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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 $[0,1]$.

Community
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InfimumMaximum
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  • 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
  • I really wish those who feel need to quote texts verbatim would either type out the quotations in Latex (with proper citings of course) or link to a larger, more hi res image . . . I can never read these included copies. Not against the practice, just the presentation. Thanks. – Robert Lewis Jul 23 '13 at 18:04
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    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
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    @ThomasAndrews I don't see where the complication is... – David Mitra Jul 23 '13 at 18:29
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    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

3 Answers3

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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 $$yn \leqslant x < yn + y \Rightarrow 0 \leqslant x - yn < y < \varepsilon $$

Pedro
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  • 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 Jul 23 '13 at 19:49
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    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
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    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
  • @InfimumMaximum See my edit. – Pedro Jul 23 '13 at 21:05
  • @InfimumMaximum Since $\lfloor k \alpha \rfloor$ is an integer, you can write $\lfloor k \alpha \rfloor = \lfloor k \alpha \pm \lfloor k \alpha \rfloor \rfloor \mp \lfloor k \alpha \rfloor$. Doing this for both $k_1 \alpha$ and $k_2 \alpha$ lets you show that $(k_1 - k_2)\alpha - \left( \lfloor k_1 \alpha \rfloor - \lfloor k_2 \alpha \rfloor \right) = (k_1 - k_2)\alpha - \lfloor (k_1 - k_2) \alpha \rfloor$. – AnonSubmitter85 Jul 23 '13 at 21:50
  • It may be that $k_2>k_1$ and ${k_2\alpha}<{k_1\alpha}$; then ${(k_2-k_1)\alpha}$ is in $[(n-1)/n,1)$ (see the comments above). But it is easy to remedy this... Actually, it might be slightly more efficient to prove (b) first. – David Mitra Jul 23 '13 at 22:00
  • @DavidMitra I got the subindices mixed. – Pedro Jul 23 '13 at 22:07
  • I missed that; but it doesn't affect my point. The difference (with a positive multiple) could either be in the first interval or the last. – David Mitra Jul 23 '13 at 22:10
  • @DavidMitra I am showing the difference of fractional parts equals the fractional part of the differences. Thus this last value is in $\left[0,\frac 1 n\right]$ as we showed above. – Pedro Jul 23 '13 at 22:12
  • Yes, your first sentence is true. But, assuming $k_2>k_1$, you can't conclude that ${k_2\alpha} >{k_1\alpha}$. It's possible that ${(k_2-k_1)\alpha}$ is negative. After modding, according to the definition in the OP, it would be in $[(n-1)/n,1)$. – David Mitra Jul 23 '13 at 22:17
  • @DavidMitra I never assumed $k_2>k_1$, what I assumed an inequality of fractional parts. Now I see what you mean: if $k_2-k_1<0$ then this isn't in the desired set. Oh, well... now I have to go. I wouldn't mind if you added the necessary corrections! – Pedro Jul 23 '13 at 22:18
  • Yes, sorry; that's what I should have said. – David Mitra Jul 23 '13 at 22:20
  • @PeterTamaroff Thank you so much! I think your solution proved (a) but didn't prove (b) itself. However, you wrote that your proof shows the numbers are dense. I'm confused about it! – InfimumMaximum Jul 24 '13 at 01:42
  • @InfimumMaximum From $(a)$ you deduce that $\inf G^+=0$ in the sense of what I added. Then you can follow the proof in $(2)$. – Pedro Jul 24 '13 at 02:14
  • @InfimumMaximum Are you sure you weren't given $\Bbb Z$ instead of $\Bbb N$? – Pedro Jul 24 '13 at 02:24
  • @PeterTamaroff I'm sure that I'm given N. – InfimumMaximum Jul 24 '13 at 03:00
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    @PeterTamaroff Peter, I think {kαmod1∣k∈N} is not a positive additive subgroup!!! – InfimumMaximum Jul 24 '13 at 15:25
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(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)

Hagen von Eitzen
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    Suppose that $\frac an \le k_1\alpha \pmod 1 \lt k_2\alpha \pmod 1\le \frac{a+1}n$, then clearly $k_2\alpha \pmod 1-k_1\alpha \pmod 1 =(k_2-k_1)\pmod 1$, but what if $k_2\lt k_1$? If $k_2\lt k_1$ then the quantity $(k_2-k_1)\pmod 1$ does not belong to the set which is to be proven to be dense. Can you please clarify this? Thanks. – Koro Jul 20 '21 at 16:53
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Hint:
Check your definition of irrational. You can, by multiplying with $10^d \in \mathbb{N}$, generate an arbitrary small $[k\alpha]$.

AlexR
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