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 , pandqa r ei n t e g e r ss u c ht h a tn ,q"!="0andq m = p n"}` . Then

Options

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

Answer

C

Explanation

Case I Given, the relation R is defined as
`R={(x,y)|x,y" are real numbers and x = wy for some rational number w"}`
Reflexive `xRximpliesx=wx`
`:." "w=1in` Rational number
So, the relation R is reflexive.
Symmetric `"xRy"cancelimplies"yRx"`
As, `0R1implies0=0.(1)"but "1R0implies1=w.(0)`,
which is not true for any rational number.
So, the relation R is not symmetric. Hence, R is not equivalence relation. Case II Now, for the relation S defined as
`S={((m)/(n),(p)/(q))}|{m,n,p" and "q" are integers such that "n,q ne0" and "qm=pn}`
Reflexive `(m)/(n)S (m)/(n)impliesmn=mn`
So, the relation S is reflexive.
Symmetric `(m)/(n)S(p)/(q)impliesmq=np`
`implies" "np=mqimplies(p)/(q)S(m)/(n)`
So, the relation S is symmetric.
Transitive `(m)/(n)S(p)/(q)" and "(p)/(q)S(r)/(s)`
`implies" "mq=np" and "ps=rq`
`implies" "mq.ps=np.rq`
`implies" "ms=nrimplies(m)/(n)=(r)/(s)`
`implies" "(m)/(n)S(r)/(s)`
So, the relation S is transitive.
Hence, the relation S is an equivalence relation.