Saturday, January 31, 2009

Another quick "2009" stuff

Let



and let




be a function satisfying






Then f has at least one fixed point, that is, there exists an element x in X such that f(x)=x.

Indeed, it is not hard to see that f is a bijection and induces a permutation of X in which all cycles are either of length 3 or 1 (those of length one correspond to the fixed points of f ). Since 2009 is not divisible by 3, there must be a cycle of length 1.