# How would you show $\mathbb{Q}=\langle \frac{1}{n!}: n\in \mathbb{N} \rangle$?

The containment of one side is obvious, but I can't see how to show $\mathbb{Q} \subseteq\langle \frac{1}{n!}: n\in \mathbb{N} \rangle$, and would prefer an answer that shows the set containment via algebraic techniques.

I am mean how to show that every rational number can be written as a finite sum of the reciprocals of factorials (or their negatives)?

That is, I am considering the rationals as an additive group and $\langle \cdot \rangle$ means the subgroup generated by the set.

• Is this a question about rings, groups, $\Bbb Z$-modules, $\Bbb Q$ vector spaces? – user2520938 Jan 4 '17 at 10:46
• If $n>b$ then $n!=b\times m$ so $\frac ab=\frac {am}{n!}$. – lulu Jan 4 '17 at 10:49
• Note: I am assuming that you meant "show that every rational number can be written as a finite sum of the reciprocals of factorials (or their negatives)" . If you meant something else, you should clarify. – lulu Jan 4 '17 at 10:50
• @lulu Yes your interpretation is correct and that is the question I was asking. – Mark Jan 4 '17 at 10:52
• Say you want $\frac{a}{b},\ a,b > 0$, then $\frac{a}{b}= a(b-1)! \cdot \frac{1}{b!}$. – Zubzub Jan 4 '17 at 11:02

It follows from the Fundamental Theorem of Arithmetic that $\Bbb Q$ is generated, as a group, by the set $$\left\{\frac1{p^k}\text{ such that p is prime and k>0}\right\}$$ (one can make this set of generators smaller, but that is irrelevant here). Then it is enough to show that each $1/p^k$ can be obtained from the $1/n!$, but then it is enough to notice that $$\frac1{p^k}=(p^k-1)!\frac1{(p^k)!}.$$
HINT: The generated subgroup also contains any finite sum of the same element $k$ times since $$\frac{k}{n!} = \underbrace{\frac{1}{n!} + \ldots \frac{1}{n!}}_{k \text{ times}}.$$