# If $\alpha$ is an irrational real number, why is $\alpha\mathbb{Z}+\mathbb{Z}$ dense in $\mathbb{R}$?

This is chapter $4$ exercise $25.b$ of Walter Rudin's Principles of Mathematical Analysis, this problem has occupied my mind for a long time, and I haven't been able to solve it, I would like to see an answer to this question.

Thanks.

• Almost a duplicate of math.stackexchange.com/questions/73262/… – lhf Nov 26 '12 at 22:48
• You can use Dirichlet approximation theorem, which essentially follows from the pigeonhole principle, to prove that fact. – Sangchul Lee Sep 5 '13 at 23:37

## 2 Answers

Hint: Let $a_n$ be the fractional part of $n\alpha$. This is $n\alpha-\lfloor n\alpha\rfloor$.

Note that because $\alpha$ is irrational, the $a_n$ are all distinct.

Therefore for any $\epsilon\gt 0$, there exist distinct $m$ and $n$ such that $|a_m-a_n|\lt \epsilon$. (There is still some work to do.)

• @Camilo This is a very good hint; perhaps what you are missing is the last part? The first two lines show you that there are infinitely many distinct numbers in $\alpha\Bbb{Z}+\Bbb{Z}$ that are within $(0,1)$. So surely some two such numbers are very close together. Once that is established, their difference is in the set too... And after that, all integer multiples of this difference... – alex.jordan Dec 19 '12 at 4:24

Here's one relatively straightforward approach to it: By a simplified version of Hurwitz's theorem, there's are infinitely many rationals $m/n$ with $\left|\alpha-\dfrac{m}{n}\right|\leq \dfrac{1}{n^2}$; multiplying by $n$ we have $\left|n\alpha-m\right|\leq \dfrac{1}{n}$ - or, defining $\beta\equiv\left|n\alpha-m\right|$, $\beta\leq\dfrac{1}{n}$. Letting $q=\left\lfloor\dfrac{1}{\beta}\right\rfloor$, the $q$ numbers $\beta, 2\beta, \ldots, q\beta$ - each of which is of the form $a\alpha+b$ for some $a, b\in\mathbb{Z}$ - then provide a cover of $(0..1)$ to 'resolution' $\frac{1}{n}$; by translation we then get the density in $\mathbb{Z}$.