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 $x = y + 2^{-n}$.$\square$