Prove that $A = \left\{ \frac{m}{p^n} \right\}$ is dense in $\mathbb{R}$ 
Prove that the set $A$, defined as follows, is dense in $\mathbb{R}$.
$$A = \left\{ \frac{m}{p^n}~|~m \in \mathbb{Z}, n \in \mathbb{N}, p \in \mathbb{N} ~~\text{with} ~~p>1 \right\}.     $$

Remark: $A$ is dense in $\mathbb{R}$ if any open interval $(a, b)$ has an element of $A$.
Observation: In the textbook where this problem is from, the convention is that $0 \notin \mathbb{N}$.
Attempt:
Let $b - a = \varepsilon$, with $\varepsilon > 0$. We have to demonstrate the existence of some $\frac{m}{p^n}$ such that $\frac{m}{p^n} < \varepsilon ~~\forall~~ \varepsilon$. So using this inequality, we have
$$ \frac{m}{\varepsilon} < p^n \tag{1} $$
$$ \left( \frac{m}{\varepsilon} \right)^{1/n} < p \tag {2} $$
Now take any $\varepsilon$ arbitrarily small. By setting $p > \left( \frac{m}{\varepsilon} \right)^{1/n}$, we have that $\frac{m}{p^n} < \varepsilon $, and therefore $\frac{m}{p^n} \in (a, b). $
Thoughts:
I feel that what I'm doing might be wrong because I just solved for one variable, $p$ in this case, then back-substituted, but $m$ and $n$ are also variables.
 A: Fix any $p \ge 2$. Then if the set $A_p=\{\frac{m}{p^n}; n \in \mathbb{N}; m \in \mathbb{Z}\}$ is dense, then so is $A$, as $A_p \subset A$. So now we show that $A_p$ is dense. To this end, it suffices to show the existence of some $m \in \mathbb{Z}$ and $n \in \mathbb{N}$ such that $a < \frac{m}{p^n} < b$. [So this is where I think your proof is convoluted OP, why does $\frac{m}{p^n} < b-a$ imply an element in $A$ i.e., of the form $\frac{m'}{p^{n'}}$ between $a$ and $b$?]
There is indeed the existence of some $m \in \mathbb{Z}$ and $n \in \mathbb{N}$ such that $a < \frac{m}{p^n} < b$. Indeed, let $n \in \mathbb{N}$ be such that $2p^{-n} < (b-a)$. [There exists such an $n$ as $p^n$ goes to infinity as $n$ does, it follows that $p^{-n}$ goes to 0 as $n$ goes to infinity. So for any $b$ and $a$ satisfying $b-a > 0$ there indeed exists an $n$ s.t. $2p^{-n} < (b-a)$.] Next, write $a=A/p^n$ and $b=B/p^n$ for some real $A$ and $B$, where $n$ is picked so that $2p^{-n} < (b-a)$. Then $(b-a)p^n = B-A > 2$. So let $m$ be an integer satisfying $A < m <B$; there is indeed such an integer $m$ because $B-A > 2$, then $\frac{m}{p^n} \in (a,b)$.
A: Just because $\frac{m}{p^n}$ is less than $\varepsilon$, that does not mean that $\frac{m}{p^n}$ is inside $(a,b)$. That only tells you that $\frac{m}{p^n}$ is inside $(0,\varepsilon)$ (assuming $m$ positive).
You should choose $p$ and $n$ such that $\frac{1}{p^n}<\varepsilon$. There are many solutions to this part, since any $n$ will give you a $p$ in a way similar to your attempt.
Then you argue that some integer multiple of $\frac{1}{p^n}$ lands inside $(a,b)$. There are many ways to do this, but one is to note that the collection of all $\left(\frac{m-1}{p^n},\frac{m}{p^n}\right]$ cover $\mathbb{R}$. (For clarity, I mean for fixed $p$ and $n$, and letting $m$ vary through $\mathbb{Z}$.) So one interval of the form $\left(\frac{m-1}{p^n},\frac{m}{p^n}\right]$ intersects with $(a,b)$. Since $(a,b)$ is longer, actually at least two adjacent $\left(\frac{m-1}{p^n},\frac{m}{p^n}\right]$ intersect with $(a,b)$, and therefore for one of those two values of $m$, $\frac{m}{p^n}$ is inside $(a,b)$. 
A: I think it would be more interesting to fix $p$, because as it is, the question is almost trivial since $A$ contains $\mathbb Q$.
Indeed let have $\dfrac uv\in\mathbb Q$ then 


*

*if $v=1$ take $m=u,\ n=0,\ p=2$ 

*if $v>1$ take $m=u,\ n=1,\ p=v$
On the other hand if you fix $p$ then you can show that $A$ is an additive subgroup of $\mathbb R$.
Let assume $n_1\le n_2\quad$ then $\quad\dfrac{m_1}{p^{n_1}}+\dfrac{m_2}{p^{n_2}}=\dfrac{m_1p^{n_2-n_1}+m_2}{p^{n_2}}=\dfrac{m_3}{p^{n_2}}$.
Now you can use that additive subgroups of $\mathbb R$ are either $\alpha\mathbb Z$ or dense : Subgroup of $\mathbb{R}$ either dense or has a least positive element?
Now since $\dfrac 1{p^n}\in A$ and $\lim\limits_{n\to\infty}\frac 1{p^n}=0$ there is no least positive element, so $A$ is dense.
