TITLE: Is there a natural bijection from $\mathbb{N}$ to $\mathbb{Q}$?
QUESTION [22 upvotes]: In a conversation where it came up that the Pythagoreans probably found an enumeration of the rational numbers I erroneously remarked that Georg Cantor found a natural bijection from $\mathbb{N}$ to $\mathbb{Q}$ with his pairing function. Is there a natural bijection bethween these sets?
Naturalness is of course not a precise criterion. But we may distinguish between degrees of naturalness and say that a bijection $f$ between $\mathbb{N}$ and $\mathbb{Q}$ is more natural than another bijection $g$ between these sets if for the identity statements $f(n)=\alpha(n)$ and $g(n)=\beta(n)$ the formula $\alpha(n)$ is lower in the arithmetical hierarchy than formula $\beta(n)$. Also, $f$ is more natural than $g$ if the formula $\alpha(n)$ is shorter than the formula $\beta(n)$.

REPLY [8 votes]: There is an interesting bijection between the factorial numbering system and the interval of rationals $[0,1)$. Factorials replace powers in the factorial system. $321_! = 3 \cdot 3! + 2 \cdot 2! + 1 = 23_{base10}$. The numerals in each position are limited by the factorial. Only 0 and 1 can be in the first position; 0, 1, or 2 in the second position; 0-3 in the third, etc.
Similarly, we can write fractions using the inverses $\frac{1}{2!}, \frac{1}{3!}, \frac{1}{4!}$. $0.123_! = \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} = \frac{23}{24}$. Every rational number has a unique finite representation in the factorial numbering system. We can count in factorial, 1, 10, 11, 20, 21, ..., and take the "inverse" to get a bijection with the rationals in $[0,1) : .1 = \frac{1}{2}, .01 = \frac{1}{6}, .11= \frac{2}{3}, .02 = \frac{1}{3}, .12 = \frac{5}{6}, .001 = \frac{1}{24}$ etc.
The factorial numbering system is just one of many product based numbering systems that have a unique finite representation for every rational number.