Cut Associated with 1 is Identity for Multiplication of Cuts

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Theorem

Let $\alpha$ be a cut.

Let $1^*$ denote the rational cut (rational) number $1$.


Then:

$\alpha 1^* = \alpha$

where $\alpha 1^*$ denote the product of $\alpha$ and $1^*$.


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
$0^*$ denotes the rational cut associated with the (rational) number $0$.
$\ge$ denotes the ordering on cuts.

We have that:

$0 < \dfrac 1 2 < 1$

and so:

$\dfrac 1 2 \notin 0^*$

but:

$\dfrac 1 2 \in 1^*$

Thus by definition of the strict ordering of cuts:

$1^* > 0^*$


Thus by definition of absolute value of $\alpha$:

$\size {1^*} = 1^*$


Thus:

$\alpha 1^* := \begin {cases}

\size \alpha \, 1^* & : \alpha \ge 0^* \\ -\paren {\size \alpha \, 1^*} & : \alpha < 0^* \end {cases}$


Let $\alpha > 0^*$.

Let $r \in \alpha 1^*$.

Then by definition of Multiplication of Positive Cuts, either:

$r < 0$

or

$\exists p \in \beta, q \in 1^*: r = p q$

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

Because $0 \le q < 1$ it follows that $p q < p$.

Thus if $r > 0$ it follows that $r < p$.

That is:

$r \in \alpha$

If $r < 0$ then $r \in \alpha$ because $0^* < \alpha$.

Thus in either case $r \in \alpha$.

That is:

$\alpha 1^* \le \alpha$


This needs considerable tedious hard slog to complete it.
In particular: tedious
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Sources

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