# uniqueness of additive inverse in a ring

###### Lemma.

Let $R$ be a ring, and let $a$ be any element of $R$. There exists a unique element $b$ of $R$ such that $a\mathrm{+}b\mathrm{=}\mathrm{0}$, i.e. there is a unique additive inverse (http://planetmath.org/Ring) for $a$.

###### Proof.

Let $a$ be an element of $R$. By definition of ring, there exists at least one additive inverse (http://planetmath.org/Ring) of $a$, call it ${b}_{1}$, so that $a+{b}_{1}=0$. Now, suppose ${b}_{2}$ is another additive inverse of $a$, i.e. another element of $R$ such that

$$a+{b}_{2}=0$$ |

where $0$ is the zero element (http://planetmath.org/Ring) of $R$. Let us show that ${b}_{1}={b}_{2}$. Using properties for a ring and the above equations for ${b}_{1}$ and ${b}_{2}$ yields

${b}_{1}$ | $=$ | ${b}_{1}+0\mathit{\hspace{1em}}\text{(definition of zero)}$ | ||

$=$ | ${b}_{1}+(a+{b}_{2})\mathit{\hspace{1em}}({b}_{2}\text{is an additive inverse of}a)$ | |||

$=$ | $({b}_{1}+a)+{b}_{2}\mathit{\hspace{1em}}(\text{associativity in}R)$ | |||

$=$ | $0+{b}_{2}\mathit{\hspace{1em}}({b}_{1}\text{is an additive inverse of}a)$ | |||

$=$ | ${b}_{2}\mathit{\hspace{1em}}\text{(definition of zero)}.$ |

Therefore, there is a unique additive inverse for $a$. ∎

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

Canonical name | UniquenessOfAdditiveInverseInARing |

Date of creation | 2013-03-22 14:13:54 |

Last modified on | 2013-03-22 14:13:54 |

Owner | alozano (2414) |

Last modified by | alozano (2414) |

Numerical id | 7 |

Author | alozano (2414) |

Entry type | Theorem |

Classification | msc 20-00 |

Classification | msc 16-00 |

Classification | msc 13-00 |

Related topic | UniquenessOfInverseForGroups |