resources:
math overflow, wikipedia, Hardy+Wright textbook chapter 3
content
We denote the Farey numbers by \(F_n\).
Proposition. 1. When you go from \(F_n\) to \(F_{n+1}\) the inserted numbers are mediants. 2. Consecutive Farey fractions \(\frac{a}{b}, \frac{c}{d}\) have difference \(\frac{1}{bd}\)
Proof. 1. There is a relatively straightforward inductive proof
- I like this constructive proof better though:
We construct the successor to \(\frac{h}{k}\) in \(F_n\). Take \(x,y\) solving \(kx-hy = 1\); they must exist because \(h\perp k\). In fact, if \(x,y\) are a solution then so is \(x+rh,y+rk\) for any \(r\in \mathbb{Z}\). So take \(r\) such that \(n-k<y\le n\).
Now, imagine that \(\frac{x}{y}\) is not the successor of \(\frac{h}{k}\) despite having \(\frac{h}{k}-\frac{x}{y}=\frac{1}{ky}\). Then we must have coprime \(h',k'\) which is between \(\frac{h}{k}\) and \(\frac{x}{y}.\)
We compute: \[\frac{x}{y}-\frac{h'}{k'} \ge \frac{1}{k'y}\] \[\frac{h'}{k'}-\frac{h}{k} \ge \frac{1}{k'k}\] We already knew \[\frac{x}{y}-\frac{h}{k} = \frac{1}{ky}.\] Combining these equations gives \[\frac{1}{ky} \ge \frac{1}{k'} (\frac{1}{y}+\frac{1}{k}) > \frac{1}{ky}\] a contradiction.
Then they give a geometrical proof :O. I’m hyped lets see this.
ok actually I didn’t understand the lattice proof right now. I will come back to it later maybe