Every positive rational number $x$ can be expressed in the form $\sum_{k=1}^n \frac{a_k}{k!}$ with $ a_k≤ k − 1$ for $k ≥ 2$ . I have this theorem which I can't prove.Please help.
"Show that every positive rational number $x$ can be expressed in the form $\sum_{k=1}^n \frac{a_k}{k!}$ in one and only one way where each $a_k$ is non-negative integer with $ a_k≤ k − 1$ for $k ≥ 2$ and $a_n>0$."
I think the ONE way is this.But I don't know that how to prove that it is the ONLY way.
Or is this not that ONE way?
Please prove it.Thank you.
 A: HINT: For existence I find it easiest to work backwards. Express the fractional part of $x$ as a fraction $\frac{c}d$ in lowest terms, and let $n$ be minimal such that $d\mid n!$. Write $x$ as $\frac{c}{n!}$, and write $c=q_0n+a_n$, where $0\le a_n<n$. Then the fractional part of $x$ is
$$x-\lfloor x\rfloor=\frac{c}{n!}=\frac{q_0n+a_n}{n!}=\frac{q_0}{(n-1)!}+\frac{a_n}{n!}\;.$$
Now write $q_0=q_1(n-1)+a_{n-1}$, where $0\le a_{n-1}<n-1$, and note that
$$\frac{q_0}{(n-1)!}=\frac{q_1(n-1)+r_{n-1}}{(n-1)!}=\frac{q_1}{(n-2)!}+\frac{r_{n-1}}{(n-1)!}\;,$$
so that you can continue the process to find in succession $a_n$, $a_{n-1}$, $a_{n-2}$, and so on to $a_1$.
You can actually organize this as an induction on $n$: if your induction hypothesis is that every rational number that can be written with a denominator that divides $(n-1)!$ has a decomposition of the desire kind, the first calculation above show that the same is true of fractions with denominators dividing $n!$.
For uniqueness use the fact that for $1\le m\le n$ we have
$$\sum_{k\ge m}^n\frac{k-1}{k!}=\sum_{k\ge m}^n\left(\frac1{(k-1)!}-\frac1{k!}\right)=\frac1{(m-1)!}-\frac1{n!}<\frac1{(m-1)!}\;;$$
this implies that if you don’t choose $a_{m-1}$ to be as large as possible, the terms with smaller denominators cannot make up the difference.
A: First, we show existence. For any rational $x$, define a sequence of integers as follows:
$a_1 = \lfloor x \rfloor$, observe that $ 0 \leq x - a_1 < 1$. If it is 0, stop.
$a_2 = \lfloor 2!(x - \frac{a_1}{1}) \rfloor$, the previous observation gives us that $a_2 \leq 1$. Observe that $ 0 \leq x - \frac{a_1}{1} - \frac{a_2}{2!} < \frac{1}{2!}$.  If it is 0, stop.
$a_3 = \lfloor 3!(x - a_1 - a_2) \rfloor$, the previous observation gives us that $a_3 \leq 2$. Obseve that $0 \leq x - \frac{a_1}{1!} - \frac{a_2}{2!} - \frac{a_3}{3!} < \frac{1}{3!} $. If it is 0, stop.
Continue in this manner to define $a_n$.   
Next, we show uniqueness. This is pretty straightforward, so I'd leave it to you.
