ProofOfLongDivision 2005-12-04 01:08:13 2006-03-05 14:34:49 Proof long division 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \begin{proof}[Proof of theorem 1] Let $a,b$ be integers, $b \ne 0$. Set \[q=\begin{cases} \left\lfloor \frac{a}{b}\right\rfloor &\text{if $b>0$} \\ -\left\lfloor \frac{a}{\lvert b\rvert}\right\rfloor &\text{otherwise}, \end{cases} \] and $r=a-q\cdot b$. Since $0\le x -\lfloor x\rfloor < 1$ for any real $x$, we get for positive $b$ \[0\le \frac{a}{b} -q =\frac{r}{b} < 1\], and for $b<0$ \[0 \le \frac{a}{\lvert b \rvert} -\left\lfloor\frac{a}{\lvert b\rvert}\right\rfloor =\frac{a}{\lvert b\rvert} +q=\frac{r}{\lvert b\rvert} <1,\] and the statement follows immediately. \end{proof} \begin{proof}[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 \[q(x)= \begin{cases} 0&\text{if $\deg(a(x)) <n$}\\ a_n&\text{if $\deg(a(x))=n$}, \end{cases}\] where $a_n$ is the leading coefficient of $a(x)$. Then $r(x)=a(x) -q(x)\cdot b(x)$ is either 0 or $\deg(r(x))<\deg(b(x))$, as desired. Now let $m \ge \deg(b(x))$. Then the degree of the polynomial \[\check{a}(x)=a(x) -a_{m+1}b(x)\cdot x^{m+1-n}\] is at most $m$. So by assumption we can write $a(x)$ as \[a(x)=b(x)\cdot(\check{q}(x)+a_{m+1}x^{m+1-n}) +\check{r}(x)\] where $\check{r}(x)$ is either 0, or its degree is $<b(x)$. \end{proof}