How can I show that the set of rational numbers with denominator a power of two form a dense subset of the reals?

Solution 1:

This follows immediately from the Archimedian property. To find a dyadic $\rm\ m/2^n \in (r,s)\ $ start at at any dyadic $\rm\ k < r\ $ (e.g. an integer) and keep taking dyadic step sizes smaller than the interval, say $\rm\ 1/2^j\ <\ s-r\:.\: $ By the Archimedean property eventually you'll land in the interval, necessarily at some dyadic rational (being a sum of such). This is a special case of the proof I explained here.

Solution 2:

Just expand the reals in base 2. With this convention, the rationals of the form $m/2^r$ will be identified with those reals whose expansion's digit is definitely $0$, i.e. those which have a "finite" expansion.

It is evident that you can get a finitely expanded real as close as you want to any fixed real $r$, just cut away all the digits after the $n$-th, whatever $n$ you choose (exactly as you do with decimal expansions!)

Solution 3:

HINT: Given $x \in \mathbb{R}$, how can you use $\frac{\left \lfloor 2^kx \right \rfloor}{2^k}$?

Here $\lfloor \cdot \rfloor$ is the floor function, which in this case sends $2^kx$ to the nearest integer less than or equal to $2^kx$.

The rational numbers whose denominator is a power of two are called dyadic rationals.