Let $$x = \frac{A}{y}-B,$$ for some $A$ and $B$. Suppose that $A >0 $ ad $B>0$. I would like to prove that if $y \in (C, D)$, for some $C$ and $D$, then $x \in (0,1)$.
To this aim, I worked in this "reverse" way:
$$0 < \frac{A}{y}-B < 1 \Rightarrow \\ B < \frac{A}{y} < 1+B \Rightarrow \\ \frac{B}{A} < \frac{1}{y} < \frac{1+B}{A}.$$
Therefore, since both $\frac{B}{A}$ and $\frac{1+B}{A}$ are positive, then also $\frac{1}{y}$ is positive, and hence $y$ is positive.
As a consequence:
$$\frac{A}{1+B} < y < \frac{A}{B}.$$
So, I pose $C = \frac{A}{1+B}$ and $D = \frac{A}{B}$ and I say:
Proposition. Consider $C = \frac{A}{1+B}$ and $D = \frac{A}{B}$, where $A > 0$ and $B>0$. Then:
Can I also "upgrade" my proposition by adding the "if and only if", i.e.,
Proposition. Consider $C = \frac{A}{1+B}$ and $D = \frac{A}{B}$, where $A > 0$ and $B>0$. Then:
$$ y \in (C, D) \Leftrightarrow x \in (0,1).$$
Since I prove it in a reverse way, then can I conclude that the "if and only if" relation holds?
Hint: To see if the equivalence holds ("iff"), check if the implication in each of your steps can be reversed.
For example, you have that $0 < A/y - B < 1 \implies B < A/y < 1 + B$. Why is that? Can you also use the same argument to show $B < A/y < 1 + B \implies 0 < A/y - B < 1$? If yes, then you have "iff" in the first step. Do the same for all the steps.