In article <email@example.com> firstname.lastname@example.org (James Harris) writes: > Paul Murray <email@example.com> wrote in message news:<firstname.lastname@example.org>... ... > > >> and I started thinking about > > >> > > >> [x] + [x + 1/k] = [2x] > > >> > > >> also in reals, with k>1, and it turns out you need x>1 as well, which > > >> I was thinking about didn't put down before. > > > > > > You can have x less than 1 as what's needed is > > > > > > xk + 1 >= k > > > > Still false. > > You're right. The requirement is that x>=1.
Still wrong. But as you apparently do not read what I write... I have posted a counterexample already (x = 5/3, k = 6). Moreover, it has already been proven that if the relation holds for a value x, it also holds for x-1. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; <a href="http://www.cwi.nl/~dik/">http://www.cwi.nl/~dik/</a>