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Math Help - How to interpret this question

  1. #1
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    How to interpret this question (induction)

    I have the following question in my textbook, it appears at the end of a section that talks about Mathematical Induction:

    Use an induction argument to show that for each natural number n, the interval (n, n+1) fails to contain any natural number.


    Not exactly sure how to structure the proof.
    Since this is induction, I think I'm going to have to deal with two consecutive intervals. Interval k and interval k+1.
    I show that (1, 2) intersected with N is empty then I assume the interval k contains no natural numbers and show that interval k+1 also contains no natural numbers? I don't know if this makes sense to me.

    Maybe I'm looking at this from the wrong angle. Are only the end points the inductive ingredients in the proof?

    Thanks
    Last edited by director; September 9th 2013 at 07:17 PM.
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  2. #2
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    Re: How to interpret this question

    Hey director.

    This seems like a pointless question (no offence), but you could use the fact that [n,n+1] = (n,n+1) + {n} + {n+1} which implies (n,n+1) = [n,n+1] \ {n} or {n+1}.
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  3. #3
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    Re: How to interpret this question

    Base case: n = 0

    (0,1) does not contain any natural number.

    Recall that (0,1) = {x in R: 0 < x < 1}

    For (0,1) to contain a natural number, we must have A = N∩(0,1) ≠ . If we proceed by way of contradiction, we suppose that A is non-empty. By the well-ordering theorem of natural numbers, A must contain a least element, k.

    We then have 0 < k < 1. Since 0 < k, we must have k = S(k'), for some natural number k', that is k = k'+1. Hence k = k'+1 < 1, implies k' < 0, a contradiction, as 0 is the smallest natural number by definition.

    Inductive step:

    Suppose that for n = m ≥ 0 , we have: (m,m+1) contains no natural numbers, and again suppose by way of contradiction that (m+1,m+2) DOES. Then there is some natural number r with m+1 < r < m+2.

    Hence m < r-1 < m+1, so r - 1 cannot be a natural number, which means we MUST have r = 0 (for if r = S(k) for some natural number k, then r = k+1, and thus r -1 = (k+1) - 1 = k, a natural number). But r > m+1 > m ≥ 0, and r cannot both be greater than AND equal to 0. Thus there is no such r, and we are finished.
    Thanks from topsquark
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  4. #4
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    Re: How to interpret this question

    Thanks to both of you Now I know how to think about similar problems.
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