This is a problem from a first chapter of a calculus textbook.
for what follows, [x]=x-floor(x).
Let x be an irrational number, and A={[xn] : n is natural}.
We need to prove that for any interval (a,b) in [0,1], there exists a c in A that belongs to the interval (or, A is dense in [0,1]).
The textbook gave clues for the proof:
1. Prove: if there exists a natural k such that [kx]<(b-a) or [kx]>1-(b-a), there exists a natural m such that [mkx] is in (a,b).
2. To prove that such a k exists, choose a natural n such that 1/n<(b-a). Then, partition [0,1] into n equal segments and look at [x], [2x], ..., [nx], [(n+1)x]. Of these points, there must b two that belong to the same segments (because there are n segments and (n+1) points).
All I could prove is that 1. is true for [kx]<(b-a) (with m=floor(a/[kx])+1). I have no idea how 2. is going to help me prove anything. If anyone can think of an answer, please tell me.
for what follows, [x]=x-floor(x).
Let x be an irrational number, and A={[xn] : n is natural}.
We need to prove that for any interval (a,b) in [0,1], there exists a c in A that belongs to the interval (or, A is dense in [0,1]).
The textbook gave clues for the proof:
1. Prove: if there exists a natural k such that [kx]<(b-a) or [kx]>1-(b-a), there exists a natural m such that [mkx] is in (a,b).
2. To prove that such a k exists, choose a natural n such that 1/n<(b-a). Then, partition [0,1] into n equal segments and look at [x], [2x], ..., [nx], [(n+1)x]. Of these points, there must b two that belong to the same segments (because there are n segments and (n+1) points).
All I could prove is that 1. is true for [kx]<(b-a) (with m=floor(a/[kx])+1). I have no idea how 2. is going to help me prove anything. If anyone can think of an answer, please tell me.