0 and qm=pm}`. Then

Options

A. S is an equivalence relation but R is not an equivalence relation
B. R and S both are equivalence relations
C. R is an equivalence relation but S is not an equivalence relation
D. neither R nor S is an equivalence relation

Answer

A

Explanation

We observe that
`(0, 2) in R`, because `0=0xx2`.
But, `(2, 0) cancelin R` because `2 ne` (Any rational number) `xx 0`
So, R is not a symmetric relation and hence it is not an equivalence relation.
We observe the following properties of relation S:
Reflexivity: Let m, n `in Z` such that `n ne 0`. Then,
`mn=nmimplies(m)/(n)=(m)/(n)implies((m)/(n),(m)/(n))inSimpliesS` is reflexive.
Symmetry: Let `((m)/(n),(p)/(q))inS`. Then,
`mq=npimpliespn=qmimplies((p)/(q),(m)/(n))inS`
So, S is symmetric.
Transivity: Let `((m)/(n),(p)/(q))inS and ((p)/(q),(r)/(s))inS`. Then,
`qm=pnand sp=rq`
`impliesqmxxsp=pnxxrqimpliesms=nrimplies((m)/(n),(r)/(s))inS`
So, S is transitive.
Hence, it is an equivalence relation.