Consider the following relations

R = {(x,y): x, y are real numbers and x = wy for some rational number w};

S = {$\left(\frac{m}{n},\frac{p}{q}\right)$: m, n, p and q are integers such that $n,q\ne 0$ and qm = pn}

We have $(x,x)\in R$ for w = 1 implying that R is reflexive

For $a\ne 0,(a,0)\notin R$ for any w; but $(0,a)\in R$. Thus R is not symmetric

Hence R is not an equivalence relation 

As $\left(\frac{{\displaystyle m}}{{\displaystyle n}},\frac{{\displaystyle m}}{{\displaystyle n}}\right)\in S$ since mn = mn.

$\therefore S$ is reflexive

$\left(\frac{{\displaystyle m}}{{\displaystyle n}},\frac{{\displaystyle p}}{{\displaystyle q}}\right)\in S\Rightarrow qm=pn\Rightarrow np=mq\Rightarrow \left(\frac{{\displaystyle p}}{{\displaystyle q}},\frac{{\displaystyle m}}{{\displaystyle n}}\right)\in S$

Thus S is symmetric

$\left(\frac{m}{n},\frac{p}{q}\right)\in S and \left(\frac{p}{q},\frac{a}{b}\right)\in S\Rightarrow qm=pn and bp=aq\Rightarrow \frac{m}{n}=\frac{p}{q} and \frac{p}{q}=\frac{a}{b}\Rightarrow \frac{m}{n}=\frac{a}{b}\Rightarrow \left(\frac{{\displaystyle m}}{{\displaystyle n}},\frac{{\displaystyle a}}{{\displaystyle b}}\right)\in S$

Thus S is transitive