MathJewels math guide · Reason about repeat-stable iteration
Repeat-Stable and Reversible Maps
Use F(F(x))=F(x) to show that every later repeat stabilizes, then use reversibility to decide whether the map must be the identity.
The big idea
A repeat-stable rule satisfies \(F(F(x))=F(x)\). After the first output \(y=F(x)\) appears, applying the rule again gives \(F(y)=y\), so every later application keeps that same output. Repeat-stability alone does not say that \(F(x)=x\): a rule may collapse several inputs onto one fixed output. A supplied undo rule changes the conclusion because an undo cannot recover two different inputs from the same output.
Worked example
Suppose a whole-number rule is \(F(x)=dx\), where \(d\) is a nonnegative whole number. Repeat-stability at input 1 gives
\[ F(F(1))=d^2=F(1)=d. \]
Thus \(d=0\) or \(d=1\). If \(d=0\), then \(F(2)=F(3)=0\), so one undo rule would have to send 0 back to both 2 and 3. That is impossible. Therefore a reversible rule of this form has \(d=1\) and \(F(x)=x\).
How children may show it
A learner may draw a chain \(x\to y\to y\to y\), write a short induction for every positive number of repeats, or exhibit two colliding inputs to disprove reversibility. These are different views of the same structure.
Common mix-up
It is easy to jump from “two passes equal one” directly to “the first pass changes nothing.” Ask, “Could several inputs land on the same fixed output?” Then ask what an undo rule would have to do with that collision.
Try it together
Test the rules \(F(x)=x\), \(F(x)=0\), and “round down to the nearest ten.” Decide which are repeat-stable, which are reversible, and which are both. For each decision, use either a repeat chain or a concrete collision.