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Background

For any positive real number α, its Simple Continued Fraction can be written as:

         ----1-----
α = a0 + a1 + -1---
              a2+...

where for all i > 0, ai are positive integers, and a0 is a non-negative integer.

If α is irrational, this continued fraction never terminates, i.e. contains infinitely many ai terms, and is denoted by: [a0;a1,a2,]. For rational α, it always terminates at some term an, and is denoted by: [a0;a1,a2,,an].

For example, 2511 = [2;3,1,2],149 = [1;1,1,4],the Golden Ratio ϕ (irrational) = [1;1,1,1,] (irrational) = [3;7,15,1,292,].

We may consider only the first few ai terms as below and call it the kth Convergent: pk∕qk = [a0;a1,a2,,ak]. For example,

p ∕q  = a
 0  0    0
p1∕q1 = a0 + 1 ∕a1
p2∕q2 = a0 + 1 ∕(a1 + 1∕a2)

If we define p1 = 1,q1 = 0, the following recurrence relations hold for all k 0:

p    = a    p + p
  k+1    k+1  k   k−1
 qk+1 = ak+1qk + qk−1                                         (2)

It is known that pk and qk are coprime for all k.

The Three Distance Theorem also involves use of integer values (qk1 + qk) for k 0. Note that the sequence of qk values is: q1 = 0,q0 = 1,q1 = a1,q2 = a2a1 + 1 etc. From (2) we also know the recurrence relation of qk. Since ai 1 for all i > 0, qk are strictly increasing with k for k 1. We also see that the sequence of (qk1 + qk) is strictly increasing with k for all k 0, starting at 1 for k = 0.

Hence, any integer N 1 either equals (qk1 + qk) for some k 0, or is between two consecutive values of this sequence. We can conclude that, any integer N 1 corresponds to a unique integer t 0 such that:

qt−1 + qt ≤ N < qt + qt+1
(3)

Note that the last endpoint of this range can be written as:

q + q   − 1 = q + (a   q + q   )− 1
 t   t+1       t    t+1 t   t−1
            = qt−1 + (at+1)qt + (qt − 1)

So, in the range of (3), (N qt1) varies from qt to (at+1)qt + (qt 1). Due to the Division Theorem, given (N qt1) and qt, there exist unique integers r and s such that:

(N − q   ) = rq  + s                                          (4)
      t−1     t
0 ≤ s ≤ qt − 1      {Division Theorem }
1 ≤ r ≤ at+1        {due to the range of (N − qt−1)}

Note that for a fixed t, as N varies in range of (3), every increase in N by qt would increase r by 1.

Thus we now know that, any integer N 1 corresponds to unique integers t, r and s as defined by (3) and (4).

Let us now move to exploring the theorem.

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