Density of $\{na\}$

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by iarnab_kundu, Dec 25, 2014, 6:50 AM

Statement: It is quite well know that for any irrational $a$ , the set $\{na\}_{n=1}^{\infty}$ is dense in $(0,1)$ .In fact it is equidistributed but that is another story.

We shall prove that using PHP.
First some reduction of the problem statement.
It is equivalent to proving that for any given rational $\dfrac{p}{q}$ and for any $\epsilon>0$ , there exists positive integers $m$ and $n$ such that $\left|na-m-\dfrac{p}{q}\right|<\epsilon$ .

Again that is equivalent to proving for any positive real $\epsilon$ there exists positive integers $m$ and $n$ such that $|na-m|<\epsilon$ .

We shall be using the binary expansion of $a$ to solve it, and at the outset we shall prove that the number $|na-m|<\dfrac{1}{2^d}$ . We shall find a suitable $n$ such that the binary expansion of $na$ has arbitrary number of tailing $0$ s after the decimal. ( $.000000\cdots$ )

First take $n=2,4, .....$ or equivalently $n=10,100,1000$ in binary. It is easy by PHP to see that there are infinitely many $0$ s in the expansion.
Now suppose there are no consecutive $0$ s. Take some number $2^sa$ . Suppose $i$ and $j$ places have $0s$ and everything place in between them is occupied by $1$ s. (Assume $i>j$ )
Then $\left(2^{s+i}+2^{s+j}\right)a$ has a lot of consecutive 0s.
So like this we can keep continuing ans hence we shall be done.

This post has been edited 1 time. Last edited by iarnab_kundu, Dec 29, 2014, 6:33 AM

Cracking the Nut Open

Density of $\{na\}$

by iarnab_kundu, Dec 25, 2014, 6:50 AM

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