# additive inverse of one element times another element is the additive inverse of their product

Let $R$ be a ring. For all $x,y\in R$

$(-x)\cdot y=x\cdot (-y)=-(x\cdot y)$

All we need to prove is that $(-x)\cdot y+x\cdot y=x\cdot (-y)+x\cdot y=0$

Now: $(-x)\cdot y+x\cdot y=((-x)+x)\cdot y$ by distributivity.

Since $(-x)+x=0$ by definition and for all $y$, $0\cdot y=0$ we get:

$(-x)\cdot y+x\cdot y=0\cdot y=0$ and thus $(-x)\cdot y=-(x\cdot y)$

For $x\cdot (-y)$, use the previous properties of rings to show that

$x\cdot (-y)+x\cdot y=x\cdot ((-y)+y)=x\cdot 0=0$

and thus $x\cdot (-y)=-(x\cdot y)$

Title | additive inverse of one element times another element is the additive inverse of their product |
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Canonical name | AdditiveInverseOfOneElementTimesAnotherElementIsTheAdditiveInverseOfTheirProduct |

Date of creation | 2013-03-22 15:43:40 |

Last modified on | 2013-03-22 15:43:40 |

Owner | cvalente (11260) |

Last modified by | cvalente (11260) |

Numerical id | 8 |

Author | cvalente (11260) |

Entry type | Theorem |

Classification | msc 16-00 |

Classification | msc 20-00 |

Classification | msc 13-00 |