Why does the additive subgroup of $\mathbb{R}$ generated by $1$ and $\sqrt{2}$ contain arbitrary small elements?

Question

Let $G\subset \mathbb{R}$ be the additive subgroup of $(\mathbb{R},+)$ defined by $G=\mathbb{Z}+\sqrt{2}\mathbb{Z}$. I want to prove that for every $\epsilon>0$ there exists an element $g_\epsilon\in G$ with $\epsilon>g_\epsilon>0$. Can anybody imagine a nice proof?

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Answer 1

Hint: $0<\sqrt 2 -1< 1$ and this group is also closed under multiplication.

Answer 2

If you know the more general result about subgroups of $(\mathbb{R},+)$ : a subgroup of $(\mathbb{R},+)$ is either dense or of the form $a\mathbb{Z}$ (see for example here). Then you see that if $\mathbb{Z}+\sqrt{2}\mathbb{Z}$ were of the form $a\mathbb{Z}$ then there will be $p,q\in \mathbb{Z}$ such that $1=aq$ and $\sqrt{2}=ap$.

Consequently, $\sqrt{2}=\frac{p}{q}$ which is absurd since $\sqrt{2}\not\in \mathbb{Q}$. So $\mathbb{Z}+\sqrt{2}\mathbb{Z}$ is dense.