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 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.