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Math Help - Help needed.

  1. #1
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    Help needed.

    Be \alpha\in\mathbb{R} and p_{0}, p_{1}, p_{2},\cdots, p_{n} distintict positive intergers. I must prove there's an interger r and indices i and j with 0\le i<j \le n such that \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 r = \lfloor\mid\alpha(p_{i} - p_{j}) - r\mid\rfloor and considering the sets [\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 \alpha\in\mathbb{R} and p_{0}, p_{1}, p_{2},\cdots, p_{n} distintict positive intergers. I must prove there's an interger r and indices i and j with 0\le i<j \le n such that \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 r = \lfloor\mid\alpha(p_{i} - p_{j}) - r\mid\rfloor and considering the sets [\frac{2}{n},\frac{3}{n}),\cdots\,[\frac{n-1}{n},1). I don't see it
    Define  \{x\} to be the fractional part of  x .

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

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

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

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

    This is done a bit sloppily but I believe it works.
    Last edited by chiph588@; July 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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