Given a ring with circumference n, if a person walks s distance per step on it, could we prove that the person will eventually step on every point? If we add more restriction so that the question became

Given a circle divided into x points each have same distance between each other. Now a person who walks s amount of point per step, if x and s are co-prime, then eventually the person would step on every points.

But that only applies on $\mathbb{Z}^+$ realm, could we extend it to $\mathbb{R}$ realm to say that if circumference is not a multiple of the distance the person walks each step, eventually the person would step through every point on the ring?
What I thought is that if circumference $n$ isn't a multiple of the step distance $s$, it means $n\mod{s}\neq 0$ and with an offset not equal to 0, I think eventually the offset would go through the whole circle hence walking though all points. But this was only obvious what the points on a circle in finite, and if we push things to continuum, I couldn't give a solid proof.

I think if written in style of sets, we could write (n is circumference, s is step distance)
$$
n,s\in\mathbb{R} \\
\mathbb{A}=\{n\cdot s, n\in\mathbb{Z}^+\} \\
\mathbb{B}=\{x\mod{n}, x\in\mathbb{A}\}\\
\text{Ask if }\mathbb{B}=[0,n)
$$
My initial reaction is a proof by contradiction. Assume the person does not step on every point. All that is needed to be proven is something to the degree of the amount stepped eventually repeats, that is it is a repeating decimal or something. This is just a rough intuition/idea; but again I think mathematics is all about ideas so I hope this gives some insight and helps.
My initial reaction is a proof by contradiction. Assume the person does not step on every point. All that is needed to be proven is something to the degree of the amount stepped eventually repeats, that is it is a repeating decimal or something. This is just a rough intuition/idea; but again I think mathematics is all about ideas so I hope this gives some insight and helps.