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