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Math Help - Number theory proof

  1. #1
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    Number theory proof

    Show that the integer n is even if and only if n - 2[n/2] = 0

    Thanks very much.
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  2. #2
    Senior Member OReilly's Avatar
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    Quote Originally Posted by suedenation View Post
    Show that the integer n is even if and only if n - 2[n/2] = 0

    Thanks very much.
    Does n - 2[n/2] = 0 means n - 2\frac{n}{2} = 0?
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  3. #3
    Senior Member OReilly's Avatar
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    Quote Originally Posted by OReilly View Post
    Does n - 2[n/2] = 0 means n - 2\frac{n}{2} = 0?
    Why LaTex isn't shown?!
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  4. #4
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    Quote Originally Posted by OReilly View Post
    Why LaTex isn't shown?!
    The software driving MHF has been upgraded and LaTeX has not
    yet been reinstalled.

    RonL
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  5. #5
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    Quote Originally Posted by OReilly View Post
    Does n - 2[n/2] = 0 means n - 2\frac{n}{2} = 0?
    It should mean that there exists a k in N, such that:

    n=2k

    It could be either the floor, ceiling of some sort of nearest integer function,
    as these are n/2 if this is an integer, and something else if not an integer.

    RonL
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  6. #6
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    Quote Originally Posted by suedenation View Post
    Show that the integer n is even if and only if n - 2[n/2] = 0

    Thanks very much.
    1. Suppose n is even, then it may be written n=2k, for some k in N.
    Then n-2[n/2]=2k-2[k], but k is in N so [k]=k, therefore n-2[n/2]=2k-2k=0.

    Which is half of the proof, we have proven that if n is even n-2[n/2]=0

    2. Suppose n-2[n/2]=0, now n is in N, so may be written n=2k+r, where
    r is in {0,1}. Then n/2=k+r/2, so [n/2]=k (assuming that [x] denotes the
    greatest integer less than x, make your own arrangements for other
    intended meanings), therefore n-2[n/2]=2k+r-2k, but this is zero, therefore
    r=0, and so n is even.

    Which is the other half of the proof, as we have proven that if n-2[n/2]=0
    then n is even.

    RonL
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    1. Suppose n is even, then it may be written n=2k, for some k in N.
    Then n-2[n/2]=2k-2[k], but k is in N so [k]=k, therefore n-2[n/2]=2k-2k=0.

    Which is half of the proof, we have proven that if n is even n-2[n/2]=0

    2. Suppose n-2[n/2]=0, now n is in N, so may be written n=2k+r, where
    r is in {0,1}. Then n/2=k+r/2, so [n/2]=k (assuming that [x] denotes the
    greatest integer less than x, make your own arrangements for other
    intended meanings), therefore n-2[n/2]=2k+r-2k, but this is zero, therefore
    r=0, and so n is even.

    Which is the other half of the proof, as we have proven that if n-2[n/2]=0
    then n is even.

    RonL
    Perhaps I'm not seeing something here, but shouldn't we also be showing that if n is odd then n - 2[n/2] is not 0? (It's not hard to do, it just seems necessary.) 2. already shows that if n - 2[n/2] = 0 then n is even so we don't have to cover any other case.

    -Dan
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by topsquark View Post
    Perhaps I'm not seeing something here, but shouldn't we also be showing that if n is odd then n - 2[n/2] is not 0? (It's not hard to do, it just seems necessary.) 2. already shows that if n - 2[n/2] = 0 then n is even so we don't have to cover any other case.

    -Dan
    No. If and only if means that if n - 2[n/2] = 0, then n is even, and if
    n is even then n - 2[n/2] = 0.

    It is a short hand for bi-implication.

    Also as a consequence if n is not even then n - 2[n/2] != 0, similarly
    if n - 2[n/2] != 0 then n is not even.

    RonL
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