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Math Help - Does 0.999... have an independent existence?

  1. #1
    RMB
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    Does 0.999... have an independent existence?

    We know that analytically 0.999... = 1.0. But does 0.999... have any sort of "independent" existence (on the real-number line or otherwise)? Here's my argument against that:

    Consider the semi-open line segment [0,1). This set of points has no greatest member. Let's say it did have a greatest member. Then surely that would equal 0.999.... Put another way, if 0.999... doesn't equal 1, then we have to conclude that 0.999... only equals 0.999..., which is to say that it's in a class by itself. But where on the real-number line is it located? Intuitively, it would have to be the point "just before" unity. But no such point exists because no two points are "next to" each other on the line. Therefore the idea that 0.999... exists is vague and has to be relegated to the realm of metaphysics.

    My argument is intuitive, not analytical. But is the conclusion correct?
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  2. #2
    Member anonimnystefy's Avatar
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    Re: Does 0.999... have an independent existence?

    Up to the end it is correct,but i don't agree with you that 0.999... doesn't exist.
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  3. #3
    Super Member TheChaz's Avatar
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    Re: Does 0.999... have an independent existence?

    There's plenty wrong with this!
    .9999.... represents the sum of an infinite geometric series. The claim is that this series converges and equals one.

    There is nothing vague about the existence of the limit in question.
    And yes, we say that two numbers that are arbitrarily "next to" each other are indeed equal.
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