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

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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## marked as duplicate by Jyrki Lahtonen, Namaste group-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 30 '14 at 13:26

• for $\epsilon=\frac{1}{2}$ can you find $g_{\epsilon}$ – user87543 Jun 30 '14 at 13:05
Hint: $0<\sqrt 2 -1< 1$ and this group is also closed under multiplication.
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.