Product of Cut with Zero Cut equals Zero Cut

Theorem
Let $\alpha$ be a cut.

Let $0^*$ denote the rational cut associated with the (rational) number $0$.

Then:
 * $\alpha 0^* = 0^*$

where $\alpha 0^*$ denote the product of $\alpha$ and $0^*$.

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 $\alpha$
 * $\size \alpha \, \size \beta$ is defined as in Multiplication of Positive Cuts
 * $\ge$ denotes the ordering on cuts.

From Absolute Value of Cut is Zero iff Cut is Zero:
 * $\size {0^*} = 0^*$

Thus:
 * $\alpha 0^* := \begin {cases}

\size \alpha \, 0^* & : \alpha \ge 0^* \\ -\paren {\size \alpha \, 0^*} & : \alpha < 0^* \end {cases}$

From Absolute Value of Cut is Greater Than or Equal To Zero Cut, we have that:
 * $\size \alpha = \beta$

where $\beta \ge 0^*$.

By definition of Multiplication of Positive Cuts:

$\beta 0^*$ is the set of all rational numbers $r$ such that either:
 * $r < 0$

or
 * $\exists p \in \beta, q \in 0^*: r = p q$

where $p \ge 0$ and $q \ge 0$

By definition of $0^*$, it is not possible for $q \ge 0$.

Thus:
 * $\beta 0^* = \set {r \in \Q: r < 0}$

Hence by definition of a cut:
 * $\beta 0^* = 0^*$

By Identity Element for Addition of Cuts and definition of negative of cut:
 * $-0^* = 0^*$

and the result follows.