# How to determine the orbits of points under the tripling map $f(x)=3x\bmod 1$?

Let $f$ be the tripling map $f(x) = 3x \mod(1)$. Determine the complete orbit of the points $\frac{1}{8}$ and $\frac{1}{72}$. Indicate whether each of these points is periodic, eventually periodic, or neither.

I've asked my professor for plenty of help on this matter, but she can't explain this too me at all and makes the whole matter even more confusing. Is there any help out there to better explain this type of problem that can lead me to the correct answer?

• What have you tried so far? In particular, what are the first few iterates of these points under the tripling map? – Semiclassical Sep 18 '14 at 4:29
• Also notice that $\frac{3^2}{72} = \frac18$. So if $\frac18$ is periodic and $\frac1{72}$ does not appear in its orbit, what do you think will happen to $\frac1{72}$? Can it be periodic? – Isomorphism Sep 18 '14 at 5:02
• Wait, you cannot determine the orbit of $\frac18$? – Did Sep 19 '14 at 10:32

I suppose $f$ is a map $\Bbb{R} \rightarrow \Bbb{R}$. First you should consider what happens to irrational resp. rational numbers. The$\mod(1)$ expression creates an equivalence relation $x\sim y \Leftrightarrow x-y \in \Bbb {Z}$. Can an irrational number become rational? (i.e can an irrational belong to the same equivalence class as a rational?). Now consider a rational number of the form $\frac{p}{q}<1$ and $\gcd(p,q)=1$. The numbers in the orbit of this number are all of the form $\frac{r}{q}<1$. Show that sooner or later a value will hit a previous value (use the pigeonhole principle). What happens after a previous value has been "hit"?
NestList[Mod[3 #, 1] &, 1/72, 10]

It's now crystal clear that $$\frac{1}{72} \rightarrow \frac{1}{24} \rightarrow \frac{1}{8} \rightarrow \frac{3}{8} \rightarrow \frac{1}{8}$$ and that you've now reached a loop.