Find the functions $f: \mathbb{Q}_{+}^{*} \rightarrow \mathbb{Q}_{+}^{*}$ such that for all $x \in \mathbb{Q}_{+}^{*}$, we have $f(x+1)=$ $f(x)+1$ and $f(1 / x)=1 / f(x)$.
Clearly, $f(n)=n$ if $n \in \mathbb{N}^{*}$, and more generally:

$$
f(x+n)=f(x)+n
$$

for $x \in \mathbb{Q}_{+}^{*}$ and $n \in \mathbb{N}$. Moreover, the behavior of $f$ with respect to the inverse allows the calculation of other values: for example, $f(2 / 3)=1 / f(3 / 2)=1 /(1+f(1 / 2))=1 /(1+1 / f(2))=2 / 3$. The idea is as follows: when we have a number greater than 1, we use the relation; if we have a number less than 1, we switch to the inverse, and we reduce it to the following case, by induction. Thus, to calculate $f(11 / 7)$, we start by calculating $f(4 / 7)$. To do this, we calculate $f(7 / 4)$, and therefore we first calculate $f(3 / 4)$. It is sufficient to calculate $f(4 / 3)$, and for that, we just need to calculate $f(1 / 3)=1 / 3$. We can then trace back the calculations and obtain $f(11 / 7)=11 / 7$. It is then sufficient to show that this procedure - which is actually the Euclidean algorithm - terminates in a finite number of steps.