- MacTutor History of Mathematics

Solution: Day 4, problem 3

Obviously the thickness and radius of the pie play no role, all the action taking place along the circumference,
which we can identify with T = R/Z.

Hence we can restate the problem more mathematically as follows:

Given two numbers

a, b

strictly between 0 and 1, show that the piecewise linear map

f

: T to T defined by

x \mapsto a - x + b

x \isin I = \mathbb{Z} + [ 0,a ]

and

x \mapsto x + b

x \notin I

, has finite order. We first show that for any

x

in T there is a positive integer

n = n(x)

with

f^{n}(x) = x

, and then show that

n(x)

is bounded. (This proves the assertion of the problem since then the least common multiple of all the

n(x)

's is a period for

f

.)

We distinguish two cases. If

x + \mathbb{Z} b

is disjoint from

I

(which can only happen if

b

is rational, since otherwise

x + \mathbb{Z} b

is dense in R/Z), then

f^{n}(x) = x + nb

for all

n

and therefore

f^{N}(x) = x

where

N

is the denominator of

b

. If not, then there are integers

n_{1} > 0

and

n_{2} ≥ 0

such that

x - n_{1}b

and

x + n_{2}b

belong to

I

. Choose them minimal.
Then

x

belongs to an orbit of

f

of length

2(n_{1} + n_{2})

, consisting of the points

x + nb (n = -n_{1} + 1, -n_{1} + 2, ... ,n_{2})

followed by the points

a - x + nb (n = -n_{2} + 1, -n_{2} + 2, ... , n_{1})

. It remains only to show that the integer

N = n_{1} + n_{2}

is bounded.
This integer has the property that there is an element

y

I

(namely,

x - n_{1}b

) such that

y + Nb

is in

I

but

y + nb

is not in

I

for

0 < n < N

. If

b

is rational, then clearly

N

is bounded by the denominator of

b

. If not, then there is a positive integer

M

such that

Mb

is congruent modulo 1 to a number

d

between 0 and

a

.

Then the sequence

y + Mb, y + 2Mb, ...

advances in steps of

d < | I |

, so must meet

I

after at
most

[(1 - a)/d]

steps, giving for

N

an upper bound

M[(1 - a)/d]

which is independent of

x