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it seems like asserting that .1111... = 1/9 is circular to proving that .999...=1. I'm not saying it's not true, but if you don't believe that .9999..=1 then why would you believe that .3333...=1/3 or .1111...=1/9 — Preceding unsigned comment added by 24.19.2.53 (talk) 05:12, 27 August 2014 (UTC)[reply]

Long division cranks out an endless sequence of threes when applied to 1/3, and an endless sequence of ones when applied to 1/9. Although neither of these are infinite series, since no-one has infinite time to carry out long division, they add plausibility to the idea of these as infinite series. Since long division is a simple procedure we learn in school, and gives the correct results for other problems, it's easier for people to believe in than more abstract procedures. -- The Anome (talk) 09:11, 27 August 2014 (UTC)[reply]

According the Dedekind cut section, a real number is a subset of the set of rational numbers, but a rational number is a kind of real number, which means that all rational numbers contain themselves which according to ZF is impossible. Blackbombchu (talk) 00:18, 12 May 2014 (UTC)[reply]

Technically speaking, a mathematician would say that there is a subset of the real numbers that is in one-to-one correspondence with the rational numbers, and that the real number addition and multiplication exactly correspond to the rational number addition and multiplication. In other words there is a set within the real numbers that is isomorphic to the rational numbers as ordered fields. So while the are different we choose not to distinguish them in most contexts. Thenub314 (talk) 00:44, 12 May 2014 (UTC)[reply]

More generally, given any set S, I can form the set , where obviously S and S' are in bijection. So while there's an obvious mapping which sends x to {x}, neither x nor {x} contain themselves as an element. Something similar happens for the rational numbers. Huon (talk) 21:22, 12 May 2014 (UTC)[reply]

This is in fact true of the binary expansions of many rational numbers, where the values of the numbers are equal but the corresponding binary tree paths are different. For example, 0.10111...₂ = 0.11000...₂, which are both equal to 3⁄4, but the first representation corresponds to the binary tree path LRLRRR... while the second corresponds to the different path LRRLLL....

Should not it say “LRLRLLL...” and “LRLLRRR…”? -- Zygmunt Zzzyzzyzkoff (talk) 18:08, 5 July 2014 (UTC)[reply]

Why should it? The current paths are correct, with "L" and "R" in the path corresponding to 0 and 1 in the binary representation, respectively. Huon (talk) 18:17, 5 July 2014 (UTC)[reply]

I thought it would work like this:

Until a color change, each segment is worth +1 or -1 (depending on whether it is Blue or Red, respectively).

Once a color change occurs, each subsequent segment (regardless of color), is worth half of the previous segment, with a +/- corresponding to the color.

Thus, the string BBRB would be worth +1+1-1/2+1/4=7/4.

Otherwise, this statement is false:

For example, the value of the Hackenbush string LRRLRLRL... is 0.010101₂... = ¹/₃.

-- Zygmunt Zzzyzzyzkoff (talk) 19:25, 5 July 2014 (UTC)[reply]

You're right, I was wrong. The L and R don't correspond to 1 and 0, but to 1 and -1. Thanks for providing the more thorough explanation. I'll correct the article. Huon (talk) 00:38, 6 July 2014 (UTC)[reply]

• I reasonably understand the surreal numbers, but I didn't know the hackenstring notation. In surreal numbers there exist distinct numbers 1 (generation 1), 1-ε, and 1+ε (both generation ω). The first is simply {0|}, the latter are {0, 1/2, 3/4, 7/8 ...|1} and {1|2, 1+1/2, 1+1/4, 1+1/8 ...}. I guess in hackenstring notation the numbers would be R, RLRRRR... and RRLLLL... - Mike Rosoft (talk) 00:01, 7 July 2014 (UTC)[reply]
• And I think I now understand the relationship between the hackenstring and set notation: L means that the number is one step smaller, i.e. put the current value in the right set; R means that the number is one step larger, i.e. put the current value in the left set. - Mike Rosoft (talk) 04:44, 7 July 2014 (UTC)[reply]
  • On the second thought, I think I have swapped the L and R symbols; i.e. L means "put the current value in the left set" (the resulting value is more than the current one), rather than "go left from the current value" (the resulting value is less than the current one). In that case: 1, 1-ε, and 1+ε are L, LRLLLL..., and LLRRRR..., respectively. (Otherwise, numbers starting with L would have been negative.) - Mike Rosoft (talk) 19:04, 9 July 2014 (UTC)[reply]

The current example in the "Cauchy sequences" section uses the sequence (1, 1⁄10, 1⁄100, 1⁄1000, ...), with a limit of 0. Could we instead provide a slightly more intuitive example using the sequence (9⁄10, 99⁄100, 999⁄1000, ...), having a limit of 1, to more closely reflect the number 0.999...? — Loadmaster (talk) 22:20, 10 July 2014 (UTC)[reply]

• The section constructs real numbers as equivalence classes of Cauchy sequences of rational numbers, with two sequences considered equal if the limit of their difference is zero. So what needs to be proven is that given the sequence 0, 0.9, 0.99, 0.999, ... (corresponding to the decimal representation 0.999...) and the constant sequence 1, 1.0, 1.00, 1.000, ... (corresponding to the decimal representation 1.000...), the limit of their difference is indeed zero - meaning that they both represent the same real number. - Mike Rosoft (talk) 03:33, 12 July 2014 (UTC)[reply]

Section Infinite series and sequences includes a graphical representation of the limit, but it's base-4 and doesn't represent the measure of "arbitrarily closeness". I think we can add a second graph representing the distance |x − xn| as a segment on the real line, and showing how it becomes closer to 1 than 1- as the series increases. This would provide a redundant, visual representation of the concept already explained in that section, which could help us visual thinkers. Diego (talk) 10:33, 15 October 2014 (UTC)[reply]

I have begun a discussion of this talk page's "Arguments" subpage at MfD, because it is being used as a forum in violation of Wikipedia policy. Lagrange613 04:18, 16 October 2014 (UTC)[reply]