3.1 Dyadic Rational Numbers
Definition 3.1.1 (Dyadic Rational Number). Let $x \in \rational$, then $x$ is a dyadic rational number if there exists $n \in \natp$ and $k \in \nat$ such that $x = k/2^{-n}$.
For each $n \in \natp$, denote $\mathbb{D}_{n} = \bracs{k/2^{-n}|k \in \integer}$. The set $\mathbb{D}= \bigcup_{n \in \natp}\mathbb{D}_{n}$ is the collection of all dyadic rational numbers.
Definition 3.1.2 (Rank). Let $q \in \mathbb{D}$, then
is the dyadic rank of $q$.
Lemma 3.1.3. Let $x \in \mathbb{D}$ with $n = \text{rk}(x) > 1$, then there exists a unique $y \in \mathbb{D}_{n - 1}$ such that $x = y + 2^{-n}$.
Proof. Since $x \in \mathbb{D}_{n}$, there exists a unique $k \in \integer$ such that $x = k/2^{-n}$. Given that $x \not\in \mathbb{D}_{n-1}$, $k$ is odd. Therefore $y = (k - 1)/2^{-n}$ is the unique element of $\mathbb{D}_{n - 1}$ with $x = y + 2^{-n}$.$\square$
Proposition 3.1.4. Let $x \in \mathbb{D}\cap [0, 1)$, then there exists a unique $M(x) \subset \natp \cap [1, \text{rk}(x)]$ such that $x = \sum_{n \in M(x)}2^{-n}$.
Proof. First suppose that $\text{rk}(x) = 1$. In which case, $x \in \bracs{0, 1/2}$, and either $M(x) = \emptyset$ or $M(y) = \bracs{1}$.
Assume inductively that the proposition holds for all dyadic rationals of rank $n$. Let $x \in \mathbb{D}_{n+1}\setminus \mathbb{D}_{n}$. By Lemma 3.1.3, there exists a unique $y \in \mathbb{D}_{n}$ such that $x = y + 2^{-n-1}$. Since $x > 0$, $x \ge 1/2^{-n-1}$, so $y \in [0, 1)$. By the inductive assumption, there exists a unique $M(y) \subset \natp \cap [1, n]$ such that $y = \sum_{k \in M(y)}2^{-k}$. In which case, $M(x) = M(y) \cup \bracs{n}$ is the desired set.$\square$
Proposition 3.1.5. Let $G$ be a commutative ordered semigroup, and $\seq{g_n}\subset G$ such that for each $n \in \natp$, $g_{n+1}+ g_{n+1}\le g_{n}$. For each $x \in \mathbb{D}\cap [0, 1)$, let $\phi(x) = \sum_{n \in M(x)}g_{n}$, then
For any $x, y \in \mathbb{D}\cap [0, 1)$ such that $x + y \in [0, 1)$, $\phi(x) + \phi(y) \le \phi(x + y)$.
For any $x, y \in \mathbb{D}\cap [0, 1)$ with $x \le y$, $\phi(x) \le \phi(y)$.
Proof. Assume without loss of generality that $x, y \in (0, 1)$. First suppose that $\text{rk}(x) = 2$ and $\text{rk}(y) \le 2$. In which case, since $G$ is commutative, it is sufficient to consider two cases: $(x, y) = (1/4, 1/4)$ and $(x, y) \in (1/2, 1/4)$. In the first case,
In the second, $\phi(x) + \phi(y) = \phi(x + y)$.
Now assume inductively that the proposition holds for all $x, y \in \mathbb{D}\cap (0, 1)$ with $\text{rk}(x) = n$ and $\text{rk}(y) \le n$. Let $x \in \mathbb{D}_{n+1}\setminus \mathbb{D}_{n}$ and $y \in \mathbb{D}_{n+1}$. By Lemma 3.1.3, there exists $x_{0} \in \mathbb{D}_{n}$ such that $x = x_{0} + 2^{-n-1}$. If $y \in \mathbb{D}_{n}$, then
by the inductive assumption. Otherwise, there exists $y_{0} \in \mathbb{D}_{n}$ such that $y = y_{0} + 2^{-n-1}$, so
by the inductive assumption.$\square$