Multiplication of Cuts Distributes over Addition

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Theorem

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

Let:

$\alpha + \beta$ denote the sum of $\alpha$ and $\beta$.
$\alpha \beta$ denote the product of $\alpha$ and $\beta$.


Then:

$\alpha \paren {\beta + \gamma} = \alpha \beta + \alpha \gamma$


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


Let $\alpha \ge 0^*$, $\beta \ge 0^*$ and $\gamma \ge 0^*$.

$\alpha \paren {\beta + \gamma}$ is the set of all rational numbers $s$ of the form:

$s = p \paren {q + r}$

such that:

$s < 0$

or:

$p \in \alpha$, $q \in \beta$ and $r \in \gamma$.


$\alpha \beta + \alpha \gamma$ is the set of all rational numbers $s$ of the form:

$s = p q + p r$

such that:

$s < 0$

or:

$p \in \alpha$, $q \in \beta$ and $r \in \gamma$.

From Rational Multiplication Distributes over Addition: $p \paren {q + r} = p q + p r$

and the result follows.

$\blacksquare$


Sources