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Thread: Math problems

  1. #61
    Saizou is offline Senior Member Always Around
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    Quote Originally Posted by Urameshi-sama View Post
    ^I don't want to do the calculus, but you could split up .999... to 9/10+9/100..., which is a convergent infinite series that approaches 1.
    Well, you're supposed to prove that the two terms are equivalent, not that one converges towards the other. While it is true that 0.999... converges towards 1, it doesn't prove that 0.999... = 1.

    It's actually possible to prove this with simple arithmetic, or by an indirect proof.

  2. #62
    adonai is offline Senior Member Community Builder
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    Quote Originally Posted by Saizou View Post
    GLORIOUS REVIVAL.

    All right y'all. Time for a screwy one.

    Prove that 0.999... = 1.

    For the record, 0.999... means an infinite amount of nines.


    1/3=.333...

    (1/3)*3=.999...

    (1/3)*3=1

    1=.999...

  3. #63
    Urameshi-sama is offline Senior Member Community Builder
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    @Sai-
    Can you just structure it as an infinite geometric series and then use the sum formula S(n) =a/(1-r)=(9/10)/(1-[1/10])=1

    Does this work?
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  4. #64
    Saizou is offline Senior Member Always Around
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    Quote Originally Posted by adonai View Post
    1/3=.333...

    (1/3)*3=.999...

    (1/3)*3=1

    1=.999...
    Correct.

    The other easy way to prove this would be to demonstrate that there cannot exist an real number X that satisfies the equation 0.999...+X=1.

    Quote Originally Posted by Urameshi=Sama
    @Sai-
    Can you just structure it as an infinite geometric series and then use the sum formula S(n) =a/(1-r)=(9/10)/(1-[1/10])=1

    Does this work?
    As far as I can see, it does work.


    If anyone else wants to post a new problem, feel free to do so.

  5. #65
    Urameshi-sama is offline Senior Member Community Builder
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    Wow, I have to get out of problem solving mode to proofing mode...I'll let you and adon post the problems. I shame this thread with my lack of insight
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