# Does this suggest a unique additive factorization on the rational numbers?

I'm reading Hardy's Course of Pure Mathematics (3rd edition) and there is an interesting exercise on page 34:

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

$$a_1+\frac{a_2}{1\cdot 2}+\frac{a_3}{1\cdot 2 \cdot 3}+\dots+\frac{a_k}{1\cdot 2 \cdot 3 \dots k'}$$

Where $a_1, a_2,\dots ,a_k$ are integers and

$$0\leq a_1, \quad 0\leq a_2 < 2, \quad 0\leq a_3 < 3 \quad \dots \quad 0\leq a_k <k$$

Does this somehow suggest that there is a unique additive factorization on the rational numbers? With $\cfrac{a_n}{n!}$ acting like prime numbers?

• No - you can have any number of primes in a multiplicative factorization, but the coefficients of your $\frac{1}{n!}$s are restricted. You can write any positive integer in base $10$ uniquely using digits from $0,\cdots,9$ and we don't call this a "factorization" either. – anon Jul 18 '15 at 0:46
• This is more like a base; it might be called "factorial base." – Qiaochu Yuan Jul 18 '15 at 1:03