dense in $\mathbb{R}$ and dense in $$ modulus $1$

I want to solve this problem given here.
The problem is:

(2) The exercise is useful to help us show the following lemma. $\{ar +b: a,b \in \mathbb Z\}$ where $r \in \mathbb Q^c$ is dense in $\mathbb R$. Its equivalent to: $\{ar : a \in \mathbb Z\}$, where $r \in \mathbb Q^c$ is dense in $[0,1]$ modulus $1$.

But, I don’t know the meaning of dense in $\mathbb{R}$ and dense in $[0,1]$ modulus $1$ as well as I don’t know what does the “modulus” mean here.

Please help.
Thank you.

Solutions Collecting From Web of "dense in $\mathbb{R}$ and dense in $$ modulus $1$"

A set $A\subseteq\Bbb R$ is dense in $\Bbb R$ if $(a,b)\cap A\ne\varnothing$ whenever $a,b\in\Bbb R$ and $a<b$; in other words, $A$ is dense in $\Bbb R$ if every non-empty open interval in $\Bbb R$ contains some member of $A$. An example of a dense subset of $\Bbb R$ is $\Bbb Q$, the set of rational numbers.

A set $D\subseteq[0,1]$ is dense in $[0,1]$ if $D\cap(a,b)\ne\varnothing$ whenever $(a,b)$ is an open interval in $\Bbb R$ such that $(a,b)\cap[0,1]\ne\varnothing$.

Finally, $A\subseteq\Bbb R$ is dense in $[0,1]$ modulus $1$ if $\{x-\lfloor x\rfloor:x\in A\}$ is dense in $[0,1]$, where $\lfloor x\rfloor$ is the greatest integer in $x$, i.e., the largest integer $n$ such that $n\le x$; $x-\lfloor x\rfloor$ is the fractional part of $x$.

A set $S$ is dense in $\mathbb{R}$ if, for any $r \in \mathbb{R}$ and any $\delta > 0$, there is a point $s\in S$ such that
$$
r – \delta < s < r+\delta
$$

“Modulus 1” means that you take the fractional part of a number. ie.
$$
s\,\text{mod}(1) = s – [s]
$$
where $[s]$ denotes the integer part of $s$.

In this case, you want to show that, for any $x\in [0,1]$ and any $\delta > 0$ there is a number $s \in S$ such that
$$
x-\delta < s-[s] < x+\delta
$$