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
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