Positive integer multiples of an irrational mod 1 are dense I'm not sure how to solve this one.
Thank you!

$2.$ For any $\alpha\in \mathbb R$ we define $$\lfloor \alpha \rfloor = \max_{n\in\mathbb Z}\{\,n\mid n\leq \alpha\,\}$$ and $$\alpha\bmod 1 = \alpha - \lfloor \alpha \rfloor$$ Let $\alpha$ be irrational.
(a) Given $n\in\mathbb N$ show that $\{\,k\alpha\bmod1\mid k\in\mathbb N\,\}\cap\left[0,\frac{1}{n}\right]\neq\emptyset$
(b) Prove that $\{\,n\alpha\bmod 1\mid n\in\mathbb N\,\}$ is dense in $[0,1]$.

 A: What I show below doesn't strictly follows the items, but still shows the numbers are dense.
STEP 1 Pick any $n\in\Bbb N$. Consider the $n+1$ distinct numbers $$x_k=k\alpha-\lfloor k\alpha\rfloor=\{k\alpha\}\; ;\;k=0,1,2,\ldots,n$$
These are $n+1$ numbers that fit into $n$ places, namely
$$\left[0,\frac 1 n\right),\ldots,\left[1-\frac{1}n,1\right)$$
By the Dirichlet's principle there must exist at least a pair of them which fall in the same interval of length $\dfrac 1n$.
STEP 2 We obtained $$\frac{k}{n} \leqslant \left\{ {{k_1}\alpha } \right\} < \left\{ {{k_2}\alpha } \right\} < \frac{{k + 1}}{n}$$
for some $k=0,\ldots,n$. We then know that $$0 < \left\{ {{k_2}\alpha } \right\} - \left\{ {{k_1}\alpha } \right\} \leqslant \frac{1}{n}$$
But note that $$\begin{align}
  \left\{ {{k_2}\alpha } \right\} - \left\{ {{k_1}\alpha } \right\} &= \left\{ {\left\{ {{k_2}\alpha } \right\} - \left\{ {{k_1}\alpha } \right\}} \right\} \cr 
   &= \left\{ {{k_2}\alpha  - {k_1}\alpha  - \left( {\left\lfloor {{k_2}\alpha } \right\rfloor  - \left\lfloor {{k_1}\alpha } \right\rfloor } \right)} \right\} \cr 
   &= \left\{ {{k_2}\alpha  - {k_1}\alpha  - {\text{integer}}} \right\} \cr 
   &= \left\{ {\left( {{k_2} - {k_1}} \right)\alpha } \right\} \end{align} $$

Now, you may as well try and prove the following:

Let $G$ be an additive subgroup of $\Bbb R$. Let $G^+$ denote the positive elements of $G$. Then 
$(1)$ If $\inf G^+=\alpha >0$, $G=\alpha\Bbb Z$
$(2)$ If $\inf G^+=0$, $G$ is dense in $\Bbb R$.

Hint
For $(1)$. Show that $\alpha\in G$. 
If not, pick $\epsilon =\alpha/2$ in the defintion of infimum, and  $g,g'\in G$ such that $\alpha  \leqslant g < g' <\alpha  + \frac{\alpha }{2}$. Look at $g'-g$.

 Then $g'-g\in G$ and $g'-g\leq \alpha/2<\alpha$ which is impossible. 

Now pick $g>0$. Look at $g-\alpha \left\lfloor {\dfrac{g}{\alpha }} \right\rfloor$. Use the definition of integer part to show it must be zero.

 Thus $g = \alpha \left\lfloor {\dfrac{g}{\alpha }} \right\rfloor  \in \alpha {\Bbb Z}$. Since opposites are in $G$ too, $(1)$ is proven.

For $(2)$, pick any $x\in \Bbb R$. We know we can find $y>0$ in $G$ with $0<y<\epsilon$. Let $n=\left\lfloor {\dfrac{x}{y}} \right\rfloor $. What can you deduce from $$n \leqslant \frac{x}{y} < n + 1\text{ ? }$$

 We have $ny\in G$ by additivity, and $$yn \leqslant x < yn + y \Rightarrow 0 \leqslant x - yn < y < \varepsilon $$

A: (a) Hint: If $\frac an\le k_1\alpha\bmod 1< k_2\alpha\bmod 1\le\frac{a+1}n$, then $(k_2-k_1)\alpha$ ...
(b) follows from (a)
A: Hint:
Check your definition of irrational. You can, by multiplying with $10^d \in \mathbb{N}$, generate an arbitrary small $[k\alpha]$.
