16B70-AdditiveInverseOfASumInARing.tex

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\pmtitle{additive inverse of a sum in a ring}
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\begin{document}
Let $R$ be a ring with elements $a, b \in R$.  
Suppose we want to find the inverse of the element $(a + b) \in R$.  
(Note that we call the element $(a+b)$ the sum of $a$ and $b$.)  
So we want the unique element $c \in R$ so that $(a + b) + c = 0$.  
Actually, let's put $c = (-a) + (-b)$ where $(-a) \in R$ is the additive inverse of $a$ and $(-b) \in R$ is the additive inverse of $b$.  
Because addition in the ring is both associative and commutative we see that
\begin{eqnarray*}
(a + b) + ((-a) + (-b)) & = & (a + (-a)) +(b + (-b))\\
 & = & 0 + 0 = 0
\end{eqnarray*}
since $(-a) \in R$ is the additive inverse of $a$ and $(-b) \in R$ is the additive inverse of $b$.  
Since additive inverses are unique this means that the additive inverse of $(a + b)$ must be $(-a) + (-b)$.  We write this as
\[ -(a + b) = (-a) + (-b). \]

It is important to note that we {\em cannot} just distribute the minus sign across the sum because this would imply that $-1 \in R$ which is not the case if our ring is not with unity.
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