Product with Ring Negative

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Theorem

Let $\struct {R, +, \circ}$ be a ring.


Then:

$\forall x, y \in \struct {R, +, \circ}: \paren {-x} \circ y = -\paren {x \circ y} = x \circ \paren {-y}$

where $\paren {-x}$ denotes the negative of $x$.


Corollary

Let $\struct {R, +, \circ}$ be a ring with unity $1_R$.

Then:

$\forall x \in R: \paren {-1_R} \circ x = -x$


Proof

We have:

\(\ds \paren {x + \paren {-x} } \circ y\) \(=\) \(\ds 0_R \circ y\) Definition of Ring Zero
\(\ds \) \(=\) \(\ds 0_R\) Ring Product with Zero
\(\ds \leadstoandfrom \ \ \) \(\ds \paren {x \circ y} + \paren {\paren {-x} \circ y}\) \(=\) \(\ds 0_R\) Ring Axiom $\text D$: Distributivity of Product over Addition


So from Group Axiom $\text G 3$: Existence of Inverse Element as applied to $\struct {R, +}$:

$\paren {-x} \circ y = -\paren {x \circ y}$


The proof that $x \circ \paren {-y} = -\paren {x \circ y}$ follows identical lines.

$\blacksquare$


Also see


Sources

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