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Revision difference : factorial base representation of fractions
Version 3 Version 2
One can represent fractions as well as whole numbers using factorials One can represent fractions as well as whole numbers using factorials
much in the same way that one has, say, a decimal representation of much in the same way that one has, say, a decimal representation of
both whole numbers and fractions. both whole numbers and fractions.
Suppose that $x$ is a rational number. For simplicity, let us assume Suppose that $x$ is a rational number. For simplicity, let us assume
that $0 < x < 1$. Then we can write that $0 < x < 1$. Then we can write
\[x = \sum_{k=2}^N {d_k \over k!}\] \[x = \sum_{k=2}^N {d_k \over k!}\]
where $0 \le d_k < k$ for some integer $N$. Unlike decimal representations where $0 \le d_k < k$ for some integer $N$. Unlike decimal representations
of fractions and, more generally representations with any fixed base, of fractions and, more generally representations with any fixed base,
factorial base representations of rational numbers all terminate. factorial base representations of rational numbers all terminate.
Let us illustrate with some simple examples: Let us illustrate with some simple examples:
\begin{eqnarray*} \begin{eqnarray*}
\frac{1}{2} &=& \frac{1}{2!} \\ \frac{1}{2} &=& \frac{1}{2!} \\
\frac{1}{3} &=& \frac{2}{3!} \\ \frac{1}{3} &=& \frac{2}{3!} \\
\frac{2}{3} &=& \frac{1}{2!} + \frac{1}{3!} \\ \frac{2}{3} &=& \frac{1}{2!} + \frac{1}{3!} \\
\frac{1}{4} &=& \frac{1}{3!} + \frac{2}{4!} \\ \frac{1}{4} &=& \frac{1}{3!} + \frac{2}{4!} \\
\frac{3}{4} &=& \frac{1}{2!} + \frac{1}{3!} + \frac{2}{4!} \\ \frac{3}{4} &=& \frac{1}{2!} + \frac{1}{3!} + \frac{2}{4!} \\
\frac{1}{5} &=& \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} \\ \frac{1}{5} &=& \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} \\
\frac{2}{5} &=& \frac{2}{3!} + \frac{1}{4!} + \frac{3}{5!} \\ \frac{2}{5} &=& \frac{2}{3!} + \frac{1}{4!} + \frac{3}{5!} \\
\frac{3}{5} &=& \frac{1}{2!} + \frac{2}{4!} + \frac{2}{5!} \\ \frac{3}{5} &=& \frac{1}{2!} + \frac{1}{3!} + \frac{3}{4!} + \frac{1}{5!}
\frac{4}{5} &=& \frac{1}{2!} + \frac{1}{3!} + \frac{3}{4!) +
\frac{1}{5!}
\end{eqnarray*} \end{eqnarray*}