# Prove that $x = a_1+\dfrac{a_2}{2!}+\dfrac{a_3}{3!}+\cdots$

If $x$ is a positive rational number, show that $x$ can be uniquely expressed in the form $$x = a_1+\dfrac{a_2}{2!}+\dfrac{a_3}{3!}+\cdots\text{,}$$ where $a_1,a_2,\ldots$ are integers, $0 \leq a_n \leq n-1$ for $n > 1$, and the series terminates.

I don't see how in the solution below we can take "$a_n \in \{0,\ldots,n-1\}$ such that $m - a_n = nm_1$ for some $m_1$".

Book's solution:

• You want $a_n \equiv m \pmod{n}$. And $\{0,1,\dotsc,n-1\}$ is a complete residue system modulo $n$. May 19 '16 at 20:15
• @DanielFischer How do we know that for the particular rational number $\dfrac{m}{n!}$ we have that $a_n \equiv m \pmod{n}$? May 19 '16 at 20:16
• We choose $a_n$ so that the congruence holds. May 19 '16 at 20:17
• @DanielFischer They say we are given a rational number $\dfrac{m}{n!}$ and so $a_n$ is unique, so we can't just pick $a_n$. May 19 '16 at 20:18
• Does this answer your question? Representation of positive rational numbers as series. Jul 18 at 16:48

It's the archmedian principal. For any two natural number $m$ and $n$ there exist a unique natural number (including 0) $k$ such that $k*n \le m < (k+1)n$.
Or in other words for any two natural numbers $m$ and $n$ there are unique natural $k$ and $a$ such that $m = k*n + a; 0\le a < n$.
Or in other words for any two natural numbers $m$ and $n$ there are unique $m-a = k*n$.
Or in other words for any $m/n!$ we can choice $a_n$ to be the unique $a_n \in \{0,..... ,n-1\}$ so that $m - a_n = n*m_1$ for some natural $m_1$. (i.e. $a_n = a$ above and $m_1 = k$ above.)