Proving that a subset $\mathscr{H}$ is dense in $\mathbb{R}$ 
Consider the subset $\mathscr{H} =\displaystyle\left\{\frac{a}{10^{b}} \ |\: a,b\in\mathbb{Z}\: \right\}$. Then is $\mathscr{H}$ dense in $\mathbb{R}$. 

To prove that $\mathscr{H}$ is dense I have to show that given $x,y \in\mathbb{R}$  there is a $m,n\in\mathbb{Z}$ such that $x < \frac{m}{10^{n}} <y$. 
Thats it I am hopelessly stuck. 
 A: We write out an argument in fairly formal style.  We show there is an integer $m$, and a positive integer $n$, such that $x\lt \frac{m}{10^n}\lt y$. 
Suppose that $x\lt y$, and let $d=y-x$. There is a positive integer $n$ such that $\frac{1}{10^n}\lt d$.  
Consider the set $S$ of integers $k$ (positive, negative, or $0$) such that $\frac{k}{10^n}\lt y$. The set $S$ is bounded above, so there is a largest integer $m$ in $S$. 
We show that $x\lt \frac{m}{10^n}\lt y$. It is obvious from the choice of $m$ that $\frac{m}{10^n}\lt y$. We show that $x\lt \frac{m}{10^n}$.
Suppose to the contrary that $\frac{m}{10^n}\le x$. Then 
$$\frac{m+1}{10^n}\le  x+\frac{1}{10^n} \lt x+d\lt y,$$ 
contradicting the choice of $m$ as the largest integer such that $\frac{m}{10^n}\lt y$. 
A: Hint: For instance, to get 0.563, choose $a=5*10^2 + 6*10^1 + 3*10^0 = 563$ and $b=3$. Generalizing this method should give you all decimal numbers with a finite number of decimals.
A: From how I've learned to do density proofs, the easist way to go about this is by expanding the interval $(x,y)$ (note I will assume from here on - w.l.o.g. ofc - that $x<y$) and forcing some number in between (one of the form $\frac{m}{10^n}$ in this case).
So let's start by expanding the interval so that we get an interval with a length greater than 1 (it's pretty easy to fit a number into an interval of length greater than 1, of any form). So let's say we find $r \in \mathbb{R}$ so that
$$
ry - rx > 1
$$
Now from this we can see that
$$
\lfloor rx \rfloor + 1 > rx
$$
and also from that first inequality we get the following
$$
ry - 1 > rx \ge \lfloor rx \rfloor \implies ry > \lfloor rx \rfloor + 1
$$
so we have found some integer $q = \lfloor rx \rfloor$ s.t.
$$
rx < q < ry
$$
Our next step is to find an exact formula for $r$, however we must be specific - we will be dividing this above inequality by $r$ so we will want $r$ to be of the form $10^m : m \in \mathbb{Z}$. I first want to note that
$$
10^m > m \, \forall \, m \in \mathbb{Z}
$$
you can use induction on that if you insist (noting that for any negative m the inequality is trivially true). Next we go back to our original inquality
$$
ry - rx > 1 \iff r > \frac{1}{y-x}
$$
So we can easily pick $r = \left\lfloor \frac{1}{y-x} \right\rfloor$ however we require $R$ to be of the form $10^m$. We now take into account that $10^m > m$ so it must be true that $$
10^{\left\lfloor \frac{1}{y - x} \right\rfloor} > \left\lfloor \frac{1}{y - x} \right\rfloor > \frac{1}{y-x}
$$
So if we choose $r = 10^{\left\lfloor \frac{1}{y - x} \right\rfloor}$ then our inequality becomes as follows
$$
10^{\left\lfloor \frac{1}{y - x} \right\rfloor} y > \left\lfloor 10^{\left\lfloor \frac{1}{y - x} \right\rfloor} x \right\rfloor + 1 > 10^{\left\lfloor \frac{1}{y - x} \right\rfloor} x
$$
Now dividing through by $10^{\left\lfloor \frac{1}{y - x} \right\rfloor}$ we end up with the following
$$
y > \frac{\left\lfloor 10^{\left\lfloor \frac{1}{y - x} \right\rfloor} x \right\rfloor + 1}{10^{\left\lfloor \frac{1}{y - x} \right\rfloor}} > x
$$
and so we have found a number of the form $\frac{a}{10^b}$ between any two given real numbers so it follows that $\mathscr{H}$ is dense in $\mathbb{R}$.
A: Hint: What do elements of $\mathscr{H}$ look like in decimal notation?
