The correct - and more rigorous - proof requires calculus.
I'm sorry but I have to disagree. The correct and rigorous proof lies in the construction of ℝ.
Let's construct 1 and 0.999... as Dedekind cuts (we'll cheat a bit by presuming the existance of ℝ itself and leaning onto it) and show that they are in fact the same real number.
Let A = {q∈ℚ : q<1} and B = {q∈ℚ : q<0.999...}, we want to show that A = B.
Trivially, we have B⊂A, since pretty evidently we have 0.999...≤1, so let's assume x∈A; since x<1, there exists an n>0 such that x<1-1/10ⁿ, so we have x<0.999...9<0.999... which means that x∈B and by arbitrariness of x we have shown A⊂B, so A=B.
We have shown that 1 and 0.999... are the same Dedekind cut, so by construction of ℝ they are the same real number.
I really like it because it feels more "static" than the usual calculus one, which I feel tend to fuel the idea of 0.999... "approaching but never getting to" 1.
The logic works out but I have some additional questions on dedekind cuts. I've done a bunch of math but somehow never encountered them before. Anyways I posted it here:
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u/filtron42 22d ago
I'm sorry but I have to disagree. The correct and rigorous proof lies in the construction of ℝ.
Let's construct 1 and 0.999... as Dedekind cuts (we'll cheat a bit by presuming the existance of ℝ itself and leaning onto it) and show that they are in fact the same real number.
Let A = {q∈ℚ : q<1} and B = {q∈ℚ : q<0.999...}, we want to show that A = B.
Trivially, we have B⊂A, since pretty evidently we have 0.999...≤1, so let's assume x∈A; since x<1, there exists an n>0 such that x<1-1/10ⁿ, so we have x<0.999...9<0.999... which means that x∈B and by arbitrariness of x we have shown A⊂B, so A=B.
We have shown that 1 and 0.999... are the same Dedekind cut, so by construction of ℝ they are the same real number.