# proof of long division

###### Proof of theorem 1.

Let $a,b$ be integers, $b\ne 0$. Set

$$q=\{\begin{array}{cc}\lfloor \frac{a}{b}\rfloor \hfill & \text{if}b0\hfill \\ -\lfloor \frac{a}{|b|}\rfloor \hfill & \text{otherwise},\hfill \end{array}$$ |

and $r=a-q\cdot b$. Since $$ for any real $x$, we get for positive $b$

$$ |

, and for $$

$$ |

and the statement follows immediately. ∎

###### Proof of theorem 2.

Let $R$ be a commutative ring with 1, and take $b(x)$ from $R[x]$, where the leading coefficient of $b(x)$ is a unit in $R$. Without loss of generality we may assume the leading coefficient of $b(x)$ is 1.

If $n$ is the degree of $b(x)$, then set

$$ |

where ${a}_{n}$ is the leading coefficient of $a(x)$. Then $r(x)=a(x)-q(x)\cdot b(x)$ is either 0 or $$, as desired.

Now let $m\ge \mathrm{deg}(b(x))$. Then the degree of the polynomial

$$\stackrel{\u02c7}{a}(x)=a(x)-{a}_{m+1}b(x)\cdot {x}^{m+1-n}$$ |

is at most $m$. So by assumption we can write $a(x)$ as

$$a(x)=b(x)\cdot (\stackrel{\u02c7}{q}(x)+{a}_{m+1}{x}^{m+1-n})+\stackrel{\u02c7}{r}(x)$$ |

where $\stackrel{\u02c7}{r}(x)$ is either 0, or its degree is $$. ∎

Title | proof of long division |
---|---|

Canonical name | ProofOfLongDivision |

Date of creation | 2013-03-22 15:36:00 |

Last modified on | 2013-03-22 15:36:00 |

Owner | Thomas Heye (1234) |

Last modified by | Thomas Heye (1234) |

Numerical id | 6 |

Author | Thomas Heye (1234) |

Entry type | Proof |

Classification | msc 11A05 |

Classification | msc 12E99 |

Classification | msc 00A05 |