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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?

M159
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1 Answers1

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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.

M159
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Apass.Jack
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  • Of course, we don't have to stick to that wonderful solution. We can prove directly $([a,b] X [c, d]) X [e,f]=[\min(ace, acf, ade,adf,bce,bcf, bde,bdf), \max(ace, acf, ade,adf,bce,bcf, bde,bdf)]=[a,b] X( [c, d] X [e,f]).$ – Apass.Jack Dec 15 '22 at 17:32