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

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Theorem

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

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


Then:

$\alpha \beta = 0^*$

if and only if:

$\alpha = 0^*$ 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:

$\alpha \beta = 0^*$

$\Box$


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 $\alpha \beta = 0^*$.

Aiming for a contradiction, suppose $r = p q$ 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.

$\blacksquare$


Sources

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