Everywhere dense set $\sqrt{n} \mod 1$

Question

Prove that set $\{ \ \{\sqrt{n}\} \ |n\in \mathbb{N} \}$ is dense in $(0,1)$. My try:

$(a,b) \subseteq (0,1) $ arbitrary. I tried to make $\{\sqrt{n}\}$ ''small'' and then multiply it by m so that it get in $(a,b)$ but i failed.

Every hint or help is appreciated.

Enough to show that ${n\sqrt 2}$ is dense. Toward that end, suppose there were a least positive term. Deduce that this would imply $\sqrt 2 \in \mathbb Q$. — lulu, Nov 05 '19 at 19:40
@user661541 I believe it is the fractional part of $\sqrt{n}$. — 79037662, Nov 05 '19 at 19:40
Forget the fractional part. Prove that the set, $S$, of all terms $a+b\sqrt 2$ with $a,b\in \mathbb Z$ is dense in $\mathbb R$. To do so, suppose that $\alpha$ was the least positive element of $S$ (seeking a contradiction). Prove that every element in $S$ is an integer multiple of $\alpha$. Derive a contradiction. — lulu, Nov 05 '19 at 19:51
Please try some of the hints I have provided. I don't want to write out a full solution. — lulu, Nov 05 '19 at 19:56
here is a duplicate question (note that that one is also a duplicate). — lulu, Nov 05 '19 at 19:58
Possible duplicate of For every irrational $\alpha$, the set ${a+b\alpha: a,b\in \mathbb{Z}}$ is dense in $\mathbb R$ — Arthur, Nov 05 '19 at 20:10
another hint: $\sqrt{n+1}-\sqrt{n}=\frac1{\sqrt{n+1}+\sqrt{n}}\to0$ as $n\to\infty$. See what happens with ${\sqrt{n}}$ as $n$ goes from $m^2$ to $(m+1)^2$ (and where $m\to\infty$). Then $\sqrt{n}$ goes from $m$ to $m+1$, so ${\sqrt{n}}$ goes from $0$ to $1$ (disregarding for a moment that the fractional part of $1$ is $0$). — Mirko, Nov 06 '19 at 01:17
@Mirko could you be more precise ,I dont see how you hint helps — Mark, Nov 08 '19 at 00:00

Mirko · Accepted Answer · 2019-11-15T00:37:54.467

There are several nice hints in the comments, I expand on mine.

Clearly $\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac1{\sqrt{n+1}+\sqrt{n}}\to0$ as $n\to\infty$.

The idea is that, given any $m$, if $n$ takes values from $m^2$ to $(m+1)^2$ then $\sqrt{n}$ takes values from $m$ to $m+1$, with some small, and more or less "uniform" step.

To illustrate, say $m=4$, and $n$ takes values from $16$ to $25$. Then we have $\sqrt{16}=4$, $\sqrt{17}\approx4.123$, $\sqrt{18}\approx4.243$, $\sqrt{19}\approx4.359$, $\sqrt{20}\approx4.472$, $\sqrt{21}\approx4.583$, $\sqrt{22}\approx4.690$, $\sqrt{23}\approx4.796$, $\sqrt{24}\approx4.899$, $\sqrt{25}\approx5$. So, the fractional part $\{n\}$ takes values $0$, $0.123$, $0.243$, $0.359$, $0.472$, $0.583$, $0.690$, $0.796$, $0.899$, $0$ (or, one could also say $1$). The point is, that the fractional part gradually changes from $0$, with small steps, to a number close to $1$, and then to $1\equiv0\mod1$. It starts with step about $0.12$, and the step decreases a bit, ending about $0.10$, but not changing that much.

It should be clear that for bigger values of $m$ the step would be smaller. More precisely, if $m$ is arbitrary (and big), and $n$ takes values $m^2,m^2+1,m^2+2,...,(m+1)^2-2,(m+1)^2-1,(m+1)^2$ then $\sqrt{n}$ takes values $v_0=m,v_1,v_2,...,v_{2m-1},v_{2m},v_{2m+1}=m+1$, so the fractional parts $\{v_0\},\{v_1\},\{v_2\},...,\{v_{2m-1}\},\{v_{2m}\},\{v_{2m+1}\}$ start at $0$ and gradually increase to values close to $1$, and the last value $\{v_{2m+1}\}$ goes back to $0$. Note that $\{v_1\}-\{v_0\}=v_1-v_0=\sqrt{m+1}-\sqrt{m}=\frac1{\sqrt{m+1}+\sqrt{m}}$ and this is the largest step within the finite sequence $v_{j}-v_{j-1}$, where $j=1,2,...,2m-1$. Note that, $\{v_{2m-1}\}-\{v_{2m-2}\}=v_{2m-1}-v_{2m-2}=$ $\sqrt{2m-1}-\sqrt{2m-2}=\frac1{\sqrt{2m-1}+\sqrt{2m-2}}$, and $1-\{v_{2m-1}\}=v_{2m}-v_{2m-1}=$ $\sqrt{2m}-\sqrt{2m-1}=\frac1{\sqrt{2m}+\sqrt{2m-1}}$ (the latter being the smallest step).

Formally, take $(a,b)\subseteq(0,1)$ arbitrary. Take $m$ big enough so that $\frac1{\sqrt{m+1}+\sqrt{m}}<b-a$. Then, in the notation from the preceding paragraph, as $n$ takes values $m^2,m^2+1,m^2+2,...,(m+1)^2-2,$ $(m+1)^2-1,(m+1)^2$, the fractional parts $\{\sqrt{m^2}\},\{\sqrt{m^2+1}\},\{\sqrt{m^2+2}\},...,$ $\{\sqrt{(m+1)^2-2}\},\{\sqrt{(m+1)^2-1}\},\{\sqrt{(m+1)^2}\}$ start at $0$ and gradually approach $1$ with steps $<b-a$, and hence (at least) one of these values would be in the interval $(a,b)$.

This is precisely what I wanted ,@Mirko thanks for help – Mark Nov 15 '19 at 00:24 — Mark, Nov 15 '19 at 00:24
You are welcome, glad to help ! – Mirko Nov 15 '19 at 00:31 — Mirko, Nov 15 '19 at 00:31

Everywhere dense set $\sqrt{n} \mod 1$

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