Mothematician post
Mothematician post
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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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.
I like how compact this one is ;)
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Not quite. The wording "equivalence classes of ... with respect to the relation R: aRb <==> lim( a_n - b_n) as ->inf" is key.
https://en.wikipedia.org/wiki/Equivalence_class
loosely, an equivalence relation is a relation between things in a set that behaves the way we want an equal sign to
For an element in a set, a, the equivalence class of a is the set of all things in the larger set that are equivalent to a.
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"Having the same age" is an equivalence relation between people.
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It is reflexive: Bob is always the same age as himself
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It is symmetric: if Bob is the same age as Sally, then Sally is the same age as Bob
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It is transitive: If Bob is the same age as Sally, and Sally is the same age as Fred, then Bob is the same age as Fred.
using symbols:
Bob ~ Bob Bob ~ Sally ⇒ Sally ~ Bob Bob ~ Sally and Sally ~ Fred ⇒ Bob ~ Fred"⇒" means "the statement on the left implies the statement on the right." When people in this thread write =>, <=, and <=> they mean ⇒, ⇐, and ⇔
An "equivalence class" is the set of all items that obey the equivalence relation with each other. So, "being 25 years old" is an equivalence class containing every person who is 25 years old. Those people might be different in every other way, but they are equivalent in that specific regard.
In their proof earlier, @Tomorrow_Farewell@hexbear.net defined two equivalence classes. Instead of "people who are 25 years old," the classes were "infinite sequences that converge to 1" and "infinite sequences that converge to 0.999...." They showed that these are the same class.
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So, under the relevant construction of the space of real numbers, every real number is an equivalence class of Cauchy sequences of rational numbers with respect to the relation R outlined in my comment. In other words, under this definition, a real number is an equivalence class that includes all such sequences that for every pair of them the relation R holds (and R is, indeed, an equivalence relation - it is reflexive, symmetric, and transitive, - that is not hard to prove).
We prove that, for the sequences (1, 1, 1,...) and (0.9, 0.99, 0.999,...), the relation R holds, which means that they are both in the same equivalence class of those sequences.
The decimals '1' and '0.999...', under the relevant definition, refer to numbers that are equivalence classes that include the aforementioned sequences as their elements. However, as we have proven, the sequences both belong to the same equivalence class, meaning that the decimals '1' and '0.999...' refer to the same equivalence class of Cauchy sequences of rational numbers with respect to R, i.e. they refer to the same real number, i.e. 0.999... = 1.
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!
Thank you.
And no, I do not notice any errors.
I should re-learn Lateχ.