# 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))

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 $. ∎

Title proof of long division ProofOfLongDivision 2013-03-22 15:36:00 2013-03-22 15:36:00 Thomas Heye (1234) Thomas Heye (1234) 6 Thomas Heye (1234) Proof msc 11A05 msc 12E99 msc 00A05