proof of long division

Proof of theorem 1.

Let $a, b$ be integers, $b\neq 0$ . 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 $0\leq x-\lfloor x\rfloor<1$ for any real $x$ , we get for positive $b$

0\leq\frac{a}{b}-q=\frac{r}{b}<1

, and for $b<0$

0\leq\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. ∎

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 $m\geq\deg(b(x))$ . 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 $<b(x)$ . ∎

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