I am attempting Miscellaneous Examples on Chapter 1, Number 2, from Hardy's Course of Pure Mathematics.


Any positive rational number can be expressed in one and only one way in the form: $a_1+\dfrac{a_2}{1\times2}+\dfrac{a_3}{1\times2\times3}+...+\dfrac{a_k}{1\times2\times3\times...\times k}$

My attempt

A positive rational number can be expressed in the form $\frac{w}{v}$ and is in lowest common terms.

First suppose that $w>v$, this implies that $a_1>0$ Likewise, suppose $w<v$, then $a_1=0$

Take the number of times that $v$ divides into $w$. This number is $a_1$. This may leave a remainder, $w_1$ and we now have $a_1+\dfrac{w_1}{v}$

Take $w_1$ and multiple it by $2$. There are two possible cases, $v$ divides evenly into $2w_1$, or there is a remainder. If the former case holds, then stop and we end with $a_1+\dfrac{a_2}{2}$ where $a_2$ is the number of times that $v$ divides into $2w_1$. If the latter is true, than we have $a_1+\dfrac{a_2}{2}+\dfrac{w_2}{v}$, where $v$ is the remainder.

Repeat this process until no remainder results from the division. This yields what Hardy had provided in the example.

My Concerns

My concerns are:

1) I am unsure of the last part of my proof. It just doesn't seem completely right to me.

2) I cannot seem to find a way to address the "expressed in one and one way only" part of the claim.

What I would like

What I would appreciate is if I could get verification on whether my proof is going in the correct direction. I would also appreciate a hint on how to make the proof complete. Please don't provide me with the answer.

Though not the primary purpose of asking this question, since I am self-teaching myself mathematics, I am still new to proof-writing. I would be grateful for any feedback on my current proof-writing skills.

  • 3
    $\begingroup$ You didn't state the problem correctly. It is not a continued fraction, but just as you wrote. Also, the $a_k$ should satisfy $0<a_k<k$. $\endgroup$ – Ron Gordon Jan 14 '13 at 4:10
  • $\begingroup$ @rlgordonma: shouldn't that be $0\le a_k\lt k$? $\endgroup$ – robjohn Jan 14 '13 at 7:13
  • $\begingroup$ @robjohn, almost right. $0 \le a_j <j$ $\forall j \in \left\{1,2, 3,\ldots,k-1 \right\}$ but $0<a_k<k$. $\endgroup$ – Ron Gordon Jan 14 '13 at 7:18
  • $\begingroup$ @rlgordonma: Okay; $0\lt a_k\lt k$ prevents extra representations by adding $0$ terms. However, to cover all rationals, we should let $a_1$ be any integer. The condition above forces $a_1=0$ and covers only $(0,1)$. $\endgroup$ – robjohn Jan 14 '13 at 7:26

You should show that after $a_1$ (which can be anything as robjohn notes), each $a_i \lt i$. Then, considering the prime factorization of $v$, show that $v$ divides evenly into $k!$ for some $k$. The lowest such $k$ will be the last term in your expression. The "expressed in one way only" comes from the division with remainder algorithm. Each $a_i$ is the quotient of a division, and the remainder goes into the rest of the $a$'s. Because the quotient is unique, so is $a_i$.


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