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I'm currently working on a problem where I need to provide a rigorous proof that multiplication among interval numbers is a associative. For those of you who haven't heard of an interval number before, here are some definitions:

We say that $x$ is an interval number on $\mathbb{R}^n$ if $x \subset \mathbb{R}^n$ and $x = [x^L,x^u]$, where $x^L \leq x^U$ and $x^L, x^U \in \mathbb{R}^n$. Here $x^L \leq x^U$ is defined component-wise, so if $x^L = (x^L_1,...x^L_n)$ and $x^U = (x^U_1,...x^U_n)$ then $x^L \leq x^U$ implies $x^L_1 \leq x^U_1, ... x^L_n \leq x^U_n$.

Given two interval numbers $x$ and $y$ on $\mathbb{R}^n$, where $x = [x^L,x^U]$ and $y = [y^L,y^U]$, their product is another interval $xy$ that has the form $[\min\{x^Ly^L,x^Ly^U,x^Uy^L,x^Uy^U\}, \max\{x^Ly^L,x^Ly^U,x^Uy^L,x^Uy^U\}]$.

Basically what I am trying to show is $x(yz) = (xy)z$ where $x,y,z \subset \mathbb{R}^n$ are interval numbers such that $x = [x^L,x^u]$, $y =[y^L,y^U]$ and $z = [z^L,z^U]$.

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    Thank you for the comments - just fixed the question.2011-04-02

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As Arturo commented, it's not really clear what you mean by inequalities and products in $\mathbb{R}^n$, so I'll assume that we are in $\mathbb{R}$.

Then it's easy if you use the fact that $xy = \{ ab : a \in x, b \in y \}$ rather than the characterization of $xy$ in terms of min and max, since $(xy)z$ and $x(yz)$ will both be equal to the interval $\{ abc : a \in x, b \in y, c \in z \}$.

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    @Berk: I don't understand that comment. Of course intervals have upper and lower $b$ounds, that's what make them intervals. The whole point is that you *don't* need to bother with the min/max $f$ormulation since the other formulation is exactly equivalent to it. (Of course, if this is some kind of assignment where you are ordered to prove it using the min/max formulation explicitly, then you will need to use it! That would lead to a tedious case-by-case analysis, I guess.)2011-04-02