# Take a square and a ray that will bounce off the walls of the square, and determine what points have an unbounded distance from the initial point.

Take a square with side length 1 and vertices at (0,0) (1,0) (0,1) and (1,1).

Now consider a ray with an initial trajectory of $$(\sqrt{2}, 1)$$ and vertex (0,0). Imagine that when this ray intersects a side of the square it bounces cleanly off of that edge and continues in that new direction.

Consider the set of points this ray will intersect if extended indefinitely. It will never hit the same place twice on the side of the square, and it will therefore bounce off of an infinite number of points.

Therefore, I can mentally conceive two primary sets of points: points whose particular distance (as measured along the entire length of the path the ray took) is unbounded (i.e. infinite) and points whose distance has a definite value.

I know for a fact there must be such points as the set of reflection points is infinite and exists in a finite space. I suspect this sort of infinity is similar to how there are an infinite number of real numbers between two integers and yet they both exist. I suspect this is a case where “arc length” is unbounded even though the two points exist.

What I want to determine is either the set of points “at infinity” in this scenario or a particular point that is “at infinity”. Preferably a proof or sufficient justification would come with such an example.

• Forgive me if I’ve used the wrong terminology. This is effectively just a random thought experiment I came up with. – user64742 Jul 9 '19 at 23:21
• There are no "infinite points." Each point is reached in finite time. Just because we can keep counting forever, it doesn't mean there are infinite positive integers. – saulspatz Jul 9 '19 at 23:26
• As a starting point, if you “unfold” the path, it’s fairly easy to see that the points on the bottom of the square touched by the path will have $x$-coordinates of the form $\{2n\sqrt2\}$ or $1-\{2n\sqrt2\}$, depending on the parity of $n$. The contact points for the top edge are similar-looking, as are those of the left and right edges, though with $1/\sqrt2$ instead of $\sqrt2$. – amd Jul 9 '19 at 23:38
• @saulspatz you can define a fractal shape such that it has two end points while meanwhile the length of the path between those points is unbounded. How is this case any different? – user64742 Jul 10 '19 at 0:38
• You don't seem to be reading what I am writing. A point is on the path if the ray gets there. To get there it must get there in finite time. That's what "get there" means. The length is unbounded because the process never stops, but that doesn't meant there is a point at infinity. Th real line is unbounded, but there are no points at infinity. – saulspatz Jul 10 '19 at 1:03

If you ignore the reflections initially, the ray passes through all the points of the form $$(a\sqrt 2, a)$$ with $$0 \le a$$. To find the point inside the square this corresponds to, separate the coordinates into integer and fractional parts, so write $$a\sqrt 2=\lfloor a\sqrt 2 \rfloor + \{a \sqrt 2 \}$$ and similarly for the vertical coordinate. If $$\lfloor a\sqrt 2 \rfloor$$ is even, the ray is at $$\{a \sqrt 2 \}$$ in the horizontal direction and traveling right. If $$\lfloor a\sqrt 2 \rfloor$$ is odd, the ray is at $$1-\{a \sqrt 2 \}$$ in the horizontal direction and traveling left. If $$\lfloor a \rfloor$$ is even, the ray is at $$\{a \}$$ in the vertical direction and traveling up. If $$\lfloor a\rfloor$$ is odd, the ray is at $$1-\{a \}$$ in the vertical direction and traveling down.