# 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.

Title | additive inverse of a sum in a ring |
---|---|

Canonical name | AdditiveInverseOfASumInARing |

Date of creation | 2013-03-22 15:45:02 |

Last modified on | 2013-03-22 15:45:02 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 11 |

Author | rspuzio (6075) |

Entry type | Theorem |

Classification | msc 16B70 |