On 24/08/2014 12:50, J.P. Grossman wrote:
I'm not sure what you mean by "bijective", but if you graph the two sequences then they form an X pattern. The numbers in the upper branches are always {n+1, ..., 2n} in some order, and the numbers in the lower branches are always {1, ..., n} in some order, so the sum |a_i - b_i| is always (n+1 + n+2 + ... + 2n) - (1 + 2 + ... + n) = n^2
More formally, [...]
Yes, this is basically the same as one of the proofs on the cut-the-knot page I linked to. It's not bad, but not quite what I was hoping for. By "bijective" I mean: We're trying to prove that one positive integer equals another. Perhaps there's a proof that makes it obvious that each of them is the size of some set, and exhibits a bijection between the two sets. -- g