1. ## 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 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

2. Originally Posted by Kichigo
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 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.

3. ## Thanks...

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