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

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Dik T. Winter

Posts: 7,899
Registered: 12/6/04
Re: Composites, and neat relation
Posted: Sep 12, 2004 7:15 PM
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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>

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