Given a circle with radius 1. Take an arbitrary starting point. Then go around the circle an infinite number of times, always drawing a point when you are one unit further. So the next point is always an arc length 1 step further. Easy to prove that all points will be different! But my question is: is each point randomly close approached? I believe this is true, but I can’t prove it!

ureq8

ureq8

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2022-08-21

Given a circle with radius 1. Take an arbitrary starting point. Then go around the circle an infinite number of times, always drawing a point when you are one unit further. So the next point is always an arc length 1 step further. Easy to prove that all points will be different! But my question is: is each point randomly close approached? I believe this is true, but I can’t prove it!

Answer & Explanation

Paola Mercer

Paola Mercer

Beginner2022-08-22Added 11 answers

As mentioned in the comments, this has to do with π being irrational. Let's assume that, and go on to prove that you'll end up arbitrarily close to any given point on the circle.

Represent the circle as the interval [ 0 , 2 π ) -- basically, a point is represented by how far counterclockwise you have to go from your starting point to get there. In your process, you're starting at 0 and adding 1 to your location every step, and when the location becomes greater than 2 π, you subtract 2 π so that it lies between 0 and 2 π.

Assume that there's some 0 x < 2 π and some ϵ > 0 for which no point in the interval ( x ϵ , x + ϵ ) is ever reached. Now, consider the first
n = π ϵ
steps, and say the locations reached (including the starting point 0) are
0 = x 0 < x 1 < x 2 < < x n < 2 π
(note that x i is not the location after i steps, but the ith smallest location in ( 0 , 2 π ) that you reach). Since they're all in [ 0 , 2 π ), there can't be a distance of greater than 2 ϵ between each pair -- otherwise,
2 π > x n = ( x n x n 1 ) + ( x n 1 x n 2 ) + + ( x 2 x 1 ) + ( x 1 x 0 ) 2 n ϵ 2 π .
So, there must be some integers a < b < n for which the ath step and the bth step are very close to each other -- within 2 ϵ. So, the (b−a)th step is within 2 ϵ of 0, since going b−a units forward from the ath location gets you to a point within 2 ϵ of the ath location. Let this location be δ units away from 0, so it is either δ or 2 π δ.
Finally, consider the points reached after k(b−a) steps for various integers k. As you add 1 to k, this point moves δ units around the circle, either clockwise (if the b−ath location is δ units clockwise from 0) or counterclockwise (if it's counterclockwise from 0). Because δ < 2 ϵ, you can't ever "jump over" the interval ( x ϵ , x + ϵ ) in this way -- you'll hit some point inside it at step k ( b a ) for some integer k. So, there is some step that drops you in ( x ϵ , x + ϵ ), as desiblack.

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