General Positivity Rule in Ordered Integral Domain

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\struct {D, +, \times}$ be an ordered integral domain, whose (strict) positivity property is denoted $P$.

Let $S \subset D$ be a subset of $D$ such that:

$\forall s \in S: \map P x$


Then the following are true:

$\displaystyle \forall n \in \N_{>0}: \forall s_i \in S: \map P {\sum_{i \mathop = 1}^n s_i}$
$\displaystyle \forall n \in \N_{>0}: \forall s_i \in S: \map P {\prod_{i \mathop = 1}^n s_i}$

where:

$\displaystyle \sum_{i \mathop = 1}^n s_i = s_1 + s_2 + \cdots + s_n$
$\displaystyle \prod_{i \mathop = 1}^n s_i = s_1 \times s_2 \times \cdots \times s_n$

That is, the ring sum and ring product of any number of elements in $D$ which have the (strict) positivity property also have the (strict) positivity property.


Corollary

Let $\map P x$ where $x \in D$.

Then:

$\map P {n \cdot x}$ and $\map P {x^n}$


Proof

Trivially true for $n = 1$, and true by definition for $n = 2$.

Suppose it is true for all $n$ up to $n = k$ for some $k \in \N$.

Then:

$\displaystyle \forall s_i \in S: \map P {\sum_{i \mathop = 1}^k s_i}$
$\displaystyle \forall s_i \in S: \map P {\prod_{i \mathop = 1}^k s_i}$


Take any $\displaystyle \sum_{i \mathop = 1}^{k + 1} s_i = \sum_{i \mathop = 1}^k s_i + s_{k + 1}$.

Then:

$\displaystyle \map P {\sum_{i \mathop = 1}^k s_i}$ and $\map P {s_{k + 1} }$

so:

$\displaystyle \map P {\sum_{i \mathop = 1}^{k + 1} s_i}$

because the (strict) positivity property holds for all $s_i \in S$.


Similarly:

$\displaystyle \map P {\prod_{i \mathop = 1}^k s_i}$ and $\map P {s_{k + 1} }$

so:

$\displaystyle \map P {\prod_{i \mathop = 1}^{k + 1} s_i}$


The result follows by the Principle of Complete Induction.

$\blacksquare$


Sources