What number follows 0.9999999... if you repeat the nines to infinity? Is it the same as one or less than one?

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  • 5 months ago

    Let me ask you a question?

    Is 0.333333... equal to 1/3 or is it *almost* 1/3?

    Almost everyone can agree if you divide out 1/3, you get 0.333333... with an unending sequence of threes. The decimal representation of 1/3 is 0.333333... and both are equivalent. Very few people say that 0.333333... is *almost* 1/3. If they do, they are wrong.

    However, when it comes to 0.999999... our intuition gets messed up and we can't make the same assertion. People start arguing that the value isn't exactly 1; that it's just a little smaller than 1 or *almost* 1. So why the different reasoning?

    The fact is, just like 0.333333... is exactly 1/3, it's also true that 0.999999... is exactly 1.

    0.333333... + 0.333333... + 0.333333... = 1/3 + 1/3 + 1/3

    0.999999... = 1

    Another pattern that might help; look at the fractions 1/9, 2/9, etc. written as decimals:

    0.1111... = 1/9

    0.2222... = 2/9

    0.3333... = 3/9 (or 1/3)

    0.4444... = 4/9

    0.5555... = 5/9

    0.6666... = 6/9 (or 2/3)

    0.7777... = 7/9

    0.8888... = 8/9

    0.9999... = 9/9 (or 1)

    Answer:

    The value of 0.9999999... is exactly 1. You can think or imply that it is the number "just before 1" but it isn't.

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  • 5 months ago

    I should probably start my answer with a question myself. What do you mean by "repeat the nines to infinity"? I understand the number 0.99999... where there is a well-defined finite number of nines, but infinity is NOT a number. Therefore, the expression 0.99999 has no meaning unless we give it one. To define what the expression 0.999...inf (or equivalently 0.999r) means, we define the sequence a_n as follows:

    a_1 = 0.9, a_2 = 0.99, a_3 = 0.999, ... or equivalently a_n = 0.999 ... n nines ... 999

    and we DEFINE 0.999r as the LIMIT of this sequence, where the limit of a sequence (assuming it has a limit) is a number L such that we can get a_n arbitrarily close to L simply by choosing a sufficiently large value for n. In the case of 0.999r, we see that no member of the sequence a_n equals 1, but we can get a number as close to 1 as we like simply by writing enough nines in the number 0.99999... Therefore, we can say that the limit of the a_n sequence is EQUAL to 1, or equivalently 0.999r = 1.

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  • 5 months ago

    It's the same as one. I can prove it.

    Let n = .99999999999...

    10n = 9.99999999999...

    -n = -.99999999999...

    ____________________

    9n = 9

    n = 1

    The next number after it is 2.

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  • 5 months ago

    ¹∕₉ = 0.1111111... which I will write as 0.1ᵣ, meaning the last digit repeats indefinitely.

    0.9999999... = 0.9ᵣ = 9 × 0.1ᵣ = 9 × ¹∕₉ = 1

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  • geezer
    Lv 7
    5 months ago

    EVENTUALLY

    0.99999999999999999... will be so close to 1 that it will be 1

    So .. to answer your question

    If 0.9999999999999999... is 1

    the number that ''follows'' 0.9999999999999999 (1) ... is 2

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  • 5 months ago

    0.99999.... to infinity =

    9/9

    Cancel down = 1

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  • TomV
    Lv 7
    5 months ago

    0.999999... is exactly equal to 1.0000...

    This is a very well known (but very non-intuitive) fact of mathematics. There are any number of videos on YouTube demonstrating the proof of that statement. Google "0.9999 YouTube" for a representative listing.

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  • 5 months ago

    0.9999999...

    = 0.9^_ (repeating decimal)

    = 1

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  • 5 months ago

    Its less than one

    But infinity doesnt exist in the real world

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  • 5 months ago

    that is the same as 1.

    If by number following, you mean next integer, that is 2

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