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Numerical Growth Formula

  1. Sep 28, 2008 #1
    hey folks, been a long time since i been here, stranded on a tropical island for a few years.. nice to be back

    to the business.

    I have a problem with some " rhetoric" i was reading today in one of the old posts. Seems to me that you all seem to think that .999 followed by about 1billion other 9's seems to equal 1. Yet some others are true naysayers and go against the flow, and say, no it is really .9999 not equal to 1.

    Now i must ask the question. .999999 can only be equal to 1 if at some point in time the last 9 in the chain, gets an increase of 1 numerical value. This I really can not deny.

    So I pose my real question. If 1/2 is 0.5 as you all say. Why is it not really 0.49999999999999...
    I can prove using a very simple equation that this is the case, and no matter how far you go, a person can not find that numerical increase to truly make it 0.5

    Now dont go all crazy and start saying, well, then any terminal decimal must be a .9999 somewhere's.

    interested in the formula??
    Last edited by a moderator: Sep 29, 2008
  2. jcsd
  3. Sep 29, 2008 #2


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    At some point in time? there is not "time" involved in a numeral. And there is no "increase of 1 numerical value". 0.99999... (note the dots: the "9"s do not end) is defined as the limit of the series [itex] 0.9\sum_{n=0}^\infity (0.9)^n[/itex], ... and that is a geometric sequence which can easily be shown to have limit 1.

    0.5 and 0.49999... and 1/2, not to mention 2/4, 3/6, ... are all different ways of writing the same number. Just as 1.0, 0.99999..., 2/2, 3/3, ... are all different ways of writing the same number.

    I hope I'm not going crazy to say, yes, any terminating decimal, say ending in the non-zero digit n, can be written as a decimal with digit n-1 followed by 9999... . Any number can be written in many different ways.
    Last edited: Sep 29, 2008
  4. Sep 29, 2008 #3


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    0.9999 = 1 - 1/10000 is not equal to 1. 0.99..(999,999,996 other 9s)..99 = 1 - 1/10^1000000000 is not equal to 1.

    0.99999... = 1.

    There is no last 9, and the value does not change over time. 0.999... is just 0.999... (which happens to equal 1, 14 - 13, and pi^0).

    0.4999... = 0.5 = 1/2.

    If by that you mean there is no e > 0 such that 0.4999... + e = 0.5, I agree. 0.5 + e > 0.5 since e > 0.

    With z and n positive integers, a terminating decimal can be represented as z * 10^-n. It's easy to see that
    z * 10^-n = (z - 1 + 0.999...) * 10^-n
    so any terminating decimal can be replaced by a form with repeating 9s. For example,
    22.3 = 22.299999...
  5. Sep 29, 2008 #4
    If you didnt like any of the previous answers, here is anotherone.
    If you imagine that for every 9 you add to any repeating digts, you are bringing it closer to the actual value, .999999.....=>1
    Then for every 9 you add, you are creating a limit and this limit approaches that value 1.
    So in essence ,you are only creating a limit, but cannot but will infinitley approach 1.
    So yes, in essence .9999999 does not equal 1, but likewise we can determin 1/0 = infinity. It is an approximation
  6. Sep 29, 2008 #5


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    If you mean "0.999..99 with some finite number of 9s does not equal 1" then I agree.

    But 0.999... = 1 exactly; it's not an approximation.
  7. Sep 29, 2008 #6
    look, there is no way to say that .9999........... equals 1, as i said, there is a limit, and when you take that limit, it does equal 1, but since we are looking at it as a real number, it does not apply
  8. Sep 29, 2008 #7


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    But .999.... IS the limit.

    Think of if this way, take as many of the 9s as want, that is go as far out the infinite string of 9s as you wish and stop. Now you have a finite number of 9s which we can all agree is less then 1, you might even say that it is "approximately" 1 but it is less then 1. Now consider the 9s that you chopped off. How many are there? The correct answer is that there is an INFINTE number of 9s, these 9s must add up to something greater then 0. At this point you have 2 unknown numbers. The difference between our "approximation" and 1 and a infinitely long string of 9s which add up to some positive number. Why is it so hard to understand that that bit represented by the infinite string of 9s is EXACTLY the difference between your "approximation" and 1.
  9. Sep 29, 2008 #8


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    .9999..... equals 1! There, I just said it so there is a way!

    What definition of base 10 numerals are you using? Every definition I know says that the number represented by the infinite decimal 0.abcde... IS the limit of the sequence 0.a, 0.ab, 0.abc, 0.abcd, 0.abcde, ... or, equivalently the limit of the sum 0.a+ 0.0b+ 0.00c+ 0.000d+ 0.0000e+ .... It makes no sense to say that 0.999... is the limit and not the number because the number IS the limit. If you disagree with that, then please tell us, as I asked before, what definition you are using.
  10. Sep 30, 2008 #9


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    I have had similar discussions with students of mine who have flatly stated that

    0.999 \dots \ne 1

    - it wasn't that they "weren't sure", it was that it was impossible. All of them had the same rationale for their position: since [tex] 0.99999\dots [/tex] has something to do with infinity, and only their God can comprehend the infinite, therefore we cannot know it, and so their position is justified.

    Could that be the current objection?
  11. Sep 30, 2008 #10
    well that is mine in a nutshell
  12. Sep 30, 2008 #11


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    It seems more than a little odd to allow an item of faith to lead one to say "mathematicians the world over have no idea what they are doing".
    (Overstated, I realize) - but what, exactly, is the rub with this bit of mathematical fact? (I use the word fact because, when push comes to shove in mathematics, it is a fact that [tex] 0.9999 \dots [/tex] is equal to one)
  13. Sep 30, 2008 #12
    I think it can be seen this way: suppose than 0.9999... and 1 are not equal, and call their (positive) difference D. Now, what is the decimal expansion of D? It should look like 0.0000... Now, if there is any non-zero digit in it, anywhere, then when you add 0.9999... you'll get a number greater than 1 (try). And if there are no non-zero digits in D, then D=0.

    However, I believe the reason why other posters have answered radically "yes, they are equal", is because the real numbers are defined like that: a real "number" is, by definition, a sequence of rationals that get closer and closer (Cauchy sequence).
    Last edited: Sep 30, 2008
  14. Sep 30, 2008 #13


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    Sorry to criticize some one who is trying to agree with me, but a real number is NOT, by definition, "a sequence of rationals that get closer and closer". One common definition is that a real number is an equivalence class of such sequences under a specific equivalence relation: {an} and {bn are equivalent if and only if the sequence {an- bn} converges to 0. (And the Wikipedia article you link to says that.)
    That is precisely the confusion that causes the question here. If a real number is the sequence 0.9, 0.99, 0.999, 0.9999, ..., then it cannot also be the sequence 1.0, 1.00, 1.00, 1.000, 1.0000, ... But the real number is NOT the sequence. Those two sequences belong to the same equivalence class- the one that in normally labled "1".
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