minus one times an element is the additive inverse in a ring

Lemma 1.

Let $R$ be a ring (with unity $\mathrm{1}$) and let $a$ be an element of $R$. Then

$$(-1)\cdot a=-a$$

where $\mathrm{-}\mathrm{1}$ is the additive inverse of $\mathrm{1}$ and $\mathrm{-}a$ is the additive inverse of $a$.

Proof.

Note that for any $a$ in $R$ there exists a unique “$-a$” by the uniqueness of additive inverse in a ring. We check that $(-1)\cdot a$ equals the additive inverse of $a$.

$a+(-1)\cdot a$$=$$1\cdot a+(-1)\cdot a,\text{ by the definition of }1$

$=$$(1+(-1))\cdot a,\text{ by the distributive law}$

$=$$0\cdot a,\text{ by the definition of }-1$

$=$$0,\text{ as a result of the properties of zero}$

Hence $(-1)\cdot a$ is “an” additive inverse for $a$, and by uniqueness $(-1)\cdot a=-a$, the additive inverse of $a$. Analogously, we can prove that $a\cdot (-1)=-a$ as well.∎