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We knew Epicycloid as a kind of trace curve of a specific point attached to a circle and rolls on another circle. But there's a limit for the ratio R/r of the radius of two circle, R and r, which is that R/r must be rational, so that the curve will loop through its origins.(Without losing generality, we may assume the circle with radius r rolls on the circle with radius R, abbr. as R-circle and r-circle for now)

What if we choose R/r to be irrational real numbers and than draw the curve?

If we choose concentric circles with radius R and R+2r centered at the R-circle. My question is, is it sufficient to imply that such curve coveres the whole region between the two circle of concentric circles?

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    The trajectory will be **dense everywhere** in the said region, that's for sure. – 2017-02-06
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    @IvanNeretin How can we see that it'll be dense everwhere? Does that implies it'll be complete? – 2017-02-06
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    The whole cycloid business is just an unnecessary complication to your problem. Look at it this way: let's simply go around the circle, marking points at regular intervals of $\alpha$, with $\alpha\over2\pi$ being irrational. Would the set of points be dense in the circle? Would it cover the whole circle? – 2017-02-06
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    I think it would be dense. Because we can degenerate this question into the non-existence of minimum within the set {αx+πy|x,y are integer}\{0}, the existence is equivalent with rationality of α/π, so if α/π irrational, the points must be dense in the circle. But intuitively I think it does not cover the whole circle, but I have no much reasons to justify this intuition. But how does that justify the density of the trajectory? I still cannot catch t it. – 2017-02-06
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    Yes, it is dense in the circle. And of course the set doesn't cover the whole circle, because the latter is _continuous_ and the former _countably infinite_. Now back to the (epi)cycloid: say, one arc of it subtends the angle of $\alpha$, so an arc starts from each of our points. – 2017-02-06
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    Oh okay, that is quite convincing actually, but how do we actually prove it in the aspect of formal way. Clearly we can show that it will not be complete (by cardinality). But for density it's seems not to be so obvious. In one dimensional circle edge we can prove it arithmetically but in 2D space how do we do the similar? – 2017-02-06
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    Cardinality doesn't work in 2D: the curve is continuous, and so is the whole region. Both density and non-completeness must be proven by relating a point inside the ring to a certain angle where you must start an arc to hit exactly that point. We don't have arcs at all angles, but we have a dense set of them. – 2017-02-06
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    Oh by cardinality I meant the turning point of an arc where it hits the edge of the circle, guess that I wasn't saying quite clear haha. So we can use the density of the origin of each arc to show the density of the trajectory, is that what you mean? – 2017-02-06
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    Yes, I think so. – 2017-02-06
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    okay I think I can figure it out now, thanks! – 2017-02-06

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Ok Now I've sort of figured it out with @IvanNeretin. And because there's no response for now I'll wrote one.

In conclusion, the trajectory will be dense but not complete in such region.

First we can show that the turning point of each arc forms a dense but not complete set on inner edge of the concentric circles. Because we can degenerate its density into the non-existence of minimum absolute value within the set {Rx+ry|x,y are integer}\{0}, the existence is equivalent with rationality of R/r, so if R/r irrational, it gives us the density as a result. It is, however, not complete because the set of such turning point is countable, but it must be uncountable if it was to cover the whole circle. This leads to contradiction and shows us that the set must be incomplete.

Because the origins of each arc is incomplete, this immediately gives us that the total trajectory cannot be complete. However, because the density of the origin of the arc, for each two arc, we can surely find another arc in between, so that if we connect each two points on that pair of arc respectively we can certainly find a cross point of the in-between arc and the connection, which is also a part of the trajectory. Thus the trajectory must be dense.