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.

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 $<b(x)$