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  1. #1
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    Help needed.

    Be $\displaystyle \alpha\in\mathbb{R}$ and $\displaystyle p_{0}, p_{1}, p_{2},\cdots, p_{n} $ distintict positive intergers. I must prove there's an interger $\displaystyle r$ and indices $\displaystyle i$ and $\displaystyle j$ with $\displaystyle 0\le i<j \le n$ such that $\displaystyle \mid\alpha(p_{i} - p_{j}) - r\mid < \frac{1}{n}$

    I've been trying to prove this using Dirichlet's principle but I only got to a few results using all possible values for $\displaystyle r = \lfloor\mid\alpha(p_{i} - p_{j}) - r\mid\rfloor$ and considering the sets $\displaystyle [\frac{2}{n},\frac{3}{n}),\cdots\,[\frac{n-1}{n},1)$. I don't see it
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Kichigo View Post
    Be $\displaystyle \alpha\in\mathbb{R}$ and $\displaystyle p_{0}, p_{1}, p_{2},\cdots, p_{n} $ distintict positive intergers. I must prove there's an interger $\displaystyle r$ and indices $\displaystyle i$ and $\displaystyle j$ with $\displaystyle 0\le i<j \le n$ such that $\displaystyle \mid\alpha(p_{i} - p_{j}) - r\mid < \frac{1}{n}$

    I've been trying to prove this using Dirichlet's principle but I only got to a few results using all possible values for $\displaystyle r = \lfloor\mid\alpha(p_{i} - p_{j}) - r\mid\rfloor$ and considering the sets $\displaystyle [\frac{2}{n},\frac{3}{n}),\cdots\,[\frac{n-1}{n},1)$. I don't see it
    Define $\displaystyle \{x\} $ to be the fractional part of $\displaystyle x $.

    What I would do is define $\displaystyle r_{ij}=\alpha p_i-\alpha p_j-\{\alpha p_i-\alpha p_j\} $ (WLOG assume $\displaystyle r_{ij}\geq0 $).

    Now, assume $\displaystyle \alpha(p_i-p_j)-r_{ij}\geq\frac1n \; \forall \; i,j $ i.e. $\displaystyle \{\alpha p_i-\alpha p_j\}\geq\frac1n $.

    If $\displaystyle \{\alpha p_i\}\geq\{\alpha p_j\} $, then $\displaystyle \{\alpha p_i-\alpha p_j\}=\{\alpha p_i\}-\{\alpha p_j\} $. This would mean we would have to have $\displaystyle n+1 $ objects: $\displaystyle \left(\{\alpha p_i\}\right)_{i=0}^n $, in the interval $\displaystyle [0,1) $ that are at least $\displaystyle \frac1n $ apart. But that would mean one element would be outside $\displaystyle [0,1) $ as $\displaystyle \frac nn>1 $.

    If $\displaystyle \{\alpha p_i\}\leq\{\alpha p_j\} $, then $\displaystyle \{\alpha p_i-\alpha p_j\}=\{\alpha p_i\}-\{\alpha p_j\}+1 $. So do what was done above, but in the interval $\displaystyle [-1,0) $.

    This is done a bit sloppily but I believe it works.
    Last edited by chiph588@; Jul 12th 2010 at 09:04 PM.
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  3. #3
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    Thanks...

    That's kind of what I was doing but got stuck in some part there. Thanks.
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