Fold a paper once and fold it every time along the existing crease. When can we have the crease formed at the center? Suppose you are folding a piece of paper along the vertical axis. If the top edge (breadth) of the paper is initially $1$ and the first fold reduces the breadth of the paper by $x \in (0,1)$ units, then every fold hereon will be along the existing vertical axis. The existing vertical axis is basically the line/crease you get after each fold.
If we are to mathematically write it, we can recursively track it as :
$x_0 = \{x,1-x\}$
$x_{i+1} = \{\min(x_0), \max(x_0)-\min(x_0)\}$
Can we conclude from here that at some point we will have the crease at the center iff $x \in \mathbb{Q} \backslash \{0\}$?
Here are my observations :

*

*Let $x$ be a non-zero rational number. Then since $x_i$ and $x_j$ for $i \neq j$ will have the same denominator, after finitely many steps, we will have $(\max - \min)(x_n) = 0$ for some $n$. Can we find this $n$?


*If $x$ is irrational, then $x_i$ will comprise of numbers of the form $a + bx$ where $a,b \in \mathbb{Z}$. This can never be $0$.
The two observations lead to my conclusion. Can someone verify if my proof is correct?
 A: I could not understand the folding picture by your description alone, but the numerical description let me figure it out. At any point, there will be a left edge, a right edge, and a folding line. Initially, the left edge is at $0$, the right edge is at $1$, and the folding line is at $x$. At each stage, you do one of two things. If the left edge is closer to the folding line than the right edge, you fold the left edge over the folding line, then the left edge and folding line switch roles. If instead the right edge is closer, do the same with the right edge.
You are correct that when $x\in \mathbb Q$, the process eventually stops. Write $x=\frac{p}{q}$, with $\gcd(p,q)=1$. The two distances to the folding line after the $n^{th}$ fold are $a_n/q$ and $b_n/q$, with $a_0=p$ and $b_0=q-p$. The sequence $(a_0,b_0),(a_1,b_1),\dots$ is just the result of applying Euclid's algorithm to $(a_0,b_0)$. Since $\gcd(a_0,b_0)=1$, this eventually reaches $(a_n,b_n)=(1,0)$, wherein the process stops with a paper of width $1/q$.
There is no formula for the number of iterations it takes. The "worst case" is when $a_0$ and $b_0$ are consecutive Fibonacci numbers; if $a_0=F_n$ and $b_0=F_{n-1}$, then the process terminates in $n-1$ steps.
For the reason you said, the process never terminates when $x$ is irrational.
