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- #1

- Mar 10, 2012

- 835

- Thread starter caffeinemachine
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- Thread starter
- #1

- Mar 10, 2012

- 835

That is not a complete question.

But guessing at what it means, I think this theorem will help.

- Jan 30, 2012

- 2,575

Equidistribution theorem is much stronger than the fact that $\{\mathop{\mbox{frac}}(n\sqrt{3})\mid n\in\mathbb{Z}^+\}$ is dense in [0, 1], which is sufficient here. This was discussed in the old forum, but there are some differences, at least at first glance. Namely, \[\begin{aligned}0&=\inf\{m\cdot1+n\cdot \sqrt{3}\mid m\cdot1+n\cdot\sqrt{3}>0\mbox{ and }m,n\in\mathbb{Z}\}\\&\le\inf\{m\cdot1+n\cdot\sqrt{3}\mid m\cdot1+n\cdot \sqrt{3}>0,m\in\mathbb{Z}^-,n\in\mathbb{Z}^+\}\\& = \inf\{\mathop{\mbox{frac}} (n\sqrt{3}) \mid n\in\mathbb{Z}^+\}\end{aligned}\] but I am not sure about equality right away.

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- #4

- Mar 10, 2012

- 835

I am so sorry. The infimum is to be proved to be equal to zero as Makarov has pointed out.That is not a complete question.

---------- Post added at 11:16 PM ---------- Previous post was at 11:05 PM ----------

Actually this can be done using Pigeon hole principle.

Denote $x_n=\text{frac}(n \sqrt{3})$

Let $n \in \mathbb{Z}^{+}$. Partition the interval $(0,1)$ into $n$ parts, viz, $(0,\frac{1}{n}),(\frac{1}{n},\frac{2}{n}), \ldots, (\frac{n-1}{n},1)$

Consider $n+1$ numbers, $x_1, x_2, \ldots, x_{n+1}$.

By PHP there exist $i,j, i \neq j$ such that $x_i,x_j \in (\frac{k}{n},\frac{k+1}{n})$.

This implies $\text{frac}(|i-j|\sqrt{3}) < \frac{1}{n}$. Thus $x_{|i-j|} < \frac{1}{n}$.

I can't think of any other proof.

- Jan 30, 2012

- 2,575

Not necessarily. Suppose $j > i$, but $\mathop{\mbox{frac}}(j\sqrt{3}) < \mathop{\mbox{frac}}(i\sqrt{3})$. Then \[\mathop{\mbox{frac}}(j\sqrt{3}-i\sqrt{3})=\mathop{\mbox{frac}}(j\sqrt{3})-\mathop{\mbox{frac}}(i\sqrt{3})+1\] so \[1-1/n<\mathop{\mbox{frac}}((j-i)\sqrt{3})<1\]By PHP there exist $i,j, i \neq j$ such that $x_i,x_j \in (\frac{k}{n},\frac{k+1}{n})$.

This implies $\text{frac}(|i-j|\sqrt{3}) < \frac{1}{n}$.

For example, $\sqrt{3}\approx1.73$ and $2\sqrt{3}\approx 3.46$, so $0 < \mathop{\mbox{frac}}(\sqrt{3})-\mathop{\mbox{frac}}(2\sqrt{3}) < 0.3$. However, $1-0.3<\mathop{\mbox{frac}}(2\sqrt{3}-\sqrt{3})<1$.

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- #6

- Mar 10, 2012

- 835

Ah! yes. My mistake.Not necessarily. Suppose $j > i$, but $\mathop{\mbox{frac}}(j\sqrt{3}) < \mathop{\mbox{frac}}(i\sqrt{3})$. Then \[\mathop{\mbox{frac}}(j\sqrt{3}-i\sqrt{3})=\mathop{\mbox{frac}}(j\sqrt{3})-\mathop{\mbox{frac}}(i\sqrt{3})+1\] so \[1-1/n<\mathop{\mbox{frac}}((j-i)\sqrt{3})<1\]

For example, $\sqrt{3}\approx1.73$ and $2\sqrt{3}\approx 3.46$, so $0 < \mathop{\mbox{frac}}(\sqrt{3})-\mathop{\mbox{frac}}(2\sqrt{3}) < 0.3$. However, $1-0.3<\mathop{\mbox{frac}}(2\sqrt{3}-\sqrt{3})<1$.

The proof changes a little bit.

We have $x_i, x_j \in (\frac{k}{n},\frac{k+1}{n})$

suppose $x_i>x_j$ then $frac((i-j) \sqrt{3}) < \frac{1}{n} \Rightarrow x_{i-j} < \frac{1}{n}$ EDIT: what if $i<j$ :O

Similarly for $x_j > x_i$

$x_i=x_j$ is not a possibility.

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- #7

- Mar 10, 2012

- 835

I don't even know whether the latter is true.

- Admin
- #8

- Jan 26, 2012

- 4,203

- Jan 30, 2012

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Well, this is not true, of course, because this set is not bounded from below... I'll return to this question when I have more time.I can only prove that $\inf \{ n \sqrt{3} : n \in \mathbb{Z} \} =0$.

- Thread starter
- #10

- Mar 10, 2012

- 835

That was a typo again... sorry:Well, this is not true, of course, because this set is not bounded from below... I'll return to this question when I have more time.

What I meant was $\inf \{ \text{frac}(n \sqrt{3}): n \in \mathbb{Z} \}=0$

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- #11

- Feb 7, 2012

- 2,807

One very constructive way to approach this is to use the continued fraction approximants to $\sqrt3$. These are alternately slightly less than, and slightly greater than, $\sqrt3.$ Take one of the approximants that is less than $\sqrt3$, for example 265/153. You can then check that $153\sqrt3\approx 265.00377...$, whose fractional part is very small.Prove that $\inf \{ \text{ frac}(n \sqrt{3}) : n \in \mathbb{Z}^+ \} = 0$ where $\text{frac}(x)$ is the fractional part of $x$

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- #12

- Mar 10, 2012

- 835

Thank you Opalg but I am ignorant about continued fractions. Anyways.. thanks for the help.One very constructive way to approach this is to use the continued fraction approximants to $\sqrt3$. These are alternately slightly less than, and slightly greater than, $\sqrt3.$ Take one of the approximants that is less than $\sqrt3$, for example 265/153. You can then check that $153\sqrt3\approx 265.00377...$, whose fractional part is very small.