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Math Help - Infinite periodic decimal.

  1. #1
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    Infinite periodic decimal.

    Guys chill,
    I know this is probably the dummest question that have been ever uploaded in this forum, but I have to know that answer!

    So if 1/3=0.3333333(and so on)
    then how can 1/3 *3 = 0.33333333333 *3 = 1
    why isn't it 0.9999999(and so on) since 3*3=9, not 10.
    I know that if we do it in a fraction we will get 1 as 1*3/3 = 3/3 = 1
    but how? is the answer really one or is it 0.9999999999999999999999999 (infinity) rounded to 1?

    thanks
    Last edited by mr fantastic; June 6th 2011 at 01:26 AM. Reason: Re-titled.
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  2. #2
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    also I read a joke that made me think
    here it is:

    An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer.
    The bartender says "I see where this is going", and pours two beers.

    so if 1+0.5+0.25+0.125 goes on to infinity, will it reach 2, the numbers get smaller as we go, hmmmm
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  3. #3
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    Quote Originally Posted by igcsestudent View Post
    Guys chill,
    I know this is probably the dummest question that have been ever uploaded in this forum, but I have to know that answer!

    So if 1/3=0.3333333(and so on)
    then how can 1/3 *3 = 0.33333333333 *3 = 1
    why isn't it 0.9999999(and so on) since 3*3=9, not 10.
    I know that if we do it in a fraction we will get 1 as 1*3/3 = 3/3 = 1
    but how? is the answer really one or is it 0.9999999999999999999999999 (infinity) rounded to 1?

    thanks
    \displaystyle \begin{align*} x &= 0.999\,999\,999\dots \\ 10x &= 9.999\,999\,999\dots \\ 10x - x &= 9.999\,999\,999\dots - 0.999\,999\,999\dots \\ 9x &= 9 \\ x&= 1\end{align*}

    Therefore \displaystyle 0.999\,999\,999\dots = 1
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  4. #4
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    The notation 0.99999... means 0.9+ 0.09+ 0.009+ ...= 0.9(1+ .1+ .01+ ...), an infinite geometric sum. It is well known that the geometric sum, \sum_{n=0}^\infty ar^n= a\sum_{n=0}^\infty r^n, as long as |r|< 1, is \frac{a}{1- r}. In this case, a= 0.9 and r= .1< 1 so the sum is \frac{0.9}{1- 0.1}= \frac{0.9}{0.9}= 1.

    There is no rounding. That "infinite decimal" is exactly equal to 1.
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