additive inverse of a sum in a ring

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

$(a+b)+((-a)+(-b))$$=$$(a+(-a))+(b+(-b))$

$=$$0+0=0$

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.