Let $p\in\mathbb{R}\setminus\mathbb{Q}.$ Show that $A=\{np-m:m\in\mathbb{Z},n\in\mathbb{N}\}$ is dense in $\mathbb{R}.$
Note that I am taking $0\notin\mathbb{N}$ by convention.
I do have an "almost complete proof" for this. I want to know how to complete my "proof", and I also want to know if there is a cleaner way to prove this statement. This is my proof:
Set $A^*=\{|np-m|:m\in\mathbb{Z},n\in\mathbb{N}\}.$ As $p$ is irrational, we know that $0\notin A^*.$ Clearly, $A^*$ is non-empty and bounded below by $0.$ So, it's infimum exists in $\mathbb{R}.$ We will first show $\inf(A^*)=0.$ Let $\varepsilon>0.$ We need to show that there exist $m\in\mathbb{Z},n\in\mathbb{N}$ such that $|np-m|<\varepsilon.$ The convergents of the continued simple fraction of $p$ produce infinitely many rationals $\frac{m}{n}$ with $|p-\frac{m}{n}|<\frac{1}{n^2},$ with $m\in\mathbb{Z}$ and $n\in\mathbb{N},$ and $\gcd(m,n)=1.$ Pick one such rational with $n>\frac{1}{\varepsilon}.$ Such a rational exists as there are only finitely many $n\in\mathbb{N}$ with $n≤\frac{1}{\varepsilon},$ and for each of these values, only finitely many values of $m$ work as we have $-\frac{1}{n}-np<-m<\frac{1}{n}-np.$ So, for this rational, $|p-\frac{m}{n}|<\frac{1}{n^2}.$ This gives $|np-m|<\frac{1}{n}<\varepsilon.$ This proves that $\inf(A^*)=0.$
Note that if $x\in A$ and $t\in\mathbb{N},$ then, $tx\in A.$
Now, let $a,b\in\mathbb{R}$ with $a<b.$ Let $\varepsilon=b-a.$
Suppose $(a,b)$ has non-empty intersection with $(0,\infty)$ as well as $(-\infty,0).$ Then, there exists some $\delta>0$ such that $(-\delta,\delta)$ is contained in $(a,b).$ We can now pick some $x\in A$ with $x\in(-\delta,\delta).$ This gives $x\in(a,b).$ Suppose $(a,b)$ has non-empty intersection with only one of $(0,\infty)$ and $(-\infty,0).$ Suppose $(a,b)$ has non-empty intersection with only $(0,\infty).$ Pick some $x\in A$ with $x\in\left(-\frac{\varepsilon}{2},\frac{\varepsilon}{2}\right).$ If $x>0,$ then for some $t\in\mathbb{N}$ we get that $tx\in(a,b),$ as each time we are going up in increments of $\frac{\varepsilon}{2}<\varepsilon.$ What to do if $x<0?$
