$\{mb + n \mid b\in \mathbb{R} \setminus \mathbb{Q} \text{ and } m, n \in \mathbb{Z} \}$ is dense in $\mathbb{R}$ I'm fairly new to analysis so this might be very simple.
I have found other pages discussing similar questions, but I haven't found any of the answers particularly helpful for me.
I've already shown that for any $k \in \mathbb{N}$
$$0 < mb + n < \frac{1}{k}$$ 
for some choice of $m$, and $n$.
My approach was using the archimedian principle to choose $k$ so that $0 < \frac{1}{k} < b - x$, where $b$ is a real number, $a < x < b$, and $x$ is an integer so then I could combine these results and add $x$ to every term
which would give that $a < mb + (n+x) < b$ thus showing the set is dense in the real numbers, but I couldn't find a valid choice of $x$ so I assume this method would not work. 
Other pages suggest showing that the set of fraction parts of $\{mb+n\}$ is dense in $[0, 1]$, but I wasn't able to follow how they reached their conclusions.
I've seen this page does support LaTex, but I've never had to use it so sorry for the fairly poor formatting.
edit: I was hoping to get some help on showing this utilizing the fact that for any $k \in \mathbb{N}$
$$0 < mb + n < \frac{1}{k}$$ 
 A: To show that $A:=\{mb+n|b \in \Bbb Q^c, m,n \in \Bbb Z\}$ is dense in $\Bbb R$, it is sufficient to proove that any real number is a limit point of A. Let $x\notin \Bbb Q$ and $\epsilon >0$. By the Archimedean Property (A. P.) there exists $k\in\Bbb N :\frac {|x|}{k} <\epsilon$. Then $x-\epsilon <x< x+\frac {|x|}{k} < x+\epsilon$, but $x+\frac {|x|}{k} =x\pm \frac {x}{k}=(k\pm1)\frac xk \in \{mb+n|b \in \Bbb Q^c, m,n \in \Bbb Z\}$, so we found an element of $A$ in the neighborhood $(x-\epsilon, x+\epsilon)\setminus \{x\}$, which means that an arbitrary irrational number is limit point of$A$. If $0\neq x\in \Bbb Q$, so there exist $μ\in \Bbb Z,ν\in \Bbb N$ such that $x=\frac μν$, by the A. P. there exists $k\in\Bbb N :\frac {|x|\sqrt 2}{k} <\epsilon$. Then $x-\epsilon <x< x+\frac {|x|\sqrt 2}{k} < x+\epsilon$, but $x+\frac {|x|\sqrt 2}{k} =x\pm \frac {x\sqrt 2}{k}=\frac {μ}{k}(\frac kν\pm \frac {\sqrt 2}{ν}) \in \{mb+n|b \in \Bbb Q^c, m,n \in \Bbb Z\}$, so we found an element of $A$ in the neighborhood $(x-\epsilon, x+\epsilon)\setminus \{x\}$, which means that an arbitrary rational number is limit point of$A$. Finally, $0$ is a limit point of $A$, because by A. P. there exists a $k \in \Bbb N: \frac {\sqrt 2}{k}<\epsilon$, so $\frac {\sqrt 2}{k}\in A\cap [(-\epsilon,\epsilon)\setminus \{0\}]$. Consequently every real number is a limit point of $A$, so $A$ is dense in $\Bbb R$.
