Proving that no periodic orbits exist for f(x) = x + q (mod 1)

Question

We have the following map: $$f(x) = x + q \quad\text{(mod 1)}$$ where $q$ is irrational. I would like to show that this map has no periodic fixed points, for any period $k$. This amounts to showing that $$f^k(x_0) = x_0$$ is never true for any fixed $x_0$. The hint of my textbook asserts that $$f^k(x_0) = x_0 + kq = x_0 \quad \text{(mod 1)}$$ Since $q$ is irrational, this amounts to saying that $kq = n$, for some integer $n$, which is clearly not possible. I am a little stuck how they got to their expression for $f^k$, though. Am I missing something obvious? It looks like, in general, it would have a nasty expansion.

Answer 1

At each iteration, $q$ is added.

After $k$ iterations, $kq$ has been added.

The $\bmod 1$ can be done after every iteration or just at the end - the result will be the same.