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Math Help - Floors and Ceiling proof (Discrete Mathematics)

  1. #1
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    Floors and Ceiling proof (Discrete Mathematics)

    ∀a ∈ Z, if a is odd then ⌊a/2⌋ is odd or ⌈a/2⌉ is odd.

    Hi,

    I'm having a little trouble proving this statement.
    I supposed a is an integer that is odd.
    Then a has the form 2k+1 for some integer k.
    Then a/2 is k + 1/2.
    Now I'm not sure how to show that the floor or ceiling is odd.

    I was thinking of saying that since k is an integer, once it is added with a strictly rational number(1/2), it would result in a non-integer, and thus be a number between two integers. Then I could say that since every consecutive integer is of opposite parity, then when you round down(floor) or round up(ceiling), it would result in 2 consecutive integers, and one of them must be odd.

    Would that be sufficient?

    Thanks in advance!
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  2. #2
    Senior Member
    Joined
    Feb 2008
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    Quote Originally Posted by Whiz View Post
    ∀a ∈ Z, if a is odd then ⌊a/2⌋ is odd or ⌈a/2⌉ is odd.

    Hi,

    I'm having a little trouble proving this statement.
    I supposed a is an integer that is odd.
    Then a has the form 2k+1 for some integer k.
    Then a/2 is k + 1/2.
    Now I'm not sure how to show that the floor or ceiling is odd.

    I was thinking of saying that since k is an integer, once it is added with a strictly rational number(1/2), it would result in a non-integer, and thus be a number between two integers. Then I could say that since every consecutive integer is of opposite parity, then when you round down(floor) or round up(ceiling), it would result in 2 consecutive integers, and one of them must be odd.

    Would that be sufficient?
    Yup, that is exactly right.
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  3. #3
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    Joined
    Jan 2010
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    Okay I wasn't sure of that one since my other questions were of a very different style.

    Thank you!
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