20-00-UniquenessOfAdditiveInverseInARing.tex

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\pmtitle{uniqueness of additive inverse in a ring}
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\begin{document}
\begin{lemma}
Let $R$ be a ring, and let $a$ be any element of $R$. There exists a unique element $b$ of $R$ such that $a+b=0$, i.e. there is a unique \PMlinkname{additive inverse}{Ring} for $a$.
\end{lemma}
\begin{proof}
Let $a$ be an element of $R$. By definition of ring, there exists at least one \PMlinkname{additive inverse}{Ring} of $a$, call it $b_1$, so that $a+b_1=0$. Now, suppose $b_2$ is another additive inverse of $a$, i.e. another element of $R$ such that 
$$a+b_2=0$$
where $0$ is the \PMlinkname{zero element}{Ring} of $R$. Let us show that $b_1=b_2$. Using properties for a ring and the above equations for $b_1$ and $b_2$
yields
\begin{eqnarray*}
b_1 &=& b_1+0 \quad \text{(definition of zero)}\\
 &=& b_1+(a+b_2) \quad (b_2 \text{ is an additive inverse of }a)\\
 &=& (b_1+a)+b_2 \quad (\text{associativity in }R)\\
 &=& 0+b_2 \quad (b_1\text{ is an additive inverse of }a)\\
 &=& b_2 \quad \text{(definition of zero)}.
\end{eqnarray*}
Therefore, there is a unique additive inverse for $a$.
\end{proof}
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