Anonymous
7/17/2025, 11:15:37 PM No.16727090
Claim: [math] 0.999_{\dots} \neq 1[/math]
Proof: We use induction. The base case is trivial: [math] 0.9 \neq 1[/math]. Next we introduce the notation that [math]0.9_n = \underbrace{0.9999999}_{n-\text{many nines}}[/math] is the decimal with n-many 9s.
Now the inductive step: we assume [math]0.9_n \neq 1[/math]. Then trivially [math]0.9_{n+1} \neq 1 [/math]. It might help to notice that [math] 1 - 0.9_{n+1} \neq 0[/math].
This implies that [math]0.9_n \neq 1 \qquad \forall n\in \mathbb{N}[/math]
Finally, we define [math] 0.999_{\dots} := \lim_{n\to\infty} 0.9_n[/math].
[math]\therefore 0.999_{\dots} \neq 1 \qquad \square [/math]
Proof: We use induction. The base case is trivial: [math] 0.9 \neq 1[/math]. Next we introduce the notation that [math]0.9_n = \underbrace{0.9999999}_{n-\text{many nines}}[/math] is the decimal with n-many 9s.
Now the inductive step: we assume [math]0.9_n \neq 1[/math]. Then trivially [math]0.9_{n+1} \neq 1 [/math]. It might help to notice that [math] 1 - 0.9_{n+1} \neq 0[/math].
This implies that [math]0.9_n \neq 1 \qquad \forall n\in \mathbb{N}[/math]
Finally, we define [math] 0.999_{\dots} := \lim_{n\to\infty} 0.9_n[/math].
[math]\therefore 0.999_{\dots} \neq 1 \qquad \square [/math]
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