This little guy craves the light of knowledge and wants to know why 0.999... = 1. He wants rigour, but he does accept proofs starting with any sort of premise.

Enlighten him.

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[–] Tomorrow_Farewell@hexbear.net 6 points 1 year ago (9 children)

I suppose I will post one myself, as I do not expect anybody else to have that one in mind.

The decimals '0.999...' and '1' refer to the real numbers that are equivalence classes of Cauchy sequences of rational numbers (0.9, 0.99, 0.999,...) and (1, 1, 1,...) with respect to the relation R: (aRb) <=> (lim(a_n-b_n) as n->inf, where a_n and b_n are the nth elements of sequences a and b, respectively).

For a = (1, 1, 1,...) and b = (0.9, 0.99, 0.999,...) we have lim(a_n-b_n) as n->inf = lim(1-sum(9/10^k) for k from 1 to n) as n->inf = lim(1/10^n) as n->inf = 0. That means that (1, 1, 1,...)R(0.9, 0.99, 0.999,...), i.e. that these sequences belong to the same equivalence class of Cauchy sequences of rational numbers with respect to R. In other words, the decimals '0.999...' and '1' refer to the same real number. QED.

[–] AernaLingus@hexbear.net 3 points 1 year ago (1 children)

Had a little trouble following this as plain text, so I wrote it up in LaTeX (it'll be a bit small if you try to read it inline--you'll probably want to tap to enlarge on mobile or open the image in a new tab/click this direct link to the image):

I tried to hew as closely to your notation as I could, but let me know if you spot any errors!

[–] Tomorrow_Farewell@hexbear.net 3 points 1 year ago* (last edited 1 year ago)

Thank you.

And no, I do not notice any errors.

I should re-learn Lateχ.

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