10c - c

Discussion in 'Physics & Math' started by BrJLa, Aug 12, 2008.

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  1. BrJLa Registered Senior Member

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    Doesn't the 10c - c method for converting repeating patterns of decimals to fractions actually assume .9¯ = 1 rather than prove it?

    If you apply 10c - c to .285714¯, it says that it's equal to 2/7. But if you multiply .285714¯ by 7/2 you get .9¯. Likewise, 10c - c says .3¯ = 1/3 when .3¯ x 3 = .9¯.

    10c - c works by assuming .9¯/x = 1/x. Or, at least, the operation can't be validated without assuming .9¯ = 1. Doesn't it follow, then, that this operation can't be used to prove .9¯ = 1? Rather, it assumes it.

    (I tried to search the forum for the subject but all the elements are either too short for a search or too common.)
     
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  3. prometheus viva voce! Registered Senior Member

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    Thats a nice easy proof that 0.9r = 1

    1/3 = 0.3r
    1/3 x 3 = 1
    0.3r x 3 = 0.9r

    therefore 0.9r = 1
     
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  5. funkstar ratsknuf Valued Senior Member

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    Look, the concept of the real number 'one' has more than one representation: 1 is one. 1.0r is another. And, yes, 0.9r is yet another. It is, if you want, a redundancy in the decimal system.

    This isn't much different from the fact that 1/2 and 2/4 also represent the same number. The fact that they don't "look the same" means nothing - they still represent the same number.
     
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  7. BrJLa Registered Senior Member

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    Aren't you proving they're equal by assuming they're equal?

    In my mind, what that shows is that long division can't be performed to an actual infinite number of places and give a true expression of a fraction. You have to assume it's a true expression to prove it's a true expression.

    1/3 can be expressed as .3 and 1/30. Or .333333 and 1/3000000. But an actual infinite expression of 1/3, without the corresponding "and 1/(3 x 10^the number of places the decimal is expressed to)" fails, because when multiplied by it's reciprocal it produces .9¯ . That doesn't prove .9¯ (a decimal point followed by an infinite number of 9's) equals 1. It states that an operation that equals 1 at any finite number of iterations would still equal 1 if it could be performed an actual infinity number of iterations.
     
  8. BrJLa Registered Senior Member

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    What I'm saying is that you're not saying anything meaningful about a decimal point followed by an actual infinite number of 9's as a distinct concept. You're talking about an operation that always equals 1 performed an actual infinite amount of times. And you're taking that it always equals 1 to prove it equals 1.
     
  9. Steve100 O͓͍̯̬̯̙͈̟̥̳̩͒̆̿ͬ̑̀̓̿͋ͬ ̙̳ͅ ̫̪̳͔O Valued Senior Member

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    No it does not.

    9.9r - 0.9r clearly = 9
    You do not need to assume 0.9r = 1 to get to this stage.

    There is however many more ways to prove 0.9r = 1
     
  10. BrJLa Registered Senior Member

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    But 10c - c can be shown in other instances to provide an invalid result, unless you assume .9¯ = 1. All you're showing is that 10c - c is consistent with itself.

    Where I'm really going is that I think "the largest real number less than 1" can be conceptualized as .9¯. But you're not allowed to do that because the concept of a largest real number less than 1 isn't "real" (even though, logically, by necessity, it would be symbolized as .9¯ - it's not as though it could end in something other than a 9). For any real number, there have to be other real numbers between it and 1 because you never reach a number that can't be added to. But this logic doesn't get applied to long division, such that you say "that's not real because you never reach a "real" place in the process where no more iterations can be performed".

    I'm open to other proofs. I'm just saying 10c - c isn't it. That, by itself, doesn't say anything meaningful about .9¯.

    (Although I do think all the other proofs fall back on this notion that the actual infinite expression of an operation can be taken as real.)
     
  11. Guest254 Valued Senior Member

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    Hi BrJLa,

    You'll be pleased to know that you're not the first person to be confused by this fact! There is an abundance of information available on the matter: I think perhaps wikipedia is a good start.

    Please Register or Log in to view the hidden image!

     
  12. BrJLa Registered Senior Member

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    I'm asking a simple question, based solely on logic. I'm not asking if .9r = 1. I'm asking if a system that can't be shown to always produce valid results unless you assume that .9r = 1 can be used to prove .9r = 1?
     
    Last edited: Aug 12, 2008
  13. BrJLa Registered Senior Member

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    And it's not just my argument. It's an extension of L.E.J. Brouwer's intuitionism, which doesn't allow infinite expressions of decimals. I wouldn't say that 10c - c isn't 9. I'd say that .3r isn't real, and certainly isn't a real expression of 1/3 since it assumes long division performed such that there can be no more iterations.
     
    Last edited: Aug 12, 2008
  14. Guest254 Valued Senior Member

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    Do you feel all the proofs in the link I provided use the fact that 0.9r=1?
     
  15. Guest254 Valued Senior Member

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    Actually, I think this is probably not the right way to go. Rather than ask you about what you don't yet understand, I should ask you about what you think you understand - hopefully this will sort out any confusion.
    Ok, here goes! If you can, would you mind defining exactly what you mean by "conceptualized" with regard the real numbers? If you don't know what the real numbers are, no probs - just ask.

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  16. BrJLa Registered Senior Member

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    I think it's easy to understand why you can't express infinity. An infinitely large number doesn't have to be 99999r. It can 1111r, and still be infinitely large, or it could take infinite other forms.

    But there is not doubt that the largest real number less than 1 - if such a number were possible - must be expressed as .9r. It could not possibly take another form.

    It can't be real. But for the exact same reason, a decimal point followed by an actual infinite number of 9s can't be real. There is no real number that is a decimal point followed by so many 9s that another 9 couldn't be added.
     
  17. iceaura Valued Senior Member

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    Am I allowed to write 1/9 ? Can I multiply it by 9 ? Can I use decimal notation and its conventions for that operation, instead of fractions ?

    If one can show consistency and establish convention, what's wrong with any notation ?
     
  18. Guest254 Valued Senior Member

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    I'll assume that your lack of response to my questions indicates you feel you don't need my help - fair enough! Good luck with your quest.

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  19. BrJLa Registered Senior Member

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    I tried. I guess I wasn't literal enough about your question.

    I said the largest real number less than 1 can be conceptualized as .9r. By "conceptualized" I meant that through logic it's possible to see the largest real number less than 1 can't take any form other than a decimal followed by 9's. It can't be a sequence that includes 2's or 4's or 8's.
     
  20. BrJLa Registered Senior Member

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    I mean "conceptualized" in the same way you might say you can conceptualize what happens when you perform long division to an infinite number of places, and drop the remainder.
     
  21. Guest254 Valued Senior Member

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    Ok, this is a good start. I shall state your conclusion clearly:

    If there is a largest number less than 1, it can be no less than 0.9r.

    Is this your position?
     
  22. BrJLa Registered Senior Member

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    Yes, that is fair.
     
  23. Guest254 Valued Senior Member

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    Ok, so now we've got that previous bit of information, could you answer the following:

    If there is a largest number less than one, it must greater than or equal to ? and less than 1.

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