mjhjmtu

A number is said to be dyadic if it has the form $frack2^n$ for some integers $k$
and $n$ in $mathbb Z^+ $. Show that the set of all dyadic numbers is countable.

Here is a proof I have (but couldn't complete),

Proof:
Let $ M_k := frack2^n: n in mathbb Z^+ $ for each $k in mathbb Z^+$

To show that $M_k$ is countable for any $k$, I tried to construct a one-to-one mapping from $M_k to mathbb N$. But I couldn't be sure if such a mapping is correct: $frack2^m mapsto m-1$ (or while mapping is the left hand side always a single variable such as x so I have to do mapping like $x mapsto ...$?). After mapping, I will show that the function(mapping) is one-to-one. Since $M_k$ is countable and the countable union of countable sets is countable, the set of dyadic numbers is countable.

The set $mathbbZ^+timesmathbbZ^+$ is countable and, by definition,$$beginarraycccmathbbZ^+timesmathbbZ^+&longrightarrow&textdyadic numbers\(k,n)&mapsto&dfrac k2^nendarray$$is surjective. Therefore, the set of dyadic numbers is either finite or countable. But since it is infinite, it is countable.

Your mapping is just fine.

Notice, however, that $0 not in mathbbZ^+$ and $0 not in mathbbN$ and thus you do not need to have $frack2^mmapsto m-1$.

Also, it is totally fine to write the mapping like you did, with

$$frack2^m mapsto m$$

If it confuses you, you can also instead define a one-to-one mapping from $mathbbN to M_k$ by

$$m mapsto frack2^m$$