How to divide two inequalities in the form $a - \epsilon < x < a + \epsilon$? I have the following inequalities
$$
a - \epsilon_x < x < a + \epsilon_x \\
b - \epsilon_y < y < b + \epsilon_y
$$
with $x,y,a,b,\epsilon_x,\epsilon_y\in {\rm I\!R}$, $|\epsilon_x|>0$ and $|\epsilon_y|>0$.
How to bound $\frac{x}{y}$?

I first tried to inverse the second inequality
$$
\frac{1}{b + \epsilon_y} < \frac{1}{y} < \frac{1}{b - \epsilon_y}
$$
I am not sure if it is correct, but I then used this answer to obtain
$$
\min\left( \frac{a-\epsilon_x}{b+\epsilon_y}, \frac{a-\epsilon_x}{b-\epsilon_y}, \frac{a+\epsilon_x}{b+\epsilon_y} \right) < \frac{x}{y} < \max\left( \frac{a+\epsilon_x}{b-\epsilon_y}, \frac{a+\epsilon_x}{b+\epsilon_y}, \frac{a-\epsilon_x}{b-\epsilon_y} \right)
$$
Can I conclude that
$$
\frac{a-\epsilon_x}{b+\epsilon_y} < \frac{x}{y} < \frac{a+\epsilon_x}{b-\epsilon_y}
$$
or should I detail every special cases depending on the sign of $a$, $b$... ?

I assume that $0\notin(b-\epsilon_y, b+\epsilon_y)$.
I tried to detail the four cases

*

*if $a > 0$ and $b > 0$ then $\frac{a-\epsilon_x}{b+\epsilon_y} < \frac{x}{y} < \frac{a+\epsilon_x}{b-\epsilon_y}$


*if $a > 0$ and $b < 0$ then $\frac{a+\epsilon_x}{b+\epsilon_y} < \frac{x}{y} < \frac{a-\epsilon_x}{b-\epsilon_y}$


*if $a < 0$ and $b > 0$ then $\frac{a-\epsilon_x}{b-\epsilon_y} < \frac{x}{y} < \frac{a+\epsilon_x}{b+\epsilon_y}$


*if $a < 0$ and $b < 0$ then $\frac{a+\epsilon_x}{b-\epsilon_y} < \frac{x}{y} < \frac{a-\epsilon_x}{b+\epsilon_y}$
But it seems different from @mathcounterexamples.net 's answer (see below).

If $0 \in (b- \epsilon_y, b + \epsilon_y)$, there is no way to bound $\frac{x}{y}$.
Otherwise you have $$\vert y \vert \gt m = \begin{cases}
b-\epsilon_y & \text{ if } b - \epsilon_y \gt 0\\
-(b + \epsilon_y) & \text{ if } b + \epsilon_y\lt 0
\end{cases}$$As $\vert x \vert \lt \vert a \vert + \epsilon_x$, you get
$$\left\vert \frac{x}{y} \right\vert \lt \frac{\vert a \vert + \epsilon_x}{m}.$$
