4 ms·

I don't think that 0.999... = 1 is actually provable. I think this and all of calculus is actually axiomatic, which has the following axiom: Given ε = 1/∞ then

by ristos 6y ago

I don't think that 0.999... = 1 is actually provable. I think this and all of calculus is actually axiomatic, which has the following axiom: Given ε = 1/∞ then: ε = 0 Am I wrong in thinking this way? It seems as though there's no way to actually truly prove that an infinite series converging towards zero actually hits zero (from a constructivist pov)

qppo 6y ago
I think the mistake in your thinking is that infinity is a value or reachable destination. It's a concept and it behaves differently than a number. Also go read Generatingfunctionology and Concrete Mathematics
- ristos 6y ago
  That's the point I'm trying to make actually. 1/infinity is an infinite series, which would be a computation that takes infinite time to compute. Saying that 0.9999...=1 is saying that the infinite series is the same as that concrete value, which don't actually know and can't prove.
  - qppo 6y ago
    I think the hiccup here is the notion that an open form expression doesn't have a closed form representation, which is not always the case. Closed forms exist for infinite series and vice versa. 1 - sum{k=1}^{\ifnty} 10^-k is one example.
SpicyLemonZest 6y ago
A mathematician would quibble with your notation, but you're basically right. The fact that 0.999... = 1 depends on the fact that 0 is the only number smaller than 1/n for every integer n. This isn't exactly an arbitrary axiom, it encodes some of our natural intuitions about what a number is, but you can construct a system known as the "hyperreal numbers" where it isn't true and 0.999... doesn't converge.
jmull 6y ago
Well, that wiki page has several proofs using various approaches. There's also a section which addresses common objections. I think you'd have to dismiss them all to make your claim (to do it well, that is).
- ristos 6y ago
  Afaict all of these rely on the axiom of Archimedes which is what I'm talking about, some of the proofs explicitly stating it and others implicitly. Unless I missed something.
karatinversion 6y ago
That's not an axiom we take; the study of real numbers takes some algebraic and order axioms, and a completeness axiom: Any non-empty set of real numbers with an upper bound has a least upper bound. We can't prove your statement Given ε = 1/∞ then: ε = 0 because it is not well-defined, but we can prove this: If ε ≥ 0 and, for every natural number n, ε < 1/n, the n ε = 0. For suppose there exists an ε which is a counter-example, i.e. ε > 0 and ε < 1/n for every natural number n. Then the set S = { x : x a real number, x > 0 and x < 1/n for every natural n}, is non-empty, and has an upper bound (e.g. 1). So it has a least upper bound, say y. In particular, y is an upper bound, so 2y is not in S. It is > 0, so there must exist a natural number N for which 2y >= 1/N. But then y/2 > 1/4N, and 4N is also a natural number. So for any element x of S, x < 1/4N < y/2; thus y/2 is an upper bound which is less than y. This is a contradiction, so the claim is proved.
- ristos 6y ago
  Why is your proof for every natural number n and not infinity (ω)? also isn't the law of excluded middle axiomatic?
deleted 6y ago
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by ristos 6y ago

qppo 6y ago

I think the mistake in your thinking is that infinity is a value or reachable destination. It's a concept and it behaves differently than a number. Also go read Generatingfunctionology and Concrete Mathematics

ristos 6y ago
That's the point I'm trying to make actually. 1/infinity is an infinite series, which would be a computation that takes infinite time to compute. Saying that 0.9999...=1 is saying that the infinite series is the same as that concrete value, which don't actually know and can't prove.
- qppo 6y ago
  I think the hiccup here is the notion that an open form expression doesn't have a closed form representation, which is not always the case. Closed forms exist for infinite series and vice versa. 1 - sum{k=1}^{\ifnty} 10^-k is one example.

SpicyLemonZest 6y ago

A mathematician would quibble with your notation, but you're basically right. The fact that 0.999... = 1 depends on the fact that 0 is the only number smaller than 1/n for every integer n. This isn't exactly an arbitrary axiom, it encodes some of our natural intuitions about what a number is, but you can construct a system known as the "hyperreal numbers" where it isn't true and 0.999... doesn't converge.

jmull 6y ago

Well, that wiki page has several proofs using various approaches. There's also a section which addresses common objections. I think you'd have to dismiss them all to make your claim (to do it well, that is).

ristos 6y ago
Afaict all of these rely on the axiom of Archimedes which is what I'm talking about, some of the proofs explicitly stating it and others implicitly. Unless I missed something.

karatinversion 6y ago

That's not an axiom we take; the study of real numbers takes some algebraic and order axioms, and a completeness axiom: Any non-empty set of real numbers with an upper bound has a least upper bound. We can't prove your statement Given ε = 1/∞ then: ε = 0 because it is not well-defined, but we can prove this: If ε ≥ 0 and, for every natural number n, ε < 1/n, the n ε = 0. For suppose there exists an ε which is a counter-example, i.e. ε > 0 and ε < 1/n for every natural number n. Then the set S = { x : x a real number, x > 0 and x < 1/n for every natural n}, is non-empty, and has an upper bound (e.g. 1). So it has a least upper bound, say y. In particular, y is an upper bound, so 2y is not in S. It is > 0, so there must exist a natural number N for which 2y >= 1/N. But then y/2 > 1/4N, and 4N is also a natural number. So for any element x of S, x < 1/4N < y/2; thus y/2 is an upper bound which is less than y. This is a contradiction, so the claim is proved.

ristos 6y ago
Why is your proof for every natural number n and not infinity (ω)? also isn't the law of excluded middle axiomatic?

deleted 6y ago

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