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.9999 = 1

  1. Apr 27, 2004 #1
    I'm sure most of you already know this. The real point of the thread is finding different ways of approaching it. You see, I have a friend who refuses to accept what seems to me to be so obvious: .9 repeating (infinite 9s after the decimal) is exactly equal to the whole number 1.

    Here are the proofs that I've used so far:

    1) 1/3 is precisely equal to .333... (to which he agreed). 3*.333... equals .999... and 3*1/3 equals 1. Therefore, since I multiplied equal numbers by the same number, they should still be equal...ergo 1 = .999...

    2) If 1 is greater than .999..., then by how much, exactly, is it greater? After all, 1 is greater than .9 by .1. It is greater than .99 by .01. It is greater than .99999999999999999999 by .00000000000000000001. And so on, and so on. However, if the number of nines is infinite, then the number of zeros preceding the 1 will be infinite, and 0.000... is obviously equal to 0, which means that 1 is greater than .999... by exactly 0.

    3) After this point, my friend revealed that he doesn't really believe in actual infinites. He thinks you will eventually reach the 1, at the end of the infinite series of 0s. Ridiculous as that may sound, I respected that as his opinion and posed this next point: If there are no infinites then the term equal to 1/3, which is usually considered .333..., would actually (eventually) terminate in a 4. I asked him if that was right. After some thought, he said "yes". So, I asked, if we multiplied this number by 2, would you logically expect to get the same number 6s as you had 3s, and then terminate in an 8 (since 4+4 is 8)? He said, "yes", not realizing that he had just implied that the infinitely repeating number .666..., which is equal to 2/3 must now terminate in an 8, instead of a 7 (as his reasoning would naturaly have assumed).


    Are there any other points I can use? I wasn't going to ask for help on this, until he talked to his math teacher, and the teacher agreed with him!! Now he's got this quaint little ipse dixe argument, "well, the math teacher said I'm right".

    Any new points or comments on my previous ones are appreciated.
     
  2. jcsd
  3. Apr 27, 2004 #2
    btw, I don't know if this was the right place to have posted this thread, but I couldn't think of a better place for it.
     
  4. Apr 27, 2004 #3
    You have provided your friends some arguments as to why .9999.... must be the same as 1, otherwise math would not be perfect. Your friend is saying math is not perfect. Countering his argument with an unsubstantiated claim that math is perfect, end of discussion, doesn't seem to do much for him.

    But you can still get him to think about the issue from a different perspective. If he doesn't accept that numbers can be infinitely small, then he must agree that there exists a minimum quantity which cannot be divided. For instance, if he thinks the smallest number possible is N, then you can ask him, how much is N/3? Is it N or is it 0? There is no known answer to that question, but if the problem is real then an answer must necessarily exist. And that means the answer can be determined with probabilities. The probability of N/3 being N is 1/3, and the probability of it being 0 is 2/3.

    So the correct answer to his problem is that the result of 3 * 0.3333...3 must also be determined by probabilities, and 1, not 0.9999...9, is by far the most likely answer.
     
  5. Apr 27, 2004 #4

    NateTG

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    There is a fundemental question that is generally skipped in the discussions of [tex]0.\bar{9} = 1[/tex], which is the assumption that [tex]0.\bar{9}[/tex] is a real (or rational) number.

    A second shortcoming is that 'bar' notation is generally not well explored. If your friend is comfortable with bar notation, the following might be interesting for him:

    [tex]1[/tex]
    [tex]=\frac{1}{9} + \frac{8}{9}[/tex]
    [tex]=0.\bar{1}+0.\bar{8}[/tex]
    [tex]=0.\bar{9}[/tex]

    Although this is, of course, identical to the [tex]3 \times \frac{1}{3}[/tex] example that you gave.

    The traditional elementary school method is:
    [tex]9 \times 0.\bar{9} = 10 \times 0.\bar{9} - (1)0.\bar{9} = 9.\bar{9} - 0.\bar{9} = 9[/tex]
    so
    [tex]9 \times 0.\bar{9} = 9[/tex]
    so
    [tex]0.\bar{9}=1[/tex]

    You can also talk about infinite sums, but getting satisfactory answers with sums would (eventually) involve invoking the completeness of the real numbers - which is not necessarily an accessible subject.
    [tex]0.\bar{9} = \sum_{i=1}^\infty \frac{9}{10^i}[/tex]

    You can ask your friend what:
    [tex]\frac{1}{1 - 0.\bar{9}}[/tex]
    is equal to. Since [tex]0.\bar{9} \neq 1[/tex] the denominator is non-zero right?
     
  6. Apr 27, 2004 #5

    Integral

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    Give this link a look
     
  7. Apr 27, 2004 #6
    [tex]x = 0.\bar{9}[/tex]
    [tex]9x = 10x - x[/tex]
    [tex]9x = 9.\bar{9} - 0.\bar{9}[/tex]
    [tex]9x = 9[/tex]
    [tex]x = 1[/tex]

    Of course, I tend to think that the idea of a value of anything repeating is illogial--after all, it's just a defect of the number system (decimal, aka base-10). The proof that I just listed relies on the fact that:
    infinity - 1 = infinity

    0.\bar{9} cannot be viewed as anything other than a defect of the number system. It is not a value, as it involves infinity.
     
    Last edited: Apr 27, 2004
  8. Apr 27, 2004 #7

    Integral

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    I don't see it that way. Just the opposite if we were UNABLE to assign a value to any infinitely repeating number that would be a defect in the number system.
    This is NOT a facet of DECIMAL numbers, it is a facet of the density of the Real number line. No matter how you choose to represent the points of the real number line there will always be numbers which require an infinite number of digits to be specified.

    The Base does not matter. for example

    [tex] .1_{10} = .00011001100... _2[/tex]

    So in binary .1 has an infinitely repeating representation which MUST be truncated in all computations. This fact was first made apparent to me years ago when I did the following loop in Applesoft Basic
    10 x= 0
    20 For i = 1 to 10
    30 x= x+ .1
    40 Print x
    50 next i

    The final number would not be 1 but .999999, in only 10 steps the round off error would become apparent.

    (Round off error has nothing to do with the fact that 1=.999....)
     
  9. Apr 27, 2004 #8
    If we allow .0...1 (sorry not sure how to code for the bar notation) to have a non-zero value, then it must be the smallest positive number. If we allow for numbers to have a minimum value then the number line ceases to be continuous as any 2 numbers no longer have an infinite range of values between them. Now we have x, x+.0...1, x+.0...2, etc as our number line. There can be no number between these values because then there would be a smaller number then .0...1 Since this causes a contradiction, we can determine that .0...1 cannot have a non-zero positive value.
     
  10. Apr 27, 2004 #9

    Hurkyl

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    If we allow this, then why is .0...01 not a smaller number? After all, if we're going to allow ourselves "decimals" with infinity + 1 digits, then why not decimals with infinity + 2 digits?

    Before we ask "what is .0...1?" we should ask "does the notation .0...1 even make sense?"


    Try asking him how many digits 1/3 has.
     
  11. Apr 27, 2004 #10
    Since there are already an infinite number of zeroes before the one, adding one more will not make a difference (property of infinity). Thats part of the problem. Giving the 1 at the end of infinity a value means a) allowing the infinity to end and b) since the value of the one is equal to 1/10^(infinity+1) we are allowing 1/infinity to have a non-zero defined positive value. Neither of these is consistent with the way which math works. The number .0...1 cannot possibly exist in any form which would allow it to be added to .99... to equal 1 (and hence prove that the two are not equal).
     
  12. Apr 28, 2004 #11

    hypnagogue

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    Mentat, you have your friend stuck on a pair of horns. If 1/3 = 0.333...4, then 3 * 1/3 = 1.000...2, contradicting 3 * 1/3 = 1. So if your friend continues to hold that 1/3 = 0.333...4, then he must believe that

    a) 1 = 1.000...2, which is just as bad (in his eyes) as the original position he was seeking to refute, or
    b) 1/3 * 3 = 1 is false, or
    c) that mathematics is not internally consistent, which essentially means that statements like 1 = 2 can be shown to be true.

    If this forces him to reject that 1/3 = 0.333...4, then you can return to the conventional arguments. If not, then your friend should have no faith in mathematics whatsoever.

    Another plan of attack would be to convert the numbers to a base where we don't get infinitely repeating decimals. For instance, [tex]3_{10} = 10_3[/tex], and [tex]1/3_{10} = 0.1_3[/tex]. So if he accepts that 1/3 = 0.333..., and therefore that 3 * 1/3 = 0.999..., then he must accept that 0.999... = 1, since in base 3, [tex]10_3 * 0.1_3 = 1_3[/tex], and [tex]1_3 = 1_{10}[/tex].
     
    Last edited: Apr 28, 2004
  13. Apr 28, 2004 #12
    Thanks for all the replies, guys :smile:. All very useful stuff (even those who seem to agree with my friend, as I can use my own rebuttals to their reasoning as added arguments against his).

    Hypna, you are absolutely right. He should indeed have no faith in math whatsoever. The problem is, his math teacher agreed with him...which means I'm going to have to talk to the math teacher first, thus removing his ability to argue from authority.
     
  14. Apr 28, 2004 #13
     
  15. Apr 29, 2004 #14
    This has bothered me for years. My own logic tells me they aren't equal, while math tells me they are. I haven't decided yet if my logic is flawed or the number system is. I never got anywhere with it when I ask a math head because they are stuck on the definitions they are taught and seem to be oblivious to anything else. I personally think it is as close to 1 as you can get without ever being 1. In any pratical use of math it can be called 1, so it really doesn't matter anyway.
     
    Last edited: Apr 29, 2004
  16. Apr 29, 2004 #15

    Hurkyl

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    Suppose that's true. Then how does [itex](0.\bar{9} + 1)/2[/itex] compare?
     
  17. Apr 29, 2004 #16
    D00d, you got served! :biggrin: Just kidding, of course. I just had to say that. :rolleyes:
     
  18. Apr 29, 2004 #17
    Also, if we assume that they are not equal and that it is "as close to one as possible without being one". Then we define must define a smallest number which represents their difference. If we assume a smallest possible number, bad things happen. Like the number line ceases to be continuous, distance ceases to exist, and other bad stuff.
     
  19. Apr 29, 2004 #18

    Integral

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    Consider this inequality.

    [tex] 1 - .1^n < x < 1+ .1^n[/tex]

    It seems clear that there is only a single number for which this is true for ALL values of n>0, x= 1 . This is a simple statement that any number added to x results in something greater then 1, or any number subtracted from x results in something less then one. I doubt that you will find many people who will argue with the truth of the statement, x=1.

    Now, in the link I posted above, I show that using simple arithmetic, involving only valid rational numbers, one can construct this inequality.

    [tex] 1- .1^n < .999.... < 1+.1^n [/tex]

    notice that is the exact inequality as above, thus we have x=.999... If the original statement is correct that the inequality can only be satisfied by 1 you must be lead to the conclusion that
    1=.999...
     
  20. Apr 29, 2004 #19

    honestrosewater

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    What if you make a distinction between 1.000... and .999..., i.e. so that they are different numbers, and then figure out "how infinite" this set of numbers is. If this new set is "more infinite" than the set of real numbers, case closed. That is, if your friend doesn't object to 1-1 correspondence, countablilty, and the various ways of comparing infinite sets ;)

    I'm not sure how to make such a distinction. Perhaps the sets of all arrangements, where an "arrangement" is an infinite string of digits and one decimal point, ex.
    1010= ...0001010.000...
    98765.4321= ...00098765.4321000...
    1= ...0001.000...
    .999...= ...000.999...

    The set of all arrangements should include all real numbers, plus two strings for each terminating decimal, but may not be more infinte than the reals. Just a thought. The answer is probably already known. Though now I find it interesting in itself :)

    The whole idea is that 1=.999... implies that there must be more representations (numerals) than numbers. This may be something your friend has not realized. And it may be a way to get him/her to take another look at the axioms and theorems. Of course, "more" becomes vague when dealing with the infinite, but...
    Happy thoughts
    Rachel
    EDIT- Nevermind, I see my mistake.
    But I would like to know more about the sets of all numerals.
     
    Last edited: Apr 29, 2004
  21. Apr 29, 2004 #20
    Does your friend deny the fact that .9999... when rounded is 1? Is that the problem he/she seems to have difficulty with?
     
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