<t>Thanks paulptli for the proof. It is quite clean. The fractional notation is very powerful.QuoteOriginally posted by: paulptlito yongge: here's a simple proof using pigeonhole principle:Statement: let x>0. For every N>1, there exist integers p and 0<q<N, such that |qx-p|<=1/NNotations:[z] = integ...

<r>Sorry, I didn't read the original post carefully. My mistake. I had thought it about for real x all along. I thought of considering the interval [0,1] as a simplification, since any x in interval [i, i+1] can be easily translated to the interval [0,1]. In fact, all of the three proofs in this thr...

<r>one counter example, for x=0, how are you able to find non-zero integers m,n such that |m*0-n|<0.01. I do not think I can find the answer.My revised proof can only cover the region (0,1] after careful thinking as I searched from the right. This is the same reason that my first proof can only cove...

<r>I read your proof, dunrewpp. I think it is good and I like it.One thing I note that in your proof, how can you guarantee that N=1-b(n,0)+b(n,1)-b(n,2)+-?.+{(-1)^(n)}*b(n,n-1) is nonzero because of so many terms, which might be canceled out.The <URL url="http://en.wikipedia.org/wiki/Dirichlet%27s_...

<t>Thanks for this, here are my answers.yes m_1=M. I will also update the my original post.Your question: Given that m_0=1, we have |m_0 x-n|=|1*x-n|=|x-n|<d0. So, what is n here? It's obvious that the only n you can be sure of to fit the inequality is that n=0. My answer: I only consider x in [0,1)...

Nice catch, I was only considering one direction. Still for alpha=1/2, it still needs modifications.QuoteOriginally posted by: paulptliyongge: what's wrong with p=1?

<t>I have edited after answering dunrewpp's questions.This could be proved by the following strategy.Assuming M is an integer such that 1/M<d<1/(M-1)denote 1/M by d0We can prove the original problem by replacing d by d0 since d0<d.Let m_0=1, we have S_0=[0,d0] satisfies that if x is in S_0, we can f...

<r>The wiki entry is not correct.I copied the wiki entry below:for any real number α and any positive integer N, there exists integers p and q such that 1 ≤ q ≤ N and counter example: For N=1, it says that , which can not be true when QuoteOriginally posted by: AVtThis follows from <URL url="http://...

<t>It is easier to solve the problem in a general setup.Let q(k,n) be the probability that the k-th person will not get in his bed with n persons and the first person being drunk.It's easy to see that q(1,n)=(n-1)/n and q(2,n)=1/n.We can derive the relationship by following which bed the first drunk...