Show that $S=\{\sqrt m - \sqrt n : m,n \in \mathbb N\}$ is dense in $\mathbb R$.

I know the definition of dense as:

A set S is dense in $\mathbb R$ if there exists $a,b \in \mathbb R$ such that $S \cap(a,b) \neq \emptyset.$

I don't understand how to proceed. Please help.

Stefan Hamcke
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5 Answers5


You could show that any real number $\alpha$ is a limit point of a sequence in $S$.

Here's one way to proceed. Consider the sequence $$ \epsilon_n = \sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n} + \sqrt{n+1}} $$ and note that $\epsilon_n \rightarrow 0$ as $n \rightarrow \infty$. Moreover, we have that for any $k \in \mathbb{N}$, $$ k \epsilon_n = \sqrt{k^2(n+1)} - \sqrt{k^2 n} \in S $$ Therefore any number of the form $k \epsilon_n$ belongs to $S$. We now show that numbers of this form are dense in $\mathbb{R}_{\geq 0}$ when $k$ ranges over the positive integers.

Fix $\alpha > 0$. For any $k > 0$ then there is some $n = n(k)$ for which $$ \frac{\alpha}{k} \in (\epsilon_{n(k)}, \epsilon_{n(k) - 1}) $$ and so $$ \alpha \in (k \epsilon_{n(k)}, k \epsilon_{n(k)-1}) $$ It suffices to check, then, that the size of this interval shrinks to zero as $k \rightarrow \infty$. It is not hard to show that $$ -1 + \left( \frac{k}{2 \alpha} \right)^2 < n(k) < 1 + \left( \frac{k}{2 \alpha} \right)^2 $$ and that $\epsilon_{n} - \epsilon_{n-1}$ goes like $\frac{1}{n \sqrt{n}}$, which implies the size of the interval decays like $1/k$.

A Blumenthal
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Suppose $a,b\in\mathbb R$ with $a<b$. We need to show that $S\cap (a,b)\ne\emptyset$. Observe that $$\sqrt{m}-\sqrt {m-1}=\frac{1}{\sqrt{m}+\sqrt{m-1}}<\frac1{2\sqrt{ m-1}}.$$ Thus for $m>\frac1{4(b-a)^2}+1$ we have $\sqrt{m}-\sqrt {m-1}<b-a$. Pick $n$ with $n>\left(b+\frac1{4(b-a)^2}+1\right)$. Then $\sqrt n -b>\frac1{4(b-a)^2}+1>1$. Let $m$ be minimal with $m>(\sqrt n-b)^2$, i.e. with $\sqrt n-\sqrt m<b$. Then $\sqrt n-\sqrt{m-1}\ge b$ and hence $\sqrt n-\sqrt m>\sqrt n-\sqrt{m-1}-(b-a)\ge a$, i.e. $$ \sqrt n-\sqrt m\in (a,b).$$

Hagen von Eitzen
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Let $x\geqslant0$. For every integer $m\gt x^2$, consider $u_m=\min\{n\geqslant1\mid \sqrt{m}-\sqrt{n}\leqslant x\}$ and $s_m=\sqrt{m}-\sqrt{u_m}$. Then $u_m$ is finite and $s_m$ is in $S$.

Furthermore, $\sqrt{m}-\sqrt{u_m}\leqslant x\lt\sqrt{m}-\sqrt{u_m-1}$ by definition of $u_m$ hence $0\leqslant x-s_m\lt t_m$ with $t_m=\sqrt{u_m}-\sqrt{u_m-1}$. Since $u_m\to\infty$ when $m\to\infty$ (can you show this?) and $t_m=1/(\sqrt{u_m}+\sqrt{u_m-1})\to0$ when $m\to\infty$ (can you show this?), $s_m\to x$ hence the closure of $S$ contains $x$.

Adapt this argument to every $x\lt0$.

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$$D=\left\{\sqrt{n+1}-\sqrt{n}:d\in\Bbb Z^+\right\}=\left\{\frac1{\sqrt{n+1}+\sqrt{n}}:d\in\Bbb Z^+\right\}\;.$$

Let $(a,b)$ be any non-empty open interval in $\Bbb R$. If $a<0<b$, then $0\in(a,b)\cap S$, so suppose that $0\le a<b$. Let $\delta=b-a$; there is a $d\in D$ such that $d<\delta$; why?

For each $k\in\Bbb Z^+$ we have

$$k\left(\sqrt{n+1}-\sqrt{n}\right)=\sqrt{k^2(n+1)}-\sqrt{k^2n}\in S\;,$$

so $\{kd:k\in\Bbb Z^+\}\subseteq S$; show that $(a,b)\cap\{kd:k\in\Bbb Z^+\}\ne\varnothing$, thereby showing that $S$ is dense in $\Bbb R^+$, the positive reals.

Once you know that $S$ is dense in $\Bbb R^+$, showing that it’s dense in the negative reals is very straightforward.

Brian M. Scott
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The result is quite clear if you write the elements of $S$ as $\sqrt{m} \left( 1 - \frac{\sqrt{n}}{\sqrt{m}}\right)$. Choose $\mu \in \mathbf{N}$ in such a way that $\frac{1}{\mu} < b - a$, $\mu > b +1$; then consider $m = \mu ^2$, $n = \nu^2$ for $\nu \in \mathbf{N}$; with these choices it becomes apparent that $S$ contains elements of the form $\mu - \frac{\nu}{\mu}$, and that for some $\nu$ one of such points belongs to the interval $\left( a, b \right)$.

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