I already have proved by induction that the equalities are true $\forall r,s\in\mathbb{Z}$
Because induction only works for statements that uses natural numbers. We first of all have to find a bijection from $\mathbb{Q}\rightarrow\mathbb{N}$ and reframe the statement.
$f:\mathbb{Q}\rightarrow\mathbb{Z}$
We are assuming that we have already reduced the rational number which we put in the function.
Then $f(\frac{x}{z}):=$
$f(\frac{1}{1}) = 1$
$f(\frac{0}{1}) = 0$
case a $\frac{x}{z}\geq 0 \rightarrow$ case 1
case b $\frac{x}{z}< 0 \rightarrow -f(-\frac{x}{z})$
case 1 $\exists k\in\mathbb{N}:\frac{x/k}{z/k}=\frac{x}{z}$ then $f(\frac{x}{z})=f(\frac{x-1}{z})$
for example: $f(\frac{3}{4})=f(\frac{2}{4})+1 =f(\frac{1}{4})+1$
case 2 $x-1 = z$ then $f(\frac{x}{z})=f(\frac{z}{x})+1$
for example: $f(\frac{5}{4})=f(\frac{4}{5})+1$
case 3 $x>z$ then $f(\frac{x}{z})=f(\frac{x}{z+1})+1$
for example: $f(\frac{7}{3})=f(\frac{7}{4})+1$
case 4 $x=1$ then $f(\frac{x}{z})=f(\frac{z-1}{x})+1$
for example: $f(\frac{1}{6})=f(\frac{5}{1})+1$
case 5 $z>x$ then $f(\frac{x}{z})=f(\frac{x-1}{z})+1$
for example: $f(\frac{3}{4})=f(\frac{2}{4})+1 =f(\frac{1}{4})+1$
This is the list of the first shortened rational numbers if we would look at the preimages:
1, 1/2, 2, 1/3, 2/3, 3/2, 3, 1/4, 3/4, 4/3, 4, 1/5, 2/5, 3/5, 4/5, 5/4, 5/3, 5/2, 5, 1/6, 5/6, 6/5, 6, 1/7, 2/7, 3/7, 4/7, 5/7, 6/7, 7/6, 7/5, 7/4, 7/3, 7/2, 7
We define
$g:\mathbb{Z}\rightarrow\mathbb{N}$
with $g(x):= 2x$ if $x$ is positve otherwise $2|x|+1$
I substitute $r$ with $(f$ o $g)^{-1}(n)$= $f^{-1}$o $g^{-1}$(n) and choose a random $s$
Inductionbase n=1
Because $g(0)=1$ and $f(\frac{0}{1}) = 0$ we have $(f$ o $g)^{-1}(1) = 0$ and
because $x^0 = 1$ the inductionbase is proved.
Can somebody help me with the inductionstep?