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A rational number $d$ is called a \emph{decimal fraction} if $10^kd$ is an integer for some non-negative integer $k$. For example, any integer, as well as rationals such as A rational number $d$ is called a \emph{decimal fraction} if $10^kd$ is an integer for some non-negative integer $k$. For example, any integer, as well as rationals such as
$$0.23123,\qquad \frac{3}{4},\qquad \frac{236}{125}$$ $$0.23123,\qquad \frac{3}{4},\qquad \frac{236}{125}$$
are all decimal fractions. Rational numbers such as $$\frac{1}{3},\qquad -\frac{227}{12}, \qquad 2.\overline{312}$$ are not. are all decimal fractions. Rational numbers such as $$\frac{1}{3},\qquad -\frac{227}{12}, \qquad 2.\overline{312}$$ are not.
There are two other ways of characterizing a decimal fraction: for a rational number $d$, There are two other ways of characterizing a decimal fraction: for a rational number $d$,
\begin{enumerate} \begin{enumerate}
\item $d$ is as in the above definition; \item $d$ is as in the above definition;
\item $d$ can be written as a fraction $\displaystyle{\frac{p}{q}}$, where $p$ and $q$ are integers, and $q=2^m5^n$ for some non-negative integers $m$ and $n$; \item $d$ can be written as a fraction $\displaystyle{\frac{p}{q}}$, where $p$ and $q$ are integers, and $q=2^m5^n$ for some non-negative integers $m$ and $n$;
\item $d$ has a terminating decimal expansion, meaning that it has a decimal representation $$a.d_1d_2\cdots d_n000\cdots$$ \item $d$ has a terminating decimal expansion.
where $a$ is an integer and $d_i$ is any one of the digits $0,\ldots,9$.
\end{enumerate} \end{enumerate}
A decimal fraction is sometimes called a \emph{decimal number}, although a decimal number in the most general sense may have non-terminating decimal expansions.
\textbf{Remarks}. Let $D\subset \mathbb{Q}$ be the set of all decimal fractions. \textbf{Remarks}. Let $D\subset \mathbb{Q}$ be the set of all decimal fractions.
\begin{itemize} \begin{itemize}
\item \item
If $a,b \in D$, then $a\cdot b$ and $a+b\in D$ as well. Also, $-a \in D$ whenever $a\in D$. In other words, $D$ is a subring of $\mathbb{Q}$. Furthermore, as an abelian group, $D$ is $2$-divisible and $5$-divisible. However, unlike $\mathbb{Q}$, $D$ is not \PMlinkname{divisible}{DivisibleGroup}. If $a,b \in D$, then $a\cdot b$ and $a+b\in D$ as well. Also, $-a \in D$ whenever $a\in D$. In other words, $D$ is a subring of $\mathbb{Q}$. Furthermore, as an abelian group, $D$ is $2$-divisible and $5$-divisible. However, unlike $\mathbb{Q}$, $D$ is not \PMlinkname{divisible}{DivisibleGroup}.
\item \item
As inherited from $\mathbb{Q}$, $D$ has a total order structure. It is easy to see that $D$ is \PMlinkname{dense}{DenseTotalOrder}: for any $a,b\in D$ with $a< b$, there is $c\in D$ such that $a<c<b$. Simply take $c=\displaystyle{\frac{a+b}{2}}$. As inherited from $\mathbb{Q}$, $D$ has a total order structure. It is easy to see that $D$ is \PMlinkname{dense}{DenseTotalOrder}: for any $a,b\in D$ with $a< b$, there is $c\in D$ such that $a<c<b$. Simply take $c=\displaystyle{\frac{a+b}{2}}$.
\item \item
From a topological point of view, $D$, as a subset of $\mathbb{R}$, is dense in $\mathbb{R}$. This is essentially the fact that every real number has a decimal expansion, so that every real number can be ``approximated'' by a decimal fraction to any degree of accuracy. From a topological point of view, $D$, as a subset of $\mathbb{R}$, is dense in $\mathbb{R}$. This is essentially the fact that every real number has a decimal expansion, so that every real number can be ``approximated'' by a decimal fraction to any degree of accuracy.
\item \item
We can associate each decimal fraction $d$ with the least non-negative integer $k(d)$ such that $10^{k(d)}d$ is an integer. This integer is uniquely determined by $d$. In fact, $k(d)$ is the last decimal place where its corresponding digit is non-zero in its decimal representation. For example, $k(1.41243)=5$ and $k(7/25)=2$. It is not hard to see that if we write $d=\displaystyle{\frac{p}{2^m5^n}}$, where $p$ and $2^m5^n$ are coprime, then $k(d)=\max(m,n)$. We can associate each decimal fraction $d$ with the least non-negative integer $k(d)$ such that $10^{k(d)}d$ is an integer. This integer is uniquely determined by $d$. In fact, $k(d)$ is the last decimal place where its corresponding digit is non-zero in its decimal representation. For example, $k(1.41243)=5$ and $k(7/25)=2$. It is not hard to see that if we write $d=\displaystyle{\frac{p}{2^m5^n}}$, where $p$ and $2^m5^n$ are coprime, then $k(d)=\max(m,n)$.
\item \item
For each non-negative integer $i$, let $D(i)$ be the set of all $d\in D$ such that $k(d)=i$. Then $D$ can be partitioned into sets $$D=D(0)\cup D(1) \cup \cdots \cup D(n) \cup \cdots.$$ Note that $D(0)=\mathbb{Z}$. Another basic property is that if $a\in D(i)$ and $b\in D(j)$ with $i<j$, then $a+b\in D(j)$. For each non-negative integer $i$, let $D(i)$ be the set of all $d\in D$ such that $k(d)=i$. Then $D$ can be partitioned into sets $$D=D(0)\cup D(1) \cup \cdots \cup D(n) \cup \cdots.$$ Note that $D(0)=\mathbb{Z}$. Another basic property is that if $a\in D(i)$ and $b\in D(j)$ with $i<j$, then $a+b\in D(j)$.
\end{itemize} \end{itemize}