Product of Cuts is Zero Cut iff Either Factor equals Zero Cut

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

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

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


 * $\alpha = 0^*$ or $\beta = 0^*$
 * $\alpha = 0^*$ or $\beta = 0^*$

where $\alpha \beta$ denotes the product of $\alpha$ and $\beta$.

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.

First we note that from Absolute Value of Cut is Zero iff Cut is Zero:
 * $\size {0^*} = 0^*$

Necessary Condition
Let $\alpha = 0^*$ or $\beta = 0^*$.

Then from Product of Cut with Zero Cut equals Zero Cut it follows directly that:


 * $\alpha \beta = 0^*$

Sufficient Condition
Let $\alpha \beta = 0^*$.

Suppose $\alpha > 0^*$ and $\beta > 0^*$.

By definition of multiplication of positive cuts:

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

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

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

We have that $\alpha \beta = 0^*$.

$r = p q$ where $p \ge 0$ and $q \ge 0$.

Then $r \ge 0$.

But then $r \notin 0^*$ by definition of $0^*$.

Thus there is no $p \in \alpha, q \in \beta$ such that $r = p q$ where $p \ge 0$ and $q \ge 0$.

It follows that $\alpha > 0^*$ and $\beta > 0^*$ cannot both be satisfied.

It follows further that for $\size \alpha \, \size \beta = 0^*$ it is not possible for $\size \alpha > 0^*$ and $\size \beta > 0^*$.

The result follows.