minus one times an element is the additive inverse in a ring1cdotAA2004-03-09 12:10:022005-11-24 11:14:40Theoremring% this is the default PlanetMath preamble. as your knowledge
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\newcommand{\Rats}{\mathbb{Q}}\begin{lemma}
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 $a$.
\end{lemma}
\begin{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 $(-1)\cdot a = -a$, {\it the} additive inverse of $a$. Analogously, we can prove that $a\cdot (-1) = -a$ as well.
\end{proof}