Are terminating decimals dense in the reals?

Yes, they are.

First, think informally (and you do this, with your sentence beginning "since"). A set $X$ is dense iff every real number $r$ can be "well-approximated" 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.

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OK, now we need to make this rigorous. Fix reals $a<b$.

• First, show that there is some $n$ such that $10^{-n}<b-a$. (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 "$n$-place decimal" which is $\le a$.

  • It may be easiest to first assume $a>0$ here.

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

Note that this works for any base, not just $10$.