α_{n} is the smallest element of the set { |a - b√3| : a, b non-negative integers with sum n}. Find sup α_{n}.

**Solution**

Answer: (√3 + 1)/2.

Evidently the elements of { |a - b√3| : a, b non-negative integers with sum n} are spaced a distance (√3 + 1) apart, so either smallest positive or the largest negative lies in the interval ( -(√3 + 1)/2, (√3 + 1)/2). Thus (√3 + 1)/2 is certainly an upper bound.

Let a_{n}, b_{n} be the pair with |a_{n} - b_{n} √3| = α_{n}. All the values (a_{n} - b_{n} √3) lie in the interval ( - (√3 + 1)/2, (√3 + 1)/2) ), so for any ε > 0, we can find two distinct values, corresponding to say m and n, which differ by less than ε. In other words α_{m-n} < ε. But now consider the sequence of values α_{m-n}, α_{2(m-n)}, α_{3(m-n)}, ... For kα_{m-n} < (√3 + 1)/2 we have α_{k(m-n)} = k α_{m-n}. So we get a sequence of increasing values with spacing less than ε. Thus the sequence gives a value within ε of (√3 + 1)/2, which is therefore the least upper bound.

© John Scholes

jscholes@kalva.demon.co.uk

1 Jan 2001