# For every irrational $\alpha$, the set $\{a+b\alpha: a,b\in \mathbb{Z}\}$ is dense in $\mathbb R$

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? –  La Belle Noiseuse 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.

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Could their difference be negative? –  André Nicolas Apr 25 '12 at 18:37
This proof needs a bit more work. The difference can be negative, and the proof only shows the existence of one limit point. It is a good start though. My comment is late; I know. I came here from another question that got marked as duplicate of this. –  Ayman Hourieh Jun 24 '13 at 0:57

Considering user29743's proof, after getting two such $i,j$ we can tell exist some $m\in\mathbb{Z}$ such that $m-\frac{1}{n}<(i-j)\alpha=k\alpha <m+\frac{1}{n}$. So now we can say for any $n$ exist $m,k \in \mathbb{Z}$ such that $|p-k\alpha|<\frac{1}{n}$. Now basically $n>k$. So from here we can conclude there exists infinitely many $q$'s such that $|\alpha-p/q|<1/q^2 \implies |q\alpha-p|<\frac{1}{q}$.

Now note as there are infinitely many $q$ so we can say partition $a,2a,3a,....,[\frac{1}{a}]a$ covers entire $(0,1)$ with length of each part being smaller enough, and any of $ia$ can be taken as a term of desired sequence. So around any ball of any $r\in (0,1)$ we'll get infinitely many terms, as a consequence its dense in $(0,1)$ as well as in $\mathbb{R}$

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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 $\mathbb{R}$

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