# minus one times an element is the additive inverse in a ring

###### Lemma 1.

Let $R$ be a ring (with unity $\mathrm{1}$) and let $a$ be an element of $R$. Then

$$(-1)\cdot a=-a$$ |

where $\mathrm{-}\mathrm{1}$ is the additive inverse of $\mathrm{1}$ and $\mathrm{-}a$ is the additive inverse of $a$.

###### Proof.

Note that for any $a$ in $R$ there exists a unique “$-a$” by the uniqueness of additive inverse in a ring. We check that $(-1)\cdot a$ equals the additive inverse of $a$.

$a+(-1)\cdot a$ | $=$ | $1\cdot a+(-1)\cdot a,\text{by the definition of}1$ | ||

$=$ | $(1+(-1))\cdot a,\text{by the distributive law}$ | |||

$=$ | $0\cdot a,\text{by the definition of}-1$ | |||

$=$ | $0,\text{as a result of the properties of zero}$ |

Hence $(-1)\cdot a$ is “an” additive inverse for $a$, and by uniqueness $(-1)\cdot a=-a$, the additive inverse of $a$. Analogously, we can prove that $a\cdot (-1)=-a$ as well. ∎

Title | minus one times an element is the additive inverse in a ring |
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Canonical name | MinusOneTimesAnElementIsTheAdditiveInverseInARing |

Date of creation | 2013-03-22 14:14:00 |

Last modified on | 2013-03-22 14:14:00 |

Owner | alozano (2414) |

Last modified by | alozano (2414) |

Numerical id | 9 |

Author | alozano (2414) |

Entry type | Theorem |

Classification | msc 16-00 |

Classification | msc 13-00 |

Classification | msc 20-00 |

Synonym | $(-1)\cdot a=-a$ |

Related topic | 0cdotA0 |