How to divide two inequalities in the form $a - \epsilon < x < a + \epsilon$?

Question

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 $\frac{x}{y}$?

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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 $a$, $b$... ?

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

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

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

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

4. 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).

Answer 1

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

$$\left\vert \frac{x}{y} \right\vert \lt \frac{\vert a \vert + \epsilon_x}{m}.$$

Answer 2

Normally you would assume that the epsilons are small enough not to make the sums and differences cross zero. You need to specify that in your answer. Then, yes, you do need to worry about the signs of $a,b$. Your result is correct if they are both positive but not otherwise. For most purposes you are interested in bounding the absolute value of $\frac xy$. If that is true, you can just take absolute values of $a,b$ and be done.