%------------------------------------------------------------------------------ % % Midy's Theorem for Periodic Decimals % % Joseph Lewittes % %------------------------------------------------------------------------------ \documentclass[12pt]{article} \textwidth= 6.5in \textheight= 9.0in \topmargin = -20pt \evensidemargin=0pt \oddsidemargin=0pt \headsep=25pt \parskip=10pt \font\smallit=cmti10 \font\smalltt=cmtt10 \font\smallrm=cmr9 \usepackage{amsmath,amssymb,amsthm} \DeclareMathOperator{\lcm}{lcm} \DeclareMathOperator{\ord}{ord} \newtheorem{theorem}{Theorem} \newtheorem*{midy}{Midy's Theorem} \newtheorem*{remark}{Remark} %\renewcommand{\baselinestretch}{2} \makeatletter %% this should really go into a style file! \renewcommand\section{\@startsection {section}{1}{\z@}% % {-3.5ex \@plus -1ex \@minus -.2ex}% % here is your vskip of 30pt: {-30pt \@plus -1ex \@minus -.2ex}% {2.3ex \@plus.2ex}% {\normalfont\normalsize\bfseries}} \renewcommand\subsection{\@startsection{subsection}{2}{\z@}% {-3.25ex\@plus -1ex \@minus -.2ex}% {1.5ex \@plus .2ex}% {\normalfont\normalsize\bfseries}} % add a point after section numbers: \renewcommand{\@seccntformat}[1]{\csname the#1\endcsname. } %\quad} \makeatother %------------------------------------------------------------------------------ \begin{document} \vspace*{-40pt} \centerline{\smalltt INTEGERS: \smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 7 (2007), \#A02} \vskip 40pt %------------------------------------------------------------------------------ %\title{\bf Midy's Theorem for Periodic Decimals} %\author{Joseph Lewittes} %\date{} %------------------------------------------------------------------------------ %\maketitle \begin{center} \uppercase{\bf Midy's Theorem for Periodic Decimals} \vskip 20pt {\bf Joseph Lewittes}\\ {\smallit Department of Mathematics and Computer Science, Lehman College (CUNY), Bronx, New York 10468}\\ {\tt joseph.lewittes@lehman.cuny.edu}\\ \end{center} \vskip 30pt \centerline{\smallit Received: 5/5/06, Revised: 1/3/07, Accepted: 1/15/07, Published: 1/23/07} \vskip 30pt \centerline{\bf Abstract} In 1836 E. Midy published at Nantes, France, a pamphlet of twenty-one pages on some topics in number theory with applications to decimals. He was the first to actually prove something about our topic. We formulate our own version, and investigate generalizations, of his main result. \pagestyle{myheadings} \markright{\smalltt INTEGERS: \smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 7 (2007), \#A02\hfill} \thispagestyle{empty} \baselineskip=15pt \vskip 30pt %------------------------------------------------------------------------------ \section{Introduction}\label{s1} %------------------------------------------------------------------------------ It is well known--and a proof will appear in our subsequent discussion--that any rational number $c/d$, with $d$ relatively prime to $10$, has a purely periodic decimal expansion of the form $I.a_1 a_2 \ldots a_n a_1 \ldots a_n a_1 \ldots$, where $I$ is an integer, $a_1$, $a_2$, $\ldots$, $a_n$ are digits, and the block $a_1 a_2 \ldots a_n$ repeats forever. The repeating block is called the {\em period} and $n$ is its {\em length}. We write the decimal as $I.\overline{a_1 a_2 \ldots a_n}$, the bar indicating the period. Consider a few examples: $1/3 = 0.\overline{3}$, $1/7 = 0.\overline{142857}$, $2/11 = 0.\overline{18}$, $1/13 = 0.\overline{076923}$, $2/13 = 0.\overline{153846}$, $1/17 = 0.\overline{0588235294117647}$, $1/37 = 0.\overline{027}$, $1/73 = 0.\overline{01369863}$. Note that when the period length is even and the period is broken into two halves of equal length which are then added, the result is a string of 9's. Thus $142 + 857 = 999$, $1 + 8 = 9$, $076 + 923 = 999$, and so on; the numerator plays no role. In each of these examples the denominator is a prime number. Try a few composite denominators: $77 = 7 \times 11$, $1/77 = 0.\overline{012987}$; $803 = 11 \times 73$, $1/803 = 0.\overline{00124533}$; $121 = 11 \times 11$, $1/121 = 0.\overline{0082644628099173553719}$. We see the property holds for $77$ and $121$ but fails for $803$. According to Dickson \cite[p. 161, footnote 19]{1}, H. Goodwyn was apparently the first to observe (in print, 1802) this phenomenon for prime denominators, based on experimental evidence. Over the past two centuries it has been rediscovered many times; it is called the `nines property' by Leavitt \cite{4} and `complementarity' by Shrader-Frechette \cite{8}. This latter reference contains a historical perspective and a bibliography of the topic. %------------------------------------------------------------------------------ In 1836 E. Midy \cite{6} published at Nantes, France, a pamphlet of twenty-one pages on some topics in number theory with applications to decimals. He was the first to actually prove something about our topic. We formulate our own version of his main result. As usual, $\gcd(a,b)$ denotes the greatest common divisor of the integers $a$, $b$. %------------------------------------------------------------------------------ \begin{midy} Let $x$ and $N$ be positive integers, with $N > 1$, $\gcd(N,10) = 1$, $\gcd(x,N) = 1$ and $1 \le x < N$. Assume $x/N = 0.\overline{a_1 a_2 \ldots a_{2k}}$ has even period length $2k$. If \vspace*{-10pt} \begin{enumerate} \item[(i)] $N$ is a prime, or \item[(ii)] $N$ is a prime power, or \item[(iii)] $\gcd(N,10^k - 1) = 1$, \end{enumerate} \vspace*{-10pt} then, for $1 \le i \le k$, \begin{equation}\label{eq1} a_i + a_{k + i} = 9. \end{equation} \end{midy} %------------------------------------------------------------------------------ Note that $(iii)$ explains the difference between $1/77$ where $k = 3$ and $\gcd(77,10^3 - 1) = 1$, which has the Midy property \eqref{eq1}, and $1/803$ where $k = 4$ and $\gcd(803,10^4 - 1) = 11$, for which \eqref{eq1} fails. %------------------------------------------------------------------------------ Various authors have given proofs of this theorem, or parts of it, most being unaware of Midy; see, for example, \cite{2}, \cite{4}, \cite{7} and \cite {8}. (As an aside, note that Rademacher and Toeplitz, \cite[p. 158]{7}, who prove the Midy property for $N$ prime, introduce the topic by saying ``we conclude ... with a property which is more amusing than significant." We leave it to the reader to decide if that comment is more significant than amusing.) Even those who do cite him do so only through Dickson's reference \cite[p. 163]{1}, undoubtedly this is due to the obscure publication of Midy's paper. Recently, Ginsberg \cite{2} extended Midy's theorem to the case where the period has length $3k$; he showed that when the period is broken into three pieces of length $k$ each and then added, the sum is again a string of $9$'s. However, his result is stated only for fractions $1/p$, $p$ a prime, and numerator restricted to be $1$. Example: $1/13 = 0.\overline{076923}$, $07 + 69 + 23 = 99$. However, note that $2/13 = 0.\overline{153846}$, $15 + 38 + 46 = 99$, $3/13 = 0.\overline{230769}$, $23 + 07 + 69 = 99$, $1/21 = 0.\overline{047619}$, $04 + 76 + 19 = 99$, all of which suggest a wider application of the result. But $4/13= 0.\overline{307692}$, $30 + 76 + 92 = 198 = 2 \times 99$. Thus here the numerator plays a role. This will be explained in Theorem \ref{th6}. %------------------------------------------------------------------------------ Eventually, I decided to actually look at Midy's paper--it is available on microfilm at the New York Public Library--and, remarkably, Midy's approach enables one to prove a general theorem that includes the above results and even more. Midy himself considered the case of period length $3k$, but he focused on the sums $a_i + a_{i + k} + a_{i + 2k}$, $1 \le i \le k$, which do not give smooth results. For example, with $1/7$ as above, $3k = 6$, $k = 2$, $1 + 2 + 5 = 8$, $4 + 8 + 7 = 19$, even though $14 + 28 + 57 = 99$. In fact, one easily sees that for period length $2k$ the two halves adding up to a string of $9$'s is equivalent to $a_i + a_{k + i} = 9$, $1 \le i \le k$, but for length $3k$ it is not so, as carrying may occur. %------------------------------------------------------------------------------ We now give a brief survey of the rest of the paper. Section \ref{s2} contains the main results as we study the Midy property in a more general setting. We consider fractions $x/N$ as before whose decimal period of length $e$ is a multiple of a given integer $d$; $e = dk$, for some positive integer $k$. If upon breaking up the period into $d$ blocks of length $k$ each, and then adding the blocks, the sum $S(x)$ is a multiple of $10^k - 1 = 99\cdots9$, a string of $k$ $9$'s, then we say $N$ has the Midy property for the divisor $d$ in base 10. Theorem \ref{th2} shows that this property depends only on $N$ and not the numerator $x$. Furthermore, a sufficient condition for the Midy property to hold is that $\gcd(N,10^k - 1) = 1$. To obtain these results we first examine in detail how the digits in the decimal expansion arise. Since it is just as easy to carry out the analysis for an arbitrary number base $B$ as it is for the decimal base $10$, we do so. Now the definition of the Midy property in base $B$ is the same as above with $10$ replaced by $B$. Theorem \ref{th3} shows that if $p$ is a prime not dividing $B$ and $e$, the order of $B \bmod p$, is a multiple of $d$, then $N = p^h$, $h \ge 1$, has the Midy property for the divisor $d$ in base $B$. In Theorem \ref{th4} we analyze when the Midy property for an integer $N$ can be deduced from its prime factorization. %------------------------------------------------------------------------------ In Section \ref{s3} we consider the ``multiplier." Namely, when $N$ has the Midy property for $B$, $d$, then for given $x/N$, we have $S(x)$ is a multiple of $B^k - 1$. Thus $S(x) = m(B^k - 1)$ for a positive integer $m$, which we call the multiplier. In general $m = m(x)$ depends on $x$, not just $N$. But in Theorem \ref{th6} we show that if $N$ has the Midy property for the divisor $2$ of $e$ then it has it for every even divisor $d$ of $e$, and the multiplier is $m = d/2$, independent of $x$. The Remark after Theorem \ref{th6} shows that our original Midy's Theorem is now a special case, $B = 10$, $d = 2$, of Theorems \ref{th2}, \ref{th3} and \ref{th6}. However, for odd divisors $d$ of $e$ the situation is quite unpredictable. In Theorem \ref{th7} we give an extended version of Ginsberg's theorem, mentioned above, showing that for $d = 3$, $e = 3k$, the multiplier is $1$ for fractions $1/N$ and $2/N$ (provided $N$ is odd) and also $3/N$ (provided $3$ does not divide $N$) except for $N = 7$. %------------------------------------------------------------------------------ \section{Base B and Midy}\label{s2} %------------------------------------------------------------------------------ Let $B$ denote an integer $> 1$ which will be the base for our numerals. The digits in base $B$, $B$-digits for short, are the numbers $0$, $1$, $2$, $\ldots$, $B - 1$. Every positive integer $c$ has a unique representation as $c = d_{n - 1}B^{n - 1} + d_{n - 2}B^{n - 2} + \ldots + d_1B + d_0$, where $n$ is a positive integer, each $d_i$ is a $B$-digit and $d_{n-1} > 0$. As in the decimal case, where $B = 10$, we write $c$ in base $B$ as the numeral $d_{n-1} d_{n-2} \ldots d_0$. When necessary to indicate the base, we write $[d_{n-1} d_{n-2} \ldots d_0]_B$. For $B = 10$ we use the usual notation. We now fix some notation. Unless otherwise noted, our variables $a$, $b$, $\ldots$ denote positive integers. $a|b$ indicates $a$ divides $b$. $B$ is the base and $N$, which will be the denominator of our fractions, is greater than $1$ and relatively prime to $B$. $N^*$ is the set $\{x | 1 \le x \le N$ and $\gcd(x,N) = 1\}$, the set of positive integers less than $N$ relatively prime to $N$. These will be the numerators of our fractions. For $x \in N^*$, $x/N$ is a reduced fraction strictly between $0$ and $1$ and we are interested in the base $B$ expansion of such a fraction. Recalling the elementary school long division process for the decimal expansion of fractions one sees that it amounts to the following. Set $x_1 = x$, let $a_1$ be the integer quotient and $x_2$ the remainder when $Bx_1$ is divided by $N$. Thus $Bx_1 = a_1N + x_2$, $0 \le x_2 < N$ and $a_1 = \lfloor Bx_1/N\rfloor$ where $\lfloor~\rfloor$ is the greatest integer, or floor, function. Continuing inductively, we obtain the following infinite sequence of equations, which we call the long division algorithm. %------------------------------------------------------------------------------ \begin{equation}\label{eq2} \begin{matrix} Bx_1 = a_1 N + x_2 \\ Bx_2 = a_2 N + x_3 \\ \vdots \\ Bx_i = a_i N + x_{i + 1} \\ \vdots \end{matrix} \end{equation} %------------------------------------------------------------------------------ Since $0 < x_1/N < 1$, $Bx_1/N < B$, $a_1 = \lfloor Bx_1/N \rfloor < B$, so $a_1$ is a $B$-digit. Also $B$ and $x_1$ are both relatively prime to $N$ so $Bx_1 \equiv x_2 \pmod N$ shows $(x_2,N) = 1$, so $x_2 \in N^*$. In the same way, for all $i \ge 1$, $a_i$ is a $B$-digit and $x_i \in N^*$. Dividing the first equation by $BN$, the second by $B^2N$, and in general the $i$th by $B^iN$ shows $x_1/N = a_1/B + a_2/B^2 + \ldots + a_i/B^i + x_{i + 1}/B^iN$. Since $0 < x_{i + 1}/B^iN < 1/B^i$ which tends to $0$ as $i \rightarrow \infty$ we have $x_1/N = \sum_{i = 1}^{\infty}a_i/B^i$ which we write as $x_1/N = 0.a_1 a_2 \ldots a_i \ldots$. This is the base $B$ expansion of $x_1/N$ ; $B$ being fixed we omit it from the notation. Reading the equations (\ref{eq2}) mod $N$ shows that for $i \ge 1$ \begin{equation}\label{eq3} x_{i+1} \equiv Bx_i \equiv B^2x_{i-1} \equiv \ldots \equiv B^ix_1 \pmod N. \end{equation} %------------------------------------------------------------------------------ Let $e$ be the order of $B \mod N$; denoted $e = \ord(B,N)$. This means $e$ is the smallest positive integer for which $B^e \equiv 1\pmod N$ and $B^f \equiv 1 \pmod N~\mbox{if and only if}~e|f$. By (\ref{eq3}), $x_{e+1} \equiv B^ex_1 \equiv x_1 \pmod N$ and $x_{i+1} \not\equiv x_1 \pmod N$ for $1 \le i < e$. Since $x_1$, $x_{e + 1}$ both belong to $N^*$, $|x_1 - x_{e + 1}| < N$, so their congruence forces $x_{e + 1} = x_1$. Then $a_{e + 1} = a_1$, $x_{e + 2} = x_2$ and in general $x_{i + e} = x_i$, $a_{i + e} = a_i$, $i \ge 1$. Thus the system (\ref{eq2}) consists of the first $e$ equations which then repeat forever. In particular, the base $B$ expansion of $x_1/N$ is periodic with length $e$ and we write it as $x_1/N = 0.\overline{a_1 a_2 \ldots a_e}$. Since $e$ depends only on $N$ and $B$, not $x_1$, we see that every fraction $x/N$ with $x \in N^*$ has period length $e$. Grouping the terms of the infinite series for $x_1/N$ into blocks of $e$ terms each, and setting $A = [a_1 a_2 \ldots a_e]_B$, produces the geometric series $\sum_{i = 1}^\infty(A/B^{ei})$ and shows $x_1/N = A/(B^e - 1)$. It may be helpful to do a simple numerical example: find the periodic expansion of $1/14$ in base 5. $N = 14$, $B = 5$, $x_1 = 1$; we don't need to know $e = \ord(5,14)$ in advance. The equations (\ref{eq2}) now are \begin{equation}\label{eq4} \begin{matrix} 5 \cdot 1 & = & 0 \cdot 14 + 5 \\ 5 \cdot 5 & = & 1 \cdot 14 + 11 \\ 5 \cdot 11 & = & 3 \cdot 14 + 13 \\ 5 \cdot 13 & = & 4 \cdot 14 + 9 \\ 5 \cdot 9 & = & 3 \cdot 14 + 3 \\ 5 \cdot 3 & = & 1 \cdot 14 + 1 \end{matrix} \end{equation} Having reached the remainder $x_7 = 1 = x_1$, we know that $e = 6$ and $1/14 = 0.\overline {013431}$ in base $5$. %------------------------------------------------------------------------------ Let $d$ be a divisor of $e$ and let $k = e/d$, $e = dk$. Break up the first $e$ equations of (\ref{eq2}) into $d$ groups of $k$ equations each. For $1 \le j \le d$, the $j$th group consists of the following $k$ equations: %------------------------------------------------------------------------------ \begin{equation}\label{eq5} \begin{matrix} Bx_{(j - 1)k + 1} = a_{(j - 1)k + 1}N + x_{(j - 1)k + 2} \\ Bx_{(j - 1)k + 2} = a_{(j - 1)k + 2}N + x_{(j - 1)k + 3} \\ \vdots \\ Bx_{jk} = a_{jk} N + x_{jk + 1} . \end{matrix} \end{equation} %------------------------------------------------------------------------------ Multiply the first equation by $B^{k-1}$, the second by $B^{k-2}$, $\ldots$, the $(k - 1)$th by $B$, and the $k$th by $B^0 = 1$ to obtain %------------------------------------------------------------------------------ \begin{equation}\label{eq6} \begin{matrix} B^k x_{(j-1)k + 1} = a_{(j-1)k + 1}B^{k-1}N + B^{k-1}x_{(j-1)k + 2} \\ B^{k-1}x_{(j-1)k + 2} = a_{(j-1)k + 2}B^{k-2}N + B^{k-2}x_{(j-1)k + 3} \\ \vdots \\ Bx_{jk} = a_{jk} N + x_{jk + 1} . \end{matrix} \end{equation} %------------------------------------------------------------------------------ In (\ref{eq6}), the rightmost term of each equation is the left side of the next equation; so replace the rightmost term of the first equation by the right side of the second equation, then replace the rightmost term of the resulting equation by the right side of the third equation, and so on. Eventually one has \begin{equation}\label{eq7} B^kx_{(j - 1)k + 1} = (a_{(j - 1)k + 1}B^{k - 1} + a_{(j - 1)k + 2}B^{k - 2} + \ldots + a_{jk})N + x_{jk + 1} . \end{equation} The quantity in parentheses is $[a_{(j - 1)k + 1} a_{(j - 1)k + 2} \ldots a_{jk}]_B$, the number represented by the base $B$ numeral consisting of the $j$th block of $k$ $B$-digits in the period; denote this number by $A_j$. So (\ref{eq7}) now becomes $B^kx_{(j - 1)k + 1} = A_jN + x_{jk + 1}$. Add these equations (\ref{eq7}) for $j = 1$, $2$, $\ldots$, $d$ to obtain \begin{equation}\label{eq8} B^k\sum_{j = 1}^d x_{(j - 1)k + 1} = N(\sum_{j = 1}^d A_j) + \sum_{j = 1}^d x_{jk + 1} . \end{equation} But both sums over $x$ are equal since $x_{dk + 1} = x_{e + 1} = x_1$, so (\ref{eq8}) may be rewritten as \begin{equation}\label{eq9} (B^k - 1)\sum_{j = 1}^d x_{(j - 1)k + 1} = N(\sum_{j = 1}^d A_j) . \end{equation} This relation between the two sums is the key to all that follows. It is convenient to define \begin{equation}\label{eq10} R_d(x) = \sum_{j = 1}^d x_{(j - 1)k + 1} ~~ {\rm and} ~~ S_d(x) = \sum_{j = 1}^d A_j . \end{equation} Call the set $\{x_1,x_{k + 1},\ldots,x_{(d - 1)k + 1}\} = \{x_{jk + 1}~|~j \bmod d\}$ the $d$-cycle of $x_1$; more generally, for any $i \ge 1$, $\{x_i,x_{k + i},\ldots,x_{(d - 1)k + i}\} = \{x_{jk + i}~|~j \bmod d\}$ is the $d$-cycle of $x_i$. For any two indices $s$ and $t$, $x_s$ and $x_t$ have the same $d$-cycle if and only if $s \equiv t \pmod k$ and for any $x \in N^*$, $R_d(x)$ and $S_d(x)$ depend only on the $d$-cycle of $x$. Of course, $R$ and $S$ depend also on $B$, $N$ and $e = dk$, but we consider these fixed for the discussion. We summarize the above as %------------------------------------------------------------------------------ \begin{theorem}\label{th1} Given positive integers $N$ and $B>1$, let $e = \ord(B,N)$ and $e = dk$, with $d$ and $k$ positive integers. Let $x \in N^*$ and $x/N = 0.\overline{a_1 a_2 \ldots a_e}$ in base $B$. Break up the period $a_1 a_2 \ldots a_e$ into $d$ blocks of length $k$ each. For $j = 1$, $2$, $\ldots$, $d$, let $A_j = [a_{(j - 1)k + 1} \ldots a_{jk}]_B$, the number represented by the base $B$ numeral consisting of the $j$th block. Let $x_1 = x$, $x_2$, $\ldots$ be the remainders in the long division algorithm {\rm (\ref{eq2})} for $x/N$. Then the following hold: \begin{equation}\label{eq11a} S_d(x) = (R_d(x)/N)(B^k - 1), \end{equation} \begin{equation}\label{eq11b} S_d(x) \equiv 0 \pmod {B^k - 1} ~~ \mbox{iff} ~~ R_d(x) \equiv 0 \pmod N. \end{equation} \end{theorem} %------------------------------------------------------------------------------ \begin{proof} (\ref{eq11a}) is just a rewriting of (\ref{eq9}) in the notation (\ref{eq10}) and then (\ref{eq11b}) is immediate. \end{proof} %------------------------------------------------------------------------------ \noindent {\bf Definition.} Let $N$, $B$, $e$, $d$, $k$ be as above. We say $N$ has the base $B$ Midy property for the divisor $d$ (of $e$) if for every $x \in N^*$, $S_d(x) \equiv 0 \pmod {B^k - 1}$. We denote by $M_d(B)$ the set of integers that have the Midy property in base $B$ for the divisor $d$. %------------------------------------------------------------------------------ \begin{theorem}\label{th2} The following are equivalent: \vspace*{-10pt} \begin{enumerate} \item[$(i)$] $N \in M_d(B)$ \item[$(ii)$] For some $x \in N^*$, $S_d(x) \equiv 0 \pmod {B^k - 1}$ \item[$(iii)$] For some $x \in N^*$, $R_d(x) \equiv 0 \pmod N$ \item[$(iv)$] $(B^e - 1)/(B^k - 1) = B^{k(d - 1)} + B^{k(d - 2)} + \ldots + B^k + 1 \equiv 0 \pmod N.$ \end{enumerate} \vspace*{-10pt} Furthermore $\gcd(B^k - 1,N) = 1$ implies $N \in M_d(B)$. \end{theorem} %------------------------------------------------------------------------------ \begin{proof} The equivalence of $(ii)$ and $(iii)$ follows from Theorem \ref{th1}. Noting (\ref{eq3}), we have $\displaystyle R_d(x) = \sum_{j = 1}^dx_{(j - 1)k + 1} \equiv ( \sum_{j = 1}^dB^{(j-1)k})x\pmod N.$ Since $\gcd(x,N) = 1$, $\displaystyle R_d(x) \equiv 0 \pmod N $ {\rm if and only if} $ \sum_{j = 1}^dB^{k(j - 1)} \equiv 0 \pmod N,$ showing $(iv)$ equivalent to $(ii)$ and $(iii)$. Now $(iv)$ is independent of $x$, so $(iv)$ is equivalent to saying $S_d(x) \equiv 0 \pmod {B^k - 1}$ for every $x \in N^*$, which, by definition, is $(i)$. For the last statement, let $F_d(t)$ be the polynomial $t^{d - 1} + t^{d - 2} + \ldots + t + 1$, so $(iv)$ amounts to $F_d(B^k) \equiv 0 \pmod N$. But $(B^k - 1)F_d(B^k) = B^e - 1 \equiv 0 \pmod N$, by definition of $e$. Thus $\gcd(B^k - 1,N) = 1$ implies $(iv)$, hence $N \in M_d(B)$, completing the proof. \end{proof} %------------------------------------------------------------------------------ Here is an example to show that $\gcd(B^k - 1,N) = 1$ is only sufficient for $N \in M_d(B)$, but not necessary. Take $B = 10$, $N = 21$, $1/21 = 0.\overline{047619}$, $e = 6$. With $d = 3$, $k = 2$, $S_3(1) = 04 + 76 + 19 = 99 \equiv 0 \pmod {10^2 - 1}$, so $21 \in M_3(10)$, but $\gcd(10^2 - 1,21) = 3 \ne 1$. %------------------------------------------------------------------------------ For a numerical illustration, take $N = 14$, $B = 5$, $e = 6$, $x = 1$, as in (\ref{eq4}) above. The period is $013431$ and the remainders $x_1$, $\ldots$, $x_6$ are $1$, $5$, $11$, $13$, $9$, $3$, respectively. With $d = 2$, $k = 3$, $S_2(1) = A_1 + A_2 = [013]_5 + [431]_5 = [444]_5 = 5^3 - 1$ and $R_2(1) = x_1 + x_4 = 1 + 13 = 14$; thus $14 \in M_2(5)$. With $d = 3$, $k = 2$, $S_3(1) = A_1 + A_2 + A_3 = [01]_5 + [34]_5 + [31]_5 = 36 \not\equiv 0 \pmod {5^2 - 1}$, $R_3(1) = x_1 + x_3 + x_5 = 1 + 11 + 9 = 21$; so $14 \not\in M_3(5)$. Note that the relation (\ref{eq11a}) holds: $36 = (21/14)(5^2 - 1)$. %------------------------------------------------------------------------------ For $d = 1$, $k = e$, we never have $N \in M_1(B)$, for this would imply $1 = R_1(1) \equiv 0 \pmod N$, which is impossible. Equivalently, $N \in M_1(B)$ says that for any $x \in N^*$, $S_1(x) \equiv 0 \pmod {B^e - 1}$. But $S_1(x) = A = [a_1 a_2 \ldots a_e]_B$, and we have seen that $A/(B^e - 1) = x/N$. So $M_1(B)$ is empty; from now on we consider only $d > 1$. For $d = e$, $k = 1$, $S_e(x)$ is $\sum_{j = 1}^ea_j$, the sum of the $B$-digits in the period. By Theorem \ref{th2}, $S_e(x) \equiv 0 \pmod {B - 1}$ if $(B - 1,N) = 1$. In particular, with $B = 10$, the period of the decimal for $x/N$ has the sum of its digits divisible by $9$ whenever $N$ is not divisible by $3$. %------------------------------------------------------------------------------ Given $B>1$ and $d>1$ we would like to describe all numbers having the base $B$ Midy property for the divisor $d$; here we make only a few observations in this direction. %------------------------------------------------------------------------------ \begin{theorem}\label{th3} If $p$ is a prime that does not divide $B$ and $e = \ord(B,p)$ is a multiple of $d$, then $p \in M_d(B)$. Then also $p^h \in M_d(B)$ for every $h > 0$. \end{theorem} %------------------------------------------------------------------------------ \begin{proof} Write $e = dk$; $k < e$ since $d > 1$, so $B^k \not\equiv 1 \pmod p$, hence $\gcd(B^k - 1,p) = 1$ and the result follows from Theorem \ref{th2}. Note that $p$ is not $2$, for if so then $B$ is odd and $B^1 \equiv 1 \pmod 2$, so $\ord(B,2) = 1$, which is not a multiple of $d$. For $p \ne 2$ it is known that $e_h = \ord(B,p^h) = ep^g$, where $g$, depending on $h$, is an integer $\ge 0$ whose exact value is not relevant here; see \cite[p. 52]{5}. Thus $e_h = dK$, where $K = kp^g$. By Fermat, $B^K = (B^k)^{p^g} \equiv B^k \pmod p$, so $\gcd(B^K - 1,p^h) \!=\! \gcd(B^k - 1,p) = 1$, and the result follows from Theorem \ref{th2}. \end{proof} %------------------------------------------------------------------------------ Suppose $p_1$, $p_2$, $\ldots$, $p_r$ are distinct primes all belonging to $M_d(B)$ and $N = p_1^{h_1} p_2^{h_2} \ldots p_r^{h_r}$, where $h_1$, $h_2$, $\ldots$, $h_r$ are positive integers. Does $N \in M_d(B)$? It turns out that the answer does not depend on the values of the the $h_i$. For $i = 1$, $2$, $\ldots$, $r$, let $\ord(B,p_i) = e_i = dk_i$, $E_i = \ord(B,p_i^{h_i}) = e_ip_i^{g_i}$, $g_i \ge 0$. Now $E = \ord(B,N) = \lcm(E_1,\ldots,E_r) = \lcm(dk_1p_1^{g_1},\ldots,dk_rp_r^{g_r})$. Set $K_i = E_i/d = k_ip_i^{g_i}$ and $K = E/d$, so $K = \lcm(K_1,\ldots,K_r)$. We need some preliminary remarks. If $q$ is a prime and $w$ a positive integer, denote by $v_q(w)$ the multiplicity of $q$ as a factor of $w$. Thus \begin{equation}\label{eq12} w = \prod_qq^{v_q(w)}, ~~ {\rm the~product~taken~over~all~prime~numbers} ~~ q, \end{equation} where almost all the exponents are $0$. For positive integers $w_1$, $\ldots$, $w_r$, \begin{equation}\label{eq13} \lcm(w_1,\ldots,w_r) = \prod_qq^{m_q}, ~~ {\rm where} ~~ m_q = \max(v_q(w_1),\ldots,v_q(w_r)). \end{equation} If $Q$ is a set of primes, denote by $Q'$ its complement in the set of all primes. Define the $Q$ part of $w$ to be $u = \prod_{q \in Q}q^{v_q(w)}$ and the $Q'$ part $y = \prod_{q \in Q'}q^{v_q(w)}$ so by (\ref{eq12}) $w = uy$. In the same way, (\ref{eq13}) says $\lcm(w_1,\ldots,w_r) = \lcm(u_1,\ldots,u_r) \lcm(y_1,\ldots,y_r).$ Returning to $N$ above, let $Q$ be the set of primes which divide $d$, and $Q'$ the complementary set. Note that each $p_i$ belongs to $Q'$, because $d|e_i \le p_i - 1 < p_i$. Finally, let $c_i$ be the the largest integer $\ge 0$ for which $d^{c_i}$ divides $k_i$; so $k_i = d^{c_i}w_i$ and $d \not| ~ w_i$. Let $u_i$ be the $Q$ part of $w_i$ and $y_i$ the $Q'$ part. Thus $K_i = k_ip_i^{g_i} = (d^{c_i}u_i)(y_ip_i^{g_i})$ is the factorization of $K_i$ into the product of its $Q$ part and $Q'$ part, and $K = \lcm(K_1,\ldots,K_r) = \lcm(d^{c_1}u_1,\ldots,d^{c_r}u_r) \lcm(y_1p_1^{g_1},\ldots,y_rp_r^{g_r}).$ Set \begin{equation}\label{eq14} U = \lcm(d^{c_1}u_1,\ldots,d^{c_r}u_r), ~~ Y = \lcm(y_1p_1^{g_1},\ldots,y_rp_r^{g_r}). \end{equation} so $K = UY$ is the factorization of $K$ into the product of its $Q$ part $U$ and $Q'$ part $Y$. %------------------------------------------------------------------------------ \begin{theorem}\label{th4} Let $p_1$, $\ldots$, $p_r$ be primes each belonging to $M_d(B)$ and $h_1$, $\ldots$, $h_r$ positive integers and $N = p_1^{h_1} \ldots p_r^{h_r}$. With the notations introduced above, $N \in M_d(B)$ if and only if \begin{equation}\label{eq15} for ~~ i = 1,\ldots,r, ~~~ U/(d^{c_i}u_i) \not\equiv 0 \pmod d. \end{equation} This condition depends only on the primes $p_1$, $\ldots$, $p_r$ and not the exponents $h_1$, $\ldots$, $h_r$. If $d$ is a prime $q$, $N \in M_q(B)$ if and only if $q$ occurs with the same multiplicity in each $e_i$: \begin{equation}\label{eq16} v_q(e_1) = v_q(e_2) = \ldots = v_q(e_r). \end{equation} \end{theorem} %------------------------------------------------------------------------------ \begin{proof} Clearly, by definition of $U$, $U/(d^{c_i}u_i)$ is an integer for each $i$. If for some $i$, $U/(d^{c_i}u_i) \equiv 0 \pmod d$ then $dd^{c_i}u_i|U$ and, since also $y_ip_i^{g_i}|Y$, it follows that $E_i = dd^{c_i}u_iy_ip_i^{g_i}|UY = K$. Hence $B^K \equiv 1 \pmod {p_i^{h_i}}$ and, in particular, $B^K \equiv 1 \pmod {p_i}$. Then $F_d(B^K) = \sum_{j = 1}^d(B^K)^{j - 1} \equiv \sum_{j = 1}^d1 \equiv d \pmod {p_i}$. Now by Theorem \ref{th2}, if $N \in M_d(B)$ then $F_d(B^K) \equiv 0 \pmod N$ implying $F_d(B^K) \equiv 0 \pmod {p_i}$, which combined with the previous congruence shows $d \equiv 0 \pmod {p_i}$ which is absurd since $d|e_i < p_i$. So the condition (\ref{eq15}) is necessary for $N \in M_d(B)$. Suppose now that (\ref{eq15}) is satisfied. Then for each $i$, $dd^{c_i}u_i \not| ~ U$, so $e_i = dd^{c_i}u_iy_i \not| ~ UY = K$ and so $B^K \not\equiv 1 \pmod {p_i}$. Thus for each $i$, $(B^K - 1,p_i) = 1$, hence $(B^K - 1,N) =1$, which, by Theorem \ref{th2}, implies $N \in M_d(B)$. This proves (\ref{eq15}) is also sufficient, and clearly (\ref{eq15}) is independent of $h_1$, $\ldots$, $h_r$. Now consider the case where $d$ is a prime number $q$, then $Q = \{q\}$ consists of the single prime $q$. Then the definition of $c_i$ as the the largest integer for which $q^{c_i}|k_i$ says $c_i = v_q(k_i)$; thus $k_i = q^{c_i}w_i$ and $q \not| ~ w_i$ so the $Q$ part $u_i$ of $w_i$ is $1$ which means $U = \lcm(q^{c_1},\ldots,q^{c_r}) = q^c$, where $c = \max(c_1,\ldots,c_r)$. Hence the conditions of (\ref{eq15}) become simply that for each $i$, $q^c/q^{c_i}$ is not divisible by $q$, so $c_i = c$ and $v_q(e_i) = v_q(qk_i) = 1 + c$. This completes the proof. \end{proof} %------------------------------------------------------------------------------ Theorem \ref{th4} was first proved by Jenkins \cite{3} in the case $d = 2$. In \cite[p. 94]{8}, the author seems to claim that if $d$ is any integer, prime or not, then $N \in M_d(B)$ if and only if all the $c_i$ are equal: $c_1 = c_2 = \ldots = c_r$. As our proof shows, this is true only when $d$ is a prime. %------------------------------------------------------------------------------ Here are numerical illustrations of some of our results, which will also show that the above claim is false. We keep the usual notations. Let $p_1 = 7$, $p_2 = 9901$, $p_3 = 19$, $B = 10$: $1/7 = 0.\overline{142857}$, $e_1 = 6$; $1/9901 = 0.\overline{000100999899}$, $e_2 = 12$; $1/19 = 0.\overline{052631578947368421}$, $e_3 = 18$. One checks easily that each $p_i \in M_d(10)$, for each $d | 6$, $d > 1$, as stated in Theorem \ref{th3}. For example, for $19$ with $d = 6$, $k = 3$, $S_6(1) = 052 + 631 + 578 + 947 + 368 + 421 = 2997 \equiv 0 \pmod {10^3 - 1}$. Note that in the setup of Theorem \ref{th4}, whenever some $h_i = 1$, then $g_i = 0$, $E_i = e_i$, $K_i = k_i$; this will be the case in what follows. Now for $p_1p_2 = 7 \times 9901 = 69307$, $E = \lcm(6,12) = 12$, $1/69307 = 0.\overline{000014428557}$. Consider, for Theorem \ref{th4}, those $d$ which divide both $6$ and $12$: $2$, $3$, which are primes, and $6$ which is not. Now $v_3(6) = 1 = v_3(12)$, so $69307 \in M_3(10)$, while $v_2(6) = 1 \ne v_2(12) = 2$, so $69307 \not\in M_2(10)$, as one also easily verifies from the period. For $d = 6$, $Q = \{2,3\}$, $K_1 = k_1 = 1$, $K_2 = k_2 = 2$, $c_1 = c_2 = 0$, $u_1 = y_1 = 1$, $u_2 = 2$, $y_2 = 1$, and (\ref{eq14}) gives $U = \lcm(1,2) = 2$, $Y = \lcm(1,1) = 1$, $K = UY = 2$. Now (\ref{eq15}) is satisfied: for $i = 1$, $2/1 \not\equiv 0 \pmod 6$; for $i = 2$, $2/2 \not\equiv 0 \pmod 6$. Thus we know $69307 \in M_6(10)$; again we verify this directly from the period. We have $S_6(1) = 00 + 00 + 14 + 42 + 85 + 57 = 198 \equiv 0 \pmod {10^2 - 1}$. For a later application we note here that $S_4(1) = 000 + 014 + 428 + 557 = 999$. %------------------------------------------------------------------------------ Now consider $p_1p_2p_3 = 7 \times 9901 \times 19 = 1316833$, $E = \lcm(6,12,18) = 36$, $1/1316833 = 0.\overline{000000759397736842864660894737601503}$. For $d = 2$, $v_2(6) = 1$, $v_2(12) = 2$, $v_2(18) = 1$ and for $d = 3$, $v_3(6) = v_3(12) = 1$, $v_3(18) = 2$, so $1316833$ is not in $M_d(10)$ for $d = 2$ and $3$--again this can be verified from the period. With $d = 6$, $K_1 = k_1 = 1$, $K_2 = k_2 = 2$, $K_3 = k_3 = 3$; none of these is divisible by $6$, so $c_1 = c_2 = c_3 = 0$. $Q = \{2,3\}$, $u_1 = 1$, $u_2 = 2$, $u_3 = 3$ while $y_1 = y_2 = y_3 = 1$, $U = \lcm(1,2,3) = 6$, $Y = 1$, $K = 6$. Now consider (\ref{eq15}): for $i = 1$, $U/(d^{c_1}u_1) = 6/1 \equiv 0 \pmod 6$, so the condition is not satisfied and $1316833 \not\in M_6(10)$. This is a counterexample to the aforementioned claim. To check this numerically, $S_6(1) = 000000 + 759397 + 736842 + 864660 + 894737 + 601503 = 3857139 \not\equiv 0 \pmod {10^6 - 1}$. In fact, $3857139/999999 = 27/7$. The other divisors of $36$ which do not arise from Theorem \ref{th4} are $d = 4$, $9$, $12$, $18$, $36$ and the reader may verify that $1316833 \in M_d(10)$ for each of these. The next theorem shows that not all of this is accidental, but that once it is known for $4$ and $9$ the result follows for their multiples $12$, $18$, $36$. %------------------------------------------------------------------------------ \begin{theorem}\label{th5} Suppose $e = \ord(B,N)$ and $d_1|d_2$, $d_2|e$. If $N \in M_{d_1}(B)$ then $N \in M_{d_2}(B)$. \end{theorem} %------------------------------------------------------------------------------ \begin{proof} An anonymous referee suggested the following simple proof. We have $$\frac{B^e - 1}{B^{k_2} - 1} = \left(\frac{B^e - 1}{B^{k_1} - 1}\right) \left(\frac{B^{k_1} - 1}{B^{k_2} - 1}\right).$$ Both fractions on the right are integers since $k_1$ divides $e$ and $k_2$ divides $k_1$. The first factor is $\equiv 0 \pmod N$ by Theorem \ref{th2}$(iv)$, because $N \in M_{d_1}(B)$. Thus the product $(B^e - 1)/(B^{k_2} - 1) \equiv 0 \pmod N$ and so $N \in M_{d_2}(B)$, again by Theorem \ref{th2}$(iv)$. \end{proof} %------------------------------------------------------------------------------ We also include our somewhat complicated proof because the result \eqref{eq18} below will be used later in the proof of Theorem \ref{th6}. %------------------------------------------------------------------------------ \begin{proof} Write $e = d_1k_1 = d_2k_2$ and set $c = d_2/d_1 = k_1/k_2$. Since $N \in M_{d_1}(B)$, $R_{d_1}(x) \equiv 0 \pmod N$ for every $x \in N^*$. By definition, $R_{d_2}(x) = \sum_{j = 0}^{d_2 - 1}x_{jk_2 + 1}$. We will show that $R_{d_2}(x) = \sum_{r = 0}^{c - 1}R_{d_1}(x_{rk_2 + 1})$, hence $R_{d_2}(x)$ is a sum of terms $\equiv 0 \pmod N$ so it is also $\equiv 0 \pmod N$ which implies $S_{d_2}(x) \equiv 0 \pmod {B^{k_2} - 1}$ and $N \in M_{d_2}(B)$. The numbers $j = 0$, $1$, $\dots$, $d_2 - 1 = cd_1 - 1$ may be written as $j = ic + r$, where $i = 0$, $1$, $\ldots$, $d_1 - 1$ and $r = 0$, $1$, $\ldots$, $c - 1$; then $jk_2 + 1 = ick_2 + rk_2 + 1 = ik_1 + rk_2 + 1$. Thus \begin{equation}\label{eq18} R_{d_2}(x) = \sum_{r = 0}^{c - 1}\sum_{i = 0}^{d_1 - 1}x_{ik_1 + rk_2 + 1}, \end{equation} and the inner sum is just $R_{d_1}(x_{rk_2 + 1})$; this completes the proof. \end{proof} %------------------------------------------------------------------------------ The basic idea here is that the $d_2$-cycle of $x$ is a union of $c$ $d_1$-cycles. %------------------------------------------------------------------------------ \vspace*{-10pt} \section{The Multiplier}\label{s3} %------------------------------------------------------------------------------ For $N \in M_d(B)$ and $x \in N^*$ we have, by definition, $S_d(x) \equiv 0 \pmod {B^k - 1}$, and more precisely, by (\ref{eq11a}), $S_d(x) = m_d(x)(B^k - 1)$ where $m_d(x) = R_d(x)/N$ is an integer, which we call the multiplier; in general it depends on both $d$ and the $d$-cycle of $x$. %------------------------------------------------------------------------------ \begin{theorem}\label{th6} If $N \in M_2(B)$ then for every even $d|e$, $N \in M_d(B)$ and $m_d(x) = d/2$ for every $x \in N^*$. \end{theorem} %------------------------------------------------------------------------------ \noindent \begin{remark} Midy's Theorem of the {\rm Introduction} now follows. For taking $B = 10$, the conditions stated there about $N$ show, by Theorems \ref{th3} and \ref{th2}, that $N \in M_2(10)$ and then this Theorem shows $m_2(x) = 1$, so $S_2(x) = 10^k - 1$, which is a string of $k = e/2$ 9's. \end{remark} %------------------------------------------------------------------------------ \begin{proof} Let $e = 2k$. By Theorem \ref{th2}$(iv)$, $B^k + 1 \equiv 0$, or $B^k \equiv -1 \pmod N$, which, by (\ref{eq3}) with $i = k + 1$, shows $x_{k + 1} \equiv -x_1 \pmod N$. But the only member of $N^*$ that is congruent to $-x_1$ is $N - x_1$, hence $x_{k+1} = N - x_1$, so $R_2(x) = x_1 + x_{k + 1} = x_1 + (N - x_1) = N$, $m_2(x) = 1$; this proves the case $d = 2$. Now say $d > 2$, $2 | d$, $d | e$, $c = d/2$, $k' = e/d$; as shown in the proof of Theorem \ref{th5} the $d$-cycle of $x$ is a union of $c$ $2$-cycles and $R_d(x) = \sum_{r = 0}^{c - 1}R_2(x_{rk' + 1}) = \sum_{r = 0}^{c - 1}N = (d/2)N$, hence $m_d(x) = d/2$. \end{proof} %------------------------------------------------------------------------------ The condition $N \in M_2(B)$ in Theorem \ref{th6} cannot be omitted. For example, we have seen--after the proof of Theorem \ref{th4}--that for $N = 69307$, $e = 12$, $N$ does not belong to $M_2(10)$ but $N$ does belong to $M_4(10)$ and $S_4(1) = 999 = (10^3 - 1)$. Thus in this case $d = 4$ is even and $m_4(1) = 1 \ne 4/2$. %------------------------------------------------------------------------------ We now study the multiplier $m_3(x)$ for $N \in M_3(B)$, $e = 3k$. Recall the result of Ginsberg \cite{2} stated in the Introduction which, in our current notation, says $m_3(1) = 1$ if $N$ is a prime. We now show that such a result holds much more extensively. %------------------------------------------------------------------------------ \begin{theorem}\label{th7} Suppose $N \in M_3(B)$, $e = 3k$. Then \vspace*{-10pt} \begin{enumerate} \item[$(i)$] $m_3(1) = 1$ \item[$(ii)$] if $N$ is odd, $m_3(2) = 1$ \item[$(iii)$] if $3 \not| ~ N$ and $N \ne 7$, $m_3(3) = 1$. \end{enumerate} \end{theorem} %------------------------------------------------------------------------------ \begin{proof} For $x \in N^*$, $R_3(x) = x_1 + x_{k + 1} + x_{2k + 1} < N + N + N = 3N$. Since $R_3(x) \equiv 0 \pmod N$, $R_3(x_1) = N$ or $2N$. If $x = 1$ or $2$, then $x_{k + 1}$, $x_{2k + 1}$ are at most $N - 1$, $N - 2$ (in some order). Thus $R_3(x) \le 2 + (N - 1) + (N - 2) < 2N$, which forces $R_3(x) = N$, $m_3(x_1) = 1$, proving $(i)$ and $(ii)$. Now take $x = 3$; $R_3(3) \le 3 + (N - 1) + (N - 2) \le 2N$, where equality holds if and only if $x_{k + 1} = N - 1$, $x_{2k + 1} = N - 2$, or $x_{k + 1} = N - 2$, $x_{2k + 1} = N - 1$. In the former case, by (\ref{eq3}), $N - 1 \equiv 3B^k$ and $N - 2 \equiv 3B^{2k} \pmod N$, so $9B^{3k} \equiv 2 \pmod N$. But $3k = e$, $B^{3k} \equiv 1 \pmod N$, so $9 \equiv 2 \pmod N$, hence $N = 7$. In the latter case the argument is the same with $N - 1$, $N - 2$ interchanged. This proves $(iii)$. \end{proof} %------------------------------------------------------------------------------ Note that $7$ really is exceptional; take, say, $B = 10$, $3/7 = 0.\overline{428571}$, $S_3(3) = 42 + 85 + 71 = 198 = 2(10^2 - 1)$, so here $m_3(3) = 2$. %------------------------------------------------------------------------------ \vspace*{-10pt} \section{Conclusion}\label{s4} %------------------------------------------------------------------------------ \vskip -10pt Midy's Theorem and its extensions deserve to be better known and certainly have a place in elementary number theory. These patterns in the decimal expansions of rational numbers provide an unexpected glimpse of the charm, and structure, of mathematical objects. Many questions and unexplored pathways remain to be investigated. %------------------------------------------------------------------------------ \vspace*{-10pt} \begin{thebibliography}{} \footnotesize %------------------------------------------------------------------------------ \bibitem[1]{1} L. E. Dickson, {\em History of the Theory of Numbers}, vol. 1, Chelsea, 1952. %------------------------------------------------------------------------------ \bibitem[2]{2} B. D. Ginsberg, Midy's (Nearly) Secret Theorem--An Extension After 165 Years, {\em College Mathematics Journal}, 35 (2004) 26--30. %------------------------------------------------------------------------------ \bibitem[3]{3} M. Jenkins, Question \#1998, {\em Mathematical Questions and Solutions}, 7 (1867) 31--32. %------------------------------------------------------------------------------ \bibitem[4]{4} W. G. Leavitt, A Theorem on Repeating Decimals, {\em Amer. Math. Monthly}, 74 (1967) 669--673. %------------------------------------------------------------------------------ \bibitem[5]{5} W. J. LeVeque, {\em Topics in Number Theory}, vol. 1, Addison - Wesley, 1956. %------------------------------------------------------------------------------ \bibitem[6]{6} E. Midy, {\em De Quelques Propri\'et\'es des Nombres et des Fractions D\'ecimales P\'eriodiques}, Nantes, 1836. %------------------------------------------------------------------------------ \bibitem[7]{7} H. Rademacher and O. Toeplitz, {\em The Enjoyment of Mathematics}, Dover Publications, 1990. %------------------------------------------------------------------------------ \bibitem[8]{8} M. Shrader-Frechette, Complementary Rational Numbers, {\em Math. Mag.}, 51 (1978) 90--98. %------------------------------------------------------------------------------ \end{thebibliography} %------------------------------------------------------------------------------ %\noindent %Department of Mathematics and Computer Science, % Lehman College, CUNY, Bronx, NY 10468 %------------------------------------------------------------------------------ %\noindent %joseph.lewittes@lehman.cuny.edu %------------------------------------------------------------------------------ \end{document} %------------------------------------------------------------------------------