# how to show this is a dense set?

I am not able to prove that this set is dense in $\mathbb{R}$. Will be pleased if you help in a easiest way, $\{a+b\alpha: a,b\in \mathbb{Z}\}$ where $\alpha\in\mathbb{Q}^c$ is a fixed irrational.

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isn't it standard notation of the set of all irrational numbers? –  Bunuelian Trick Apr 25 '12 at 6:33
it's a little funny, because $\{a + b\alpha: a, b\in \mathbb{Z}, \alpha \in \mathbb{Q}^c\} = \mathbb{Q}^c$ –  user29743 Apr 25 '12 at 6:36
I think OP intended that $\alpha$ is fixed. –  Ragib Zaman Apr 25 '12 at 6:36
This set contains all irrationals: any irrational $x$ is of this form with $a=0, b=1, \alpha=x$. Perhaps you intended to ask the more interesting question "If $\alpha$ is a fixed irrational, then is $\{a+b\alpha : a, b \in \mathbb{Z} \}$ dense in $\mathbb{R}$?" –  Chris Eagle Apr 25 '12 at 6:37
@countinghaus: No, it's $\mathbb{Q}^c \cup \mathbb{Z}$. –  Chris Eagle Apr 25 '12 at 6:38
I will write $\{x\}$ to mean the fractional part of $x$, i.e. for $x$ minus the floor of $x$. What we need to show is that we can get arbitrarily close to $0$ by taking $\{m\alpha\}$ for varying integers $m$. Note that, because $\alpha$ is irrational, $\{m\alpha\} \neq \{m'\alpha\}$ for $m \neq m'$.
Let's show that we can get within $1/n$ of $0$ for an arbitrary positive integer $n$. Divide up the interval $[0, 1]$ into $n$ closed intervals of length $1/n$. We have $n+1$ distinct quantities $0, \{\alpha\}, \{2\alpha\}, \ldots, \{n\alpha\}$.
By the pigeonhole principle, two of these, say $\{i\alpha\}$ and $\{j\alpha\}$ with $i > j$, lie in the same closed interval $[k/n, (k+1)/n]$, and so their difference, which is $\{(i - j)\alpha\}$, is closer than $1/n$ to $0$; as $n$ was arbitrary, we're done.