For all $e\ge 0,z>0,e,z\in\Bbb Q$, there exists $m\in \Bbb Z,n\in\Bbb N$ such that $e<\dfrac{m}{2^n}<e+z$.
Does my proof look fine or contain logical gaps and flaws? I'm pleased to receive your suggestion!
My attempt:
It suffices to prove that for all $e\ge 0$ and for all $1>z>0$, there exist $m\in \Bbb Z,n\in\Bbb N$ such that $$e<\dfrac{m}{2^n}<e+z$$
We have three cases in total.
- $e \in \Bbb Z$
Let $n'=\min\{n\in\Bbb N\mid\dfrac{1}{2^{n}}<z\}$. Then $e+\dfrac{1}{2^{n'}}$ has a form of $\dfrac{m}{2^n}$ and $e<e+\dfrac{1}{2^{n'}}<e+z$.
- $e<a<e+z$ for some $a\in Z$
Let $n'=\min\{n\in\Bbb N\mid\dfrac{1}{2^{n}}<z\}$. Then $a+\dfrac{1}{2^{n'}}$ has a form of $\dfrac{m}{2^n}$ and $e<a+\dfrac{1}{2^{n'}}<e+z$.
- $a<e<e+z\le a+1$ for some $a\in Z$
Let $n'=\min\left\{\min\{n\in\Bbb N\mid \dfrac{1}{2^n}\le e-a\} ,\min\{ n\in\Bbb N\mid\dfrac{1}{2^n}<z\}\right\}$. Notice that $n'\ge 1$.
It follows that $\dfrac{1}{2^{n'}}\le e-a$ and $\dfrac{1}{2^{n'}}<z$ and $\dfrac{1}{2^{n'-1}}> e-a$ and $\dfrac{1}{2^{n'-1}}\ge z$.
Thus $e-a < \dfrac{1}{2^{n'-1}} = \dfrac{2}{2^{n'}}=\dfrac{1}{2^{n'}}+\dfrac{1}{2^{n'}}<(e-a)+z$.
Hence $e<\dfrac{2}{2^{n'}}+a<e+z$ where $\dfrac{2}{2^{n'}}+a$ has a form of $\dfrac{m}{2^n}$.
Find $n$ with $1 < z×2^n$.
Find $m$ with $e×2^n < m \le e×2^n + 1$.