1. The Clock Problem

I have a blank clock face and a single hand which moves in jumps of one radian. The hand starts at angle 0. For an arbitrary line scratched on the surface at angle alpha, and starting from 0, will the hand ever come within a small angle epsilon of the line? After how many jumps?

2. Originally Posted by Hartlw
I have a blank clock face and a single hand which moves in jumps of one radian. The hand starts at angle 0. For an arbitrary line scratched on the surface at angle alpha, and starting from 0, will the hand ever come within a small angle epsilon of the line? After how many jumps?

EDIT: By intelligible I mean to someone who has taken math through calculus.
Finally got it, what a lulu.

Use the principle of Archimedes (Ref Rosenlicht or Shilov):

For any real a and integer N, K exists st

K $\displaystyle \leq$ Na< K+1, or

0 $\displaystyle \leq$ a - K/N < 1/N

So we can come to within 1/N of a after K steps.

Mathematically, this also proves that e^iK is dense on the unit circle.

3. Originally Posted by Hartlw
Finally got it, what a lulu.

Use the principle of Archimedes (Ref Rosenlicht or Shilov):

For any real a and integer N, K exists st

K $\displaystyle \leq$ Na< K+1, or

0 $\displaystyle \leq$ a - K/N < 1/N

So we can come to within 1/N of a after K steps.

Mathematically, this also proves that e^iK is dense on the unit circle.
WRONG!!!!! (MADE THE SAME MISTAKE TWICE) All I have done is made the obvious statement that K exists st after K steps of 1/N each I will be within 1/N of a.

Given K I can write:

K/2pi = M + b/2pi , 0 <b <2pi , (b can't be zero or 2pi because that would mean pi is rational)

but given "a" I can't find K so that "b" is within $\displaystyle \epsilon$ of a.

EDIT Sorry about bumping someone off the top of the line with this. I give up.

4. Here is an observation that may help.

Suppose $\displaystyle \epsilon > 0$ is given. Choose n sufficiently large that $\displaystyle 2 \pi / n < \epsilon$. Partition the circle into n arcs

$\displaystyle [0, 2 \pi /n), [2 \pi /n, 4 \pi /n), [4 \pi /n, 6 \pi /n), \dots , [2 (n-1) \pi /n, 2 \pi)$.

Consider the angles 1, 2, ..., n , n+1 (radians). No two of these angles can be the same (remember that angles that differ by a an integer multiple of $\displaystyle 2 \pi$ are equivalent), because if j and k were equivalent as angles we would have $\displaystyle j-k = 2 m \pi$ for some integer m, contradicting the irrationality of $\displaystyle \pi$.

Since there are n arcs and n+1 angles, by the Pigeonhole Principle one of the arcs must contain at least two angles, say j and k. Let's say j < k. Define N = k-j. Then N is an angle less, in absolute value, than the width of the arc, $\displaystyle 2 \pi / n$, i.e. $\displaystyle N = 2 m \pi + \theta$, where m is an integer and $\displaystyle | \theta | < 2 \pi /n < \epsilon$.

Now consider the multiples of N; what can you say about them as angles?

5. Originally Posted by Hartlw
I have a blank clock face and a single hand which moves in jumps of one radian. The hand starts at angle 0. For an arbitrary line scratched on the surface at angle alpha, and starting from 0, will the hand ever come within a small angle epsilon of the line? After how many jumps?
Each 7 steps of 1 you advance .7168
7(1) = 2pi + .7168

Each 9 increments of .7168 you advance .1680
9(.7168) = 2pi + .1680

Each 38 increments of .1680 you advance .1008
38(.1680) = 2pi + .1008

63(.1008) = 2pi + .0672
94(.0672) = 2pi + .0336
187(.0336) = 2pi + .0001

The general pattern is:
7/2pi = 1/N1 + r1, N1=1, r1<1
r1/2pi = 1/N2 + r2, N2>N1, r2<r1
r2/2pi = 1/N3 + r3, N3>N2, r3<r2
.........
any rNi = 0 implies pi rational (substitute and work backwards).

Ni=>infinity, rNi=>0

So the circle is divided into aritrarily small increments as the hand steps around and we can come as close as we please to the line scribed on the face.