How many distinct equivalence classes does this equivalence on rationals have? Let $$A = \{ r\in \mathbb Q \mid \exists p\in \mathbb Z,\text{ and $q\in \mathbb Z$, with $p$ even and $q$ odd, and $r = p/q$} \}$$
For example, $A$ contains such $2/9, 16/(-34)$, and $4$. $A$ does not contain $9/10$, or $-15$ , or $18/32$.
Let $\mathbb Q$ be the set of all rational numbers and let the relation ~ be defined by $x \sim y$ if and only if $x − y ∈ A$.
And I need help for:

How many distinct equivalence classes does ~ have$?$ Describe them.

I think it's an infinite set, but professor told us it's finite. How could it be finite?
 A: This isn't a complete answer, as I can't classify all equivalence classes.
Your professor isn't right. The equivalence relation which you have provided in fact has infinitely many equivalence classes. Take for example $\frac{1}{2^n},n\in\Bbb N$. I claim these numbers are pairwise inequivalent under this relation.
Indeed, suppose $n<m$. Then we have $\frac{1}{2^n}-\frac{1}{2^m}=\frac{2^{m-n}-1}{2^m}$ which has odd numerator and even denominator, and hence can't be represented as $\frac{\text{even}}{\text{odd}}$.
Let me remark that $\frac{1}{2^n}$ don't generate all equivalence classes, as, for example, $\frac{1}{6}$ is not equivalent to any of these (indeed, $\frac{1}{6}-\frac{1}{2^n}=\frac{2^{n-1}-3}{3\cdot 2^n}$).
A: $\newcommand{\Q}{\mathbb{Q}}$Let me try and describe all the equivalence classes.
Write $\alpha \in \Q$ as
$$
\alpha = \frac{a}{b},
$$
with $a, b$ coprime integers.
If $a$ is even, then $\alpha \in A$, so it is in the class of $0$.
All $\alpha$ such that both $a, b$ are odd are in the class of $1$, as
$$
\alpha - 1 = \frac{a}{b} - 1 = \frac{a - 1}{b} \in A.
$$
We are left with the $\alpha$ with $a$ odd and $b$ even. Let
$$
\alpha = \frac{a}{2^{e} s}, 
\qquad
\beta = \frac{b}{2^{f} t},
$$
with $a, b, s, t$ odd, and $e \ge f > 0$.
Then
$$
\alpha - \beta
=
\frac{a t - 2^{e - f} b s}{2^{f} s t}.
$$
If $e > f$, then this is not in $A$, so that $\alpha$ and $\beta$ are not equivalent. If $e = f$, then
$$
\alpha - \beta
=
\frac{a t - b s}{2^{f} s t}
$$
is in $A$ if and only if
$$\tag{cond}
2^{f+1}\ \text{divides}\ a t - b s.
$$ 
Now given
$$
\alpha = \frac{a}{2^{f} s}, 
\qquad
\beta = \frac{b}{2^{f} t},
$$
with $\alpha$ fixed, what are the $\beta$ that are equivalent to $\alpha$? We need (cond) to hold. Choose $t$ to be an arbitrary odd integer, then we have
$$
b s \equiv a t \pmod{2^{f+1}},
$$ 
which has the unique solution modulo $2^{f+1}$
$$\tag{bnaught}
b_{0} = s^{-1} a t \pmod{2^{f+1}},
$$
where $a^{-1}$ denotes the inverse of the (odd) integer $a$ modulo $2^{f+1}$. It follows that $\beta$ is equivalent to $\alpha$ if and only if
$$
\beta = \frac{b_{0} + k 2^{t+1}}{2^{f} t}
$$
where $t$ is an arbitrary odd integer, $k$ is an arbitrary integer, and $b_{0}$ is determined from (bnaught).
This seems to me very close to a description of the classes. I'll try to come back to this later.
