1
$\begingroup$

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:

enter image description here

$\endgroup$
11
  • 1
    $\begingroup$ You want $a_n \equiv m \pmod{n}$. And $\{0,1,\dotsc,n-1\}$ is a complete residue system modulo $n$. $\endgroup$ May 19 '16 at 20:15
  • $\begingroup$ @DanielFischer How do we know that for the particular rational number $\dfrac{m}{n!}$ we have that $a_n \equiv m \pmod{n}$? $\endgroup$ May 19 '16 at 20:16
  • $\begingroup$ We choose $a_n$ so that the congruence holds. $\endgroup$ May 19 '16 at 20:17
  • $\begingroup$ @DanielFischer They say we are given a rational number $\dfrac{m}{n!}$ and so $a_n$ is unique, so we can't just pick $a_n$. $\endgroup$ May 19 '16 at 20:18
  • 2
    $\begingroup$ Does this answer your question? Representation of positive rational numbers as series. $\endgroup$
    – Buraian
    Jul 18 at 16:48
0
$\begingroup$

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.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.