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I am having trouble proving the statement:

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

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Consider rational numbers really close to $\sqrt{2}$. –  Ted Oct 17 '11 at 5:45
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$S\cap (0,\epsilon)\ne\emptyset\quad\forall\epsilon>0$ only says $S$ has a limit point at $0$, not at every $x\in\mathbb{R}$... –  anon Oct 17 '11 at 5:55
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3 Answers

up vote 21 down vote accepted

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.

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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 7:54
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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. In higher dimensions this can be replaced with Kronecker's density theorem. IOW I feel that this exercise is more about Diophantien approximation than real analysis, but to each their own :-) –  Jyrki Lahtonen Oct 17 '11 at 13:24
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Supose 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 $m$ suchthat $1=n\alpha$ 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.

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This answer should get more votes. It is more technical than Jyrki's, but is works for every irrational number and not only $\sqrt{2}$. –  KotelKanim Oct 17 '11 at 8:39
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I have a question: In the proposed answer by Dr. Mariano Suárez-Alvarez, how to show that it's always possible to find 2 different numbers s and t between $\alpha$ and (1+1/4)$\alpha$? I know one can always find one such number by the definition of inf., but what about the other? –  Yang Jan 30 '12 at 3:51
    
@Yang, pick one, call it $t$, such that $\alpha<t\leq(1+\tfrac14)\alpha$ and then pick another, call it $s$, such that $\alpha\leq s<t$. –  Mariano Suárez-Alvarez Jan 30 '12 at 3:59
    
(By the way, just call me Mariano, please! :) ) –  Mariano Suárez-Alvarez Jan 30 '12 at 4:00
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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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