Any positive rational number can be expressed in one and only one way in the form …

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

Question

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.

• You didn't state the problem correctly. It is not a continued fraction, but just as you wrote. Also, the $a_k$ should satisfy $0<a_k<k$. – Ron Gordon Jan 14 '13 at 4:10
• @rlgordonma: shouldn't that be $0\le a_k\lt k$? – robjohn Jan 14 '13 at 7:13
• @robjohn, almost right. $0 \le a_j <j$ $\forall j \in \left\{1,2, 3,\ldots,k-1 \right\}$ but $0<a_k<k$. – Ron Gordon Jan 14 '13 at 7:18
• @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 $(0,1)$. – 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 $a_i$.