TITLE: For a defined set $M$ (see problem) do there exist $a,b$ natural numbers so that $a,ab+1 \in M$
QUESTION [1 upvotes]: Let $\rho \in \mathbb{R}\setminus \mathbb{Q}$ be a irrational nuber, and
let $\varepsilon>0$ be arbitrarily small. Define $M=\{m \in \mathbb{N}: \exists k \in \mathbb{N}\hbox{ so that} |\rho m -k |<\varepsilon\}$. Do there exist $a,b \in \mathbb{N}$ so that $a,ab+1 \in M$?

REPLY [6 votes]: You probably want to allow $k\in\mathbb{Z}$, not just $k\in\mathbb{N}$, in the definition of $M$. Otherwise $M$ can be empty for $\rho<0$.
With this modification, the answer to your question is yes. In fact, $M$ is infinite and for any nonzero $a\in M$ there is $b\in\mathbb{N}$ such that $ab+1\in M$. To see this, we shall use the elementary fact that the multiples of any irrational number are dense modulo $1$. 
By the mentioned fact, $M$ is infinite. Let $a\in M$ be nonzero. Then $\tau=a\rho$ is irrational, so using the mentioned fact again, there exists $b\in\mathbb{N}$ such that $b\tau$ lies in $(-\rho-\epsilon,-\rho+\epsilon)$ modulo $1$. Then $(ab+1)\rho=b\tau+\rho$ lies in $(-\epsilon,\epsilon)$ modulo $1$, i.e. $ab+1\in M$.
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