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.
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.
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.
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 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.
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.
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.
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.