Are terminating decimals dense in the reals? I have read that rationals are dense in the reals, but I've been curious if the terminating decimals are as well. Since, in applied math, all effectively &quot;real&quot; numbers are approximated with terminating decimals, I would expect the terminating decimals to have this property, and if not I'm curious why. Here I'm defining the terminating decimals like this:
$$D = \{\frac{m}{10^n} \in \mathbb{Q} : m \in \mathbb{Z} \wedge n \in \mathbb{N}_0\}$$.
Yes, they are.
First, think informally (and you do this, with your sentence beginning &quot;since&quot;). A set $X$ is dense iff every real number $r$ can be &quot;well-approximated&quot; by an element of $X$. For terminating decimals, this approximation amounts to simply ... truncating at a certain point! E.g. $\pi$ is the limit of the sequence $$3.0000..., 3.1000..., 3.1400..., ...$$ of reals with terminating decimal expansions.

OK, now we need to make this rigorous. Fix reals $a&lt;b$.

First, show that there is some $n$ such that $10^{-n}&lt;b-a$. (This is actually the most substantial step, but at the same time may be the intuitively clearest one.)

Let $X=\{z\cdot 10^{-n}: z\in\mathbb{Z}\}$. Show that $X\cap (-\infty,a]$ has a greatest element - that is, there is a largest &quot;$n$-place decimal&quot; which is $\le a$.

It may be easiest to first assume $a&gt;0$ here.


Calling that number &quot;$\alpha$,&quot; what can you say about $\alpha+10^{-n}$?


Note that this works for any base, not just $10$.
Suppose $a$ and $b$ are terminating decimals. We will find a terminating decimal that lies between them, namely  $\frac{a+b}2$.  For example, between $1.7$ and $0.0001$ is $\frac{1.7+0.0001}2 = 0.850005$.
First, as you said, write $$a=\frac A{10^i}, b=\frac B{10^j}.$$
Without loss of generality, $i\le j$. Then write $$a=\frac{A\cdot 10^{j-i}}{10^j}.$$
Now $$\frac{a+b}2=   \frac{A\cdot10^{j-i} + B}{2\cdot 10^j} =
 \frac{5A\cdot 10^{j-i} + 5B}{10^{j+1}} $$
is a terminating decimal between $a$ and $b$.