Lemma 1   Let $R$ be a ring (with unity $1$ ) and let $a$ be an element of $R$ . Then $$(-1)\cdot a = -a$$ where $-1$ is the additive inverse of $1$ and $-a$ 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 $a$ . \begin{eqnarray*} a+(-1)\cdot a &=& 1\cdot a + (-1)\cdot a, \quad \text{ by the definition of }1\\ &=& (1+ (-1))\cdot a, \quad \text{ by the distributive law}\\ &=& 0\cdot a,\quad \text{ by the definition of }-1\\ &=& 0, \quad \text{ as a result of the properties of zero} \end{eqnarray*}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.