Density in $\mathbb R$ Let $A=\{a+b\sqrt{2}\mid a,b\in\mathbb Z\}$ and let $Y\subset A$ with $x\in Y$ iff $x\in [0,1]$.
I'm trying to prove that $A$ is dense in $\mathbb R$ and already noticed that it is enough to show that $Y$ is dense in $[0,1]$. Could any one point me in a direction? 
I have already looked at a proof of the density of $\mathbb Q$ in $\mathbb R$, where the core of the proof consisted of showing that between any two reals there is a rational. This was done by multiplying the reals with a natural number until the difference between them became bigger than one and noticing that you can find a rational between the two multiplied reals and divide that by the amount the reals got multiplied with and you get a rational between the two reals.
This method does not work identically in this case, because dividing an element in $A$ by a natural number won't always yield another element in $A$.
I will use this, when I have a proof, to prove that the quotient $\mathbb R/{\sim}$ with $r\sim s$ if $r-s=a+b\sqrt{2},\;\exists a,b \in \mathbb Z$ is not hausdorff. The density of every eq. clas in this quotient implies that the quotient topology must be the trivial one, hence the quotient cannot be hausdorff.
 A: Note that $0\lt \sqrt{2}-1\lt 1$, and that any integer power of $\sqrt{2}-1$ is of the form $a+b\sqrt{2}$ where $a$ and $b$ are integers.
Given any $\epsilon\gt 0$, we can find an integer $q$ such that $0\lt (\sqrt{2}-1)^q\lt \epsilon$. Now consider the numbers $n(\sqrt{2}-1)^q$, as $n$ ranges over the integers, positive, negative, and $0$. It is not hard to show that for any real number $x$, there is an integer $n$ such that the distance from $n(\sqrt{2}-1)^q$ to $x$ is $\lt \epsilon$.     
A: Another approach is to use this application of the pigeonhole principle to show that the numbers $b\sqrt2-\lfloor b\sqrt2\rfloor$ for $b\in\Bbb Z$ are dense in $[0,1]$: clearly $\{b\sqrt2-\lfloor b\sqrt2\rfloor:b\in\Bbb Z\}\subseteq A$. The same argument works with any irrational in place of $\sqrt2$.
A: You can approximate $\sqrt{2}$ as closely as you want with rational numbers $-a/b$.  If $\dfrac{-a}{b}$ is close to $\sqrt{2}$, then $a+b\sqrt{2}$ is close to $0$.  If you can make $a+b\sqrt{2}$ as close as you want to $0$ be suitable choice of $a,b\in\mathbb Z$, then there is no neighborhood of $0$ that excludes all nonzero numbers of the form $a+b\sqrt{2}$.  If the set in question were discrete, then it would have only finitely many members in every bounded interval.
