proof of long division

Proof of theorem 1.

Let $a, b$ be integers, $b\eq 0$ . 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 $<b(x)$ .∎