# Proving That Multiplication among Interval Numbers is Associative

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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What is the order in $\mathbb{R}^n$? That is, how do you define $a\leq b$ for $a,b\in\mathbb{R}^n$? (I know how to do it in $\mathbb{R}$, but there are several possible definitions for $\mathbb{R}^n$, which is why I ask). And how do you define the product of elements of $\mathbb{R}^n$? Also: you should take care not to use both $U$ and $u$, lest people be confused on whether $x^u$ and $x^U$ are the same or not. –  Arturo Magidin Apr 1 '11 at 3:31
Also, the question has "commutative" instead of "associative" in the first paragraph. –  Carl Brannen Apr 1 '11 at 5:37
In interval arithmetic, multiplication is componentwise. Same for inequalities, $a\le b$ for $a,b\in R^n$ means $a_i \le b_i$ for $i=1,\ldots,n$. –  Jitse Niesen Apr 1 '11 at 8:17
Thank you for the comments - just fixed the question. –  Elements Apr 2 '11 at 4:10

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 $\{ abc : a \in x, b \in y, c \in z \}$.
This also works in $\mathbb{R}^n$. –  Jitse Niesen Apr 1 '11 at 8:18