User:Dfeuer/Multiplying Compatible Relationship by Zero-Related Element

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Note: this theorem generalizes the sixth property of ordered rings and a strict total version for ordered integral domains to compatible relations.

Theorem

Let $(R,+,\circ)$ be a ring with zero $0_R$.

Let $\mathcal R$ be a relation compatible with $(R,+,\circ)$.

Let $x,y,z\in R$ such that:

$x \mathrel{\mathcal R} y$
$0_R \mathrel{\mathcal R} z$

Then:

$(x \circ z) \mathrel{\mathcal R} (y \circ z)$
$(z \circ x) \mathrel{\mathcal R} (z \circ y)$

Corollary

Let $(R,+,\circ,<)$ be a strictly ordered ring with zero $0_R$.

Let $x,y,z \in R$ such that:

$x < y$
$0 < z$


Then

$x \circ z < y \circ z$
$z \circ x < z \circ y$


Proof

Since $x \mathrel{\mathcal R} y$, by the definition of compatibility:

$0_R \mathrel{\mathcal R} (y + (-x))$

By the definition of compatibility:

$0_R \mathrel{\mathcal R} ((y + (-x)) \circ z)$
$0_R \mathrel{\mathcal R} (z \circ (y + (-x)))$

Since $\circ$ distributes over $+$:

$0_R \mathrel{\mathcal R} (y \circ z + (-x) \circ z)$
$0_R \mathrel{\mathcal R} (z \circ y + z \circ (-x))$

By Product with Ring Negative:

$-((-x) \circ z) = x \circ z$
$-(z \circ (-x)) = z \circ x$

Thus by compatibility:

$(x \circ z) \mathrel{\mathcal R} (y \circ z)$
$(z \circ x) \mathrel{\mathcal R} (z \circ y)$