Can we find a subsequence of { n^{1/3} - m^{1/3} **:** n, m = 0, 1, 2, ... } which converges to √2?

**Solution**

Answer: yes.

There is obviously nothing special about √2. In fact a little thought suggests that the set { n^{1/3} - m^{1/3} **:** n, m integers } is dense in the reals.

To prove this note that we can find arbitrarily small values because certainly for n > 64, we have (n^{1/3} + 1/n)^{3} < n + 3/4 + 3/1024 + 1/64^{3} < n + 1. Also if α is in the set, then so are all integral multiples of α, because (n^{3}N)^{1/3} - (n^{3}M)^{1/3} = n( N^{1/3} - M^{1/3}). So by taking a suitable multiple of a value less than ε we can get within ε of any desired value.

© John Scholes

jscholes@kalva.demon.co.uk

1 Jan 2001