Multiplication of Cuts is Associative

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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.

$\blacksquare$


Sources