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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 $n \in \mathbb{N}$

Martin Sleziak
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Guest_000
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  • answer is given for $\alpha z$ where $z$ is an integer case. – Guest_000 Aug 19 '14 at 16:35
  • It follows from the fact that the points of an equidistributed sequence of an arbitrary interval [a,b] form a dense subset of [a,b]. – TheOscillator Aug 19 '14 at 16:54
  • I do not know whether they form an "equidistributed sequence". – Guest_000 Aug 19 '14 at 16:59
  • http://math.stackexchange.com/q/843763 – Jonas Meyer Aug 19 '14 at 17:07
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    Ah, right, I misread Pedro Tamaroff's answer, sorry about that. (You still can get the desired result from that with relatively little work, but we should find one that does it for $n\in\mathbb{N}$.) – Daniel Fischer Aug 19 '14 at 17:10
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    You wrote: that this is not a duplicate of the mentioned question but you did not mention any question. You also wrote that it is not a duplicate as it does not carry a correct. I don't think that the answers have anything to do with whether it is the same questions or not. And, most importantly, using all caps is very impolite. – Martin Sleziak Aug 21 '14 at 07:33

5 Answers5

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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 $[0,1]$.

Martin Sleziak
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  • $;2k;$ intervals? I think you meant $;k;$ intervals? –  Dec 21 '14 at 13:45
  • I have two doubts: where was used that $;\alpha;$ is irrational? And why the multiples $;k(b-a)\alpha;$ have to be in all the intervals? Thanks you –  Dec 21 '14 at 14:16
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    @AntoineNemesioParras I have edited my posts and pointed out explicitly where the assumption that $\alpha$ is irrational is used. About your second question: Since ${(b-a)\alpha}$ is very small (smaller than $1/k$, we cannot skip the whole interval by moving to the next multiple. – Martin Sleziak Dec 21 '14 at 15:11
  • Hi Martin! I was wondering can you show the proof why if you take all the multiples $n(b-a)\alpha$, then in each of the $k$ intervals must be at least one of the values ${n(b-a)\alpha}$.? – RFZ Jul 25 '21 at 17:18
  • @ZFR Some kind of explanation is in the previous comment and also in the answer. I have posted something about this in chat - we can continue there if needed. – Martin Sleziak Jul 27 '21 at 13:36
  • You wrote that ${(b-a)\alpha}\in [0,1/k]$ or ${(b-a)\alpha}\in [1-1/k,1]$. But it depends on whether ${a\alpha}<{b\alpha}$ or ${a\alpha}>{b\alpha}$, respectively. Right? – RFZ Jul 29 '21 at 19:26
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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 $k$.

Andrew Dudzik
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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 $(0 , \frac{1}{k})$.

N. S.
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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.

user62223
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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 $i$, $1-i\epsilon \in (0,1/N)$.

ronno
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  • Why is enough that $;S;$ has an element in $;(0, 1/N);$ ? I don't understand this. Thank you –  Dec 21 '14 at 14:17
  • @AntoineNemesioParras Because then any interval of length $\ge 1/N$ (included in $[0,1]$) will contain some multiple of that element. Now note that any open subset of $[0,1]$ contains some such interval for large enough $N$. – ronno Dec 21 '14 at 14:21
  • Thank you so much for your kind answer, but I still am struggling to understand: I already have that $;,{m\alpha}\in (0,1/N);$ for some positive integer $;m;$ , but how from this it follows that for any $;x\in [0,1];$ there is some multiple of $;\alpha;$ s.t. it is close (say, upt to $;1/N;$ from $;x;$ ? I just can't see it, sorry. Thanks. –  Dec 22 '14 at 03:09