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Math - Proving

  1. Sep 9, 2005 #1

    Meh

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    Prove that the square of an odd integer is always of the 8k + 1, where k is an integer. Any help would be appreciated.
     
  2. jcsd
  3. Sep 9, 2005 #2

    VietDao29

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    An even integer can be written as [tex]2n, \ n \in \mathbb{Z}[/tex]
    And an odd integer can be written as [tex]2n + 1, \ n \in \mathbb{Z}[/tex]
    For example : 5 = 2 * 2 + 1 (n = 2), 7 = 2 * 3 + 1 (n = 3), 13 = 2 * 6 + 1 (n = 6).
    So the quare of an add integer can be written as: [tex](2n + 1) ^ 2 = 4n ^ 2 + 4n + 1[/tex]
    You will try to arrange [itex]4n ^ 2 + 4n + 1[/itex] into 8k + 1. Since you have '+ 1' in both sides, in fact, you just need to arrange [itex]4n ^ 2 + 4n[/itex] into 8k.
    Can you go from here?
    Viet Dao,
     
    Last edited: Sep 9, 2005
  4. Sep 9, 2005 #3

    Meh

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    Hmm what does that "Z" in "n belongs to ..." mean? Not to fimilar with that sign :P
     
  5. Sep 9, 2005 #4

    VietDao29

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    [itex]Z[/itex] is the set of all integers, ie: [itex]\mathbb{Z} = \{..., \ -3, \ , -2, \ , -1, \ 0, \ 1, \ 2, \ 3, \ ... \}[/itex]
    Do you get it now?
    Viet Dao,
     
  6. Sep 9, 2005 #5

    Meh

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    Still don't get it >.< Sorry.
     
  7. Sep 9, 2005 #6

    VietDao29

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    [tex]n \in \mathbb{Z}[/tex] means that n is an integer.
    And because:
    [tex]\mathbb{Z} = \{..., \ -3, \ -2, \ -1, \ 0, \ 1, \ 2, \ 3, \ ... \}[/tex]
    So:
    [tex]n \in \{..., \ -3, \ -2, \ -1, \ 0, \ 1, \ 2, \ 3, \ ... \}[/tex]
    That means n can take any value from that set, n can be -3, or -5, or 9, or 10, or 14, or 600, or -1004, ...
    Viet Dao,
     
  8. Sep 9, 2005 #7

    Meh

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    I understood that, just not the how to solve >.<
     
  9. Sep 9, 2005 #8

    VietDao29

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    An odd integer: [tex]2n + 1, \ n \in \mathbb{Z}[/tex]
    So the square of an odd integer is: [tex](2n + 1) ^ 2 = 4n ^ 2 + 4n + 1[/tex].
    You are going to prove [tex]\forall n \in \mathbb{Z},\ \exists k \in \mathbb{Z} \ | \ 4n ^ 2 + 4n + 1 = 8k + 1[/tex]
    + 1 is in both sides, so you are going to prove:
    [tex]\forall n \in \mathbb{Z},\ \exists k \in \mathbb{Z} \ | \ 4n ^ 2 + 4n = 8k[/tex]
    Note that:
    [itex]4n ^ 2 + 4n = 4n(n + 1)[/itex]
    What can you say about the product of two successive integers? ie : n * (n + 1).
    Can you go from here?
    Viet Dao,
     
  10. Sep 10, 2005 #9

    Meh

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    Ah, alrite I can go from there. Thanks Viet Dao.
     
  11. Sep 10, 2005 #10
    hi, I'm wondering, isn't [tex]\forall k[/tex]? cause k is stated to be an integer? just making sure..
     
  12. Sep 10, 2005 #11

    VietDao29

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    It can't be for all k, because:
    k = 4:
    8 * 4 + 1 = 32 + 1 = 33, and 33 is not a square of any integer.
    k = 5:
    8 * 5 + 1 = 40 + 1 = 41, and 41 is not a square of any integer.
    Viet Dao,
     
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