Proving that $m+n\sqrt{2}$ is dense in R

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.