We all love natural numbers. Let’s take a sequence of real numbers 0 = a0 < a1 < a2 < …and corresponding intervals An = [an, an+1]. We call an interval An lucky if it contains a natural number:Here the intervals A1, A2, A4, A5 are lucky and marked in green. We don’t want intervals to be too lucky and therefore insist that intervals are shorter than unit length. We also want each natural number to know where it belongs, so we don’t want our sequences to contain any natural numbers.
To increase happiness, we consider two such sequences an and bn shown on top of each other at height +1 and -1, respectively. Corresponding intervals An and Bn are upper and lower edges of quadrilaterals Qn.
In this particular example the sequences are linear, namely an = n(𝛗-1) and bn= n(2-𝜑), where 𝜑 is the Golden Ratio once more. We note that either An or Bn is lucky, but never both. So each quadrilateral has precisely one lucky edge.
In order to understand how this can possibly be true, we consider more generally two increasing sequences an and bn such that an + bn = n, and such that intervals are shorter than 1. Then the average sequence is cn = (an + bn)/2 = n/2, which we include in the next figure at height 0 between the sequences an and bn.
As already noted, no interval An or Bn can contain two natural numbers. If both contain a natural number, we can join them by a red segment within Qn which intersects the height 0 line in a point whose x-coordinate is of the form n/2, where n is a natural number. But all these points with half-integral coordinates are already taken by the (blue) sequence cn. So at most one of the intervals An or Bn can be lucky.
Now suppose that neither An or Bn are lucky. Then both intervals An and Bn are contained in integer intervals of length 1, marked as green dots to the left and right of the orange quadrilateral. The endpoints average to points with integral coordinates (green crosses) that are 1 unit apart and outside the orange quadrilateral. But both blue crosses on the orange quadrilateral have half integral coordinates, which can’t be. Thus either An or Bn is lucky.
As promised, this proves Raleigh’s Theorem: If irrational 𝛂, 𝛃 > 0 satisfy 1/𝛂 + 1/𝛃 = 1, then 𝛂n = [n 𝛂] and 𝛃n = [n 𝛃] are complementary, i.e. each natural number occurs in precisely one of the two sequences.
To see this, let a=1/𝛂 and b=1/𝛃, an = n a and bn= n b. Now observe that An contains a natural number k if and only if
an < k < an+1 if and only if n< k 𝛂 < n+1 if and only if n = [k 𝛂] (and likewise for Bn).
The argument in terms of intervals is a mere geometric formulation of the standard proof of Raleigh’s theorem but gives a more general result that is not so apparent in the purely algebraic version.