For an irrational number $a$ the fractional part of $na$ for $n\in\mathbb N$ is dense in $[0,1]$ How to prove that the $\{$ fractional part of $n\alpha\mid n \in \mathbb{N}$ $\}$ is dense in $[0,1]$ for an irrational number $\alpha$. 
NOTICE that $n$ is in $\mathbb{N}$
Also notice that this is not a duplicate of the mentioned question as it does not carry a correct answer and the partially correct answer in the said question is given for integer multiple case, NOT for $n \in \mathbb{N}$ 
 A: Your main issue is to get around the positivity of $n$.
So lets change the standard proof. By the pigeon hole principle you can find some $m \in \mathbb Z$ so that $mx = k+ y$ with $k \in \mathbb Z$ and $y \in (0 , \frac{1}{k})$.
If $m >0$, you are happy.
If $m <0$, then show that there exists some $l \in \mathbb N$ so that $ly \in (\frac{k-1}{k}, 1)$. Show that the fractional part of $-lmx$ is in $(0 , \frac{1}{k})$.
A: Let's start with along the lines of the standard proof.
Let us divide $[0,1]$ into $k$ intervals of length $1/k$; i.e. $[0,1/k]$, $[1/k,2/k]$, $[2/k,3/k]$, etc.
Now by Dirichlet principle there are two numbers $a\ne b$ such that  $\{a\alpha\}$, $\{b\alpha\}$ which are in the same interval.
If $b>a$, then $(b-a)$ is a positive integer and either  $\{(b-a)\alpha\}\in [0,1/k]$ or $\{(b-a)\alpha\}\in[1-1/k,1]$.
Since $\alpha$ is irrational, $\{(b-a)\alpha\}$ is non-zero. (The number $(b-a)\alpha$ cannot be an integer.)
Now if we take all multiples $n(b-a)\alpha$, $n\in\mathbb N$, then in each of the $k$ intervals must be at least one of the values $\{n(b-a)\alpha\}$. (We go either upwards from $[0,1/k]$, or downwards from the last interval, but we can never skip an interval.)
This implies that the set of all multiples is dense in $[0,1]$.
A: Pick any $k\in\mathbb{N}$.  By the pigeonhole principle, there are two multiples of $\alpha$ whose fractional part lie within $1/k$ of each other.  Taking the difference, there is a multiple of $\alpha$ with (positive) fractional part $<1/k$.
It follows that every $x\in [0,1]$ is within $1/k$ of some $\{n\alpha\}$, for any $k$.
A: We gonna prove that the set A = {{nα}, n is natural} is dense in [0,1] by contradiction.
The fact that A is dense in [0,1] is equivalent with the following statement:
for any ϵ > 0 there is N(ϵ) such that {N(ϵ)α} < ϵ. 
We gonna assume the contrary i.e. there is ϵ > 0 such that for any natural number n we have {nα} > ϵ. Now, choose any natural number N sufficient large such that 1/N < ϵ. We consider now the numbers {α}, {2α}, ..., {Nα}, {(N+1)α}.
We have N+1 distinct numbers in interval [0,1), so there are two of them, say
{pα} and {qα} such that 0 < | {pα} - {qα} | < 1/N. Assume p < q. We have
two cases here: {pα} < {qα} or {pα} > {qα}.
Case 1. p < q and {pα} < {qα}.  Then {(q-p)α} < 1/N < ϵ which contradicts our assumption.
Case 2. p < q and {pα} > {qα}.  We have 1 > {(q-p)α} > 1 - 1/N. Let q-p = R. Since p, q are from {1,2, ..., N+1} and p < q we have that R = q-p is from the set {1,2, ..., N}. Our number N was chosen arbitrary such that 1/N < ϵ, so for any natural number N such that 1/N < ϵ we can find a natural number R = R(N) < N + 1 such that {Rα} > 1 - 1/N. 
Now fix again a natural number N such that 1/N < ϵ. Consider the numbers
1 - {α}, 1 - {2α}, ..., 1 - {Nα}. Let E be the minimum of this numbers. We have E > 0 and 1/2 E < 1 - {kα} for any k = 1..N. We can choose a natural number Q sufficient large that 1/Q < 1/2 E and Q > N. For this natural number Q we can choose R(Q) from the set {1, 2, ..., Q} such that {R(Q)α} > 1-1/Q. We have {R(Q)α} > 1 - 1/Q > 1 - 1/2 E > {kα} for any k=1..N. First notice that R(Q) is not in the set {1,2,.., N} but R(Q) is in the set {1,2, .., Q}. So Q+1 > R(Q) > R(N) and {R(Q)α} > 1 - 1/Q > {R(N)α} > 1 - 1/N. Now consider the number S = R(Q) - R(N). Obvious S is a positive natural number and {Sα} = {R(Q)α} - {R(N)α} < 1/N. This is again a contradiction with our initial assumption.
qed.
A: Let $S = \{\{n \alpha\} | n \in \mathbb{N} \}$. First note that it is enough to show that $S$ intersects $(0,1/N)$ for arbitrarily large $N \in \mathbb{N}$. By the usual pigeonhole argument, for some $k \in \mathbb{N}$, $\{k\alpha\}$ is in either $(0,1/2N)$ or $(1-1/2N,1)$. Now, in the first case we are done. In the second case, let $\{k\alpha\} = 1-\epsilon$. Then since $\{x+y\} = \{\{x\}+\{y\}\}$, for an integer $i \ge 1 $ such that $i \epsilon <1$, $\{ik\alpha\} = \{i\{k \alpha\}\} = 1- i\epsilon$. Now for a suitable choice of $i$, $1-i\epsilon \in (0,1/N)$.
