S is a finite set of collinear points. Let k be the maximum distance between any two points of S. Given a pair of points of S a distance d < k apart, we can find another pair of points of S also a distance d apart. Prove that if two pairs of points of S are distances a and b apart, then a/b is rational.

**Solution**

*Hard.*

Since we are only interested in ratios, we may take k = 1 and the points to have coordinates x_{1} = 0, x_{2} ... , x_{n-1}, x_{n} = 1. Let these points generate a vector space V of dimension m over Q.

If m = 1, then every x_{i} is a rational multiple of x_{n} and hence rational, so the ratio of any two distances is rational.

Suppose that m > 1. Take a basis b_{1}, b_{2}, ... , b_{m} for V as follows. Take b_{1} = 1. Let b_{2} = x_{t}, for some irrational x_{t}, then extend {b_{1}, b_{2}} to a basis. Then each x_{i} is a unique rational linear combination of the b_{j}. In particular, x_{1} = 0, x_{t} = b_{2} and x_{n}= b_{1}. Now define a linear map f from V to Q as follows. Let f(b_{2}) = r b_{2}, where r is rational and sufficiently large that f(b_{2}) > 1, and f(b_{i}) = b_{i}/2 for all other i.

Now suppose we take two distinct points in S, we can write them as ∑ r_{i}b_{i} and ∑s_{i}b_{i}, where all r_{i} and s_{i} are rational. Hence their images under f are r r_{2}b_{2} + ∑ r_{i}b_{i}/2 and r s_{2}b_{2} + ∑ s_{i}b_{i}/2, where the summations exclude i = 2. These must be unequal, otherwise we would have a linear combination of b_{i} with rational coefficients which was zero.

So there is a unique point x in S with f(x) maximum and a unique point y in S with f(y) minimum. Then f(x - y) is greater than f(d) where d is any other difference x_{i} - x_{j}. But every difference except x_{n} - x_{1} and x_{1} - x_{n} occurs at least twice, so x and y must be the endpoints. Now f(x_{1}) = 0 and f(x_{n}) = 1/2, so f(x - y) = 1/2. But f(x_{t} - x_{1}) = f(x_{t}) = f(b_{2}) > 1. Contradiction.

© John Scholes

jscholes@kalva.demon.co.uk

5 Feb 2002