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by AbsolutValu on April 8th, 2009

AbsolutValu

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Mathematicians, I think that .999999... (infinite 9s) is equivalent to 1.0. Do you agree?

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Answers. 15 helpful answers below.

  • by bobbinhood on April 8th, 2009

    bobbinhood

    Yes, they are equivalent. The argument I have seen for it is the following:

    Consider:
    1/3 = 0.3333...
    2/3 = 0.6666...
    3/3 = 0.9999...

    But we know 3/3 = 1.
    Therefore, 0.9999... = 1


    There are several proofs for this identity here: http://en.wikipedia.org/wiki/0.999...#Proofs
    The first one is similar to (and better than) the argument I showed. You should be able to follow the second one easily as well. Your ability to understand the rest of them would depend on how much math you have taken.

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  • by Tinkerbell on April 9th, 2009

    Tinkerbell

    The key here is the infinite 9s.

    You are correct and here is a simple proof

    X = 0.999999999999...
    10x = 9.99999999999999...
    10X - X = 9.99999999999... - 0.99999999999...
    9X = 9
    X = 1

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  • by Charger on April 8th, 2009

    Charger

    What is one third of one? It's .33333333... (infinite three.)

    Now, three times that is .99999999..... (infinite 9s.)

    Therefore, .999999999.....(infinite 9s)=1.

    QED.

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  • by dandan on April 9th, 2009

    dandan

    Another way to prove this (use a calculator if you need to):

    1/9 = 0.1111111111111111111111 ......
    2/9 = 0.2222222222222222222222 ......
    3/9 = 0.3333333333333333333333 ......
    ...
    ...
    ...
    8/9 = 0.8888888888888888888888 .......

    9/9 = 0.9999999999999999999999 ........

    But, as you know 9/9=1

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  • by Mosexy on April 8th, 2009

    Mosexy

    It's limit would be 1.

    But that's not the same as saying it is 1 because 1 is a whole number.

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  • by hudini23 on April 8th, 2009

    hudini23

    no. its always gunna be the smallest amount below 1.0 no matter how many 9's...

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  • by KiwiBruce on April 8th, 2009

    KiwiBruce

    No, Ive worked with chemists, that last decimal point always counts.

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  • by arnold on May 4th, 2009

    arnold

    yup you are right, but it's not a matter of opinion.

    0.99... (with n 9s) = 1 - (1 / 10) ^ n

    so

    0.99... (with unlimited 9s) = lim(n) 1 - (1 / 10) ^ n

    which indeed equals to 1, taking into account the behaviour of

    a ^ n

    where 0 < a < 1

    there's no blurry stuff in math :)

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  • by nutmegcollector on April 8th, 2009

    nutmegcollector

    Yes, if you round off to 2 digits, it's 1.0; 3 digits 1.00; 4 digits 1.000

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  • by Anonymous T on August 17th, 2009

    Anonymous T

    It will come to You more easily to accept
    this fact, when You understand that
    0.9999... (which is not formal) or more
    formally 0,(9) is just a symbol.

    It's not 0 dot and nines stacking up
    one after another. What You must understand
    is that in maths You have many notations
    that formally are nonsense - a thought
    abbreviation -, but by the mere fact,
    that "everyone" (i.e. all mathematicians
    and nobody else) understand this thought
    abbreviation, nobody discusses it.
    It's convenient.
    Everyone knows what it stands for and
    how it would look to formally write it.
    This thought abbreviation is made so
    that proves don't take eons longer.

    What it actually means is:
    lim n -> +oo 9/10 + 9/(10^2) + 9/(10^3) + 9/(10^4) + ... (again it's informal)

    More formally it would be:
    a(n) = 9/10 + 9/(10^2) + ... + 9/(10^n) (yet again informal - this would stand for 0.9999...999 with
    n nines)

    Even more formally:
    a(n) = Sum[k=1 to n] (9/(10^k))

    Now we can finally give the definition of 0.(9)

    0.(9) = lim n->+oo a(n)
    or
    0.(9) = lim n->+oo {Sum[k=1 to n] (9/(10^k))}

    That's pretty lot of text for such an easy
    thing, don't You think? Yeah. That's why
    mathematicians don't do that stuff on paper
    or the rain forests might seize to exist. ;)

    So what does this lim actually mean?

    Here we go again. Formal definition
    of a limit:
    lim n->+oo a(n) = g <--- g is the limit
    <=>
    For every epsilon > 0, there
    exists a natural N, such that
    for every n > N we have:
    |a(n) - g| < epsilon

    -----------------------

    Hence, if You want to prove that
    0.(9) = 1
    You have to prove that:

    For every epsilon > 0, there
    exists a natural N, such that
    for every n > N we have:
    |Sum[k=1 to n] (9/(10^k)) - 1| < epsilon

    The number under the modulus
    is negative. We clear the modulus
    and add minus.

    1 - Sum[k=1 to n] (9/(10^k)) < epsilon
    <=>
    1/(10^n) < epsilon <--- that's because 1 - 0.999...99
    is actually 0.000....001 (n-1 zeros and 1)
    <=>
    1/epsilon < 10^n

    ----

    Soooo... to prove 0.(9) = 1,
    we have to prove that:

    For every epsilon > 0, there
    exists a natural N, such that
    for every n > N we have:
    1/epsilon < 10^n

    We notice that for big enough n's
    10^n > n. Hence, it's enough if we
    prove 1/epsilon < n
    to prove 1/epsilon < 10^n

    -----

    Soooo... to prove 0.(9) = 1,
    it's ENOUGH to prove that:

    For every epsilon > 0, there
    exists a natural N, such that
    for every n > N we have:
    1/epsilon < n

    How to do that?
    Just point to such an N.

    epsilon is > 0, so 1/epsilon is a finite
    positive number. The N we are looking
    for could be for example
    round(1/epsilon) + 1,
    since
    1/epsilon < round(1/epsilon) + 1.

    -----

    There we have it.
    For every epsilon > 0 we found
    a natural N = round(1/epsilon) + 1,
    such that for every n > N we have:
    1/epsilon < n

    So 0.(9) is indeed equal 1.

    -----

    Ehhhem.
    Breaks Your heart doesn't it? ;)

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  • by xprofessor on June 3rd, 2009

    xprofessor

    Yes. Mathematicians say that an infinite sequence of 9's after the decimal point -- or notation to indicate such -- is equal and equivalent to the real number that we call 1 -- just two names for the same number. One of them is much more complicated to write down than the other.

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  • by royal77 says hello friend on May 2nd, 2009

    royal77 says hello friend

    .999999999999999... is close enough to 1.0 that it is alright to equal it to 1.0

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  • by arnold on June 4th, 2009

    arnold

    the two numbers are exactly the same. this is a duplicate question, you can check the proof here:

    http://www.answerbag.com/q_view/1392961

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  • by Beany on June 4th, 2009

    Beany

    Of course they're the same

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  • by robertbrozewicz on March 6th, 2011

    robertbrozewicz

    Approximately yes, but it is not equal to 1 as 1 is 1, but 0.9999999999999.... tends to one but seems that never reaches it. We can say that 1 is its limit. We normally just don't care and we use precision notation that is we don't need to worry about let us say 1 million of 9s because this does not make sense in everyday calculations, so we accept it as equal to 1.

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More Questions. Additional questions in this category.

You're reading Mathematicians, I think that .999999... (infinite 9s) is equivalent to 1.0. Do you agree? - which can also be phrased in the following ways:

  • Do you believe that .999~ is the EXACT same (that is not approaching, not approximate, not near, not rounded up) as 1? If not, why not?

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