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.999 does not equal 1 because

  1. Nov 6, 2009 #1
    Represent .999... by (1-10-n), limit as n>infinity.
    (1-10-n)10n, limit as n>infinity = 1/e (binomial expansion)
    (1+10-n)10n, limit as n>infinity = e
    (1)10n, limit as n>infinity=1

    I think .999... is a transcendental number that can in almost every case be treated as equal to 1. Any solution to a finite algebraic equation that approximates 1 but is less than 1 will always be less than .999...
    If .999... is transcendental then so is every other recurring decimal because they are arbitrarily close to a fraction but not equal to that fraction.

    The above equation seems to converge with the number of digits of accuracy approximately equal to n.
  2. jcsd
  3. Nov 6, 2009 #2
    If you need proof that .999...=1, take a look at the infinite series that represents .999....
  4. Nov 6, 2009 #3
    If 0.99... is not equal to 1, then there's definitely a number between the two. Can you find one such number?
  5. Nov 6, 2009 #4


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    It doesn't work like that. You're saying

    [tex]\mbox{Let}~0.\overline{9} \equiv \lim_{n \rightarrow \infty} (1 - 10^{-n}).[/tex]

    First, the limit of that expression is certainly 1 - no doubt about that. It's not the way one tends to define a decimal expansion but let's use it anyways. You then consider

    [tex]\lim_{n \rightarrow \infty} (1 - 10^{-n})^{10^n}[/tex]
    which is NOT equal to

    \lim_{n \rightarrow \infty} (0.\overline{9})^{10^n}.[/tex]

    The proper expression given your proposed definition for [itex]0.\overline{9}[/itex] would be

    [tex]\lim_{n \rightarrow \infty} (0.\overline{9})^{10^n} = \lim_{n \rightarrow \infty} (\lim_{m \rightarrow \infty} (1 - 10^{-m}))^{10^n},[/tex]

    which is a different thing altogether - there are two limits involved, instead of just the one as in your original post. This makes a very big difference. Note also that you cannot exchange the order of the limits here -that would also be an invalid operation.
  6. Nov 6, 2009 #5
  7. Nov 6, 2009 #6
    I agree with Mute. The point is that you cannot make the single limit you had first into a double limit with a second occurence of the same infinity.
  8. Nov 6, 2009 #7
    This has been done to death. Google is your friend.
  9. Nov 6, 2009 #8
    Correct. If this Wikipedia article doesn't convince you, then probably nothing ever will.
  10. Nov 7, 2009 #9
    [tex]\lim_{n \rightarrow \infty} (1 - 10^{-n})^{10^n}=1/e[/tex]
    has a definite value other than 1. Try some values of n in a calculator.
    Is this expression in error?
    Does this expression not indicate that
    [tex]\lim_{n \rightarrow \infty} (1 - 10^{-n})[/tex] has a value other than 1 in the above expression?
    What then is the value of [tex]\lim_{n \rightarrow \infty} (1 - 10^{-n})[/tex] in the above expression?
  11. Nov 7, 2009 #10


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    [itex]1^{+\infty}[/itex] is an indeterminate form, just like [itex]0/0[/itex], [itex](+\infty) / (+\infty)[/itex], and [itex](+\infty) - (+\infty)[/itex].

    If you can prove 0.999... is unequal to 1, then you have found an inconsistency in the arithmetic of integers.
  12. Nov 7, 2009 #11


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    To the OP:

    What is the decimal representation of 1/3?
    Do you accept that 1/3 = 0.333...?

    If so, multiply x3 on both sides of that equation and see what you get.
  13. Nov 7, 2009 #12

    If 0.999999... is questioned to be equal to 1, then why would one accept that 0.33333333.... equals 1/3? :smile:

    I think the answer to ".999... does not equal 1 because..." should be that the number system has not been specified, as pointed out here:


  14. Nov 7, 2009 #13


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    That occurred to me. Hence my initial question, "What is the decimal representation of 1/3?" Most people are okay with saying 0.333... in answer to that, and don't question it.
    However, the perception that 0.999... looks differently than 1.000... is what distinguishes the two examples in many people's minds.
  15. Nov 7, 2009 #14


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    I don't understand -- do we not have an agreed-upon number system?
  16. Nov 7, 2009 #15


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    The author of that PDF is writing it pretty much specifically to justify the 0.999... != 1 hypothesis. The main loophole he exploits is that we haven't given a fully precise definition of the number system we are teaching them -- and so he's filling in the missing details in a... peculiar... fashion.

    That is not to say there is no merit in non-standard analysis -- but it's like the author is specifically trying to prevent students from understanding what an infinite decimal means.
  17. Nov 9, 2009 #16
    Most people have a terrible understanding of the real numbers. It isn't usually until college (more likely, never) that people learn about the reals as the completion of the rational number system. The vast majority of people thing an irrational number's defining characteristic is that the decimal expansion doesn't repeat. And they believe that you can add two irrational numbers together in the same way you add rationals, using the algorithm they learned in elementary school -- despite the fact that it obviously doesn't terminate!

    The peculiar thing isn't that people don't understand numbers, it's that they so fiercely stick to their broken notions of numbers. They won't even entertain the idea that what they learned was incomplete. After all, business and science are both done with rational numbers! And there's a strange belief (in both math and science) that there's only one correct way these kinds of things can work. But that's the beauty of math! You can make up the rules to be whatever you want as long as you strictly adhere to them.
    Last edited by a moderator: Nov 9, 2009
  18. Nov 9, 2009 #17


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    Most people do not need a good understanding of real numbers. Real numbers are quite obviously a completion of the rationals, they are the numbers between rational like sqrt(2). It is perfectly valid to take nonterminating decimal expansion as a defining characteristic of irrational numbers. Irrationals are aded the same way as rationals, or at least an infinite process analog. I think the trouble is these people (though most of them know .(9)=1 and just want to start stupid threads) do not realize a infinite decimal expansion is inherently a real number. Arguing .(9) is not real is like arguing 17/19 is not rational. One could (and these crack pots do) devise a number system different than the reals (and having horrible properties) but represented by infinite decimal expansions in which two representations that represent the same real number represent different crackpot numbers. The two sides of this "debate" are not even talking about the same thing. We have
    Crackpot: In the crackpot number system clearly .(9)!=1.(0)
    !Crackpot: In the real number system clearly .(9)=1.(0)
    They are both right.
  19. Nov 9, 2009 #18


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    No, that expression is perfectly correct.

    No, it absolutely does not indicate that. [itex]\lim_{n \rightarrow \infty} (1 - 10^{-n}) = 1[/itex]. Your error is the following: consider the expression


    If I take the limit [itex]n \rightarrow \infty[/itex] this becomes [itex](1)^{10^m} = 1[/itex] for any finite value of m. If I instead keep n fixed at some finite value and take [itex]m \rightarrow \infty[/itex] then because [itex]1-10^{-n} < 1[/itex], the expression tends to zero. If I set [itex]m = n[/itex] and then take the limit [itex]n \rightarrow \infty[/itex] I get the value of 1/e. You're mixing up these limiting processes which gives you different results depending on how you take the limit.
  20. Nov 9, 2009 #19


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    I'm possibly nitpicking a bit, but:
    . an infinite decimal expansion is a real number only because we have chosen to use decimal notation to express real numbers
    . we only use the left-finite expansions
  21. Nov 10, 2009 #20
    Most real numbers are not formally describable (the set of formally describable numbers must obviously be countable), so you can question if real numbers really exist.
  22. Nov 10, 2009 #21


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    A common assertion, but with subtle precision issues!

    If "really exist" is to mean the mathematical existence predicate, then your hypothesis here is irrelevant.

    However, if you are saying something about the real world, then the implicit conclusion you made no longer follows.

    e.g. Skolem's paradox demonstrates that, while internally to a (first-order) model of set theory the real numbers are not countable, that need not remain true in an external context. In particular, there exists a countable model of (first-order) set theory.
  23. Nov 10, 2009 #22
    That's really akin to asking whether the geometrical line, infinite sets, ordered pairs, functions or relations really exist. Do ideas exist?
  24. Nov 10, 2009 #23


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    Did you really mean to say that? If so, then .636363636363... and .222222... are irrational, despite the fact that they can be represented respectively as 7/11 and 2/9, which are clearly rational numbers.
  25. Nov 11, 2009 #24
    Thank you for trying to explain this to me.
    Clearly I am looking at the case where m=n for limits. What is the rule for separating limits? For example xsin(1/x) limit x->infinity or sin(y)/y limit y->0 would be undefined if you separate the limits. Or indeed my above examples would be undefined if you seperate the limits.

    by taking m=n, I am trying to make an example where [itex]\lim_{n \rightarrow \infty} (1 - 10^{-n}) != 1[/itex]. (albiet by an infinitesimal amount.) Are you saying this is not allowed by some rule or convention? Please could you explain that rule?
    Duncan Irvine.
  26. Nov 11, 2009 #25
    I disagree with [itex]1^{+\infty}[/itex] is indeterminate.
    If my initial equations are correct then [itex]1^{+\infty}[/itex] is bounded by e, 1/e.
    If the equations are changed to
    [tex]\lim_{n \rightarrow \infty} (1 + x(10^{-n}))^{10^n}=e^x[/tex]

    Then for x=+1 and -1, [itex]1^{+\infty}[/itex] has a value between e^x and e^-x.
    limit as x->0 gives a value between e^+0 and e^-0=e^0=1.

    An interesting variant is:
    [tex]\lim_{n \rightarrow \infty} (1 + i(pi/2)(10^{-n}))^{10^n}=i[/tex]

    An infinitesimal imaginary component can also be expanded into a complex number.

    Duncan Irvine.
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