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Like the title says:

$$ x \in \Bbb R : x \in [x_{min}, x_{max}] $$ $$ y \in \Bbb R : y \in [y_{min}, y_{max}] $$

What is the range of:

$$ x \over y $$

?

I find this hard to reason about because the division is discontinuous.

I can identify up to 49 different combinations of possibilities that I would need to analyze one by one... 7 for each variable: the cases where the variable is entirely negative, negative or 0, exactly 0, 0 or positive, entirely positive, from negative to positive.

For instance let's take the four cases where neither $x$ nor $y$ pass by 0, are somewhat simple:

  1. $$ x_{max} \lt 0 \wedge y_{max} \lt 0 \Rightarrow [{x_{min} \over y_{max}}, {x_{max} \over y_{min}}] $$

  2. $$ x_{min} \gt 0 \wedge y_{min} \gt 0 \Rightarrow [{x_{max} \over y_{min}}, {x_{min} \over y_{max}}] $$

  3. $$ x_{max} \lt 0 \wedge y_{min} \gt 0 \Rightarrow [{x_{min} \over y_{min}}, {x_{max} \over y_{max}}] $$

  4. $$ x_{min} \gt 0 \wedge y_{max} \lt 0 \Rightarrow [{x_{max} \over y_{max}}, {x_{min} \over y_{min}}] $$

These are probably the simplest cases, since they're away from the discontinuity of the division. Yet I'm getting lost with the signs. And many many more cases are still missing.

I assumed this might have been a standard question in interval arithmetic. So I was hoping there could be a simple precooked solution.

Helloer
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  • Have you tried any examples? What are your thoughts on the matter? What do you know about division? – Steven Stadnicki Jan 09 '22 at 17:33
  • Please edit to include your efforts. Also, you should indicate whether or not you are including cases in which $0$ is an admissible value for $y$ and, if you are, how you wish to handle that. – lulu Jan 09 '22 at 17:34
  • It would be useful if you support your question with your attempts. – Mostafa Ayaz Jan 09 '22 at 17:37
  • OK, I'll try to add some reasoning I made – Helloer Jan 09 '22 at 17:39
  • I listed an upper bound for a number of cases that I should think about independently from each other: 49 (realistically, the number of interesting cases may be something around 20 or so).

    I also showed the range I computed for some of these cases. But I'm pretty sure something is wrong as I'm getting lost with the signs...

    I don't know if my approach is the correct one, but it'd be tough to find the solution this way. I was hoping this might have been a common question in interval arithmetic, so maybe there was a simple solution?

    – Helloer Jan 09 '22 at 18:03
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    If the range of $y$ includes zero, then the range of $x/y$ is unbounded above and below. So you may as well assume $0<\min y<\max y$ (and deal with negative $y$ by symmetry). – Gerry Myerson Jan 09 '22 at 18:31

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