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Math Help - .9 repeated does not equal 1

  1. #1
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    .9 repeated does not equal 1

    I know it is generally accepted and proven that 0.9 repeated is equal to 1. However I got an idea for a proof against it, and decided to write the proof to see where I run into a flaw or contradiction. As of now I am unable to see the flaw in my logic. Can somebody else run through it and tell me where it is? I apologize for the complex notations I added.

    Hypothesis: 0.9 repeated does not equal 1


    Proof:by mathematical induction


    Let the notation 0.(x) mean zero with x amount of 9's after the decimal point and the notation 0.(#) also mean zero with some amount of 9's after the decimal place dictated by the number in place of #


    Let the notation 0.#[x1] #[x2] #[x3] mean that the digit in # is in the decimal place as defined by x1, x2, or x3. For example, 0.5[3] 6[5]1[6] is 0.005061


    Let the zero'th decimal place be the ones decimal place.


    Let P(x) be the predicate 0.(x) is less than 1 if x >= 0


    Step1: Prove P(0)
    P(0)= 0.(0) is less than 1.
    As defined above, 0.(0) = 0
    0 is less than 1.
    Thus P(0) is true.


    Step2: Prove that if P(n) is true, then P(n+1) is true.
    By P(n) we know that 0.(n) is less than 1.
    We must show that 0.(n+1) is less than 1.
    0.(n+1) is equal to 0.(n) + 9 X 10 ^ -(n+1) by separating out the last digit added.
    We must prove the inequality 0.(n) + 9 X 10 ^ -(n+1) < 1
    =9 X 10 ^ -(n+1) < 1 0.(n) by subtracting 0.(n) from each side
    =0 < 1 0.(n) - 9 X 10 ^ -(n+1) by adding 9 X 10 ^ -(n+1) to each side.


    But 1 0.(n) - 9 X 10 ^ -(n+1)
    =0.1[n] - 9 X 10 ^ -(n+1)
    =0.1[n+1]


    So we have 0 < 0.1[n+1] which is true because 0.1[n+1] is a positive number and is greater than 0.

    Therefore we have proven that if P(n) is true, then P(n+1) is true.


    By mathematical induction we have shown that P(n) is true for all integers n >= 0. There fore P(positive infinity) is true or in other words, a 0 with an infinite number of 9's after it is less than 1.
    Last edited by betabob2000; May 11th 2012 at 11:47 AM.
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    Re: .9 repeated does not equal 1

    I think I see one potential flaw. By mathematical induction i have proved this over the interval x = [0, infinity) and this interval does not include infinity (which of course is a concept, not a number.)
    Therefore my predicate cannot be evaluated at infinity, only at numbers with arbitrarily large magnitudes. What do you guys think?
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    Re: .9 repeated does not equal 1

    You are confusing the statement "0.(n) < 1 for infinitely many n" with the statement "0.(∞) < 1". 0.(n) is a terminating decimal whereas 0.(∞) isn't.
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    Re: .9 repeated does not equal 1

    I think what you are saying was similar to what I replied just above. Thank you for the clarification.

    The basis of the proof is that at no point does adding another 9 on make it one, but we can never reach 0.9 repeated by this method, whether by induction or not.
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    Re: .9 repeated does not equal 1

    Yes, we can. 0.(n) is \sum_{r=1}^n\frac9{10^r} whereas 0.(\infty) is \sum_{r=1}^\infty\frac9{10^r} the latter is, by definition, the limit of the sequence of partial sums \sum_{r=1}^n\frac9{10^r}. A sequence of real numbers less than 1 can indeed converge to 1.
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    Re: .9 repeated does not equal 1

    Quote Originally Posted by Sylvia104 View Post
    You are confusing the statement "0.(n) < 1 for infinitely many n" with the statement "0.(∞) < 1". 0.(n) is a terminating decimal whereas 0.(∞) isn't.
    I will add to the above. You proved the for every n \sum\limits_{k = 1}^n {\frac{9}{{{{10}^k}}}}  < 1.

    BUT \sum\limits_{k = 1}^\infty  {\frac{9}{{{{10}^k}}}}  = {\lim _{n \to \infty }}\sum\limits_{k = 1}^n {\frac{9}{{{{10}^k}}}}  = 1
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    Re: .9 repeated does not equal 1

    very cool. I see now how that works. I thought the point of induction is that it proved the predicate all the way up to infinity, but I can see now that this is not necessarily the case.
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  8. #8
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    Re: .9 repeated does not equal 1

    This problem reminds me of the paradoxes of motion posed by the ancient Greek philosopher Zeno of Elea twenty-five centuries ago.
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