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.
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 $\{t_n+\rho\}$.