Say you have the set $A = \{r\in\mathbb{Q}:\exists\,q,p\in\mathbb{Z},$ with $p$ odd and $q$ even, and $r=\frac{p}{q}\}$, and a relation $R$ on $\mathbb{Q}$ where for $x,y\in A$, then $xRy$ if $x-y\in A$.
I started proving that the relation is an equivalence relation on $\mathbb{Q}$, and proving relfexivity was easy:
Let $x\in\mathbb{Q}$. Then for $p,q\in\mathbb{Z},\,q\neq0,\,x=\frac{p}{q}$, and $x-x=\frac{p}{q}-\frac{p}{q}=0$. Since $0=\frac{0}{1}$, then $0\in A$, and $R$ is reflexive.
Proof of symmetry:
Let $x,y\in\mathbb{Q}$. Then, for $p,q,r,s\in\mathbb{Z}$, with $q,s\neq0$, $x=\frac{p}{q}$ and $y=\frac{r}{s}$. Suppose $xRy$. Then $ps-qr=2k, k\in\mathbb{Z}$ and $qs=2l+1, l\in\mathbb{Z}$. Algebraically manipulating the first statement, we see that $qr-ps=2(-k)$, and since $-k\in\mathbb{Z}$, then $qr-rs$ is even. Then we can see that $y-x=\frac{qr-ps}{qs}\in A$, meaning $yRx$, making $R$ symmetric.
However, using similar methods to prove transitivity, I get caught up. I find that, given $p,q,r,s,m,n\in\mathbb{Z}$, the rational numbers $x=\frac{q}{p}$, $y=\frac{r}{s}$, and $z=\frac{m}{n}$ (with none of the denominators 0, of course).
Since we suppose $xRy$ and $yRz$, then we know that $ps-qr$ and $rn-sm$ are even, and $qs$ and $sn$ are odd.
So $x-z=\frac{q}{p}-\frac{m}{n}$.
What's the easiest way to finish this proof?