I am having trouble proving the statement:

Let $$S = \{m + n\sqrt 2 : m, n \in\mathbb Z\}$$ Prove that for every $\epsilon > 0$, the intersection of $S$ and $(0, \epsilon)$ is nonempty.

Rodrigo de Azevedo
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5 Answers5


Hint: $|\sqrt2 -1|<1/2$, so as $n\to\infty$ we have that $(\sqrt2-1)^n\to ?$ In addition to that use the fact that the set $S$ is a ring, i.e. closed under multiplication and addition.

Jyrki Lahtonen
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    If you are unfamiliar with the language of rings, you can just compute $(\sqrt2-1)^n$ using the binomial formula to verify that it is, indeed, in the set $S$. – Jyrki Lahtonen Oct 17 '11 at 07:54
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    If $\sqrt2$ is replaced with a generic irrational real number $\alpha$, then we can use the pigeon hole principle to prove the same result as [Dirichlet's approximation theorem.](http://en.wikipedia.org/wiki/Dirichlet%27s_approximation_theorem) In higher dimensions this can be replaced with [Kronecker's density theorem](http://en.wikipedia.org/wiki/Kronecker%27s_theorem). IOW I feel that this exercise is more about Diophantine approximation than real analysis, but to each their own :-) – Jyrki Lahtonen Oct 17 '11 at 13:24
  • sir , if the set is $\{m+n\sqrt{2} : m,n\in\mathbb{N}\}$ , then is it still dense in $\mathbb{R}$ ? I think then it is dense in $[0,\infty)$ . Am I right,sir? – A learner Nov 07 '20 at 16:22
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    @Alearner The set $\{m+n\sqrt2\mid m,n\in\Bbb{N}\}$ has only finitely many elements in any interval of the form $[0,x]$, because $m\le x$ and $n\le x/\sqrt2$. Therefore it cannot be dense anywhere. – Jyrki Lahtonen Nov 07 '20 at 17:13
  • thank you sir, for the reply. – A learner Nov 07 '20 at 17:22

Suppose not, so that

there exists an $\varepsilon>0$ such that $(0,\varepsilon)\cap S=\emptyset$. $\qquad\qquad\qquad(\star)$

It follows that $\alpha=\inf S\cap(0,+\infty)$ is a positive number.

  • The choice of $\alpha$ and its positivity implies that

the one and only element of $S$ which is in $[0,\alpha)$ is $0$.

  • I claim that $\alpha\in S$. Indeed, suppose not and let $\alpha=\inf S\cap(0,+\infty)$. The hypothesis implies that $\alpha>0$, and the choice of $\alpha$ implies that there exists elements $s$, $t\in S\cap(0,+\infty)$ such that $$\alpha\leq s<t\leq(1+\tfrac14)\alpha.$$ Then $u=t-s$ is an element of $S$ (because $S$ is closed under addition) such that $0<u\leq\tfrac14\alpha<\alpha$. This is absurd so we must have $\alpha\in S$, as I claimed.

  • Let $s\in S\cap(0,+\infty)$ and let $n=\lfloor s/\alpha\rfloor$ be the largest integer which is less than $s/\alpha$. Then $n\alpha\leq s<(n+1)\alpha$, so that $0\leq s-n\alpha <\alpha$. This tells us that $s-n\alpha$, which is an element of $S$, is in $[0,\alpha)$. The choice of $\alpha$ implies that we must then have $s-n\alpha=0$, that is, $s=n\alpha$. We conclude that every positive element of $S$ is an integer multiple of $\alpha$.

  • In particular, since $1\in S$ and $\sqrt2\in S$, there exists integers $n$ and $m$ such that $1=n\alpha$ and $\sqrt2=m\alpha$. But then $\sqrt2=\frac{\sqrt2}{1}=\frac{m\alpha}{n\alpha}=\frac mn\in\mathbb Q.$ This is absurd, and we can thus conclude that $(\star)$ is an untenable hypothesis.

Mariano Suárez-Álvarez
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Taking a step into generalization, it is true that every additive subgroup $G$ of $\mathbb R$ is either discrete or dense. This can be proved by considering $\alpha = \inf \{ x \in G : x>0 \}$. Then $G$ is discrete iff $\alpha >0$, in which case $G=\alpha \mathbb Z$. In your case, Jyrki's suggestion implies that $\alpha=0$ and so $S$ is dense.

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Let$\epsilon>0$,Let $x\in\mathbb{R}$ $x-m-\epsilon<x-m+\epsilon$ between any two reals there are infinitely many rationals. $\frac{x-m-\epsilon}{\sqrt{2}|q|}<\frac{p}{q}<\frac{x-m+\epsilon}{\sqrt{2}|q|}$ therefore $$(x-\epsilon,x+\epsilon)$$ contains the member of the given set.

Rayees Ahmad
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    I'm afraid this argument is not correct. There are, indeed, infinitely many rationals between $x-m-\epsilon$ and $x-m+\epsilon$, but your next claim is that there is an integer $p$ between $(x-m-\epsilon)/\sqrt2$ and $(x-m+\epsilon)/\sqrt2$. This does not follow. And it is obvious not true for some $x,m,\epsilon$. – Jyrki Lahtonen Dec 17 '20 at 06:46

we know: $\sqrt{2}-1<1 $ so $ (\sqrt{2}-1)^n \rightarrow 0$ so for any $\epsilon > 0$ there exist $N$ such that $n\geq N \Rightarrow (\sqrt{2}-1)^n < \epsilon$

so: $(\sqrt{2}-1)^N < \epsilon$, but we know: \begin{align} (\sqrt{2}-1)^N = \sum_{i=0}^N{2^{i/2}(-1)^{N-i}} = \sum_{i \text{ Odd}} + \sum_{i \text{ Even}} {2^{i/2}(-1)^{N-i}} \end{align} but for Even i's above sum is an integer and for Odd i's is equal to $m\sqrt{2}$ for some integer $m$, so: $(\sqrt{2}-1)^N = n+m\sqrt{2}$ for some integer's $m,n$, but $0<(\sqrt{2}-1)^N < \epsilon$

so for any $\epsilon > 0$, exist $s\in S , s=m+n\sqrt{2}$ such that $s \in (0,\epsilon)$ as desired.

Siamak Amo
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