# law of signs under multiplication in a ring

###### Lemma 1.

Let $R$ be a ring with unity, which we denote by $\mathrm{1}$. For all $x\mathrm{,}y\mathrm{\in}R$:

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

where $\mathrm{-}x$ denotes the additive inverse of $x$ in $R$.

###### Proof.

Here we use the fact $(-1)\cdot a=-a$ for all $a\in R$. First, we see that:

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

since, clearly, the additive inverse of $-a$ is $a$ itself.

Hence:

$$(-x)\cdot (-y)=(-1)\cdot x\cdot (-1)\cdot y=(-1)\cdot (-1)\cdot x\cdot y=x\cdot y$$ |

where we have used several times the associativity of $\cdot $ and the fact that $(-1)\cdot x=x\cdot (-1)=-x$. ∎

Title | law of signs under multiplication in a ring |
---|---|

Canonical name | LawOfSignsUnderMultiplicationInARing |

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

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

Owner | alozano (2414) |

Last modified by | alozano (2414) |

Numerical id | 10 |

Author | alozano (2414) |

Entry type | Derivation |

Classification | msc 20-00 |

Classification | msc 16-00 |

Classification | msc 13-00 |

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

Related topic | Ring |