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B 0.999… = 0 ?

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  1. Jul 1, 2017 #1
    Hi folks, my first post here. I was researching that guy who claims to be the smartest person in the world (Chris Langan) and found thread here… seemed like a good place to ask my question, or submit my observation.

    So… I'm thinking about 1/9, and I visualize a graph where the ones are slowly converging to zero as they become smaller through the "expanding" regress. And I sat and pondered the argument on wikipedia. 9*1/9 = 1. The convergence proof, right?

    But I thought, if you're going to assume convergence using the same rules, then why is not the first step, 1/9 = 0, before you multiply it nine times. There isn't a ratio that gives you 0.9…, you have to multiply it to get there, so by order of operations, the steps that get you to .9… should actually get you to zero, if you use the assumption of convergence.

    How am I wrong?
     
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  3. Jul 1, 2017 #2

    Orodruin

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    If you are going to reference outside material, it is customary to provide a link so that people have a chance to know what you are talking about. Please provide the appropriate link for context.

    Probably because 1/9 is not equal to zero. Without knowing what "first step" you are referring to, it is impossible to answer in any other way.
     
  4. Jul 1, 2017 #3

    mfb

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    There is no convergence to study here. 9*1/9=1 by definition of 1/9.
    Why should 1/9 be 0? It cannot be 0.


    Decimal representations are ways to write a number. They are not what defines a number!
    0.1=1/10=5/50 is all the same number - written in three different ways.
     
  5. Jul 1, 2017 #4
    1/9 has to be zero, because it converges to zero, ONLY IF .999… converges to 1. It's using the same exact logic. Simply saying 0.1… doesn't equal zero is not an argument, especially when you argue that 0.9… converges to 1 using the same exact logic.

    I stated that the argument in question was found among the 20 or so arguments on the wikipedia page: https://en.wikipedia.org/wiki/0.999…

    All you have to do is think through the steps… 0.1 gets smaller and smaller and smaller, getting closer and closer and closer to zero if you plot the regress on a graph, so if you assume convergence, as you do with other sequences, 0.1… converges at zero. It seems obvious to me.

    Mentor note for the benefit of readers of this thread. The OP, @Seasons, has many misconceptions about how real numbers work, almost too many to count.
    "1/9 has to be zero" -- It should be obvious that this can't possibly be true.
    "...because it converges to zero" -- 1/9 is a number, a constant that doesn't change -- convergence doesn't enter into the discussion.
    "ONLY IF .999… converges to 1" -- .999... is also a constant, that happens to be exactly equal to 1. No convergence here, either.
    "Simply saying 0.1… doesn't equal zero is not an argument" -- It is a fact that is (almost) universally recognized that 0.1... is not equal to zero.
    "0.1 gets smaller and smaller" -- No, 0.1 is a constant that doesn't change.

    Several members have responded in rest of this thread, attempting to correct the OP's misconceptions, but the misconceptions are so numerous, not all of them have been addressed.
     
    Last edited by a moderator: Jul 3, 2017
  6. Jul 1, 2017 #5

    Orodruin

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    No it doesn't. 1/9 is a well defined rational number. Assuming that you are at high-school level, I would say that this error is mainly based on lack of knowledge of what "converges to" means.

    It is not the same logic. You can easily find a number between 0.1... and 0 (take 0.05 for example) and so the limit cannot be zero. The corresponding thing would be 0.0... = 0, which is obvious.

    No it doesn't. You are thinking of the series 0.1, 0.01, 0.001, ... which has the limit zero but is not the same thing as 0.11..., which is the limit of the series 0.1, 0.11, 0.111, ...
     
  7. Jul 1, 2017 #6
    That's where my potential confusion is!!

    if you add .1, .01, .001 = .111 correct?

    so how are they different? I don't understand this part
     
  8. Jul 1, 2017 #7

    Orodruin

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    Yes, but this is not what the series is. You are confusing the limit of the series with the sum of the elements in the series. The series 0.9, 0.09, 0.009, ... converges to zero. If you make a series where the nth entry is the sum of that series' first n elements you get the series 0.9, 0.99, 0.999, ... - which is a different series that converges to one.
     
  9. Jul 1, 2017 #8

    mfb

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    What is different from what?
    0.1+0.01+0.001=0.111, correct.
    Note that 0.11 is larger than 0.1 and 0.111 is larger than 0.11.
     
  10. Jul 1, 2017 #9

    Ok, this is my issue. There is no such thing as a ratio that gives you 0.999…

    It can ONLY be arrived upon by a sum. Either 0.333…*3 0r 0.111…*9

    So in one sentence you said I'm confusing the limit with the sum, yet, you can only use a sum to get to the limit. I'm not fully out of my mental problem with this.
     
  11. Jul 1, 2017 #10
    Let me think about that… it seems obvious, but I think the rules that make that number might cause a contradiction - which are the rules of summing.

    What I mean to suggest, is that the same rules used to make 0.999… equal to 1, appear to make 0.111… equal to 0. Almost as if, the mental hurdle that people have with the 0.999… issue, might be something that facilitates an additional mental hurdle for the implications of those rules applied to the number 1; namely, summing and convergence.

    Is it the case that 0.9, 0.09, 0.009… equals zero in the same sense 0.1, 0.01, 0.001… equals zero, as was also stated earlier. Argghh… its hurting my brain. Maybe I need to take all this in!
     
  12. Jul 1, 2017 #11

    Orodruin

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    Yes there is: 2/2 = 0.999... because 0.999... = 1.

    You have to understand which real number is intended when we write it on decimal form. By a number on the form ##0.a_1a_2a_3\ldots## where the ##a_k## are digits, we mean the limit of the series ##0.a_1##, ##0.a_1a_2##, ##0.a_1a_2a_3##, ##\ldots## or, equivalently, the sum
    $$
    \sum_{n = 1}^\infty \frac{a_n}{10^n}.
    $$
    In the case of 0.999..., this sum is equal to one.

    A different way of stating this is that for any distinct real numbers, it is possible to find a number (even a rational number) which is between them. You can find no such number for 0.999... and 1 and therefore they must be the different representations of the same real number.
     
  13. Jul 1, 2017 #12

    Orodruin

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    No they don't. You are applying them differently even if you do not see it. The corresponding thing in the 0.999... case would be to claim that it converges to zero because the series 0.9, 0.09, 0.009, ... converges to zero - which is not true. The value of 0.999... is the limit of the series 0.9, 0.99, 0.999, ... not the limit of the series 0.9, 0.09, 0.009. Just as 0.111... is the limit of the series 0.1, 0.11, 0.111, ... not the limit of the series 0.1, 0.01, 0.001, ...
     
  14. Jul 1, 2017 #13

    mfb

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    They do not. They make 0.000..... = 0.

    0.1111.... is well above 0.
    0.1 is larger than 0 and the sequence 0.1, 0.11, 0.111 ... is increasing. How can you expect to end up at 0?
     
  15. Jul 1, 2017 #14

    Ok here! You're using a lot of implied steps that I'm puzzling about to make the claim that 2/2 = 0.999… Everyone knows that there is no actual ratio that spits out repeating 9's, the only method actually used is summing lower ratios in decimal form, and then making the equality from there.

    So the question is, if you're going to sum those ratios, why can't you sum other ratios?

    There's also no ratio that gives you 0.000…1, only a summing sequence, just like 0.999…

    So, if you're going to use summing sequences to make decimals that have no ratios, then it begs the question of why 0.111… (ratio) doesn't equal the summing sequence of 0.1, 0.01, 0.001… (no ratio) The sum gets greater (0.111…) as the summing sequence gets smaller (0.1, 0.01, 0.001) - how are they not equal? You're doing the same thing for 1 (ratio) and 0.999… (no ratio). ?? Because you've proven a ratioless number equals a ratio number here, how is not the ratioless number of 0.000…1 not equal to the sum of 0.1, 0.01, 0.001… not equal to 0.111?

    I'm also thinking about the argument that there is no number between 0.999… and 1, but the is between say, 0 and 0.111… : I want to ponder this further, because I think I might be able to make the argument that anything between them solves as them (one or the other / both beings equal)
     
  16. Jul 1, 2017 #15

    Orodruin

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    It does equal the sum of the sequence. The sum of the sequence is not equal to the limit of the sequence!
     
  17. Jul 1, 2017 #16
    Ok! This is the second time you've said this to me. So I should consider this quite seriously. The point I was trying to make is that this method is defining sums as limits!!! The method of 0.999… = 1 is using only sums to define the limit. Yet you make the distinction that I'm being confused when I do this for numbers like 1/9. Defining sums as limits that is.
     
  18. Jul 1, 2017 #17

    Orodruin

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    The limit of a (Cauchy) series ##\{x_n\}## is defined as the number ##x## such that ##|x-x_n|## is arbitrarily small as long as ##n## is large enough. An infinite sum can be defined by letting
    $$
    y_n = \sum_{k=1}^n x_n
    $$
    and taking the limit of the series ##\{y_n\}##. If ##\{x_n\}## is a series, there is no reason to believe that it converges to the same number as the series ##\{y_n\}## as defined above, since the series are different.
     
  19. Jul 1, 2017 #18

    Mark44

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    @Orodruin, you are misusing some standard terminology here. "Limit of the series" is usually stated as the "limit of the sequence of partial sums of the series." This limit, if it exists, is the sum of the series.
    Infinite series: ##\sum_{n = 0}^\infty a_n##
    Sequence of terms: ##\{a_0, a_1, a_2, \dots, a_n, \dots \}##
    Sequence of partial sums: ##\{S_0, S_1, S_2, \dots, S_N, \dots \}##, where ##S_N = \sum_{n = 0}^N a_n##
    If the limit as N grows without bound exists in the sequence of partial sums, this limit is the sum of the series.
    What you wrote is a sequence. This sequence (not series) converges to zero, but the series (.9 + .09 + .009 + ...) converges to 1.
    Again, what you wrote here is a sequence.
     
  20. Jul 1, 2017 #19

    Orodruin

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    Yes, sorry. I meant to write Cauchy sequence and sequence ##\{x_n\}## etc.

    Edit: Come to think of this. I think this is a translation issue I have been misusing for quite some time. Always room to teach old dogs to sit.
     
  21. Jul 1, 2017 #20

    Mark44

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    0.000...1 is not a number. The ellipsis ('...') means that the 0's continue indefinitely, which implies that there can't be a 1 digit "out there" somewhere. If you specify that the 1 is at a specific position, then that is meaningful, but without other context, 0.000...1 is not.
    There are two sequences involved here.
    1) {.1, .01, .001, ...} -- This is the sequence of terms.
    2) (.1, .11, .111, ...} -- This is the sequence of partial sums of the series ##\sum_{k = 1}^\infty \frac 1 {10^k}##
    The second sequence is produced by adding the terms in the first sequence. I.e., .1 = .1, .11 = .1 + .01, .111 = .1 + .01 + .001, and so on.
     
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