Given $ \frac {p} {q} $, can you achieve to paint the point C such that $\frac{AC}{CB} = \frac {p} {q}$ with permitted operations

Question

A certain segment $ AB $ has only its ends painted. Matías can perform a series of operations of the following type:

1) Choose two points $ X, Y $ painted and paint the midpoint of $ XY $.

2) Choose two points $ X, Y $ painted, a positive integer $ n $ and paint the point $ Z $ of the segment $ XY $ that satisfies $\frac {XZ} {ZY} = \frac {n} {n +1}$.

Matías affirms that given any fraction $ \frac {p} {q} $, with $ p, q $ positive integers, he can achieve to paint the point C on the segment AB such that $ \frac {AC} {CB} = \frac {p} {q} $. Is it true what Matías says?

Answer 1

Let me remove the geometry from the problem for you: Identify the segment with the unit interval $[0,1]$. The given construction steps are then as follows; starting with the numbers $0$ and $1$:

1. Take two constructed numbers $x$ and $y$ and construct $\tfrac{x+y}{2}$.
2. Take two cosntructed numbers $x$ and $y$ and a positive integer $n$, and construct $\frac{(n+1)x+ny}{2n+1}$.

Now the question is: Given positive integers $p$ and $q$, can you construct $\tfrac{p}{p+q}$?