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Homework Help: I am stubborn. Prove to me that 0.999 = 1

  1. Feb 3, 2008 #1
    1. The problem statement, all variables and given/known data
    This is not homework, this is me being stubborn. I have seen the proofs, I know that it is professionally accepted, but I still just don't agree.

    2. Relevant equations
    1 = 2
    1 + 1 = 3 for very large values of 1
    Squareroot of anything will lead to very sharp trees.

    3. The attempt at a solution
    One example of a proof:
    1/3 = 0.333...
    2/3 = 0.666...
    3/3 = 0.999...
    3/3 = 1
    Therefore 0.999... = 1
    My issue with this is, I don't consider 3/3 to be 0.999... I consider 3/3 to be 1/1 = 1. The reason this seems illusionary is because 2/3 isn't 0.666. THIS proof, I accept:
    0.999 * (1/3) = 0.333
    0.999 * (2/3) = 0.666
    0.999 * (3/3) = 0.999

    However, when you say that 2/3 = 0.666..., this is the same as 0.666 +(2/3)*0.001 (Or 0.667 if you round). However, when you denote the ellipsis after 0.666... to signify that the 6's go on forever, the fact is that 6 >= 5, therefore no matter how many 6's you have, it will still be over that (For example, 0.666... > 0.666). Therefore, 3/3 is 0.999+(3/3)*0.001, which = 1.

    Another proof says that:
    c = 0.999...
    10c = 9.999... (My issue lies with this line right here, I'll explain below)
    10c - c = 9.999... - 0.999...
    9c = 9
    c = 1
    0.999... = 1

    The problem here? Multiplying by 10 in this case is the same as just adding 9c. However, when you add 9c, you have already made the ASSUMPTION that c = 1 (Therefore when you add 9c, you just add 9), while first off assuming that c = 0.999...

    To demonstrate how I see this as a flaw, here's a proof that 10 = 5
    c = 10
    10c = 55 (See the assumption here? I assumed that c already equaled 5, and then just added 9c to each side, but that contradicts the first statement, rendering the proof invalid.)
    10c - c = 55 - 10
    9c = 45
    c = 5
    10 = 5

    And as a third note, this proof here:
    My problem with this is that the limit of something IS NOT that something, unless it can be done directly.

    The limit of 2n+5 as n approaches 10 = 25. This is fine.

    However, say we have function F (Too lazy to find a valid function for this). F has the variable n in it, and when you plug in n = 0, the whole thing breaks apart (Divide by zero, or squareroot of negative thing). However, due to factoring, L'hopal's rule, or some other trick, we can figure out the limit anyway.

    The limit of F as n approaches 0 = 2.
    The problem here? As n APPROACHES 0, the value of the function APPROACHES 2.
    The LIMIT equals 2. The value of F when n = 0, is not 2.

    Unless the question can be done without limits at all (Such as 2n+5), the limit will only tell you what the value WOULD be, IF it existed. Using the proof above shows that as 0.999 adds on more 9s, the value APPROACHES 1.. this is fine. But it does not equal.
    Last edited: Feb 3, 2008
  2. jcsd
  3. Feb 3, 2008 #2


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    It is good that you are stubborn and refuse to accept such facts without proof. I think the problem lies with the concept of "infinity", as I conclude from quotes like
    (which is, by the way, that we mean by the ellipsis and which is true: 2 / 3 is strictly greater than 0.666 ( = 0.6660000000.....)).

    Note that
    -- c = 10
    -- 10c = 55
    is flawed because 55 is not equal to 10 x 10.
    On the other hand,
    -- c = 0.999...
    -- 10c = 9.999...
    can be obtained by multiplying both sides by 10, in the same fashion as
    -- c = 12.34
    -- 10c = 10(12.34) = 123.4

    Next, I can always subtract the first equation from the second, as in;
    -- c = 12.34
    -- 10c = 123.4
    -- 9c = (10c - c) = (123.4 - 12.34) = 111.06
    They key in the 0.99... = 1 proof is indeed that "Multiplying by 10 in this case is the same as just adding 9c." as you write.

    But if you're still not satisfied, what about a mathematically rigorous argument?

    [tex]a_n = \sum_{i = 1}^n \left( 9 \times 10^{-n} \right)[/tex]
    The limit
    [tex]\lim_{n \to \infty} a_n = a[/tex]
    has a rigorous meaning, for it is defined as: for each number [itex]\epsilon > 0[/itex] there exists an integer N such that [itex]|a_n - a| < \epsilon[/itex] for all [itex]n \ge N[/itex].
    Now you can try to prove that
    [tex]\lim_{n \to \infty} a_n = 1[/tex]
    and hopefully you also see that this limit just means adding more and more nines to the end of 0.99999....
    Last edited: Feb 3, 2008
  4. Feb 3, 2008 #3
    I edited my post to address the limit-of-the-sum proof. For the sake of not spamming, refer to the original post for my response to that one.
  5. Feb 3, 2008 #4


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    What, then, is your definiition of 0.9999....?

    The only definition I know of an decimal number with an infinite number of digits, say [itex]0.a_1a_2a_3\cdot\cdot\cdot a_n\cdot\cdot\cdot[/itex] is that it is the infinite sum [itex]\sum_{n=1}^\infty a_n/10^n[/itex].

    No one is claiming "the limit of something IS that something" but that your decimal 0.99999.... IS the limit of 0.9+ 0.09+ 0.009+ .. BY DEFINITION of decimal numeration.

    And no one said it was. The point is that while it is true that the sequence 0.9, 0.99, 0.999, 0.9999, .... approaches 1, it's limit is 1. And the number 0.99999... is, again, by definition of the base 10 numeration system, the limit not the sequence!

    [/quote]Unless the question can be done without limits at all (Such as 2n+5), the limit will only tell you what the value WOULD be, IF it existed. Using the proof above shows that as 0.999 adds on more 9s, the value APPROACHES 1.. this is fine. But it does not equal.[/quote]
    You can't answer it without limits because limits are what infinite decimals are.

    Once again, the number 0.999999.... is not the sequence 0.9, 0.99, 0.999, ..., it is the limit of that sequence.

    If you disagree with that, then I will ask again, what is your definition of infinite decimals in the base 10 numeration system?
  6. Feb 3, 2008 #5
    I think I see something here.
    'cause, I can accept that the limit of this approaches 1... and if that's what the definition of 0.999... is (That it is a notation to express an otherwise unexpressable limit), then that I can see.

    But as far as 0.999... meaning just a number where the 9's go on forever... not agreed.

    What's the difference? One (The limit) says what it would equal, if such a thing could be reached. The other is a flat concept, which doesn't equal 1, far as I can see.

    If the elipsis notation is just another way to say the limit, then we agree.
  7. Feb 3, 2008 #6


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    OK, I'm going to do one more attempt to (re)state a proof. After that I give up, as the point of a prove is to convince someone that a statement is true because the proof shows it, while you seem to be convinced that the proofs are flawed because the statement is false.

    I'll enumerate my steps, so you can more easily refer to them
    1. Let us agree that 0.999... is the notation for a zero with infinitely many nines behind the decimal
    2. Let us agree that 0.999... is (by definition because it can be expressed in decimal form) a real number -- if you disagree the whole discussion is vacuous anyway
    3. Let us agree that we can define the distance between two real numbers x and y as d(x, y) = |x-y| and that two real numbers x and y are the same if and only if d(x, y) = 0.
    4. Since [itex]0.999\ldots \le 1[/itex], there is a number [itex]\epsilon \ge 0[/itex] such that |1 - 0.999...| = 1 - 0.999... = [itex]\epsilon[/itex].
    5. Now by basic arithmetic, add 0.999... to both sides of this equation, getting 1 = 0.999... + [itex]\epsilon[/itex].
    6. If you reject the statement 0.999... = 1, you automatically claim that [itex]\epsilon > 0[/itex]. For example, you say that [itex]\epsilon = 0.001[/itex]. Then I add the numbers and show to you that
      Code (Text):

      0.0010000000....  +
      which I hope you agree is not equal to one -- in fact, it is larger so the epsilon you gave me was too large.
    7. Since I can play this trick for any positive non-zero epsilon you give me, and you accepted that 0.999.... <= 1 so epsilon cannot be negative, wouldn't you have to agree that epsilon must be exactly equal to zero?

    It's not otherwise inexpressible, we can use the symbol 1 (with its usual meaning) :tongue: But let's not get caught in circular arguments :smile:

    You are making it sound like L'Hopitals rule is inferior to plugging in the value. So you don't consider
    [tex]\lim_{x \to 0} \frac{\sin x}{x} = 1[/tex]
    by L'Hopitals rule to be obtained in a agreeable/favourable/valid way?

    It cannot, because, as HallsOfIvy explained, 0.999... is a limit by definition.

    By the way, how about this statement?
    The average of 0.99999(infinitely many nines) and 1.00000(infinitely many zeroes)1 is 1. Mathematically,
    \tfrac12 \left[ \left( \lim_{n \to \infty} 9 \times 10^{-n} \right) + \left( 1 + \lim_{n \to \infty} 10^{-n} \right) \right] =
    \lim_{n \to \infty} \tfrac12 \left[ 9 \times 10^{-n} + \left( 1 + 10^{-n} \right) \right] = 1[/tex].
    Last edited: Feb 3, 2008
  8. Feb 3, 2008 #7


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    First off, I will point out that the decimal numerals ...000.999... and ...0001.000... are different.

    There are two primary ways to define the notion of a decimal number.

    Method 1
    (1) Define the real numbers.
    (2) For each (left-terminating) decimal numeral S, define it's value to be the value

    [tex]S = \sum_{n = -\infty}^{+\infty} S_n b^n[/tex]

    where b is i + i + i + i + i + i + i + i + i + i, and i denotes the multiplicative identity. (i.e. b = 10, but I'm trying to avoid using decimal numbers to say that)

    In this method, it's easy to see that the infinite sums that define the values of ...0001.000... and ...000.999... are equal, and so those numerals are equal when interpreted as decimal numbers.

    Method 2
    Provide 'algorithms' for directly computing all elementary arithmetic operations and relations on decimal numbers.

    In this method, the algorithm for computing equality explicitly states that
    Sx999... = Sy000...​
    (where S is any left-infinite string of symbols, x is a decimal digit less than 9, and y is the digit after x)

    If you apply the algorithm for computing equality to
    ...0001.000... = ...000.999...​
    you will calculate that the equation is true.
  9. Feb 3, 2008 #8


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    No, the sequence "approaches" 1. The limit is 1. The limit of a sequence is a number- it doesn't "approach" anything.

    Then I can only suggest that you review the concept of limit. If you agree that the limit of a sequence is a number, then you agree that [itex]0.\overline{999}[/itex] is equal to 1.

    I used a different notation for a repeating decimal because I want to emphasise that the "ellipsis" is not the crux of the matter- it is the definition of "decimal notation" that says the number is a limit.
  10. Feb 3, 2008 #9
    If Goldenwind hasn't accepted the all the proofs presented to him already, perhaps he wants proof that the reals is a linear continuum (i.e. for every distinct points x and y, there exists z such that x < z < y). Is that what you are looking for?

    Because there is no number between 0.999999... and 1 so they must be equal ASSUMING that each real number has no immediate successor.
  11. Feb 3, 2008 #10


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    That's the definition of "densely ordered"; the property of being a continuum is far stronger.
  12. Feb 3, 2008 #11


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    "Properties of the reals", including being a continuum, is not relevant. Every number involved is a rational number.
  13. Feb 3, 2008 #12
    something I always remember in schoolwork:

    if you really don't get a proof after considerable work, and it doesn't look like a big problem, then you're probably just missing some definition or theorem.
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