# uniqueness of division algorithm in Euclidean domain

Theorem. Let $a,b$ be non-zero elements of a Euclidean domain^{} $D$ with the Euclidean valuation $\nu $. The incomplete quotient $q$ and the remainder $r$ of the division algorithm

$$ |

are unique if and only if

$\nu (a+b)\leqq \mathrm{max}\{\nu (a),\nu (b)\}.$ | (1) |

Proof. Assume first (1) for the elements $a,b$ of $D$. If we had

$$ |

and ${r}^{\prime}\ne r$, ${q}^{\prime}\ne q$, then the properties of the Euclidean valuation (http://planetmath.org/EuclideanValuation) and the assumption^{} yield the of inequalities

$$ |

which is impossible. We must infer that ${r}^{\prime}-r=0$ or ${q}^{\prime}-q=0$. But these two conditions are equivalent^{} (http://planetmath.org/Equivalent3). Thus the division algorithm is unique.

Conversely, assume that (1) is not true for non-zero elements $a,b$ of $D$, i.e.

$$\nu (a+b)>\mathrm{max}\{\nu (a),\nu (b)\}.$$ |

Then we obtain two repsesentations

$$b=0(a+b)+b=1(a+b)-a$$ |

where $$ and $$. Thus the incomplete quotient and the remainder are not unique.

Title | uniqueness of division algorithm in Euclidean domain |
---|---|

Canonical name | UniquenessOfDivisionAlgorithmInEuclideanDomain |

Date of creation | 2013-03-22 17:53:00 |

Last modified on | 2013-03-22 17:53:00 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 7 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 13F07 |

Related topic | KrullValuation |

Related topic | Quotient |

Defines | incomplete quotient |