Multiplication of Cuts Distributes over Addition

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

Let:
 * $\alpha + \beta$ denote the sum of $\alpha$ and $\beta$.
 * $\alpha \beta$ denote the product of $\alpha$ and $\beta$.

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

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 $\gamma \ge 0^*$.

$\alpha \paren {\beta + \gamma}$ 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 $r \in \gamma$.

$\alpha \beta + \alpha \gamma$ is the set of all rational numbers $s$ of the form:
 * $s = p q + p r$

such that:
 * $s < 0$

or:
 * $p \in \alpha$, $q \in \beta$ and $r \in \gamma$.

From Rational Multiplication Distributes over Addition: $p \paren {q + r} = p q + p r$

and the result follows.