Evidence of alternative calculation of fractions

Question

I found the following iterative calculation method for fractions by chance and would like to prove its validity. It reads: \begin{equation} \frac{n}{m} = \frac{f\left(9\cdot \frac{n}{m}\right)}{10}, \hspace{20pt} f(r) = f(k+p) = k + \frac{f(10p+k)}{10}, \hspace{20pt} n,m, k \in \mathbb{N}, \hspace{10pt} r,p \in \mathbb{R}_+, \hspace{10pt} p < 1 \end{equation} Here is an example: \begin{equation} \frac{7}{18} = \frac{f\left(9\cdot \frac{7}{18}\right)}{10} = \frac{f(3+0.5)}{10} = \frac{3}{10} + \frac{f(8)}{100} = \frac{3}{10} + \frac{8}{100} + \frac{f(8)}{100} \end{equation} And because obviously the function $f$ of a natural number $k \in \mathbb{N}$ is equal to the "period of this number", so \begin{equation} f(k) = k.\bar{k} = \frac{10\cdot k}{9}, \end{equation} finally follows for my example \begin{equation} \frac{7}{18} = 0.3\bar{8} \end{equation}

Answer 1

Okay, I'm stupid. Simply choose a linear function for $f$ as an approach: $f(x) = ax+b$. If you now use this, you get \begin{equation} f(k+p) = a(k+p) + b = k + \frac{a(10p+k)+b}{10} = k + \frac{f(10p+k)}{10} \end{equation} By changeover follows: \begin{equation} \left(a-1-\frac{a}{10}\right)k + \left(a-a\right)p + b - \frac{b}{10} = 0 \end{equation} Since this must apply to all k and p, it follows immediately: $a = \frac{10}{9}, \hspace{5pt} b = 0$. Since this must apply to all k and p, it follows immediately: $a = \frac{10}{9}, \hspace{5pt} b = 0$. And also this then fulfills the equality to fraction: \begin{equation} \frac{n}{m} = \frac{f\left(9\cdot\frac{n}{m}\right)}{10} = \frac{\frac{10}{9}\left(9\cdot\frac{n}{m}\right)}{10} = \frac{n}{m} \end{equation}