Is there any numerical representation in which each rational has only one representation?

Question

In positional representations, there are always some rational numbers which have multiple representations. For example, in base 10, 1 can be written as 1 or as $0.\overline{9}$. Do there exist any numerical representations in which all rationals have exactly one representation?

Answer 1

The fundamental theorem of arithmetic states that every natural number greater than $1$ can be written as a unique product of powers of prime numbers. Extending this to the rationals simply requires us to allow the negative powers of prime factors. For example, $$60=2^2\times3\times5$$ and $$\frac{64}{375}=2^6\times 3^{-1}\times5^{-3}.$$

We know that these are unique due to the monotone nature of exponentials.

Answer 2

You can express rational numbers uniquely by terminating continued fractions. Every rational number can be expressed as a finite sequence of integers $[a_0, a_1, \ldots, a_n]$ containing at least one term, where $a_i \in \Bbb{Z}$ for $i \ge 0$, and $a_i > 0$ for $i > 0$. We associate $[a_0, a_1, \ldots, a_n]$ with

$$\huge{ a_0 + \frac{1}{a_1 + \frac{1}{a_2 + \frac{1}{\ddots_{a_{n-1}+\frac{1}{a_n}}}}}}.$$

The way you convert a rational number to its unique continued fraction is you let $a_0$ be the floor function of the given rational. Subtract $a_0$ from this number, leaving you the fractional part, a number in $[0, 1)$. If the number is $0$, then stop. If the number is positive, then compute its reciprocal to obtain a number greater than $1$. You then start over, letting $a_1$ be the floor of this new number, etc.

As an example, consider $\frac{67}{21}$. The floor of this number is $3$, so we subtract $\frac{67}{21} - 3 = \frac{4}{21}$. We reciprocate this fractional part, which produces $\frac{21}{4}$. We let $a_1$ be the integer part of this number, i.e. $5$, leaving us a fractional part $\frac{1}{4}$. Reciprocating this gives us $4$, which has an integer part of $4$ (so $a_2 = 4$), and a fractional part of $0$, telling us to stop. So, $\frac{67}{21}$ is represented by $[3, 5, 4]$, i.e. $$\frac{67}{21} = 3 + \frac{1}{5 + \frac{1}{4}}.$$ Note that the denominator of the fraction under consideration is always strictly decreasing, since the fractional part is always less than $1$, so the process will terminate for any rational number.