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 $k ≥ 2$ .

Question

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.

Do you know how to show that a number base representation is unique? This can be done in a similar way. I think you might need $x\leq 1$, unless you're allowing for $a_1$ to be any integer. — Calvin Lin, Sep 22 '13 at 04:17
I once wrote a proof here: https://blogs.ams.org/mathgradblog/2017/09/20/real-numbers-base-factorials-by-product/ — Behnam Esmayli, Feb 27 '22 at 06:55

Brian M. Scott · Accepted Answer · 2013-09-22T04:46:16.290

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.

@Sush: You’re welcome. – Brian M. Scott Sep 22 '13 at 04:55 — Brian M. Scott, Sep 22 '13 at 04:55

Calvin Lin · Answer 2 · 2020-03-14T15:14:33.217

0

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.

edited Mar 14 '20 at 15:14

answered Sep 22 '13 at 04:25

Calvin Lin

68,864

Please show the uniqueness! I don't know how to show that. – Silent Sep 22 '13 at 04:33
I think there is a typo here: for if $x = 1.9$ then $x-a_1=0.9$ and $a_2 = 1$ so $x-a_1-a_2 < 0$ – Peanut Mar 14 '20 at 11:03
@Calvin Lin can you add a quick explanation for uniqueness? – Peanut Mar 14 '20 at 16:11
@Dude Brian's approach works, and that's the way I would do it. – Calvin Lin Mar 14 '20 at 17:47

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 $k ≥ 2$ .

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