Is the set $\{\sqrt n - \lfloor\sqrt n\rfloor : n\in \mathbb Z^+\}$ dense in $[0,1]$?

Question

I was reviewing basic Analysis and thought of this question:

Is the set $\{\sqrt n - \lfloor\sqrt n\rfloor : n\in \mathbb Z^+\}$ dense in $[0,1]$?

Lest we have any ambiguity with notational stuff, $\lfloor \cdot \rfloor$ denotes the integer part of a real number i.e., for $x\in\mathbb R$, $\lfloor x\rfloor := \max\{n\in\mathbb Z\mid n\leq x\}$.

The function $f:\mathbb Z^+\to[0,1]$ defined by $f(n)=\sqrt n - \lfloor\sqrt n\rfloor$ is injective when it is restricted to the prime numbers. I showed this using contradiction.

Let $p_1$ and $p_2$ be distinct prime numbers such that $f(p_1)=f(p_2)$. It would mean that $\sqrt {p_1} + \sqrt{p_2}$ is an integer. (contradiction!)

Thus, the set in our discussion has infinitely many points.

Suppose $p_0$, $p_1$, $\ldots$ is an enumeration of prime numbers. Using pigeon-hole principle, for every $n\geq 1$, there exists primes $p_i$ and $p_j$ with $0\leq i<j\leq n$ such that $$ |f(p_i)-f(p_j)|< \frac{1}{n}$$

I am not sure, how to proceed after this.

Answer 1

This follows by a direct application of Fejer's theorem:

Theorem: Let $(f(n):n\in\mathbb{N})$ be a sequence of real numbers such that $\Delta f(n)=f(n+1)-f(n)$ is monotone in $n$. If $$\lim_{n\rightarrow\infty}\Delta f(n)=0,\quad\text{and}\quad \lim_{n\rightarrow\infty}n|\Delta f(n)|=\infty,$$ then the sequence $(f(n):n\in\mathbb{N})$ is uniformly distributed $\mod 1$.

This means that for any interval $(a,b]\subset (0,1)$ \begin{align} \lim_{n\rightarrow\infty}\frac{1}{n}\sum^n_{k=1}\mathbb{1}_{(a,b]}(f(n))=b-a\tag{0}\label{zero} \end{align}

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Let $f(n)=\sqrt{n}$, $n\in\mathbb{N}$. Notice that

1. $\Delta f(n)=f(n+1)-f(n)=\frac{1}{\sqrt{n+1}+\sqrt{n}}$ is monotone in $n$,
2. $\lim_{n\rightarrow\infty}\Delta f(n)=0$ and
3. $\lim_{n\rightarrow\infty} n|\Delta f(n)|=\lim_{n\rightarrow\infty} n\Delta f(n)=\infty$.

For a nice introduction to uniform distribution $mod 1$ and some functional results (including Fejer's theorem), see Kuipers, L. and Niederreiter, H., Uniform Distributions of Sequences, Dover Publications Inc. Chapter 1.

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Some comments about Fejer's theorem:

The result stated above proofs much more than mere density of the sequence $(\sqrt{n}-\lfloor \sqrt{n}\rfloor: n\in\mathbb{N})$ but that in the limit, the points are uniformly distributed over $(0,1)$ (this what \eqref{zero} means). This phenomena is known as equidistribution and has a rich history. Perhaps the most basic result on this subject, due to Weyl, is the following (see for example e1 and e2)

Theorem (Weyl): The sequence $(x_n:n\in\mathbb{N})\subset \mathbb{R}$ is uniformly distributed $\mod 1$ (that is $(x_n-\lfloor x_n\rfloor:n\in\mathbb{N})$ satisfies \eqref{zero}) if and only iff $$\lim_{N\rightarrow\infty}\frac1N\sum^N_{n=1}e^{2\pi ikx_n}=0$$ for any integers $k\neq0$.

Many deep results in equidistribution of a sequence lead to probing that Weyl criteria holds. Fejer's result above is of this type. Here is a sketch of a proof of Fejer's. I recommend the OP takes a look at the reference I provided in above.

Using integration by parts, one can show that for all $t\in\mathbb{R}$ \begin{align} |e^{2\pi it}-1-2\pi it|\leq 2\pi^2t^2 \end{align} Form this, we obtain that \begin{align} |e^{2\pi k if(n+1)}-e^{2\pi k if(n)} -2\pi k e^{2\pi k if(n)}|\leq 2\pi^2 k^2|\Delta f(n)|^2 \end{align} As a consequence \begin{align} \big|\frac{e^{2\pi ik f(n+1)}}{\Delta f(n+1)}-\frac{e^{2\pi ik f(n)}}{\Delta f(n)} - 2\pi ik fe^{2\pi ik f(n)}\Big|\leq \Big|\frac{1}{\Delta f(n+1)}-\frac{1}{\Delta f(n)}\Big| + 2\pi^2 k^2|\Delta f(n)|\tag{1}\label{one} \end{align} Since \begin{align} \sum^N_{n=1}e^{2\pi ik f(n)}&=\frac{1}{2\pi ik}\sum^N_{n=1}\Big(2\pi ik e^{2\pi ik f(n)}-\frac{e^{2\pi ik f(n+1)}}{\Delta f(n+1)}-\frac{e^{2\pi ik f(n)}}{\Delta f(n)}\Big) +\\ &\qquad \qquad \frac{e^{2\pi i f(N+1)}}{\Delta f(N+1)} -\frac{e^{2\pi i f(1)}}{\Delta f(1)}, \end{align} \eqref{one} and the monotonicity of $\Delta f(n)$ yields \begin{align} \frac1N\Big|\sum^N_{n=1}e^{2\pi ik f(n)}\Big|&\leq \frac1{\pi |k|}\left(\frac{1}{(N+1)\Delta f(N+1)}+\frac{1}{(N+1)\Delta f(1)}\right)+\\ &\qquad\qquad \frac{\pi |k|}{N}\sum^N_{n=1}|\Delta f(n)| \end{align} The assumptions in Fejer's theorem imply that the conditions in Weyl's criteria hold.

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Final remark:

Fejer's theorem can also be used to imply the u.d. $mod 1$ of the sequence $(a n^p\log^q n:n\in\mathbb{N})$ with $0<p<1$ and $\tau\in\mathbb{R}$.

Answer 2

Subdivide the interval $[0,1]$ in subintervals of with $\dfrac1m$, giving the bounds $\dfrac km$ with $k=0,1,\cdots m$. I claim that for every $k$, $n$ equal to $m^2+2k+1$ is such that $\sqrt n-\lfloor \sqrt n\rfloor$ belongs to the subinterval $\left[\dfrac km,\dfrac{k+1}m\right]$.

Indeed, $$\frac km\le \sqrt{m^2+2k+1}-m\le \frac{k+1}m.$$

is equivalent to

$$2k+\frac{k^2}{m^2}\le 2k+1\le 2k+2+\frac{(k+1)^2}{m^2}.$$

So for any real in $[0,1]$ you will find a subinterval as small as you want that contains that real as well as some $\sqrt n-\lfloor \sqrt n\rfloor$.

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E.g. approximate $0.371$ to less than $0.001$. We take $m=1000, k=371$ and $n=1000^2+2\cdot371+1$.

$$\sqrt{1000743}-1000=0.37143\cdots$$

Answer 3

Pointing at this question, let's prove the following

Lemma. Let $(u_n)_{n\in\mathbb{N}}$ be an increasing sequence such that $$u_n \to \infty, n\to\infty\tag{1}$$ and $$u_{n+1}-u_n \to 0,n\to\infty\tag{2}$$ Then $\left\{u_n-\lfloor u_n \rfloor : n\in \mathbb Z^+ \right\}$ is dense in $[0,1]$

Let's establish one important property, from $(2)$ and the fact that $(u_n)_{n\in\mathbb{N}}$ is increasing, $\forall \varepsilon>0$ $\exists N_{\varepsilon}\in \mathbb{N}$: $$u_n<u_{n+1}<u_n +\varepsilon \tag{3}$$ for $\forall n>N_{\varepsilon}$.

We need to show that $\forall a,b$ real numbers s.t. $0 < a < b <1$ there $\exists u_n$ from the sequence s.t. $a<\{u_n\}<b$, where $\{\}$ is the fractional part.

Proof by contradiction. Assume $(u_n)_{n\in\mathbb{N}}$ is not dense in $[0,1]$, which means there $\exists a,b$ real numbers s.t. $0 < a < b <1$ s.t. for $\forall n\in\mathbb{N}$:

• $0\le\{u_n\} \le a$ or
• $b\le\{u_n\}$

We have 3 cases to check ...

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Case 1. $0\le\{u_n\} \le a$ for $\forall n > N_{\varepsilon}$.

From $(3)$ $$u_n<u_{n+1}<u_n +\varepsilon = \left \lfloor u_n\right \rfloor +\left\{u_n\right\}+\varepsilon \le \left \lfloor u_n\right \rfloor +a+\varepsilon \Rightarrow \\ 0 < u_{n+1} - \left \lfloor u_n\right \rfloor < a+\varepsilon$$ Now $0<a<1$ and we can "adjust" $\varepsilon$ s.t. $a+\varepsilon <1$ leading to $$0 < u_{n+1} - \left \lfloor u_n\right \rfloor <1$$ Or, because the sequence is increasing $$0 \le \left \lfloor u_{n+1}\right \rfloor - \left \lfloor u_n\right \rfloor \le u_{n+1} - \left \lfloor u_n\right \rfloor <1$$ Leading to $\left \lfloor u_{n+1}\right \rfloor = \left \lfloor u_n\right \rfloor$ for $\forall n > N_{\varepsilon}$. This contradicts the $(1)$ above.

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Case 2. $b\le\{u_n\}$ for $\forall n > N_{\varepsilon}$.

From $(3)$ and $b\le\{u_n\} \Rightarrow b\le\{u_{n+1}\}$ $$\left \lfloor u_{n+1}\right \rfloor +b \le \left \lfloor u_{n+1}\right \rfloor +\left\{u_{n+1}\right\}=u_{n+1} < u_n + \varepsilon$$

We can "adjust" $\varepsilon$ s.t. $b-\varepsilon >0$, leading to

$$0< b-\varepsilon < u_n - \left \lfloor u_{n+1}\right \rfloor \Rightarrow \left \lfloor u_{n+1}\right \rfloor < u_n$$

The sequence is increasing, thus $$0 \le \left \lfloor u_{n+1}\right \rfloor - \left \lfloor u_n\right \rfloor < \left\{u_n\right\} < 1$$

Leading to $\left \lfloor u_{n+1}\right \rfloor = \left \lfloor u_n\right \rfloor$ for $\forall n > N_{\varepsilon}$. This contradicts the $(1)$ above.

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Case 3. $(u_n)_{n\in\mathbb{N}}$ is alternating between $0\le\{u_n\} \le a$ and $b\le\{u_n\}$ infinitely many times. If it doesn't alternate infinitely many times, then the sequence gets stuck in the case 1 or case 2 from some $n$ onwards.

As such, the sequence has a subsequence $(u_{n_{k}})_{k\in\mathbb{N}}$ s.t.

• $0\le \{u_{n_k}\} \le a$ and
• $b\le \{u_{n_k +1}\}$

From $(3)$ and because the sequence is increasing $$0<u_{n_{k}+1} -u_{n_k} <\varepsilon \iff \\ 0< \left \lfloor u_{n_k +1}\right \rfloor - \left \lfloor u_{n_{k}}\right \rfloor +\{u_{n_{k}+1}\}-\{u_{n_{k}}\}< \varepsilon$$ or $$0<b-a \le \{u_{n_k +1}\} - \{u_{n_k}\} \le \left \lfloor u_{n_k +1}\right \rfloor - \left \lfloor u_{n_{k}}\right \rfloor +\{u_{n_{k}+1}\}-\{u_{n_{k}}\}< \varepsilon$$ or $$0<b-a < \varepsilon$$ This contradicts the fact that $\varepsilon$ can be "adjusted" to be arbitrarily small. Technically, it contradicts $(2)$ above.

This concludes the proof.

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Now, let's apply the lemma to the $u_n=\sqrt{n}$

• $(\sqrt{n})_{n\in\mathbb{N}}$ is increasing
• $\sqrt{n} \to\infty, n\to\infty$
• $\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}} \to 0, n\to\infty$

As a result, $\left\{\sqrt{n}-\lfloor \sqrt{n} \rfloor : n\in \mathbb Z^+ \right\}$ is dense in $[0,1]$.

Answer 4

Let $r$ any real in $[0,1]$ and $\epsilon>0$, the allowed approximation error. WLOG $\epsilon=\dfrac1m$ with $m$ integer.

We set $k=[mr]$ (rounding to the nearest integer). Now consider $n=m^2+2k$. We have $m^2\le n<(m+1)^2$ so that $\lfloor\sqrt n\rfloor=m$. Then

$$\begin{align}|\sqrt n-\lfloor\sqrt n\rfloor-r|&=\left|m\sqrt{1+2\frac k{m^2}}-m-r\right|\\&=\left|\frac km-\frac{k^2}{4m^3}+\frac{k^3}{8m^5}-\cdots-r\right|\\&<\frac1m=\epsilon\end{align}$$ using a Taylor development.

The final inequality is justified by the fact that the rounding error does not exceed $\dfrac12$, and the other terms contribute little: $\dfrac{k^2}{4m^3}<\dfrac\epsilon4, \dfrac{k^3}{8m^5}<\dfrac{\epsilon^2}8,\cdots$