1. ## 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

2. Originally Posted by Kichigo
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.

3. ## Thanks...

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