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Re: Composites, and neat relation
Posted:
Sep 13, 2004 3:54 AM


In article <3c65f87.0409121727.2ef5c7a8@posting.google.com>, James Harris wrote: > jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0409121043.5988edb8@posting.google.com>... >> So someone pointed out that there's the trivial relation >> >> [x] + [x + 1/2] = [2x] where you're in reals, >> >> 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. Counterexample (similar to one posted already): x = 1.75, k = 100 xk + 1 = 176 k = 100 => xk + 1 >= k
[x] + [x + 1/k] = 2 [2x] = 3 => !([x] + [x + 1/k] = [2x])



