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2
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'long division'
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| Title of object: |
long division |
| Canonical Name: |
LongDivision |
| Type: |
Theorem |
| Created on: |
2005-03-22 14:28:16 |
| Modified on: |
2005-03-22 15:54:21 |
| Classification: |
msc:00A05, msc:12E99, msc:11A05 |
| Defines: |
dividend, remainder |
Revision comment (for changes between this and next version):
| Changes for correction #6379 ('a'). |
Preamble:
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\usepackage{amssymb}
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%\usepackage{psfrag}
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\newcommand{\Nats}{\mathbb{N}}
\newcommand{\Ints}{\mathbb{Z}}
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Content:
In this entry we treat two cases of long division.
\section{Integers}
\begin{thm}[Integer Long Division]
For every pair of integers $a, b\neq 0$ there exist a unique integers $q$ and $r$ such that:
\begin{enumerate}
\item $a=b\cdot q + r,$
\item $0\leq r < |b|$.
\end{enumerate}
\end{thm}
\begin{exa}
Let $a=10$ and $b=-3$. Then $q=-3$ and $r=1$ correspond to the long division:
$$10=(-3)\cdot(-3)+1.$$
\end{exa}
\begin{defn}
The number $r$ as in the theorem is called the remainder of the division of $a$ by $b$. The numbers $a,\ b$ and $q$ are called the dividend, divisor and quotient respectively.
\end{defn}
\section{Polynomials}
\begin{thm}[Polynomial Long Division]
Let $R$ be a commutative ring with non-zero unity and let $a(x)$ and $b(x)$ be two polynomials in $R[x]$, where the leading coefficient of $b(x)$ is a unit of $R$. Then there exist unique polynomials $q(x)$ and $r(x)$ in $R[x]$ such that:
\begin{enumerate}
\item $a(x)=b(x)\cdot q(x) + r(x),$
\item $0\leq \deg(r(x)) < \deg b(x)$ or $r(x)=0$.
\end{enumerate}
\end{thm}
\begin{exa}
Let $R=\Ints$ and let $a(x)=x^3+3$, $b(x)=x^2+1$. Then $q(x)=x$ and $r(x)=-x+3$, so that:
$$x^3+3=x(x^2+1)-x+3.$$
\end{exa}
\begin{exa}
The theorem is not true in general if the leading coefficient of $b(x)$ is not a unit. For example, if $a(x)=x^3+3$ and $b(x)=3x^2+1$ then there are no $q(x)$ and $r(x)$ {\bf with coefficients in} $\Ints$ with the required properties.
\end{exa} |
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