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 $\struct {R, +, \circ}$ be a ring with zero $0_R$.
Let $\RR$ be a relation compatible with $\struct {R, +, \circ}$.
Let $x, y, z \in R$ such that:
- $x \mathrel \RR y$
- $0_R \mathrel \RR z$
Then:
- $\paren {x \circ z} \mathrel \RR \paren {y \circ z}$
- $\paren {z \circ x} \mathrel \RR \paren {z \circ y}$
Corollary
Let $\struct {R, +, \circ, <}$ 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 $x \mathrel \RR y$, by the definition of compatibility:
- $0_R \mathrel \RR \paren {y + \paren {-x} }$
By the definition of compatibility:
- $0_R \mathrel \RR \paren {\paren {y + \paren {-x} } \circ z}$
- $0_R \mathrel \RR \paren {z \circ \paren {y + \paren {-x} } }$
Since $\circ$ distributes over $+$:
- $0_R \mathrel \RR \paren {y \circ z + \paren {-x} \circ z}$
- $0_R \mathrel \RR \paren {z \circ y + z \circ \paren {-x} }$
By Product with Ring Negative:
- $-\paren {\paren {-x} \circ z} = x \circ z$
- $-\paren {z \circ \paren {-x} } = z \circ x$
Thus by compatibility:
- $\paren {x \circ z} \mathrel \RR \paren {y \circ z}$
- $\paren {z \circ x} \mathrel \RR \paren {z \circ y}$