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Thread: how does addition and multipication of numbers with repeating decimals work?

  1. #1
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    Question how does addition and multipication of numbers with repeating decimals work?

    i just started pre-calculus and in the fundamentals it says about rational numbers and how their decimal representation repeat if they are rational. now what happened:
    using the windows calculator i divided 1 by 3 i get 0.33... and when i multiplied it by three the answer was 1. i don't get it.
    i was expecting 0.99.... to be the answer.

    now my question:

    how does addition of numbers with repeating decimals like 0.33... work that multiplying it to 3 gives me 1 and not 0.99...
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  2. #2
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    Re: how does addition and multipication of numbers with repeating decimals work?

    there's a bit of a trick dealing with infinitely repeating decimal numbers

    the question is what does $0.9999\dots$ equal

    let $x = 0.9999\dots$

    $10x-x = 9x$

    $x = \dfrac{10x-x}{9}$

    $10x = 9.9999\dots$

    $10x - x = 9$

    $x = \dfrac{9}{9} = 1$
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  3. #3
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    Re: how does addition and multipication of numbers with repeating decimals work?

    Yes, 0.9999..., extending infinitely is exactly equal to 1 but I don't think that is quite what mohsentux is asking. The "Windows calculator" shows numbers to 32 decimal places and, like any modern calculator, retains one or two more decimal places internally.

    Dividing 1 by 3 gives 0.333333333333333333333333333333333 and multiplying that by 3 gives 0.9999999999999999999999999999999999. But those additional, "internal" but undisplayed decimal places are "99" so the calcutor rounds up and displays 1.000.
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