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# 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

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

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## Mathematics Subject Classification

16B70*no label found*

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