🧵 is this a valid proof?

Anonymous at Mon, 15 Jul 2024 10:03:56 UTC No. 16282201

let [math]x \in \mathbb R_+, \displaystyle D_n(x) = \sum^n_{k=1} \mathtt d_k(x) 10^{-k} \in \mathbb Q[/math] where [math]d_k(x)[/math] is the [math]k[/math]-th decimal of [math]x[/math]

for all [math]\epsilon > 0[/math], let [math]p= \lfloor -\log10(\epsilon) \rfloor[/math] (so that [math]10^{-p} \ge \epsilon[/math])
then for all [math]\displaystyle n \ge p, \underbrace{(x-\lfloor x \rfloor)}_{0.d_1(x)d_2(x)\cdots} - \underbrace{D_n(x)}_{0.d_1(x)\cdots d_n(x)} = 0.\underbrace{0 \cdots 0}_{n\text{'s } 0} d_{n+1}(x)\cdots \le 10^{-n} \le 10^{-p} \le \epsilon[/math]
therefore [math]\displaystyle \lim_{n \to +\infty} D_n(x) = x - \lfloor x \rfloor \in \mathbb R[/math]

let [math]0 \le x < y[/math] in [math]\mathbb R[/math], let [math]\displaystyle q_n = \frac{\lfloor x \rfloor + D_n(x) + \lfloor y \rfloor + D_n(y)}{2} \in \mathbb Q[/math],
using the previous property of [math]D_n[/math], we can show that
[math]\frac{x+y}{2} - q_n \le 10^{-n+1}[/math], therefore, [math]\displaystyle \lim_{n \to +\infty} q_n = \frac{x+y}{2}[/math]

therefore [math]\bar{\mathbb Q} = \mathbb R[/math]

Is my proof valid? what do you think?