# Negative of Subring is Negative of Ring

## Theorem

Let $\struct {R, +, \circ}$ be an Ring.

For each $x \in R$ let $-x$ denote the ring negative of $x$ in $R$.

Let $\struct {S, + {\restriction_S}, \circ {\restriction_S}}$ be a subring of $R$.

For each $x \in S$ let $\mathbin \sim x$ denote the ring negative of $x$ in $S$.

Then:

$\forall x \in S: \mathbin \sim x = -x$

## Proof

Let $i_S: S \to R$ be the inclusion mapping from $S$ to $R$.

Let $x \in S$.

Then:

 $\displaystyle \mathbin \sim x$ $=$ $\displaystyle \map {i_S} {\mathbin \sim x}$ as $\mathbin \sim x \in S$ $\displaystyle$ $=$ $\displaystyle -x$ Group Homomorphism Preserves Inverses

$\blacksquare$