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Math Help - What's wrong with this proof...

  1. #16
    Bar0n janvdl's Avatar
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    Let me teach you some of the basics of a geometric series...

    The difference between a sequence and a series is:

    Series:  S_n = 1 + 2 + 4 + 8 + 16 + ...
    Sequence:  S_n = 1 ; 2 ; 4 ; 8 ; 16 ; ...

    A geometric series is formed by multiplying each term by a constant.

    For example:  S_n = 1 + 2 + 4 + 8 + 16 + ...
    Could also be written as:  S_n = (1)(2^0) + (1)(2^1) + (1)(2^2) + (1)(2^3) + (1)(2^4) + ...

    The n-th term is equal to:  ar^{n - 1}

     a is the starting term, the first term (a = 1 in the example).
     r is the common ratio (r = 2 in the example).

    You can determine  r with the following formula:  r = \frac{T_2}{T_1}

    In series and sequences, the Sigma-sign is very important.

     \sum^{3}_{n = 0} (n + 1) = (0 + 1) + (1 + 1) + (2 + 1) + (3 + 1) = 10

    Now i do hope that's all...
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  2. #17
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    There is nothing wrong with the proof, it is in fact true that .9999.... = 1.


    Note: If this thread starts getting "philosophical" , just be warned, you know I hate that stuff and I will close it.
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  3. #18
    MHF Contributor

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    The basic fact you need to recall is: \sum\limits_{k = J}^\infty  {ar^k }  = \frac{{ar^J }}{{1 - r}}\quad \text{if}\quad \left| r \right| < 1.

    Example: \sum\limits_{k = 5}^\infty  {3\left( {\frac{4}{5}} \right)^k  = \frac{{3\left( {4/5} \right)^5 }}{{1 - \left( {4/5} \right)}}}

    Here is the way a so-called repeating decimal works.
     .9\overline 9  = .9 + .09 + .009 + .0009 +  \cdots
    .9\overline 9  = \frac{9}{{10}} + \frac{9}{{100}} + \frac{9}{{1000}} + \frac{9}{{10000}} +  \cdots
     .9\overline 9  = \frac{9}{{10^1 }} + \frac{9}{{10^2 }} + \frac{9}{{10^3 }} + \frac{9}{{10^4 }} +  \cdots
    .9\overline 9  = \sum\limits_{k = 1}^\infty  {9\left( {\frac{1}{{10}}} \right)^k }  = \frac{{\left( {9/10} \right)}}{{1 - \left( {1/10} \right)}} = 1
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  4. #19
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    Hello, jtl!

    There's nothing wrong with that proof.

    Let N \;=\;0.99999\cdots \;=\;\frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots

    This is a geometric series with first term a = \frac{9}{10} and common ratio r = \frac{1}{10}
    . . Its sum is: . N \;=\;\frac{\frac{9}{10}}{1-\frac{1}{10}} \;=\;\frac{\frac{9}{10}}{\frac{9}{10}} \;=\;1

    Therefore: . 0.99999\cdots \;=\;1

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Another proof

    We have: . N \;=\;0.99999\cdots

    Multiply by 10: . 10N \;=\;9.99999\cdots
    . . Subtract N: . . N \;=\;0.99999\cdots

    And we have: . 9N \:=\:9\quad\Rightarrow\quad N \:=\:1

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  5. #20
    jtl
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    Quote Originally Posted by ThePerfectHacker View Post
    There is nothing wrong with the proof, it is in fact true that .9999.... = 1.


    Note: If this thread starts getting "philosophical" , just be warned, you know I hate that stuff and I will close it.
    Don't worry, with my level of math, I don't think I'll last in a philosophical battle.
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  6. #21
    jtl
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    Thanks everyone, Soraoban and Plato and others, it's clear now.
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