# Positive integer multiples of an irrational mod 1 are dense [duplicate]

I'm not sure how to solve this one. Thank you!

$2.$ For any $\alpha\in \mathbb R$ we define $$\lfloor \alpha \rfloor = \max_{n\in\mathbb Z}\{\,n\mid n\leq \alpha\,\}$$ and $$\alpha\bmod 1 = \alpha - \lfloor \alpha \rfloor$$ Let $\alpha$ be irrational.

(a) Given $n\in\mathbb N$ show that $\{\,k\alpha\bmod1\mid k\in\mathbb N\,\}\cap\left[0,\frac{1}{n}\right]\neq\emptyset$
(b) Prove that $\{\,n\alpha\bmod 1\mid n\in\mathbb N\,\}$ is dense in $[0,1]$.

• Please clarify your title – Zeta.Investigator Jul 23 '13 at 17:57
• Pigeonhole Principle, applied to the "holes" $[0,1/n), [1/n,2/n),\ldots,[(n-1)/n,1)$. – David Mitra Jul 23 '13 at 18:01
• The pigeonhole principle approach works better for $\mathbb Z$ rather than $\mathbb N$, because if the ordering issue. @DavidMitra – Thomas Andrews Jul 23 '13 at 18:05
• @ThomasAndrews I don't see where the complication is... – David Mitra Jul 23 '13 at 18:29
• I recall there is a simple way, but can't remember! The problem arises if $p\lt q$ and the fractional part of $q\alpha$ is a bit smaller than the fractional part of $p\alpha$. Then the difference is close to $1$, not $0$. By then multiplying by a suitable $k$, we can bring the fractional part down to small positive. But there must be a nicer very short way. – André Nicolas Jul 23 '13 at 18:39

What I show below doesn't strictly follows the items, but still shows the numbers are dense.

STEP 1 Pick any $n\in\Bbb N$. Consider the $n+1$ distinct numbers $$x_k=k\alpha-\lfloor k\alpha\rfloor=\{k\alpha\}\; ;\;k=0,1,2,\ldots,n$$

These are $n+1$ numbers that fit into $n$ places, namely

$$\left[0,\frac 1 n\right),\ldots,\left[1-\frac{1}n,1\right)$$

By the Dirichlet's principle there must exist at least a pair of them which fall in the same interval of length $\dfrac 1n$.

STEP 2 We obtained $$\frac{k}{n} \leqslant \left\{ {{k_1}\alpha } \right\} < \left\{ {{k_2}\alpha } \right\} < \frac{{k + 1}}{n}$$

for some $k=0,\ldots,n$. We then know that $$0 < \left\{ {{k_2}\alpha } \right\} - \left\{ {{k_1}\alpha } \right\} \leqslant \frac{1}{n}$$

But note that \begin{align} \left\{ {{k_2}\alpha } \right\} - \left\{ {{k_1}\alpha } \right\} &= \left\{ {\left\{ {{k_2}\alpha } \right\} - \left\{ {{k_1}\alpha } \right\}} \right\} \cr &= \left\{ {{k_2}\alpha - {k_1}\alpha - \left( {\left\lfloor {{k_2}\alpha } \right\rfloor - \left\lfloor {{k_1}\alpha } \right\rfloor } \right)} \right\} \cr &= \left\{ {{k_2}\alpha - {k_1}\alpha - {\text{integer}}} \right\} \cr &= \left\{ {\left( {{k_2} - {k_1}} \right)\alpha } \right\} \end{align}

Now, you may as well try and prove the following:

Let $G$ be an additive subgroup of $\Bbb R$. Let $G^+$ denote the positive elements of $G$. Then

$(1)$ If $\inf G^+=\alpha >0$, $G=\alpha\Bbb Z$

$(2)$ If $\inf G^+=0$, $G$ is dense in $\Bbb R$.

Hint

For $(1)$. Show that $\alpha\in G$.

If not, pick $\epsilon =\alpha/2$ in the defintion of infimum, and $g,g'\in G$ such that $\alpha \leqslant g < g' <\alpha + \frac{\alpha }{2}$. Look at $g'-g$.

Then $g'-g\in G$ and $g'-g\leq \alpha/2<\alpha$ which is impossible.

Now pick $g>0$. Look at $g-\alpha \left\lfloor {\dfrac{g}{\alpha }} \right\rfloor$. Use the definition of integer part to show it must be zero.

Thus $g = \alpha \left\lfloor {\dfrac{g}{\alpha }} \right\rfloor \in \alpha {\Bbb Z}$. Since opposites are in $G$ too, $(1)$ is proven.

For $(2)$, pick any $x\in \Bbb R$. We know we can find $y>0$ in $G$ with $0<y<\epsilon$. Let $n=\left\lfloor {\dfrac{x}{y}} \right\rfloor$. What can you deduce from $$n \leqslant \frac{x}{y} < n + 1\text{ ? }$$

We have $ny\in G$ by additivity, and $$yn \leqslant x < yn + y \Rightarrow 0 \leqslant x - yn < y < \varepsilon$$

• why do we need to suppose α>0? what if α<0? – InfimumMaximum Jul 23 '13 at 19:43
• @InfimumMaximum If you take fractional parts without sign (as I guess you do) then there is no problem, since $\{x\}$ is periodic with period $1$: it is $y=x$ extended periodically from $[0,1]$. – Pedro Tamaroff Jul 23 '13 at 19:49
• Thank you! Following your suggestion, I supposed x_k1 and x_k2 fit into the same interval. then, WLG, suppose x_k1 > x_k2. Then if we set (k_1-k_2) as x, (a) is satisfied. – InfimumMaximum Jul 23 '13 at 19:58
• However, the problem is if I can just say [k_1α] - [k_2α] > [k_1-k_2]α. I'mt not sure how to prove this. – InfimumMaximum Jul 23 '13 at 19:59
• @PeterTamaroff Peter, I think {kαmod1∣k∈N} is not a positive additive subgroup!!! – InfimumMaximum Jul 24 '13 at 15:25

(a) Hint: If $\frac an\le k_1\alpha\bmod 1< k_2\alpha\bmod 1\le\frac{a+1}n$, then $(k_2-k_1)\alpha$ ...

(b) follows from (a)

Hint:
Check your definition of irrational. You can, by multiplying with $10^d \in \mathbb{N}$, generate an arbitrary small $[k\alpha]$.

• No, that is not the definition of irrational. There are plenty of irrationals for which this is not true. – Thomas Andrews Jul 23 '13 at 18:07
• $.01001000100001\cdots$. – David Mitra Jul 23 '13 at 18:07
• Sorry, that was for transcendent numbers. However multiplication by $10^d$ should still give the desired sizes. – AlexR Jul 23 '13 at 18:34