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Topic:
Composites, and neat relation
Replies:
18
Last Post:
Sep 14, 2004 3:38 AM




Re: Composites, and neat relation
Posted:
Sep 12, 2004 7:15 PM


In article <3c65f87.0409121043.5988edb8@posting.google.com> jstevh@msn.com (James Harris) writes: > 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.
Have you a proof for that? Did you apply the null test to that proof? What do you think of the following proof: Suppose: [x] + [x + 1/k] = [2x] Trivially: [x + 1] = [x] + 1 So: [x] = [x  1] + 1, [x + 1/k] = [(x  1) + 1/k] + 1, [2x] = [2(x  1)] + 2. And so: [x  1] + [(x  1) + 1/k] = [2(x  1)]. The conclusion is that if the statement holds for x, it also holds for x  1. Think a bit about the actual condition you need for that relation to be true. As a hint, it has to do with the value of 'x  [x]', and not of the value of x relative to some number.  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>



