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Prove that k is even

  1. Oct 4, 2009 #1
    1. The problem statement, all variables and given/known data

    a) Let k be any integer. Prove that if k3 is even, then k is even.


    2. Relevant equations



    3. The attempt at a solution

    Proof by contradiction:

    If the hypothesis is true, then k3 cannot be even if k is odd.

    Assume k is odd:
    k = 2n + 1, such that n is any integer.
    k3 = (2n + 1)3
    k3 = (2n +1)(2n +1)(2n +1)
    k3 = 2(k3 + 5k2 + 3k) + 1
    Now, (k3 + 5k2 + 3k) is any integer.
    Therefore k3 is odd if k is odd, hence if k3 is even if k is even.
     
  2. jcsd
  3. Oct 4, 2009 #2
    Looks okay to me, unless you were trying to do a proof by contradiction. Your current proof is one of a contrapositive.
     
  4. Oct 4, 2009 #3
    My bad, last statement is iffy. I think it should read if k cubed is even, then k is even
     
  5. Oct 5, 2009 #4
    Ok thanks
     
  6. Oct 5, 2009 #5

    Mark44

    Staff: Mentor

    The line above doesn't make sense to me. If you multiply out the right side, you get 8n3 + 12n2 + 6n + 1 = 2(4n3 + 3n + 3) + 1. This is clearly an odd integer, which means that k3 is an odd integer.
    Another approach is to prove the statement directly.
    Assume that k3 is even.
    Then k3 = 2m for some integer m.
    Because k occurs to the third power, there must be three factors of 2 on the right.
    So k3 = 2*2*2*n for some integer n.
    Again, because k occurs to the third power, n must have three equal factors, say n = p3.
    Hence, k3 = 2*2*2*p*p*p, from which you can easily show that k = 2p, an even integer.
     
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