Can you move a clock's hand by a constant, such that the hand hits all angles

Question

My younger brother asked me if rotating a clock's hand by $\pi$ degrees (or any irrational number) would mean that the clock's hand would eventually point at all angles.

Initially I though that could be correct. After all, $\pi \cdot n$ is irrational. Thus the hand will never return to the initial position ($0$ or $n \cdot 360$). However, by the same logic, the hand will never hit any rational numbers, thus it won't point at all angles. Of course, moving the hand by a rational numbers of degrees won't hit any irrational angles either. It doesn't seem to be possible to hit all real angles.

But what if we where only interested in rational angles? Can we prove whether there is any constant $\alpha$ by which we can move the clock's hand and eventually point at any rational angle?

I think this could be rephrased as asking whether for $\forall n \in \mathbb{N} \quad \alpha * n \mod 360 \neq 0$. Aka, you just need to prove that the hand never returns to the starting position. Is that correct? If so, I'm not sure how to prove it.

Answer 1

First, as you may have surmised, you can't hope to hit more than countably many distinct angles, so it's not possible to hit all angles. But the rational numbers are a countable set....

Nonetheless, you cannot hit them all by taking multiples of a fixed angle $\alpha$ mod $2\pi$. A slightly fancy way to say this is, a rational angle $\alpha$ generates a cyclic subgroup $\alpha\mathbf{Z} = \{n\alpha : \text{$n$ an integer}\}$ of the additive group of reals. This subgroup is clearly not the set of all rational numbers, since there is no smallest positive rational, but there is a smallest positive integer multiple of $\alpha$. But the mapping from real numbers to points on the unit circle that sends a real number $\theta$ to the point making angle $\theta$ with some fixed position (say $12$ o'clock) is injective on the set of rationals: Two real numbers map to the same clock location if and only if they differ by an integer multiple of $2\pi$, and distinct rationals never do. Since there are rationals not in $\alpha\mathbf{Z}$, there are rational angles not hit if we increment our position by whole steps of size $\alpha$.

The preceding argument presumes that one turn of a circle is $2\pi$ units (namely, radians). What if instead we use degrees, so a full turn is a whole number? That turns out not to help: If our angle $\alpha$ is irrational, then we don't hit any rational angles. If on the other hand $\alpha = p/q$ is rational, then some integer multiple of $\alpha$ is an exact number of whole turns. (If we measure angle in degrees, for example, then $360q\alpha = 360p$ is $p$ whole turns. That means only finitely many angles are represented as integer multiples of $\alpha$.

What this all means is no, we cannot hit every rational angle by taking multiples of a single angle, whether or not a full turn is rational or irrational.