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For any positive real number α, its Simple Continued Fraction can be written as:
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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, 25∕11 = [2;3,1,2],14∕9 = [1;1,1,4],the Golden Ratio ϕ (irrational) = [1;1,1,1,…],π (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,
If we define p−1 = 1,q−1 = 0, the following recurrence relations hold for all k ≥ 0:
It is known that pk and qk are coprime for all k.
The Three Distance Theorem also involves use of integer values (qk−1 + qk) for k ≥ 0. Note that the sequence of qk values is: q−1 = 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 (qk−1 + qk) is strictly increasing with k for all k ≥ 0, starting at 1 for k = 0.
Hence, any integer N ≥ 1 either equals (qk−1 + 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:
| (3) |
Note that the last endpoint of this range can be written as:
So, in the range of (3), (N −qt−1) varies from qt to (at+1)qt + (qt − 1). Due to the Division Theorem, given (N −qt−1) and qt, there exist unique integers r and s such that:
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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