inf{frac{n*sqrt(3)} : n in integer}=0

caffeinemachine

Well-known member
MHB Math Scholar
Prove that $\inf \{ \text{ frac}(n \sqrt{3}) : n \in \mathbb{Z}^+ \}$ where $frac(x)$ is the fractional part of $x$

Plato

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MHB Math Helper
Prove that $\inf \{ \text{ frac}(n \sqrt{3}) : n \in \mathbb{Z}^+ \}$ where $frac(x)$ is the fractional part of $x$
That is not a complete question.
But guessing at what it means, I think this theorem will help.

Evgeny.Makarov

Well-known member
MHB Math Scholar
The OP misses "= 0," which is present in the thread title.

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.

caffeinemachine

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

---------- 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.

Evgeny.Makarov

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MHB Math Scholar
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 $\text{frac}(|i-j|\sqrt{3}) < \frac{1}{n}$.
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, $1-0.3<\mathop{\mbox{frac}}(2\sqrt{3}-\sqrt{3})<1$.

caffeinemachine

Well-known member
MHB Math Scholar
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, $1-0.3<\mathop{\mbox{frac}}(2\sqrt{3}-\sqrt{3})<1$.
Ah! yes. My mistake.
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.

caffeinemachine

Well-known member
MHB Math Scholar
Well after more fiddling around I can only prove that $inf \{ n \sqrt{3} : n \in \mathbb{Z} \} =0$. I am not able to show that $inf \{ n \sqrt{3} : n \in \mathbb{Z}^+ \} =0$.
I don't even know whether the latter is true.

Ackbach

Indicium Physicus
Staff member
Just as a little $\LaTeX$ hint: use the backslash for infinum instead of just inf. You get '$\inf$' versus '$inf$'.

Evgeny.Makarov

Well-known member
MHB Math Scholar
I can only prove that $\inf \{ n \sqrt{3} : n \in \mathbb{Z} \} =0$.
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.

caffeinemachine

Well-known member
MHB Math Scholar
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.
That was a typo again... sorry:
What I meant was $\inf \{ \text{frac}(n \sqrt{3}): n \in \mathbb{Z} \}=0$

Opalg

MHB Oldtimer
Staff member
Prove that $\inf \{ \text{ frac}(n \sqrt{3}) : n \in \mathbb{Z}^+ \} = 0$ where $\text{frac}(x)$ is the fractional part of $x$
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.

caffeinemachine

Well-known member
MHB Math Scholar
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.
Thank you Opalg but I am ignorant about continued fractions. Anyways.. thanks for the help.