# 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

 $\displaystyle(a+b)+((-a)+(-b))$ $\displaystyle=$ $\displaystyle(a+(-a))+(b+(-b))$ $\displaystyle=$ $\displaystyle 0+0=0$

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.

Title additive inverse of a sum in a ring AdditiveInverseOfASumInARing 2013-03-22 15:45:02 2013-03-22 15:45:02 rspuzio (6075) rspuzio (6075) 11 rspuzio (6075) Theorem msc 16B70