Product of Cut with Zero Cut equals Zero Cut

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Theorem

Let $\alpha$ be a cut.

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


Then:

$\alpha 0^* = 0^*$

where $\alpha 0^*$ denotes 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.

$\blacksquare$


Sources

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