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Math Help - function over Z

  1. #1
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    function over Z

    Is there any function f:Z->Z, with f(f(n))=-n for every integer n?
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  2. #2
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    Quote Originally Posted by takis1881 View Post
    Is there any function f:Z->Z, with f(f(n))=-n for every integer n?
    I thought about this for a little while, and as far as I can tell the answer is yes:

    f(n)=\begin{cases}2n\quad \ n\ \text{is odd} \\ -\frac{n}{2}\quad n\ \text{is even}\end{cases}

    Ah, posted too soon. Obviously the above does not work when n is divisible by 4. The answer might be no.. I'll keep thinking about it.
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  3. #3
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    Well I'm not faring too well with this problem, but perhaps this will be of use: Assuming such function exists (which could be the start of a proof by contradiction, or the start of a way to find such an f), f(f(f(n))) can be thought of as either

    Code:
    f(  f(f(n))  ) = f(-n)   or   f(f(  f(n)  )) = -f(n)
    So f(-n) = -f(n), meaning f(n) is an odd function.

    Edits: Ah ok so I think the answer is yes, and there is a simple function we can define.

    Here is how I got there.

    Suppose f(n) = m.

    Then from above we know f(-n) = -f(n) = -m.

    We also know f(m) = -n.

    And consequently f(-m) = n.

    So we partition the integers into groups of four, basically, (not counting 0) and the simplest is {0}, {-2,-1,1,2}, {-4,-3,3,4}, ...

    So we can write

    f(n) = \begin{cases}0\quad\quad\quad\ \ n = 0 \\ n+1\quad\ \ n\ \text{odd and positive} \\ n-1\quad\ \ n\ \text{odd and negative} \\ -n+1\quad n\ \text{even and positive}\\-n-1 \quad n\ \text{even and negative}\end{cases}

    which can probably be written in a much more concise manner, anyway that should do the trick.
    Last edited by undefined; May 30th 2010 at 08:41 PM.
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  4. #4
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    Yes, i found that f is an odd function and f(0)=0. But for the time being, i can't either prove that f does not exist, or find f, if it does exist. Another thing that may help is that if f(a)=b,
    then f seems to send a -> b -> -a -> -b -> a.
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  5. #5
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    Please see my edited post above.
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  6. #6
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    I found a function like this too, but it's not an odd one, as f should be.
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  7. #7
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    Quote Originally Posted by takis1881 View Post
    I found a function like this too, but it's not an odd one, as f should be.
    We can show f is odd as follows:

    We need to show that f(n)=-f(-n). Consider 5 cases.

    Case 1: n = 0

    This is trivial.

    Case 2: n\ \text{odd and positive}

    f(n)=n+1

    -f(-n)=-((-n)-1)=n+1

    Cases 3-5: Similar.
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  8. #8
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    Yes, you're right!! I mistakenly considered f(-n)=-n+1. Thank you very much for your help!! Greetings from Greece.
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