Multiplication of Cuts is Associative

Theorem
Let $\alpha$, $\beta$ and $\gamma$ be cuts.

Let $\alpha \beta$ denote the product of $\alpha$ and $\beta$.

Then:
 * $\paren {\alpha \beta} \gamma = \alpha \paren {\beta \gamma}$

Proof
By definition, we have that:


 * $\alpha \beta := \begin {cases}

\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\ -\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\ -\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\ \size \alpha \, \size \beta & : \alpha < 0^*, \beta < 0^* \end {cases}$ where:
 * $\size \alpha$ denotes the absolute value of $\alpha$
 * $0^*$ denotes the rational cut associated with the (rational) number $0$
 * $\ge$ denotes the ordering on cuts.

Let $\alpha \ge 0^*$, $\beta \ge 0^*$ and $\gamma \ge 0^*$.

$\paren {\alpha \beta} \gamma$ is the set of all rational numbers $s$ of the form $s = \paren {p q} r$ such that $s < 0$ or $p \in \alpha$, $q \in \beta$ and $r \in \gamma$.

Similarly, $\alpha \paren {\beta \gamma}$ is the set of all rational numbers $s$ of the form $s = p \paren {q r}$ such that $s < 0$ or $p \in \alpha$, $q \in \beta$ and $r \in \gamma$.

From Rational Multiplication is Associative we have that:
 * $\paren {p q} r = p \paren {q r}$

Thus we have that:
 * $\size \alpha \paren {\size \beta \, \size \gamma} = \paren {\size \alpha \, \size \beta} \size \gamma$

and the result follows.