TITLE: For an irrational number $\alpha$, prove that the set $\{a+b\alpha: a,b\in \mathbb{Z}\}$ is dense in $\mathbb R$
QUESTION [22 upvotes]: I am not able to prove that this set is dense in $\mathbb{R}$. Will be pleased if you help in a easiest way. $A=\{a+b\alpha: a,b\in \mathbb{Z}\}$ where $\alpha$ is a fixed irrational number.

REPLY [10 votes]: Now I'll prove a general result.
Claim: Any additive subgroup of $\mathbb{R}$ is either discrete or dense in $\mathbb{R}$
Proof: Let $V$ be a discrete subgroup of $\mathbb{R}$. So there is a interval of form 
$(-a,+a)$ which contains finitely many points of $V$. So there exist a $v\in V$ such that
$|v|\leq |x|$ for all $x\in V$. 
So now, $v\mathbb{Z} \subset V$. Suppose there exist $c\in V$ but not in $v\mathbb{Z}$.Then after dividing $c$ by $v$ we'll get remainder of smaller mod value than $v$. Which contradicts property of $v$. So, $V=v\mathbb{Z}$. Hence we can say any discrete subgroup of $\mathbb{R}$ is of form $v\mathbb{Z}$. 
So any non discrete subgroup is of form $\sum_{i=1}^{n} v_i \mathbb{Z}$. Now we need to show this is dense in $\mathbb{R}$. 
This is equivalent to show $\mathbb{Z}+\sum_{i=2}^{n} v_i\mathbb{Z}$ is dense. And this clearly follows from above result, i.e $\mathbb{Z}+\alpha \mathbb{Z}$ is dense in $\mathbb{R}$