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In this question, we prove that $\{a+b\sqrt{2}\mid a,b\in\mathbb{Z}\}$ is dense in $\mathbb{R}$ by proving that is it dense around $0$.

Why is that enough to prove that it is dense on $\mathbb{R}$ ?

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Hippalectryon
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    This isn't an answer, but a direct, nice and simple demonstration of the final result: http://math.stackexchange.com/a/90180/40061 . The key point is the fact that S is an additive group. – dan Jan 05 '15 at 13:20

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Because then you can take the integer multiples $k(a+b\sqrt2)=ka+kb\sqrt2$ to fill the rest of the line.

user2345215
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    But how are you sure that this will fill (i.e. be dense on) all the line ? – Hippalectryon Jan 05 '15 at 00:12
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    @Hippalectryon If $a+b\sqrt 2\in (-\varepsilon,\varepsilon)\setminus{0}$, then the integer multiples fill all intervals of lenght larger than $\varepsilon$. – user2345215 Jan 05 '15 at 00:14