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Assume that $a$ and $b$ are positive integers, and consider the arithmetic progression $\{{s_n}=(an+b):n=1,2,3,\cdots\}$. Let $\alpha$ be an irrational number, and consider the fractional part ${r_n} = \{{s_n}\alpha \}$ defined by $${r_n} = \{{s_n}\alpha \} = {s_n}{\alpha}-[{s_n}{\alpha}].$$ Is the sequence $\{{r_n}:n=1,2,3,\cdots\}$ dense in $(0,1)$? I tried giving a proof using the the pigeonhole principle dividing the unit interval into segments of width $1/ka$ but get stuck at the crucial point when two such distinct fractional parts inhabit one such segment! Is this result always true? How can I modify the standard proof for the case $(a,b)=(1,0)$?

I attach my attempted proof at the request of Mr. Lin (see below). First I give my attempt at a standard proof for context.

Consider for a positive integer $k$ the sequence of of mutually disjoint segments $$\{ [{\frac {j}{k}},{\frac {j+1}{k}}):0\leq j \leq (k-1)\}.$$ Then for $n>k$, the sequence of fractional parts $$\{ \{j{\alpha}\}:1\leq j\leq n \}$$ has more elements than the $k$ segments. So by the pigeon-hole principle, at least one segment should contain two distinct fractional parts-say $\{i{\alpha}\},\{j{\alpha}\}$ for $i<j$. Then the fractional part $\{(j-i){\alpha}\}$ satisfies $$\{(j-i){\alpha}\} < {\frac {1}{k}}.$$ (To see that this is so, use the definition of the fractional part to check that $\{(j-i){\alpha}\} \leq \{j{\alpha}\}-\{i{\alpha}\}.$)

This proves that to every positive integer $k$, there corresponds a large enough positive integer $n_k$ such that $$\{{n_k}{\alpha}\} < {\frac {1}{k}}.$$ Now assume that $x\in (0,1)$ is chosen arbitrarily, and consider an arbitrarily small open segement $(x-{\epsilon},x+{\epsilon}).$We can choose $k$ large enough and $n_k$ so that $$\{j{n_k}{\alpha} \}\in[{\frac {j}{k}},{\frac {j+1}{k}})\subset (x-{\epsilon},x+{\epsilon}).$$ This proves that the sequence of fractional parts of positive integeral multiples of the irrational number $\alpha$ must be dense in $(0,1)$.

In generalizing this proof to treat arithmetic progressions, my problem was this. In this proof I can simply multiply $\{{n_k}{\alpha} \}$ by the positive integer $j$ to get to the desired fractional part $\{j{n_k}{\alpha} \}$. How do I achieve this if I am only allowed to multiply fractional parts by terms $an+b$ drawn from an arithmetic progression? This is where I was stuck.

student
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    Can you state your PP proof? It should be pretty direct from there, so please clarify how you got stuck. (IE You should run into the same issues as showing that $ { n \alpha }$ is dense in $(0,1)$. So if they were resolved previously, they should be fine now. ) – Calvin Lin Feb 07 '22 at 18:31
  • $\alpha$ is replaced by $a({\alpha})$ which remains irrational and plays no role in the proof, and $b{\alpha}$ is a constant (it is not being multiplied by positive integers $n$) so the density of the sequence $({\rm mod})1$ should be "translation invariant". The standard argument proves that the sequence ${na{\alpha}}({\rm mod}1)$ is dense in $(0,1)$ for irrational $\alpha$ and a positive integer $a$. The sequence $(a{\alpha}n+b{\alpha})({\rm mod}1)$ is dense in $(0,1)$ if and only if the sequence $(a{\alpha}n)({\rm mod}1)$ is dense in $(0,1)$ – student Feb 07 '22 at 21:24
  • I was not thinking clearly! The (hinting) answer below helped. This problem came up as I need this fact in another proof! Do you want me to reproduce my attempted proof? – student Feb 07 '22 at 21:29
  • To me, it would be helpful for you to reproduce your attempted proof. – Calvin Lin Feb 08 '22 at 15:46
  • Sure! I think it would be best to write the proof for the standard result, and alongside this indicate my attempted modification of this proof and where I got I got stuck! I will try to do this by editing the post above. – student Feb 08 '22 at 18:55
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    Ah, I wasn't expecting you to "prove for $n\alpha$ then hope that you can find a subsequence of $ (an+b)\alpha$ to show density". In that instance, you're right that it can be hard to find a $(an+b)\alpha$ that must lie in that interval, since we have very little control over what we're multiplying by. $\quad$ Can you work with $ (an+b)\alpha$ directly? Once we get to $ 0 < an_k \alpha < 1/k$, can you show that the sequence will eventually fall in each $ [ j/k, (j+1) / k ] $ interval? – Calvin Lin Feb 09 '22 at 00:08
  • This is what Mr. Stadnicki sugested. Replace $\alpha$ by $a{\alpha}$ but then we still have $b{\alpha}$ to worry about. The indirect approach seems to be to apply the above argument for the irrational number $a{\alpha},$ and then argue that translating the sequence ${na{\alpha}}$ by the constant $b{\alpha}$ (i.e. adding $b{\alpha}$ to each term) and then considering fractional parts does not affect the density of the fractional parts of the resulting sequence in $(0,1)$. This is the hard part! Do you have direct approach? – student Feb 09 '22 at 03:35
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    Do you agree with "Suppose we were walking with steps of size $< 1/k$ along the number line, then we will eventually fall into a size $\geq 1/k$ hole?" That's all that's involved here. $\quad$ For Steven's solution, translating a dense sequence by a fixed constant (and taking mod 1) still results in a dense sequence. That should be quite obvious, by looking at the corresponding intervals like they suggested. Can you elaborate on why that's the hard part? – Calvin Lin Feb 09 '22 at 15:43
  • I agree with your statement Calvin. I worry (probably a blind spot) that if $b{\alpha}$ is irrational, translating each term of the sequence by an irrational number can mess things up. Am still working on convincing myself that as we translating the sequence by an irrational constant it shouldn’t matter. This is my difficulty. I am sure I am just not seeing things clearly! – student Feb 10 '22 at 20:14
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    Can you write it up, at least for the rational constant first since you seem to believe that it works? Then you can see if the rational condition is used at all. If it isn't, then the proof would hold for irrational constants too. if it is, maybe there is a way to fix it. – Calvin Lin Feb 10 '22 at 20:33
  • Useful suggestion! Clearly integer translates are irrelevant in this context! – student Feb 11 '22 at 05:36

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Converting my comment into a proper (hinting) answer: Start by considering $\beta=a\alpha$ rather than $\alpha$. It's irrational iff $\alpha$ is, and since we can write $r_n=\{(an+b)\alpha\}$ $=\{(a\alpha)n+b\alpha\}$ $=\{\beta n+b\alpha\}$, we immediately get rid of one of our degrees of freedom. To get rid of the other, consider an arbitrary sequence $\langle t_n:n\in\mathbb{N}\rangle$ and arbitrary $\rho\in\mathbb{R}$ and see if you can relate the density of $\{t_n\}$ in $(0,1)$ to that of $\{t_n+\rho\}$.

Steven Stadnicki
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