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Math Help - Proof by induction

  1. #1
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    Proof by induction

    hi the link to this question is here http://i223.photobucket.com/albums/d...f/HPIM0535.jpg
    Question 8. I think I was OK in the first part with proving 2n(2n+1) is always odd.
    However I am lost with part b p^n. I'm not sure if the answer written on the scan is correct.
    Thanks
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  2. #2
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    Let p be an odd integer.
    for n=1 p^1=p which is odd.
    Suppose p^k is odd.
    Now p^{k+1}=(p^k)p which by part a is odd.

    So the answer on the sheet was a little off.
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  3. #3
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    Quote Originally Posted by minicooper58 View Post
    hi the link to this question is here http://i223.photobucket.com/albums/d...f/HPIM0535.jpg
    Question 8. I think I was OK in the first part with proving 2n(2n+1) is always odd.
    However I am lost with part b p^n. I'm not sure if the answer written on the scan is correct.
    Thanks
    Am I missing something? 2n(2n+1) is always even, not odd. n(2n+1) is an integer so 2n(2n+1) is 2 times an integer- an even number.
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  4. #4
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    The PDF said proving that the product of two odd integers is odd. I'm assuming he meant (2m+1)(2n+1). Though, what he wrote in the thread is also written on the side of the PDF... interesting.
    Last edited by MattMan; July 1st 2010 at 04:57 PM.
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  5. #5
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    The exercise in the photocopied page says to show that the product of two odd numbers is odd.

    That is expressed:

    Show (2n+1)(2k+1) is odd.

    And it's trivial and no induction needed:

    (2n+1)(2k+1) = 2n2k + 2n + 2k +1 = 2(2nk + n + k) + 1 is odd
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  6. #6
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    Part b proof by induction

    hi I apologise for confusing the written answer on the PDF. Is two odd numbers written as (2n-1)(2n+1) acceptable as an answer for part a)
    For part b) How can I use this answer for the induction proof of p^n.
    Thanks
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  7. #7
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    Quote Originally Posted by minicooper58 View Post
    Is two odd numbers written as (2n-1)(2n+1) acceptable as an answer for part a)
    No, it's not. You're not being asked to prove something about the special case of two consecutive odd numbers, but rather about ANY pair of odd numbers. And I gave the answer just above.

    Quote Originally Posted by minicooper58 View Post
    For part b) How can I use this answer for the induction proof of p^n.
    Suppose p is odd. Show p^n is odd. Induction on n:

    n=1. Then p^n = p is odd.

    Suppose p^n is odd. Show p^(n+1) is odd.
    p^(n+1) = (p^n)*p. But we have p^n is odd, and p is odd, and the product of two odd numbers is odd, so p^(n+1) is odd.
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  8. #8
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    Quote Originally Posted by minicooper58 View Post
    hi the link to this question is here http://i223.photobucket.com/albums/d...f/HPIM0535.jpg
    Question 8. I think I was OK in the first part with proving 2n(2n+1) is always odd.
    This is not the product of 2 odd numbers as 2n is even.

    Let n be odd.

    Then n\pm2k is also odd.

    P(k)

    n\left(n\pm2k\right)=odd

    P(k+1)

    n\left(n\pm2(k+1)\right)=odd

    Show that P(k) being true will cause P(k+1) to be true

    Proof

    n\left(n\pm2k\right)=n^2\pm2kn=odd ?

    n\left(n\pm2(k+1)\right)=n^2\pm2kn\pm2n

    If P(k) is true, then n^2\pm2kn is odd,

    therefore since 2n is even and odd+even=odd, so P(k+1) is true if P(k) is.

    Then test for an initial value, such as n=1.

    There's no need to use induction for part (a) but it can be used nevertheless.



    Part (b)

    p=odd

    P(k)

    p^k=odd

    P(k+1)

    p^{k+1}=odd

    Try to show that P(k) being true causes P(k+1) to be true

    Proof

    p^{k+1}=(p)\left(p^k\right)=(odd)(odd) if P(k) is true

    Test for an initial value
    Last edited by Archie Meade; July 2nd 2010 at 03:59 AM.
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