Whether Functions With A Periodic-Like Property Are Constant 
Call a function $f(x)$ defined over $\mathbb R \setminus \{0\}$ “drunken periodic” if $$\forall k \in \mathbb Z \setminus \{0\}, q \in (0,1), f(kq)=f(q)$$Are all drunken periodic functions constant?

This question came from a math class I was taking (Honors Precalculus and AP Calculus BC)- we were discussing periodicity of functions, and the teacher asked for a rigorous definition of periodicity in a function (assuming that that function is defined over $\mathbb R$). One of the other students (presumably) misspoke in trying to make rigorous the intuitive definition, and multiplied the period to get the equality, rather than adding it. I thought this was an interesting class of function to analyze, but didn’t get anywhere- I may simply not have the tools necessary. I call it drunken periodic not to insult the student who made the mistake, but only because I find that name amusing.

First of all, there's no reason to restrict to $q\in(0,1)$ in your definition; you may as well just say $q\in\mathbb{R}$. Next, call two real numbers $x$ and $y$ rational-equivalent if there's some $r\in\mathbb{Q}\setminus \{0\}$ (yes, I appreciate the confusion in notation here!) such that $x=ry$. Note that rational-equivalence is a proper equivalence relation (this is a good thing to show!), so we can partition $\mathbb{R}$ into equivalence classes. Now, as long as $f()$ is constant on each equivalence class, then it's drunken-periodic but not constant. As a simple example, consider $f(x)=1$ if there's some $q\in\mathbb{Q}$ such that $x=q\sqrt{2}$, and $f(x)=0$ otherwise.
Trickier but worthwhile exercise: if $f()$ is continuous, then it is constant...
