Product of intervals is associative I am stuck trying to prove an operation being associative.
Let $ M = \lbrace [a, \ b] : a, \ b \in \mathbb{R} \rbrace $ be the set of all intervals on $ \mathbb{R} $ and $ X : M^2 \rightarrow M $ the product given by
$$ 
[a, \ b] X [c, \ d] = [\min(ac, ad, bc, bd), \ \max(ac, ad, bc, bd)] 
$$
Then, how can I prove this is associative?
I have founded an old question about a generalization of this to $ \mathbb{R}^n $, but it is not solved. There is an answer, but I really don't understand it... I think it really not solves the problem. And if it does, I don't understand how: Proving That Multiplication among Interval Numbers is Associative
Here is what I have tried:
\begin{align}
    ([a, \ b] X [c, \ d]) X [e, \ f]
    &= [\min(P), \ \max(P)] X [e, \ f] \ \ \mbox{ being } P = \lbrace ab, ad, bc, bd \rbrace \\
    &= [\min(P'), \max(P')] \ \ \mbox{ being } P' = \lbrace e \cdot \min(P), \ e \cdot \max(P), \ f \cdot \min(P), \ f \cdot \max(P)  \rbrace
\end{align}
On the other hand,
\begin{align}
    [a, \ b] X ([c, \ d] X [e, \ f])
    &= [a, \ b] X [\min(Q), \ \max(Q)] \ \ \mbox{ being } Q = \lbrace ce, cf, de, df \rbrace \\
    &= [\min(Q'), \max(Q')] \ \ \mbox{ being } Q' = \lbrace a \cdot \min(Q), \ a \cdot \max(Q), \ b \cdot \min(Q), \ b \cdot \max(Q)  \rbrace
\end{align}
I try to simplify this by using the following property:
$$
    a \cdot \min(A)
    = \begin{cases}
        \min(a \cdot A) \mbox{ if } a \geq 0 \\
        \max(a \cdot A) \mbox{ if } a < 0
    \end{cases}
$$
where $ a \cdot A = \lbrace a \cdot x : x \in A \rbrace $. The same works for the $ \max $ function, by changing it for $ \min $ if $ a < 0 $. Then no matters if $ a, \ b, \ e $ and $ f $ are positive or not, the following holds (because inside $ P' $ and $ Q' $ we have always both, $ x \cdot \max $ and $ x \cdot \min $):
$$
P' = \lbrace \min(e \cdot P), \ \max(e \cdot P), \ \min(f \cdot P), \ \max(f \cdot P) \\
Q' = \lbrace \min(a \cdot Q), \ \max(a \cdot Q), \ \min(b \cdot Q), \ \max(b \cdot Q) 
\rbrace
$$
But I am not sure how to continue. Any hints?
 A: Let me explain the wonderful solution by Hans Lundmark.
All variables represent real numbers.
Recall that an interval $[a,b]$ is the set $\{x: a\le x\le b\}$. We can rephrase the definition of $X$ as

Let $ M = \{\{x: a\le x\le b\}: a, \ b \in \mathbb{R} \rbrace $ be the set of all intervals on $ \mathbb{R} $ and $ X : M^2 \rightarrow M $ the product given by
$$ 
\{x: a\le x\le b\} X \{x: c\le x\le d\}\\= \{x: \min(ac, ad, bc, bd)\le x\le\max(ac, ad, bc, bd) \} 
$$

Here is a fundamental observation.
$$\min(ac, ad, bc, bd)\le x\le\max(ac, ad, bc, bd) \\ \iff \exists a\le y\le b,\, c\le z\le d\text{ such that } x=yz.$$
So we can rephrase the definition of $X$ again as

$ X : M^2 \rightarrow M $ is the product given by $ IXJ= \{yz: y\in I\land z\in J\}$.

So, $$(IXJ)XK=\{(xy)z: x\in I, y\in J, z\in K\},\\
IX(JXK)=\{x(yz): x\in I, y\in J, z\in K\}.$$
Since multiplication of real numbers is associative, i.e., $(xy)z=x(yz)$, we have $(IXJ)XK=IX(JXK)$, which is the associativity of $X$.

What remains to prove is the fundamental observation above. I will leave that easy and important step for you to verify.
