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Math Help - Induction Questions

  1. #1
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    Induction Questions

    I have quite a few questions and so i just made it an image. Also attached.
    Induction Questions
    Only need help with questions 2 and 5

    Oh and so far my lecturer has taught well-ordering, strong induction and simple induction. But I could only follow simple induction... the other two I'm quite clueless about >< Though tell me which method is best for each question.

    Attempts
    Question 2
    I have no clue to how to start it..
    all i've done is
    24 = 7 + 7 + 5 + 5
    25 = 5 + 5 + 5 + 5 + 5
    26 = 7 + 7 + 7 + 5
    27 = 7 + 5 + 5 + 5 + 5
    28 = 7 + 7 + 7 + 7
    29 = 7 + 7 + 5 + 5 + 5
    no idea what to do next

    Question 5
    Show that n/t - 1/(q+1) is positive and numerator is less than n
    where t = nq + r with 0 < r < n

    (get common denominator then expand and simplify)
    n/t - 1/(q+1)
    = n(q + 1)/[t(q+1)] - t/[t(q+1)]
    = [n(q+1) - t] / [t(q+1)]
    = [nq - t + n] / [t(q+1)]

    t = nq + r
    nq - t = -r

    hence n/t - 1/(q+1)
    = [n-r] / [t(q+1)]

    from 0 < r < n
    n > r therefore n - r > 0 (proved that numerator is positive)
    and since r > 0 then n - r < n (proved that numerator is < n)

    I'm not sure where to go from here

    Please someone help me however you can..
    Thank you in advance!!
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  2. #2
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    Quote Originally Posted by moocav View Post
    I have quite a few questions and so i just made it an image. Also attached.
    Induction Questions
    Only need help with questions 2 and 5

    Oh and so far my lecturer has taught well-ordering, strong induction and simple induction. But I could only follow simple induction... the other two I'm quite clueless about >< Though tell me which method is best for each question.

    Attempts
    Question 2
    I have no clue to how to start it..
    all i've done is
    24 = 7 + 7 + 5 + 5
    25 = 5 + 5 + 5 + 5 + 5
    26 = 7 + 7 + 7 + 5
    27 = 7 + 5 + 5 + 5 + 5
    28 = 7 + 7 + 7 + 7
    29 = 7 + 7 + 5 + 5 + 5
    no idea what to do next


    This was already practically solved some weeks ago......**sigh**...take any natural number n>29\Longrightarrow \exists k\in\mathbb{N}\,\,s.t.\,\,\,24\leq n-5k\leq 29 (why? Think!) , so n-5k can be written as a sum of 5's and 7's...and voila!



    Question 5
    Show that n/t - 1/(q+1) is positive and numerator is less than n
    where t = nq + r with 0 < r < n

    (get common denominator then expand and simplify)
    n/t - 1/(q+1)
    = n(q + 1)/[t(q+1)] - t/[t(q+1)]
    = [n(q+1) - t] / [t(q+1)]
    = [nq - t + n] / [t(q+1)]

    t = nq + r
    nq - t = -r

    hence n/t - 1/(q+1)
    = [n-r] / [t(q+1)]

    from 0 < r < n
    n > r therefore n - r > 0 (proved that numerator is positive)
    and since r > 0 then n - r < n (proved that numerator is < n)

    I'm not sure where to go from here

    Please someone help me however you can..
    Thank you in advance!!

    ...so by the inductive hypothesis the number \frac{n}{t}-\frac{1}{q+1} can be written as a sum of no more than n-1 fractions with 1 as numerator, and thus...(!!!)

    Tonio
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