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Martin Sleziak
Martin Sleziak
1:26 PM
10
A: For an irrational number $a$ the fractional part of $na$ for $n\in\mathbb N$ is dense in $[0,1]$

Martin SleziakLet's start with along the lines of the standard proof. Let us divide $[0,1]$ into $k$ intervals of length $1/k$; i.e. $[0,1/k]$, $[1/k,2/k]$, $[2/k,3/k]$, etc. Now by Dirichlet principle there are two numbers $a\ne b$ such that $\{a\alpha\}$, $\{b\alpha\}$ which are in the same interval. If $b>

I have two doubts: where was used that $\;\alpha\;$ is irrational? And why the multiples $\;k(b-a)\alpha\;$ have to be in all the intervals? Thanks you — user177692 Dec 21 '14 at 14:16
@AntoineNemesioParras I have edited my posts and pointed out explicitly where the assumption that $\alpha$ is irrational is used. About your second question: Since $\{(b-a)\alpha\}$ is very small (smaller than $1/k$, we cannot skip the whole interval by moving to the next multiple. — Martin Sleziak Dec 21 '14 at 15:11
Hi Martin! I was wondering can you show the proof why if you take all the multiples $n(b-a)\alpha$, then in each of the $k$ intervals must be at least one of the values $\{n(b-a)\alpha\}$.? — ZFR yesterday
@ZFR There is a brief explanation already in the answer, but let me try add a bit more details here.
We know that $(b-a)\alpha<1/k$.
Let us take one of the intervals, and let us assume that it does not contain any multiple of $(b-a)\alpha$. (We want to get a contradiction.)
We have the last multiple $x=n(b-a)\alpha$ before this interval. And the next multiple $y=(n+1)(b-a)\alpha$ after this interval.
Then their difference must be larger than the length of the interval - we get that $y-x = (b-a)\alpha>1/k$. And this is a contradiction.
 

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